Pick the one you are sitting. You get the sections it covers and
nothing else — the full guide material, figures and all — then questions
on exactly those sections. Coverage follows the department's own Fall 2026
lesson calendar.
§13.1 · §13.2 · §13.3 · §13.3b · §13.4 · §12.1 · §13.5 — and nothing else. Read them, then do the check-yourself questions inside each one with the answer folded away.
A vector is a displacement: how far and which way. It does not care where it starts. Two arrows of the same length pointing the same way are the same vector.
the four things you do constantlya=⟨a1,a2,a3⟩∣a∣=a12+a22+a32a+b=⟨a1+b1,a2+b2,a3+b3⟩ca=⟨ca1,ca2,ca3⟩
unit vector — same direction, length exactly 1u=∣a∣a
Vector from P to Q is PQ=⟨q1−p1,q2−p2,q3−p3⟩ — head minus tail. Getting this backwards is the single most common Unit 1 slip.
Algorithm · build a vector of prescribed length in a given direction
Get the direction as any vector a pointing that way.
Normalise: divide by its own length, u=a/∣a∣. Now it has length 1.
Scale: multiply by the length you want. v=Lu.
Normalise first, scale second. Doing it in the other order gives the wrong length.
Check yourself
What is ∣⟨3,−4⟩∣?
Answer
9+16=5.
Write the unit vector in the direction of ⟨0,3,4⟩.
Answer
Length is 5, so ⟨0,53,54⟩. Normalise first, always.
What is the vector from P(1,2) to Q(4,6)?
Answer
PQ=⟨3,4⟩ — head minus tail, Q−P. Backwards is the classic slip.
Three number lines at right angles. A point needs three numbers instead of two — and distance is still just Pythagoras, applied twice.
drag to spin
Spin it. The three brown steps and the one magenta arrow are the same information — the arrow is just the shortcut.
distance between two points∣P1P2∣=(x2−x1)2+(y2−y1)2+(z2−z1)2
sphere · centre (h,k,l), radius r(x−h)2+(y−k)2+(z−l)2=r2
Why Pythagoras twice: walk along x, then along y — that diagonal on the floor is Δx2+Δy2. Now go up by Δz. That floor-diagonal and the vertical rise are a right triangle too, so square-and-add again.
Algorithm · recognise a sphere from a messy equation
Group. Put all the x terms together, all the y terms, all the z terms. Move the constant to the right.
Complete the square in each variable: x2+ax→(x+2a)2−4a2.
Balance. Every −4a2 you created must be added to the right-hand side too.
Read it off. Centre is (h,k,l) from the three brackets, radius is right side. If the right side is negative there is no sphere; if it is zero you get a single point.
Worked example
x2+y2+z2−4x+6y−2=0.
group
(x2−4x)+(y2+6y)+z2=2
complete
(x−2)2−4+(y+3)2−9+z2=2
balance
(x−2)2+(y+3)2+z2=15
Centre (2,−3,0), radius 15.
One equation in 3-D is a surface, not a curve.x=3 is a whole plane. x2+y2=9 is an infinite cylinder — the z is free, so the circle gets extruded forever.
Check yourself
Distance from (1,0,2) to (1,4,5)?
Answer
0+16+9=5. Pythagoras, twice.
In 3-D, what does x2+y2=4 describe?
Answer
An infinite cylinder of radius 2 around the z-axis — z is free, so the circle gets extruded forever. One equation in 3-D is a surface, not a curve.
Centre and radius of (x−1)2+y2+(z+2)2=9?
Answer
Centre (1,0,−2), radius 3. The sign inside the bracket flips: (z+2)2 means z=−2.
Shine a light straight down on b. The shadow it casts along a is the whole idea. Long shadow → big number. No shadow → zero. Shadow pointing backwards → negative.
Drag the angle. a·b = (length of a) × (length of the shadow). Same direction → positive. Square corner → zero. Opposite → negative.
two formulas, one numbera⋅b=a1b1+a2b2+a3b3anda⋅b=∣a∣∣b∣cosθ
The left one is how you compute it. The right one is what it means. Setting them equal is how every angle problem gets solved.
the three facts that mattercosθ=∣a∣∣b∣a⋅ba⊥b⟺a⋅b=0a⋅a=∣a∣2
Algorithm · angle between two vectors
Dot them componentwise: a⋅b.
Find both lengths∣a∣ and ∣b∣.
Divide:cosθ=∣a∣∣b∣a⋅b.
Arccos.θ=cos−1(⋅), always in [0,π].
Sign check before you compute: positive dot ⇒ acute, zero ⇒ right angle, negative ⇒ obtuse.
direction cosines — the angles a vector makes with each axiscosα=∣a∣a1,cosβ=∣a∣a2,cosγ=∣a∣a3,cos2α+cos2β+cos2γ=1
Those three cosines are the components of the unit vectora/∣a∣ — which is exactly why they square to 1.
work — the physics reason dot products existW=F⋅d=∣F∣∣d∣cosθ
Check yourself
Compute ⟨1,2,3⟩⋅⟨4,−1,0⟩.
Answer
4−2+0=2.
If a⋅b<0, is the angle acute or obtuse?
Answer
Obtuse. Negative dot means the shadow points backwards.
These two get confused constantly, and the difference is simple: one is a number (how long the shadow is) and the other is an arrow (the shadow itself, lying along b).
Same picture, two different answers. comp is the tick mark on a ruler. proj is the arrow you could actually add to another vector.
scalar projection of a onto b — "component of a along b"compba=∣b∣a⋅ba number
vector projection of a onto bprojba=(∣b∣2a⋅b)b=(compba)∣b∣ba vector
How to never mix them up. The vector projection is literally the scalar projection times the unit vector of b. So: get the number first, then attach a direction to it. One ∣b∣ in the denominator = number. Two = vector, because the second one is busy normalising b.
Algorithm · project a onto b
Dot. Compute a⋅b.
Decide which you want. Does the answer need a direction? If not, stop after step 3.
Scalar: divide by ∣b∣ once. Done — that is compba, and it may be negative.
Vector: multiply the scalar answer by the unit vector b/∣b∣. Equivalently, divide by ∣b∣2 and multiply by b — never by ∣b∣.
Projecting onto b always lands on the b line. If your answer is not a multiple of b, you projected the wrong way round.
Worked example
a=⟨3,4,0⟩, b=⟨4,0,3⟩. Both have length 5.
dot
a⋅b=12+0+0=12
scalar
compba=512=2.4
vector
projba=2512⟨4,0,3⟩=⟨2548,0,2536⟩
check
its length is (48/25)2+(36/25)2=2560=2.4 ✓ matches the scalar answer
orthogonal decomposition — split a into "along b" plus "perpendicular to b"a=parallel to bprojba+perpendicular to b(a−projba)
This split is not a curiosity — it is reused three more times in this course: distance from a point to a line (13.5b), splitting acceleration into aT and aN (14.3), and the flux/circulation split of a vector field (Unit 4).
Trap
projba=projab. Read the subscript as "onto". Swapping them gives a vector pointing in a completely different direction. And the scalar projection keeps its sign — if θ is obtuse, compba is negative, and that negative is the answer, not an error.
Check yourself
a⋅b=10 and ∣b∣=5. What is compba?
Answer
10/5=2. Divide by ∣b∣once — that gives the number.
Same numbers. Is projba a number or a vector?
Answer
A vector: 2510b. Divide twice — the second ∣b∣ is busy normalising.
If a⊥b, what is projba?
Answer
The zero vector. No shadow — the light is shining straight along b.
scalar triple product — volume of the boxV=∣a⋅(b×c)∣=a1b1c1a2b2c2a3b3c3
Reading it:b×c gives the base area as a vector standing on the base. Dotting with a takes a's shadow on that stand-up direction — which is the height. Area × height = volume. If the triple product is zero, the box is flat, so the three vectors are coplanar.
Check yourself
∣a∣=3, ∣b∣=4, and they meet at 90°. What is ∣a×b∣?
Answer
3⋅4⋅sin90°=12 — the biggest the tile can get.
What is a×a?
Answer
0. Zero angle, flat tile, no area.
You computed a×b. How do you check it in ten seconds?
Answer
Dot it with a. You must get 0. This catches the missing minus on the j component every time.
A Cartesian equation tells you where a curve is. A parametrization
tells you how a point moves along it — where it starts, which way it goes,
how fast, and how many times round. Those extra facts are the whole reason the rest
of this course uses parametrizations for everything: curves in space, surfaces,
line integrals, flux. This is where they start.
a plane curve, parametrizedx=f(t),y=g(t),a≤t≤b
Think of t as time and (f(t),g(t)) as the position of a bug. Sweep t
from a to b and the bug draws the curve. Different bugs can draw the same picture.
Both bugs draw the same ellipse. The orange one goes
counterclockwise once; the green one goes clockwise, twice as fast, twice around.
They disagree about direction, speed, and how many laps. Eliminating the parameter keeps the picture and throws
all three away — which is exactly the information line integrals care about.
the three you must know coldsegment P→Q:circle, centre (h,k), radius R:ellipse, semi-axes a,b:r(t)=P+t(Q−P),0≤t≤1x=h+Rcost,y=k+Rsint,0≤t≤2πx=h+acost,y=k+bsint,0≤t≤2π
All three run counterclockwise as written, and all three start at
the rightmost point when t=0. Swap sin and cos, or negate one of them, and you
change the starting point or the direction — never the picture.
Algorithm · eliminate the parameter
Look for the trig identity first. If you see cost and sint, solve each
equation for the trig function and use cos2t+sin2t=1. Do not try to solve for t.
Otherwise solve the easier equation for t — usually the linear one —
and substitute into the other.
Carry the range across. Ask what values x and y actually take as t runs
over its interval. The Cartesian equation on its own will usually describe more
curve than you have.
Say the direction out loud. Plug in t=a, the midpoint, and t=b, and mark
the three points in order. That is the orientation, and no Cartesian equation records it.
Steps 3 and 4 are where the marks are. Step 1 and 2 are algebra you already have.
Worked example
Describe the curve x=2cost,y=3sint, 0≤t≤π.
identity
cost=2x and sint=3y, so
4x2+9y2=1 — an ellipse with semi-axes 2 and 3.
range
On 0≤t≤π, sint≥0, so y≥0.
Only the upper half of the ellipse.
direction
t=0⇒(2,0); t=2π⇒(0,3);
t=π⇒(−2,0). Right, up, left — counterclockwise.
answer
The upper half of 4x2+9y2=1,
traced once counterclockwise from (2,0) to (−2,0).
Trap
“Eliminate the parameter” is a lossy operation.x=t,y=t2 and
x=t2,y=t4 both give y=x2, but the first is the whole parabola and the second is
only its right half, traced in and back out. If a question mentions direction, a starting
point, or a number of laps, the Cartesian equation cannot answer it — keep the
parametrization.
Where this reappears: 13.5 writes a line as r(t)=r0+tv,
which is the segment recipe with the range removed. 14.1 is this section with a third
coordinate bolted on. 17.2 integrates along one of these, and the orientation you were
told to keep track of here is what decides the sign of the answer.
Check yourself
Parametrize the segment from (−1,4) to (3,2) on 0≤t≤1.
Answer
x=−1+4t,y=4−2t. Start, plus t times
the displacement ⟨4,−2⟩.
What curve is x=1+3cost,y=−2+3sint, 0≤t≤2π?
Answer
The circle of radius 3 centred at
(1,−2), once counterclockwise from (4,−2): (x−1)2+(y+2)2=9.
x=sint,y=sin2t. Cartesian equation, and how much of it?
Answer
y=x2, but only the arc with
−1≤x≤1, and the point retraces it back and forth forever.
Two parametrizations give the same Cartesian equation. Name two things that
can still differ.
Answer
Orientation, speed, starting point, and
how much of the curve is covered — any of these. Only the point set is shared.
Quiz 1 covers the lines half of 13.5 only. Planes are Lesson 3, on Quiz 2.
A line needs a point and a direction to walk. A plane needs a point and a direction to avoid. That is the entire difference.
Tilt n and the whole plane tilts with it. That single arrow is the plane's identity.
line through P0 with direction v=⟨a,b,c⟩r(t)=r0+tv⎩⎨⎧x=x0+aty=y0+btz=z0+ctax−x0=by−y0=cz−z0
Vector form, parametric form, symmetric form — the same line, written three ways. Symmetric form is just the parametric equations each solved for t. If a direction component is 0 you cannot divide by it: write that coordinate as a separate condition, e.g. 2x−1=5z−4,y=3.
plane through P0 with normal n=⟨a,b,c⟩n⋅(r−r0)=0⟺a(x−x0)+b(y−y0)+c(z−z0)=0⟺ax+by+cz=d
Free information: in ax+by+cz=d, the coefficients are the normal vector. You can read n=⟨a,b,c⟩ straight off any plane equation without doing anything.
Algorithm · plane through three points P, Q, R
Make two vectors in the plane:PQ and PR.
Cross them:n=PQ×PR. It sticks out of the plane by construction.
Pick any one of the three points and write n⋅(r−r0)=0.
Expand and check that all three original points satisfy it.
Algorithm · classify two lines in space
Compare directions. If v1∥v2 (one is a multiple of the other) the lines are parallel — identical if a point of one satisfies the other, otherwise distinct.
Otherwise set them equal — and use different parameters, t for one and s for the other. Same-letter parameters is the classic wrong answer.
Solve two of the three equations for t and s.
Test in the third. Consistent ⇒ they intersect (plug back in for the point). Contradiction ⇒ they are skew — never meeting, not parallel.
Angle between two planes
Angle between their normals: cosθ=∣n1∣∣n2∣∣n1⋅n2∣.
Line where two planes meet
Direction is n1×n2. For a point, set one variable to 0 and solve the two equations.
Parallel planes
Normals are multiples of each other. Same normal + different d ⇒ parallel and distinct.
Check yourself
What is the normal vector of the plane 2x−y+3z=7?
Answer
⟨2,−1,3⟩ — read straight off the coefficients, no work needed.
Line through (1,0,0) with direction ⟨2,3,−1⟩?
Answer
r(t)=⟨1+2t,3t,−t⟩.
Two lines have non-parallel directions and no common point. What are they?
Answer
Skew. They never meet and they are not parallel — only possible in 3-D.
STEP 02
The official problems
11 problems
Straight from the department's own study guide, for exactly these lessons — worked, with the answer held back until you commit. Do these first.
The projection of u onto v is projvu=∣v∣2u⋅vv — a vector *along* v.
u⋅v=(2)(1)+(−1)(2)+(3)(2)=2−2+6=6.
∣v∣2=1+4+4=9.
projvu=96⟨1,2,2⟩=32⟨1,2,2⟩=⟨32,34,34⟩.
TrapDividing by ∣v∣ instead of ∣v∣2. That gives the scalar projection compvu=36=2, which is a number, not a vector. Read which one the question wants.
2
Find the area of the triangle with vertices P(1,0,0), Q(0,2,0), and R(0,0,3).
no calculator, same as the exam
Answer27answer recomputed · sympy
Two edge vectors from P: PQ=⟨−1,2,0⟩ and PR=⟨−1,0,3⟩.
PQ×PR=⟨2⋅3−0⋅0,0⋅(−1)−(−1)⋅3,0−(−2)⟩=⟨6,3,2⟩.
∣⟨6,3,2⟩∣=36+9+4=7 — that is the parallelogram area.
A triangle is half of it: area=27.
TrapForgetting the 21. The cross product's length is the parallelogram, and 7 is sitting right there looking like an answer.
3
If u=⟨3,−1,2⟩ and v=⟨1,4,−1⟩, find u×v.
no calculator, same as the exam
Answer⟨−7,5,13⟩answer recomputed · sympy
u×v=i31j−14k2−1.
i: (−1)(−1)−(2)(4)=1−8=−7.
j: −[(3)(−1)−(2)(1)]=−(−5)=5 — note the sign flip on the middle term.
k: (3)(4)−(−1)(1)=12+1=13. So u×v=⟨−7,5,13⟩.
TrapThe minus on the j component. Dropping it gives ⟨−7,−5,13⟩, and the sanity check catches it instantly: u⋅(u×v) must be 0.
4
For what value of c are the vectors ⟨2,c,−1⟩ and ⟨3,−6,c⟩ orthogonal?
no calculator, same as the exam
Answerc=76answer recomputed · sympy
Orthogonal means the dot product vanishes.
⟨2,c,−1⟩⋅⟨3,−6,c⟩=6−6c−c=6−7c.
6−7c=0⇒c=76.
TrapMissing that c appears in both vectors, so the last term (−1)(c) also carries a c. Treating it as a constant gives c=1.
5
Find the angle between u=⟨1,1,0⟩ and v=⟨0,1,1⟩.
no calculator, same as the exam
Answerθ=3πanswer recomputed · sympy
cosθ=∣u∣∣v∣u⋅v.
u⋅v=0+1+0=1, and ∣u∣=∣v∣=2.
cosθ=21⇒θ=3π.
TrapAnswering 6π: cos−121=3π, not 6π. 6π is where the *sine* is 21.
The line through (1,3,−1) and (3,1,2) intersects the plane x+y+z=6 at what point?
no calculator, same as the exam
Answer(3,1,2)answer recomputed · sympy
Direction: v=⟨3−1,1−3,2−(−1)⟩=⟨2,−2,3⟩.
Line: r(t)=⟨1+2t,3−2t,−1+3t⟩.
Substitute into x+y+z=6: (1+2t)+(3−2t)+(−1+3t)=3+3t=6⇒t=1.
r(1)=⟨3,1,2⟩ — the second given point happens to lie on the plane.
TrapSign errors in the direction vector. Notice the x and y parameters cancel, so the whole equation collapses to 3+3t; if yours does not collapse, recheck v.
2
Find the equation of the plane containing (2,1,−1) and parallel to the vectors ⟨1,0,3⟩ and ⟨2,1,0⟩.
no calculator, same as the exam
Answer−3x+6y+z+1=0answer recomputed · sympy
Two vectors parallel to the plane give the normal through a cross product.
Point-normal form at (2,1,−1): −3(x−2)+6(y−1)+1(z+1)=0.
Expand: −3x+6+6y−6+z+1=0⇒−3x+6y+z+1=0.
Trap'Parallel to the vectors' does not mean the vectors are normals. They lie *in* the plane; their cross product is the normal.
3
Find the line of intersection between the planes x+y=2 and y+z=3.
no calculator, same as the exam
Answerx=2+t,y=−t,z=3+tanswer recomputed · sympy
Two equations, three unknowns — one free parameter, which is the line.
Let y=−t. Then x+y=2 gives x=2+t, and y+z=3 gives z=3+t.
x=2+t,y=−t,z=3+t.
Check the direction against n1×n2=⟨1,1,0⟩×⟨0,1,1⟩=⟨1,−1,1⟩ — same direction, as it must be.
TrapAny parametrisation is correct; yours may look nothing like this one. Verify by plugging back into both plane equations rather than comparing to a printed answer.
4
Find the angle between the planes x+y+z=1 and x−y=0.
no calculator, same as the exam
Answerθ=2π (planes are perpendicular)answer recomputed · sympy
The angle between planes is the angle between normals: n1=⟨1,1,1⟩, n2=⟨1,−1,0⟩.
n1⋅n2=1−1+0=0.
Zero dot product ⇒θ=2π: the planes are perpendicular.
TrapReading coefficients off a plane written as x−y=0 and forgetting the missing z is a coefficient of 0, not of 1.
5
For what value of a do the lines r1(t)=⟨1+t,2t,3+t⟩ and r2(s)=⟨3−s,s+1,2s+a⟩ intersect?
no calculator, same as the exam
Answera=2answer recomputed · sympy
Intersection needs a single point, so use different parameters t and s and solve all three components.
x: 1+t=3−s⇒t=2−s. y: 2t=s+1.
Substitute: 2(2−s)=s+1⇒4−2s=s+1⇒s=1, hence t=1.
z: 3+t=2s+a⇒3+1=2+a⇒a=2.
TrapUsing t for both lines. Two lines can cross at a point they reach at different times; forcing one parameter finds only the cases where they arrive together, and reports skew for lines that do meet.
′restored
Find the distance from the point (1,−2,3) to the plane 2x−y+2z=4.
no calculator, same as the exam
Official answer35checked · official answer wrong
Correct answer2recomputed · sympy
D=a2+b2+c2∣ax0+by0+cz0−d∣ with n=⟨2,−1,2⟩.
Numerator: ∣2(1)−(−2)+2(3)−4∣=∣2+2+6−4∣=6.
Denominator: 4+1+4=3.
D=36=2.
TrapForgetting to move d to the same side. The formula measures ax+by+cz−d, so a plane written as 2x−y+2z=4 contributes −4, not +4.
The official answer is wrongThe commented-out source gives 35. The numerator is ∣2+2+6−4∣=6, not 5, so the distance is 36=2. This may be exactly why the problem was commented out.
Commented out on the official page, so it does not appear there. The skill is still on the syllabus, so it is restored here.
STEP 03
More practice
24 questions
Mine, on those sections only. Untimed, no calculator — the quiz doesn't allow one either.
0/24 answered
112.1 · Eliminating the parameternew
The curve x=2cost,y=3sint, 0≤t≤2π, is
Answer A
Solve each equation for the trig function: cost=2x and sint=3y.
Then cos2t+sin2t=1 gives 4x2+9y2=1.
The denominators are the squares of the amplitudes: 22=4 under x, 32=9 under y.
TrapSwapping the denominators. The number multiplying cost belongs under x, squared — not under y. Option B is that swap.
Which parametrization traces the line segment from (1,2) to (5,−4) exactly once as t runs from 0 to 1, starting at (1,2)?
Answer A
The straight-line recipe is r(t)=r0+t(r1−r0): start, plus t times the displacement.
r1−r0=⟨5−1,−4−2⟩=⟨4,−6⟩.
So x=1+4t, y=2−6t. At t=0 that is (1,2); at t=1 it is (5,−4). ✓
TrapOption D traces the same segment, correctly and exactly once — but backwards, from (5,−4) to (1,2). Direction is part of the answer. Option E covers the right points at the wrong speed.
cost=2x−3 and sint=−2y+1. Squaring and adding, (x−3)2+(y+1)2=4 — a circle of radius 2 centred at (3,−1).
Now the direction. t=0⇒(5,−1), the rightmost point. t=2π⇒(3,−3), below the centre.
Right, then down, then left: that is clockwise.
TrapReading the minus sign as a change of shape. Negating sint leaves the circle identical and only reverses the orientation. Radius is 2, not 4 — the 4 is r2.
TrapSubtracting the wrong way round. P−Q has the same length, so the error hides here — but it reverses the direction of every line you build from it later.
The equation x2+y2+z2−6x+4y−12=0 describes a sphere. Find its centre and radius.
Answer A
Group and complete the square, balancing the right side each time:
(x2−6x)+(y2+4y)+z2=12⇒(x−3)2−9+(y+2)2−4+z2=12.
(x−3)2+(y+2)2+z2=25. Centre (3,−2,0), radius 5.
TrapSign flip on the centre. (x−3)² means the centre coordinate is +3, not −3.
a⋅b=3+2+4=9 and ∣b∣2=1+4+4=9.
projba=∣b∣2a⋅bb=99⟨1,2,2⟩=⟨1,2,2⟩.
The scalar projection is the different object compba=∣b∣9=39=3 — option A, the planted distractor.
TrapDividing by |b| once (a number) when the question asked for the vector. Once = number, twice = vector.
For a=⟨4,3,0⟩ and b=⟨0,1,1⟩, find the scalar projection compba.
Answer A
a⋅b=(4)(0)+(3)(1)+(0)(1)=3 and ∣b∣=2.
compba=∣b∣a⋅b=23=232.
TrapOption C is projba — the vector projection, which divides by ∣b∣2 and then multiplies by b. The question asked for a number. Option B is that same computation with the b forgotten.
L1:x=1+t,y=2−t,z=3+2t and L2:x=2+2s,y=1−2s,z=6+4s. These two lines are
Answer A
Directions: v1=⟨1,−1,2⟩ and v2=⟨2,−2,4⟩=2v1. Parallel directions, so the lines cannot be skew or intersecting.
Now decide parallel-and-distinct versus identical: is L2's point (2,1,6) on L1? From x: 1+t=2⇒t=1, giving (2,1,5) on L1 — not (2,1,6).
Same direction, different point ⇒ parallel and distinct.
TrapStopping at “the directions are proportional, so it's the same line.” Parallel lines share a direction; identical lines also share a point. You must test a point.
Find symmetric equations for the line through (2,−1,4) parallel to v=⟨3,5,−2⟩.
Answer A
Start from the parametric form x=2+3t,y=−1+5t,z=4−2t and solve each for t:
t=3x−2, t=5y−(−1)=5y+1, t=−2z−4.
Setting them equal gives the symmetric form. The point is subtracted on top; the direction goes underneath.
TrapThe point's coordinate is −1, so the numerator is y−(−1)=y+1 — option E drops that. Option D swaps the roles: it puts the direction on top and the point underneath.
L1:x=1+t,y=2+3t,z=−1+2t and L2:x=2+s,y=−1+2s,z=3−s. These two lines are
Answer A
Directions ⟨1,3,2⟩ and ⟨1,2,−1⟩ are not multiples of each other, so the lines are not parallel. Either they meet or they are skew.
Set the x's equal: 1+t=2+s⇒t=1+s. Set the y's equal: 2+3t=−1+2s, so 3(1+s)−2s=−3, giving s=−6 and t=−5.
Now test the third equation: z1=−1+2(−5)=−11 but z2=3−(−6)=9. They disagree, so no single point is on both lines.
Not parallel and never meeting ⇒ skew.
TrapSolving the first two equations, finding a solution, and declaring the lines intersect. Two equations in two unknowns almost always have a solution — the third equation is the actual test.
Quiz 2 covers the planes half of 13.5. Lines were Lesson 2, on Quiz 1.
A line needs a point and a direction to walk. A plane needs a point and a direction to avoid. That is the entire difference.
Tilt n and the whole plane tilts with it. That single arrow is the plane's identity.
line through P0 with direction v=⟨a,b,c⟩r(t)=r0+tv⎩⎨⎧x=x0+aty=y0+btz=z0+ctax−x0=by−y0=cz−z0
Vector form, parametric form, symmetric form — the same line, written three ways. Symmetric form is just the parametric equations each solved for t. If a direction component is 0 you cannot divide by it: write that coordinate as a separate condition, e.g. 2x−1=5z−4,y=3.
plane through P0 with normal n=⟨a,b,c⟩n⋅(r−r0)=0⟺a(x−x0)+b(y−y0)+c(z−z0)=0⟺ax+by+cz=d
Free information: in ax+by+cz=d, the coefficients are the normal vector. You can read n=⟨a,b,c⟩ straight off any plane equation without doing anything.
Algorithm · plane through three points P, Q, R
Make two vectors in the plane:PQ and PR.
Cross them:n=PQ×PR. It sticks out of the plane by construction.
Pick any one of the three points and write n⋅(r−r0)=0.
Expand and check that all three original points satisfy it.
Algorithm · classify two lines in space
Compare directions. If v1∥v2 (one is a multiple of the other) the lines are parallel — identical if a point of one satisfies the other, otherwise distinct.
Otherwise set them equal — and use different parameters, t for one and s for the other. Same-letter parameters is the classic wrong answer.
Solve two of the three equations for t and s.
Test in the third. Consistent ⇒ they intersect (plug back in for the point). Contradiction ⇒ they are skew — never meeting, not parallel.
Angle between two planes
Angle between their normals: cosθ=∣n1∣∣n2∣∣n1⋅n2∣.
Line where two planes meet
Direction is n1×n2. For a point, set one variable to 0 and solve the two equations.
Parallel planes
Normals are multiples of each other. Same normal + different d ⇒ parallel and distinct.
Check yourself
What is the normal vector of the plane 2x−y+3z=7?
Answer
⟨2,−1,3⟩ — read straight off the coefficients, no work needed.
Line through (1,0,0) with direction ⟨2,3,−1⟩?
Answer
r(t)=⟨1+2t,3t,−t⟩.
Two lines have non-parallel directions and no common point. What are they?
Answer
Skew. They never meet and they are not parallel — only possible in 3-D.
All four of these are the same move: build a vector from the thing to the point, then keep only the part that is perpendicular to whatever you're measuring from.
point P1 to the plane ax+by+cz+d=0D=a2+b2+c2∣ax1+by1+cz1+d∣=compnP0P1
Read it as a scalar projection onto the normal. Take any point P0 on the plane, form P0P1, and measure its shadow on n — the part of the trip that actually left the plane.
point P to the line through Q with direction vD=∣v∣∣QP×v∣
Why the cross product: ∣QP×v∣=∣QP∣∣v∣sinθ is the area of the parallelogram they span. Area ÷ base = height, and the height is the perpendicular distance. Same as ∣QP−projvQP∣ — the leftover from 13.3b.
two skew linesD=∣v1×v2∣(r2−r1)⋅(v1×v2)
Algorithm · distance between two skew lines
Find the common perpendicular:n=v1×v2. It is at right angles to both lines at once.
Pick one point on each line, P1 and P2, and form P1P2.
Project onto n: the distance is compnP1P2. Everything not along n can be walked off along the lines, so it doesn't count.
Equivalent framing: slide the two lines into two parallel planes with normal n, then measure plane-to-plane.
Two parallel planes
Take any point on one, use the point-to-plane formula on the other.
Line parallel to a plane
Take any point on the line, use point-to-plane.
Sanity check
Every distance is ≥0. Absolute value bars are not decoration.
Check yourself
Distance from the origin to x+2y+2z=9?
Answer
1+4+4∣0+0+0−9∣=39=3.
Why does point-to-line use a cross product?
Answer
∣QP×v∣ is the parallelogram's area. Area ÷ base = height, and the height is the perpendicular distance.
Section 13.6 has two halves, and the first one is
nearly free. A cylinder is what you get when a variable is missing from
the equation. Missing means unconstrained: that coordinate may be anything at all, so
the curve drawn in the other two variables is dragged along the whole missing axis.
A cylinder in this course is not necessarily round — it is any curve times a line.
a missing variable is a swept axisx2+z2=9(no y)
In the xz-plane that is a circle of radius 3. In space, nothing
constrains y, so every height y=k gives the same circle. Stack them and you
get an infinite tube whose axis is the y-axis.
The equation never mentions y, so y is free. The circle repeats at
every y — the straight lines along the tube are the rulings, and they
run parallel to the axis of the variable that went missing.
Algorithm · name a cylinder
Spot the missing variable. One of x, y, z does not appear. If all
three appear, it is not a cylinder — go to the quadric test below.
Read the curve formed by the two that remain, in their own coordinate
plane: circle, ellipse, parabola, hyperbola, or a pair of lines.
Sweep it along the missing variable's axis. That axis is the direction of
the rulings.
Name it after the curve. Circular, elliptic, parabolic or hyperbolic
cylinder — then say which axis the rulings are parallel to. Both halves are
expected in the answer.
y=−z2 has no x: a parabola in the yz-plane opening toward
−y, swept along the x-axis. A parabolic cylinder, rulings parallel to the
x-axis.
Trap
x2+z2=9 is a circle in two dimensions and a tube in three.
The question decides which by telling you the ambient space, and 13.6 is a chapter
about surfaces in R3 — so the answer is the tube. Answering
“a circle of radius 3” is the single most common way to drop this
question.
Now the other half. Every quadric is decided by signs and squares. Slice it with flat planes and see what falls out — circles, parabolas, or hyperbolas.
drag to spin
Fast test: count the minus signs. Zero → ellipsoid. One → one sheet. Two → two sheets. Right side is 0 instead of 1 → cone. One variable unsquared → paraboloid.
Algorithm · identify a quadric by its traces
Standard form first. Get it to a2x2±b2y2±c2z2=1,0, or = a single linear variable.
Count the minus signs among the squared terms: 0⇒ ellipsoid, 1⇒ hyperboloid of one sheet, 2⇒ hyperboloid of two sheets.
Right side is 0? It's a cone.
One variable not squared? It's a paraboloid — elliptic if the two squared terms have the same sign, hyperbolic (a saddle) if opposite.
Confirm with traces. Set z=k and see what curve you get, then x=k, then y=k. Ellipse / parabola / hyperbola tells you the rest.
The axis of a cone, paraboloid, or hyperboloid is the variable that behaves differently from the other two.
Algorithm · complete the square into standard form
Group by variable and move the bare constant to the right-hand side.
Factor out the coefficient of each squared term, so the bracket opens
with a bare x2, y2 or z2.
Complete each square by adding (b/2)2inside the bracket.
Because the bracket is multiplied by that factored-out coefficient, what you must
add to the other side is a⋅(b/2)2, not (b/2)2. This is where the
arithmetic goes wrong.
Divide so the right side is 1 (or 0).
Now count the minus signs with the algorithm above. Shifting a surface
never changes what it is — only where it sits.
worked · 4x2−y2+z2+8x=04(x2+2x)−y2+z24(x2+2x+1)−y2+z24(x+1)2−y2+z2(x+1)2−4y2+4z2=0=0+4⋅1=4=1
One minus sign, right side 1: a hyperboloid of one sheet, axis
the y-axis, centred at (−1,0,0). The +8x only moved it one unit along x; the
surface was never anything else.
Trap
The one-sheet hyperboloid has one minus and is connected — you can walk around the waist. The two-sheet has two minuses and a gap: there is no surface at all for small ∣z∣. Test it by setting z=0 and seeing whether you get a real curve or an impossibility.
Check yourself
In space, what is x2+z2=9?
Answer
No y, so y is free: a circular cylinder of radius 3 with rulings parallel to the y-axis. In the plane it would be a circle; in R3 it is a tube.
Identify y=−z2 in three dimensions.
Answer
No x: a parabolic cylinder, the parabola y=−z2 in the yz-plane opening toward −y, swept along the x-axis.
Put 9x2+4y2+36z2−18x=27 into standard form and identify it.
Answer
9(x2−2x)=9(x−1)2−9, so 9(x−1)2+4y2+36z2=36, i.e. 4(x−1)2+9y2+z2=1. No minus signs → an ellipsoid centred at (1,0,0).
Identify x2+y2−z2=1.
Answer
One minus sign → hyperboloid of one sheet (connected, like a cooling tower).
Identify z=x2−y2.
Answer
One variable unsquared, opposite signs → hyperbolic paraboloid, the saddle.
Identify z2=x2+y2.
Answer
Right side is 0 when rearranged → cone.
STEP 02
The official problems
5 problems
Straight from the department's own study guide, for exactly these lessons — worked, with the answer held back until you commit. Do these first.
AnswerHyperboloid of two sheets (axis along y)prose answer · not machine-checkable
Write it with the positive term first: 4y2−x2−9z2=1.
One positive square, two negative, equals +1 — that is the two-sheet signature.
Trace at y=0: −x2−9z2=1, impossible. Nothing at the origin, so it cannot be one connected sheet.
Hyperboloid of two sheets, axis along y — the axis is always the variable whose square is positive.
TrapOne sheet vs two sheets is decided by the count of minus signs, not by which variable is isolated. One minus = one sheet (connected); two minuses = two sheets.
2
Identify the surface defined by x2+4x+y2−2y+z2=−4. (Hint: complete the square.)
no calculator, same as the exam
AnswerSphere centered at (−2,1,0) with radius 1answer recomputed · sympy
Complete the square in x and y: x2+4x=(x+2)2−4 and y2−2y=(y−1)2−1.
(x+2)2−4+(y−1)2−1+z2=−4.
(x+2)2+(y−1)2+z2=1.
Sphere, centre (−2,1,0), radius 1.
TrapMoving the completion constants the wrong way. The −4 and −1 come out on the left, so they get added to the right: −4+4+1=1.
3
Identify the surface: z=4x2+9y2.
no calculator, same as the exam
AnswerElliptic paraboloid opening upwardprose answer · not machine-checkable
One variable (z) appears to the first power, the other two are squared with the same sign.
That is the paraboloid family. Both coefficients positive ⇒ opens upward.
Traces z=k>0 are ellipses 4x2+9y2=k; traces x=0 and y=0 are upward parabolas.
Elliptic paraboloid opening upward.
TrapCalling it a cone. A cone has z2, not z; the first power is what makes it a paraboloid.
4
Describe the trace of 4x2+9y2−z2=1 in the plane z=2.
TrapStopping at 4x2+9y2=5. A trace is named by its standard form, and the denominators change once you divide through.
5
Identify the surface: x2+4y2=9z2.
no calculator, same as the exam
AnswerElliptic cone with axis along the z-axisprose answer · not machine-checkable
Rewrite as x2+4y2−9z2=0 — equal to zero, not 1.
Zero is the cone signature. Traces z=k=0 are ellipses that grow linearly with ∣k∣; the trace z=0 is the single point at the origin (the vertex).
Elliptic cone with axis along the z-axis.
TrapReading it as a hyperboloid. Same sign pattern — the right-hand side is the entire difference: =1 one sheet, =−1 two sheets, =0 the cone between them.
STEP 03
More practice
42 questions
Mine, on those sections only. Untimed, no calculator — the quiz doesn't allow one either.
0/42 answered
113.5 · Planes
Find an equation of the plane through (1,0,0), (0,2,0) and (0,0,4).
Answer B
PQ=⟨−1,2,0⟩,PR=⟨−1,0,4⟩.
n=PQ×PR=⟨8,4,2⟩∥⟨4,2,1⟩.
Through (1,0,0): 4(x−1)+2y+z=0⇒4x+2y+z=4.
Fast check: the intercept form 1x+2y+4z=1 clears to the same thing.
TrapSign slip on the j component of the cross product — verify n·PQ = 0.
The planes x+y+z=1 and x−2y+3z=4 meet in a line. Which point lies on it?
Answer A
A point on the line has to satisfy **both** equations, so pick a convenient slice: set z=0.
That leaves x+y=1 and x−2y=4. Subtracting, 3y=−3, so y=−1 and x=2.
The point is (2,−1,0). Check both: 2−1+0=1 and 2+2+0=4.
With the direction n1×n2=⟨5,−2,−3⟩ from the companion question, you now have the full line.
TrapTesting against only one of the two planes. Every option here satisfies at least one of them — the line is the *intersection*, so both have to hold.
Find an equation of the plane through A(1,0,0), B(0,2,0) and C(0,0,4).
Answer A
Two vectors in the plane: AB=⟨−1,2,0⟩ and AC=⟨−1,0,4⟩.
n=AB×AC=⟨2⋅4−0⋅0,0⋅(−1)−(−1)⋅4,0−(−2)⟩=⟨8,4,2⟩, or ⟨4,2,1⟩ after dividing by 2.
Through A(1,0,0): 4(x−1)+2y+z=0, so 4x+2y+z=4.
TrapReading the intercepts straight off as coefficients gives x+2y+4z=1 — option B. The intercept form is 1x+2y+4z=1; the denominators are the intercepts, not the numerators.
Two planes are orthogonal exactly when their normals are orthogonal.
n1=⟨2,1,−1⟩. For option A, n=⟨1,1,3⟩ and n1⋅n=2+1−3=0. ✓
The others: C gives 6−1−1=4, D gives 4+1−1=4, E gives 2+2−1=3 — none zero. B and F have normals proportional to n1, so those planes are parallel to the original, not perpendicular.
TrapOptions B and F look different but are the same plane family scaled by 2 and −1. Proportional normals mean parallel; orthogonal normals mean perpendicular.
Find the plane through (1,0,2) perpendicular to the line x=3−t,y=2+4t,z=1+t.
Answer A
A plane perpendicular to a line has the line's direction as its normal.
The direction is v=⟨−1,4,1⟩.
−1(x−1)+4(y−0)+1(z−2)=0, so −x+4y+z−1=0, that is −x+4y+z=1.
Check: −(1)+4(0)+2=1. ✓
TrapUsing the line's point(3,2,1) as the normal gives option C. The point tells you where the line is; the direction tells you which way it goes, and only the direction can be a normal here.
Without doing any work, what is a normal vector to the plane 5x−2y+z=9?
Answer A
In ax+by+cz=d the coefficients **are** the normal: n=⟨a,b,c⟩.
So n=⟨5,−2,1⟩. The constant d locates the plane in space; it says nothing about which way the plane faces.
Any nonzero multiple works too — ⟨−5,2,−1⟩ is equally normal, just pointing the other way.
TrapDragging d into the vector, giving ⟨5,−2,9⟩. The normal has three components and the equation has four numbers; d is the one that is not a direction.
What is the angle between the planes x+y=1 and x+z=1?
Answer A
The angle between planes is the angle between their normals: n1=⟨1,1,0⟩, n2=⟨1,0,1⟩.
cosθ=∣n1∣∣n2∣∣n1⋅n2∣=221=21.
θ=60∘.
TrapReading the missing variable as a zero angle. Neither normal is a multiple of the other, so the planes genuinely cross — a missing variable makes a plane parallel to an axis, not parallel to another plane.
The line r(t)=⟨2t,t,−t⟩ and the plane 3x−y+5z=2. Which is true?
Answer A
Test the direction against the normal: v=⟨2,1,−1⟩, n=⟨3,−1,5⟩, and n⋅v=6−1−5=0.
A zero dot product means the line never climbs away from the plane — so it is parallel to it, or lying in it. Two cases, and the dot product cannot tell them apart.
Settle it with one point. At t=0 the line is at the origin, and 3(0)−0+5(0)=0=2, so the origin is not on the plane.
Parallel and disjoint.
TrapStopping at n⋅v=0 and answering "parallel" without checking a point — a line lying *inside* the plane passes that same test. One substitution separates them.
Count the minus signs among the squared terms: exactly one, and the right side is 1.
One minus ⇒hyperboloid of one sheet — the connected cooling-tower shape.
Check with a trace: setting z=0 gives the ellipse 4x2+9y2=1, a real curve,
so the surface exists at the waist and is not in two pieces.
TrapTwo minus signs would be two sheets; right side 0 would be a cone.
What is the trace of the cone z2=x2+y2 in the plane z=3?
Answer A
Substitute z=3: 9=x2+y2, a circle of radius 3.
The vertical traces are different: setting y=0 gives z2=x2, i.e. the pair of lines z=±x —
which is why the surface is a cone and not a paraboloid.
TrapReading the radius as z² = 9 instead of √9 = 3.
y never appears, so this is a cylinder and the rulings run along the **y-axis**.
The curve being swept is x2−z2=1 in the xz-plane: a hyperbola opening along ±x, in two branches.
So it is a **hyperbolic cylinder** — two curved sheets, both infinite in y.
TrapMistaking it for a hyperboloid. A hyperboloid has all three variables squared; here one is missing entirely, which is the signature of a cylinder no matter what curve is being swept.
Three squared terms, one of them negative — and the right-hand side is **0**, not 1. Zero on the right is the cone case.
Confirm with traces: z=k gives x2+y2=4k2, a circle whose radius grows linearly with ∣k∣ and pinches to a point at k=0.
The odd-one-out variable, z, is the axis.
TrapReading the single negative sign and answering "one sheet". The minus-sign count only decides between the hyperboloids once the right-hand side is 1; 0 overrides it and gives the cone.
Identify the surface 4x2+y2+9z2=36, and give its x-intercepts.
Answer A
Divide by 36: 9x2+36y2+4z2=1.
Three squared terms, none negative, right side 1 — an **ellipsoid**.
Intercepts are the denominators' square roots: x=±3, y=±6, z=±2. Unequal, so it is not a sphere.
TrapReading the intercept off the original coefficient rather than the standard form. In 4x2+⋯=36 the x-intercept is 36/4=3, not 6 — you must divide through first.
Every square of a real number is ≥0, so the left-hand side is never negative.
No real (x,y,z) can satisfy it: the graph is the **empty set**.
Three cases worth holding together: =9 is a sphere, =0 is the single point at the origin, and <0 is nothing at all. The right-hand side alone separates them.
TrapAnswering "a single point". That is the =0 case. A negative right-hand side is strictly emptier, and it is the one that looks most like a typo in an exam and most often gets "corrected" into a sphere.
2413.6 · The same equation in different dimensionsnew
What does the equation x=2 describe in R3?
Answer A
Count the free variables. x is pinned at 2; y and z appear nowhere, so both are free to be anything.
Two free variables sweep out a two-dimensional object: a **plane**, perpendicular to the x-axis and passing through (2,0,0).
The same equation says something different in each dimension — a **point** on the number line, a **vertical line** in the plane, a **plane** in space. The equation never changes; the ambient space does.
This is the idea the whole section runs on: a variable that is absent is a variable that is free, and every free variable is a direction you sweep along.
TrapAnswering "a line" out of habit from two-variable algebra, where x=2 really is a line. The question has to tell you the ambient space, and in this chapter it is always R3 unless it says otherwise.
2513.6 · A paraboloid that is not about the z-axisnew
Identify the surface x2+z2=−y.
Answer A
Two variables are squared (x and z) and one is linear (y) — that is a **paraboloid**, and the linear variable is always the axis. Here the axis is y, not z.
The two squared terms share a sign, so the cross-sections are circles: elliptic (in fact circular), not a saddle.
Which way does it open? Set y=c: x2+z2=−c. That needs −c≥0, so c≤0 — the surface only exists for y≤0 and opens along −y.
At y=0 the trace is x2+z2=0, the single point at the vertex.
TrapAssuming every paraboloid stands on the z-axis because that is how they are always drawn. Read which variable is unsquared — that one is the axis.
For the surface x2+y2=−z, which values of z carry no points at all?
Answer A
Slice at z=k: x2+y2=−k.
The left side is a sum of squares, so it is never negative. A trace exists only when −k≥0, i.e. k≤0.
So there is **nothing above the xy-plane**: the paraboloid opens downward from its vertex at the origin. At k=0 the trace is the single point (0,0,0); below that the traces are circles of radius −k, growing as you descend.
Same reasoning as the two-sheet hyperboloid — ask which slices force a negative sum of squares, and those are the empty ones.
TrapReading the minus sign as "reflected, so it exists everywhere". The minus sign is exactly what makes half of space empty. Test a slice rather than picturing the surface.
In the plane, the curve x2−y2=k changes character with k. Which description is right?
Answer A
k>0: dividing by k puts x2 over a positive number, so the curve meets the x-axis at x=±k and never the y-axis — it opens **left and right**.
k=0: x2−y2=0 factors as (x−y)(x+y)=0, the **two lines** y=x and y=−x. This is the degenerate member of the family.
k<0: the roles swap and the hyperbola opens **up and down**, asymptotic to the same two lines.
This is exactly the family of horizontal traces of the saddle z=x2−y2, which is why the saddle's z=0 trace is a pair of crossed lines.
TrapTreating k=0 as "empty" or "a point". A difference of squares can vanish without either square vanishing — that is what makes the asymptotes part of the family rather than something outside it.
The surface x2+y2=z2 is a circular cone. What does it actually look like?
Answer A
Solve for z: z=±x2+y2. Both signs are real for every (x,y), so the surface has an upper half and a lower half.
Trace at z=k: a circle of radius ∣k∣ — so the circles shrink to a point at k=0 and grow linearly in both directions.
The two **nappes** therefore touch at exactly one place, the origin. The equation z=x2+y2 would be only the top half; the squared form is both.
Contrast the two-sheet hyperboloid, whose pieces are separated by a genuine gap and never touch.
TrapConfusing the cone with the two-sheet hyperboloid. Both come in two pieces, but the cone's pieces meet at a point (=0 on the right) while the hyperboloid's are held apart by an empty band (=1 on the right).
Two squared variables with opposite signs, plus one linear variable.
Along y=0: z=x2, a parabola opening up. Along x=0: z=−y2, a parabola opening down.
Up one way and down the other is exactly a saddle — the hyperbolic paraboloid.
TrapReading “hyperbolic” as “hyperboloid”. A hyperboloid has three squared terms and a constant; a hyperbolic paraboloid has two squared terms and one linear one.
For the surface z=x2−y2, what is the trace in the plane z=0?
Answer A
Set z=0: x2−y2=0, so (x−y)(x+y)=0.
That is y=x or y=−x — a pair of intersecting lines.
For z=k=0 the traces really are hyperbolas, opening along x when k>0 and along y when k<0. z=0 is the level where the hyperbola degenerates into its own asymptotes.
TrapAnswering “hyperbola” because every other horizontal trace is one. The one level that breaks the pattern is the one the question asks about.
One plus and two minuses ⇒ hyperboloid of two sheets, and the axis is the positive variable: z.
Sanity check on the traces: z=k needs 9k2−1≥0, so ∣k∣≥3. There is no surface at all between z=−3 and z=3 — that gap is the two separate sheets.
TrapConfusing this with 4x2+y2−z2=36, which has two pluses and is one sheet. Count the signs, then check whether some slab has no points in it: an empty gap means two sheets.
Group the x terms and factor out the 9: 9(x2−2x)+4y2+36z2=27.
Complete the square inside: 9[(x−1)2−1]+4y2+36z2=27, so 9(x−1)2+4y2+36z2=36.
Divide by 36: 4(x−1)2+9y2+z2=1.
Three squared terms, all positive, right side 1 — an **ellipsoid**, shifted one unit along x.
TrapAdding 1 to the right-hand side instead of 9. The bracket is multiplied by 9, so completing the square inside it adds 9⋅1 to the left. Getting this wrong changes the size but not the type, which is exactly why it survives a sanity check and still loses the mark.
A surface has an **ellipse** for every horizontal trace z=k, and a **hyperbola** for every trace x=k and y=k. What is it?
Answer A
Ellipses at *every* height z=k means the surface is present for all z and closes up horizontally — that rules out the two-sheet hyperboloid, which is empty for small ∣z∣, and the ellipsoid, which stops existing past its poles.
Hyperbolas in the two vertical directions means one squared term carries the opposite sign: a2x2+b2y2−c2z2=1.
That is the **hyperboloid of one sheet** — the cooling-tower shape, connected, with a waist you can walk around.
TrapA cone also gives hyperbolas vertically, but its horizontal traces shrink to a single point at the vertex rather than staying ellipses at every height. "Every k" is the word doing the work in this question.
4x2+9y2−z2=1 is a hyperboloid of one sheet. What does the surface become if the right-hand side is changed to 0?
Answer A
The right-hand side controls the size of the waist. At z=0 the one-sheet hyperboloid gives 4x2+9y2=1, an ellipse.
With 0 on the right, z=0 forces 4x2+9y2=0, whose only real solution is the origin. The waist has pinched shut.
Every other trace is still a hyperbola or an ellipse, so the surface is the **cone** the hyperboloid was hugging all along — the cone is the boundary case between the one-sheet and two-sheet families.
TrapReading "=0" as "empty". Zero on the right is the cone; it is a *negative* right-hand side with all-positive squared terms that gives you nothing at all.
4013.6 · Where a two-sheet hyperboloid is emptynew
For the surface −x2−y2+4z2=36, which values of z carry no points of the surface at all?
Answer A
Fix z=k: −x2−y2=36−4k2, i.e. x2+y2=4k2−36.
A circle needs a non-negative right-hand side, so 4k2≥36, giving k2≥9 and ∣k∣≥3.
For ∣z∣<3 the trace would need a negative radius squared — there is nothing there. That gap between z=−3 and z=3 is what makes it **two sheets**.
TrapTesting z=0, finding nothing, and concluding the surface is empty everywhere. Two-sheet hyperboloids are always empty at the origin; that is the definition, not a contradiction.
In three dimensions, y2+z2=4 is a circular cylinder. Its rulings — the straight lines running along its length — are parallel to which axis?
Answer A
x does not appear in the equation, so x is unconstrained: any x at all satisfies it.
The circle y2+z2=4 therefore repeats at every x, and stacking those copies sweeps the curve along the **x-axis**.
General rule: the missing variable names the axis the rulings follow.
TrapPicking z out of habit, because the cylinders drawn in most textbooks stand upright on the z-axis. Read the equation instead — the axis is whichever letter is absent.
Lessons 5–7 contain exactly one new idea, and every formula on the quiz can be re-derived from it in under a minute. Twenty minutes here and you are not memorising anything.
Step 0 · Two things you already have
You already know both halves of this chapter. They have just never been introduced to each other.
A function takes a number and hands you back a number. f(t)=t2: feed it 3, get 9.
A vector is a list of numbers that means a location or a direction. ⟨1,3,5⟩ is "1 across, 3 back, 5 up".
Glue them: a vector function takes a number and hands you back a list. Feed it a time, get a position.
That is the only new object in Lessons 5–7. Everything else is Calc I applied to it.
Step 1 · A bug, with actual numbers
No notation yet. A bug flies around. At time t it is at
x=t2,y=3t,z=5.
Make a table, because a table is impossible to misunderstand:
t
x=t2
y=3t
z=5
where it is
0
0
0
5
(0,0,5)
1
1
3
5
(1,3,5)
2
4
6
5
(4,6,5)
3
9
9
5
(9,9,5)
Join those dots and you get a curve. The curve is the trail; the function is the flight. Two different bugs can trace the same trail at different speeds, which is why the function carries more information than the picture.
Notice z never moves. The bug stays at height 5 the whole time, so the trail lies flat in a horizontal plane. You just read a geometric fact off a slot — that is most of what 14.1 asks you to do.
Step 2 · Derive the derivative (do not take my word for it)
How fast is the bug going at t=1? You know how to answer this for one function: the derivative is the limit of rise over run. Do the same thing and watch what happens.
Between time t and a moment h later, the bug moves from r(t) to r(t+h). The change is the subtraction:
r(t+h)−r(t).
Subtracting two triples means subtracting each slot — that is just what subtracting vectors is. So the change is
⟨x(t+h)−x(t),y(t+h)−y(t),z(t+h)−z(t)⟩.
Now divide by h. Dividing a vector by a number divides each slot, so each slot becomes its own rise-over-run:
⟨hx(t+h)−x(t),hy(t+h)−y(t),hz(t+h)−z(t)⟩.
Finally let h→0. A point in space is close to another exactly when it is close in all three coordinates, so the limit of a triple is the triple of the limits — and each of those is an ordinary Calc I derivative.
r′(t)=⟨x′(t),y′(t),z′(t)⟩Slot by slot is not a convention. It is forced. There was never an opportunity for the slots to interact.
Which is why the same is true of limits, integrals and continuity: every one of them is built out of subtracting, dividing and taking limits, and all three go slot by slot. That single argument covers most of Lesson 5 and all of Lesson 6.
For the bug: x′=2t, y′=3, z′=0. At t=1 the velocity is ⟨2,3,0⟩.
Step 3 · Why that arrow points along the curve
Worth thirty seconds, because it is the reason the word tangent shows up everywhere.
r(t+h)−r(t) is the arrow from where the bug is now to where it will be. It cuts across the curve like a chord across a circle.
Dividing by a positive number h stretches or shrinks that arrow but cannot rotate it. So the direction survives the division untouched.
Now shrink h. The far endpoint slides back toward the near one, and the chord pivots until it is just grazing the curve.
So r′ points exactly the way the bug is heading, and its length says how fast. Direction and magnitude in one object — that is what "velocity" means.
Step 4 · Speed, and why arc length is not a new formula
The length of r(t+h)−r(t) is (nearly) the distance travelled in those h seconds. Divide by h and you have distance over time.
Speed =∣r′(t)∣. Not a definition handed down — it is distance / time, in the limit.
For the bug: ∣⟨2t,3,0⟩∣=4t2+9, which at t=1 is 13.
And now arc length for free. If you travel at speed ∣r′∣ for a tiny time dt, you cover ∣r′∣dt of ground. Add up every tiny piece:
L=∫abr′(t)dt.
That is distance = rate × time, summed. If you ever blank on it, rebuild it from that sentence in five seconds.
Step 5 · Unit tangent: direction with the size stripped off
Sometimes you want only the heading, with speed thrown away. Dividing any vector by its own length always gives length 1 — same trick as 13.1.
T(t)=∣r′(t)∣r′(t)
Check it at t=1: r′=⟨2,3,0⟩ has length 13, so T=⟨132,133,0⟩, whose length is 134+139=1. Good.
T is a signpost, not a speedometer. If a question asks for T and your answer does not have length 1, you skipped the division.
Step 6 · Why acceleration leans into the bend
This is the one result that looks like it needs intuition and actually needs two lines of algebra.
Suppose the speed never changes — a car going round a roundabout at a steady 30. Then ∣v∣ is constant, so ∣v∣2 is constant, and ∣v∣2=v⋅v.
Differentiate both sides. The left is a constant, so it gives 0. The right uses the product rule for dot products, which is the ordinary product rule:
0=dtd(v⋅v)=v′⋅v+v⋅v′=2a⋅v.
So a⋅v=0: acceleration is perpendicular to velocity whenever speed is constant.
Which makes physical sense. Acceleration along your direction of travel would speed you up or slow you down. If your speed is not changing, all of it must be spent turning — pointing sideways, into the bend.
Circular motion is exactly this case, which is why r=⟨acost,asint⟩ always has a aimed at the centre. Differentiate it twice and you get a=−r — literally pointing back at the origin.
Step 7 · Projectiles are Newton plus the integrals you just built
Purdue's Lesson 7 does projectile motion, and none of it is new. Gravity pulls down at a constant rate and nothing pushes sideways:
a(t)=⟨0,0,−9.8⟩.
Integrating undoes differentiating, slot by slot for the same reason as before. Once:
v(t)=⟨0,0,−9.8t⟩+C,and at t=0 this is C, so C=v(0).
Twice, the same way, giving r(t)=⟨0,0,−4.9t2⟩+tv(0)+r(0).
The constant of integration is a vector — three separate constants — and the initial condition is what pins it down. Every t dies at t=0, so whatever survives isC.
Now every projectile question is just reading the z slot, because only z has gravity in it:
Max height — height stops rising when z′=0. Solve for t, put it back in z.
Time of flight — it lands when z=0.
How far — take that landing time and read the x and y slots, which stayed linear.
Same three questions every time.
Every symbol on the quiz, mapped to the bug
Now the notation is just shorthand for moves you have already made:
Symbol
Out loud
The bug
r(t)
position — where it is
⟨t2,3t,5⟩
r′(t)=v(t)
velocity — heading and how fast
⟨2t,3,0⟩
∣v(t)∣
speed — one number
4t2+9
r′′(t)=a(t)
acceleration — how the heading changes
⟨2,0,0⟩
T(t)
unit tangent — heading only
v/∣v∣
∫rdt
undo a derivative, plus a vector constant
slot by slot
The arrow on top only ever means "this is a list of three, not one". The prime is still a derivative. t is still time. There is no third meaning hiding anywhere.
The one distinction the quiz will punish
Velocity is a vector. Speed is its length. Velocity knows which way you are going; speed is a single number that has forgotten.
So "find the speed at t=π/2" means differentiate, substitute, then take the square root of the sum of the squares. If your answer to a speed question still has angle brackets around it, you stopped one step early. That is the single most common lost mark in 14.3.
The jingle — now that it means something
Slot by slot, then reassemble. Prime once for velocity, twice for acceleration. Speed is the length — take the root or lose the mark. The constant is a vector; t=0 finds it.
You are ready when you can rebuild these, not recall them
Why differentiating goes slot by slot — the subtract, divide, take-the-limit argument
Why r′ points along the curve (the chord that pivots)
Why speed is ∣r′∣, and why arc length is just rate × time summed
Why a⊥v when speed is constant — the v⋅v trick
How to build T from v, and why its length is 1
How to get position from acceleration, and why C has three components
The three projectile questions and which slot answers each
Why a domain question is "where is every slot defined at once"
Anything there you cannot derive out loud in a sentence is exactly what the problems below are for. Start with the official ones.
STEP 01
The material
3 sections
§14.1 · §14.2 · §14.3 — and nothing else. Read them, then do the check-yourself questions inside each one with the answer folded away.
Differentiate every slot. That is the whole rule. The result is tangent to the curve, pointing the way the bead is heading.
Derivative of position is velocity. Derivative of velocity is acceleration. Exactly Calc I — just done to all three coordinates at once.
derivative and unit tangentr′(t)=⟨f′(t),g′(t),h′(t)⟩T(t)=∣r′(t)∣r′(t)
product rules — note which product each one keepsdtd[u⋅v]=u′⋅v+u⋅v′dtd[u×v]=u′×v+u×v′
Trap
The cross-product rule must keep its order: u′×v, never v×u′. Cross products anticommute, so swapping flips the sign of your whole answer.
A fact worth memorising: if ∣r(t)∣ is constant, then r⋅r′=0 — the velocity is perpendicular to the position. That is why anything moving on a sphere has velocity tangent to the sphere, and it's the reason T′⊥T in 14.3.
integral — antidifferentiate each slot, one vector constant∫r(t)dt=⟨∫f,∫g,∫h⟩+C
Check yourself
r(t)=⟨t2,t3⟩. Find r′(1).
Answer
⟨2t,3t2⟩ at t=1 is ⟨2,3⟩.
If ∣r(t)∣ is constant, what is r⋅r′?
Answer
0. Constant length means the velocity is perpendicular to the position — why anything moving on a sphere stays tangent to it.
Quiz 3 covers 14.3 only — velocity, acceleration, circular and projectile motion. The T, N, B frame and the components of acceleration are 14.5, on Quiz 4 and later.
Sit on a roller coaster. T is where your nose points, N is the way you're pressed into your seat, and B is out through your shoulder. All three travel with you.
drag to spin
The frame rides the curve instead of standing still. Acceleration always splits into a T part (changing speed) and an N part (changing direction).
the moving frameT=∣r′∣r′N=∣T′∣T′B=T×N
N is guaranteed perpendicular to T because T has constant length 1 — the fact from 14.2. B is perpendicular to both because that's what a cross product does.
acceleration always splits two waysa=aTT+aNN,aT=dtd∣v∣=∣r′∣r′⋅r′′,aN=κ∣v∣2=∣r′∣∣r′×r′′∣
Read it physically.aT is the gas pedal — it changes how fast you go. aN is the steering wheel — it changes where you're pointed. There is never a B component: you are never accelerated sideways out of your own turn.
Algorithm · decompose acceleration
Compute r′ and r′′.
Speed:∣r′∣.
aT: dot them and divide by the speed, ∣r′∣r′⋅r′′.
aN: cross them and divide by the speed, ∣r′∣∣r′×r′′∣.
Check:aT2+aN2=∣r′′∣2. If that fails, one of them is wrong.
Dot for the gas pedal, cross for the steering wheel. Same two products, one divisor.
Projectile motion is this with a=⟨0,0,−g⟩: integrate once for v (constant of integration = initial velocity), again for r (constant = initial position).
Check yourself
What does aT measure?
Answer
The gas pedal — how fast your speed is changing. It is zero at constant speed, even in a hard turn.
Which product gives aN, dot or cross?
Answer
Cross.aN=∣r′∣∣r′×r′′∣. Dot for the gas pedal, cross for the steering wheel.
STEP 02
The official problems
11 problems
Straight from the department's own study guide, for exactly these lessons — worked, with the answer held back until you commit. Do these first.
TrapUsing t again for the tangent line. The line has its own parameter; reusing t makes you solve on the curve instead of on its tangent.
3
Find the unit tangent vector T(0) for r(t)=⟨cost,sint,t⟩.
no calculator, same as the exam
AnswerT(0)=⟨0,21,21⟩answer recomputed · sympy
r′(t)=⟨−sint,cost,1⟩, so r′(0)=⟨0,1,1⟩.
∣r′(0)∣=0+1+1=2.
T(0)=⟨0,21,21⟩.
TrapSkipping the normalisation. T is a unit vector by definition; r′ alone is not one here.
4
Find the speed of a particle with position r(t)=⟨et,e−t,2t⟩ at t=0.
no calculator, same as the exam
AnswerSpeed =2answer recomputed · sympy
r′(t)=⟨et,−e−t,2⟩; at t=0 this is ⟨1,−1,2⟩.
Speed =1+1+2=4=2.
TrapDifferentiating e−t to e−t. The chain rule puts a minus in front, though here it squares away — the sign matters for direction, not for speed.
5
If r(t)=⟨t,t2,t3⟩, compute r′(t)×r′′(t).
no calculator, same as the exam
Answer⟨6t2,−6t,2⟩answer recomputed · sympy
r′=⟨1,2t,3t2⟩ and r′′=⟨0,2,6t⟩.
i: (2t)(6t)−(3t2)(2)=12t2−6t2=6t2.
j: −[(1)(6t)−(3t2)(0)]=−6t.
k: (1)(2)−(2t)(0)=2. So r′×r′′=⟨6t2,−6t,2⟩.
TrapThis exact vector is the numerator of the curvature formula κ=∣r′∣3∣r′×r′′∣ — recognise it and Lesson 8 gets shorter.
6
A particle travels along r(t)=⟨3cost,4sint,0⟩. Find the speed at t=2π.
no calculator, same as the exam
AnswerSpeed =3answer recomputed · sympy
r′(t)=⟨−3sint,4cost,0⟩.
At t=2π: ⟨−3,0,0⟩.
Speed =3.
TrapAssuming an ellipse is traversed at constant speed. It is not — the speed here is 9sin2t+16cos2t, which runs between 3 and 4. Only the circle has constant speed under this parametrisation.
The official set has a hole here.
Lesson 7 covers projectile motion under gravity, and uniform circular motion — and not one of the official problems above tests it.
It is still examinable, so the questions below are not optional this time — they are the only place those topics get rehearsed.
STEP 03
More practice
16 questions
Mine, on those sections only. Untimed, no calculator — the quiz doesn't allow one either.
0/16 answered
114.1 · Space curves
Which curve lies on the cylinder y2+z2=9 and advances at a constant rate along the cylinder\'s axis?
Answer B
The cylinder y2+z2=9 constrains y and z; its axis is the x-axis, so x must be the free coordinate.
Option B: y2+z2=9cos2t+9sin2t=9 ✓ and x=t advances at constant rate ✓.
A and E put the circle in x,y (wrong cylinder). D fails the identity. F has radius t, not 3.
TrapReading the cylinder’s axis off the wrong pair of variables.
Which function traces the circle of radius 3 centred on the z-axis in the plane z=5, counterclockwise as seen from above?
Answer A
Staying in the plane z=5 forces the third component to be the constant 5.
Radius 3 centred on the axis gives x=3cost, y=3sint; then x2+y2=9. ✓
(+cos,+sin) runs counterclockwise viewed from +z looking down.
TrapOption B has z=t — that is a helix, not a plane curve. Option D reverses the orientation. Option E shifts the centre to (3,3) and shrinks the radius to 1.
Find parametric equations for the tangent line to r(t)=⟨t,t2,t3⟩ at the point (2,4,8).
Answer A
First find the parameter value at that point. x=t=2, and indeed t2=4, t3=8. So t=2.
r′(t)=⟨1,2t,3t2⟩, so r′(2)=⟨1,4,12⟩ — the direction.
Line through (2,4,8) with direction ⟨1,4,12⟩: x=2+s,y=4+4s,z=8+12s.
TrapOption C uses r′ with t never substituted, effectively ⟨1,2,3⟩. Option B swaps the point and the direction. Option F evaluates 3t2 as 3⋅2=6 instead of 3⋅4=12.
r(t) is differentiable and ∣r(t)∣ is constant. What is r(t)⋅r′(t)?
Answer A
Use the product rule for the dot product: dtd[r⋅r]=r′⋅r+r⋅r′=2r⋅r′.
But r⋅r=∣r∣2, which is a constant, so its derivative is 0.
Hence 2r⋅r′=0, so r⋅r′=0.
Geometrically: a curve of constant distance from the origin lies on a sphere, and its velocity must be tangent to that sphere — perpendicular to the radius.
TrapThis is the fact behind uniform circular motion: constant speed means the acceleration is perpendicular to the velocity, which is the same statement applied to v instead of r.
A projectile leaves the origin with velocity ⟨20,0,20⟩ m/s under acceleration ⟨0,0,−10⟩ m/s². How far horizontally does it travel before returning to z=0?
Answer C
Integrate twice: z(t)=20t−5t2, x(t)=20t.
z=0⇒5t(4−t)=0⇒t=4 s (rejecting t=0, the launch).
Horizontal distance x(4)=20(4)=80 m.
TrapUsing t = 2 s (the time to the top) instead of the full flight time.
For r(t)=⟨4cos3t,4sin3t⟩, which statement is true?
Answer A
v=⟨−12sin3t,12cos3t⟩, so ∣v∣=12sin23t+cos23t=12 — constant.
a=⟨−36cos3t,−36sin3t⟩=−9r(t).
A negative multiple of the position vector points straight back at the origin. (This is centripetal acceleration.)
Also v⋅a=0, so the acceleration is perpendicular to the velocity, never parallel to it.
Trap“Constant speed” is not “no acceleration” — option F. The direction is changing the whole time, and changing direction is acceleration. Note also that the speed is 12, not the radius 4: the 3 inside the trig functions multiplies it.
A particle has acceleration a(t)=⟨0,2,6t⟩, with v(0)=⟨1,0,0⟩ and r(0)=⟨0,1,2⟩. Find r(1).
Answer A
Integrate once: v(t)=⟨0,2t,3t2⟩+C1. Since v(0)=⟨1,0,0⟩, C1=⟨1,0,0⟩, so v(t)=⟨1,2t,3t2⟩.
Integrate again: r(t)=⟨t,t2,t3⟩+C2. Since r(0)=⟨0,1,2⟩, C2=⟨0,1,2⟩.
r(t)=⟨t,t2+1,t3+2⟩, so r(1)=⟨1,2,3⟩.
TrapTwo integrations mean two vector constants, and each is fixed by its own initial condition. Losing C1 drops the x motion entirely; losing C2 gives ⟨1,1,1⟩.
A ball is launched from a height of 15 m with initial velocity ⟨20,0,10⟩ m/s, under acceleration ⟨0,0,−10⟩ m/s2. When does it hit the ground?
Answer A
Only the vertical component matters. Integrating twice, z(t)=15+10t−5t2 (initial height, plus initial vertical velocity, minus 21gt2).
Set z=0: 5t2−10t−15=0, so t2−2t−3=0, that is (t−3)(t+1)=0.
The positive root is t=3 s.
TrapDropping the initial height and solving 10t−5t2=0 gives t=2 — option B. The launch height is what keeps it in the air longer. And the coefficient is 21g=5, not g=10.
Lesson 8 runs 14.4 and 14.5 as far as Theorem 14.5. The later part of 14.5 is not on this quiz.
A curve is hard to measure; a straight stick is easy. Chop the curve into a million tiny straight sticks, measure each, add them up.
Distance = speed × time, repeated for every instant. That's all the integral is doing.
arc length = speed integrated over timeL=∫ab∣r′(t)∣dt=∫ab(dtdx)2+(dtdy)2+(dtdz)2dt
Algorithm · reparametrize by arc length
Compute the speed∣r′(t)∣.
Build the arc-length functions(t)=∫0t∣r′(u)∣du — how far you've walked by time t.
Invert it: solve s=s(t) for t in terms of s.
Substitute back: r(t(s)). Now the parameter is distance, so the speed is exactly 1.
Only clean when the speed is constant or very simple — which is exactly when this gets assigned.
Curvature asks how small is the circle you're hugging. A tight little circle means a sharp turn; a huge circle looks almost straight.
A straight line hugs a circle of infinite radius, so its curvature is 0. A circle of radius R has curvature 1/R everywhere.
three formulas for the same κ — pick by what you're givenκ=dsdT=∣r′(t)∣∣T′(t)∣=∣r′(t)∣3∣r′(t)×r′′(t)∣
Use the third one
Almost always. It needs only r′ and r′′ — no messy normalising.
Use the second one
When the problem already handed you T(t).
For a plane curve y=f(x)
κ=(1+(f′)2)3/2∣f′′∣
Sanity values: a straight line has κ=0. A circle of radius a has κ=1/a everywhere. The radius of curvature is ρ=1/κ — the radius of that best-fitting circle.
Check yourself
Speed of ⟨3cost,3sint,4t⟩?
Answer
9+16=5, constant. So length over [0,2π] is just 10π.
Curvature of a circle of radius 2?
Answer
κ=1/2. Curvature is one over the radius of the hugging circle.
Slice the hill with flat sheets of glass at heights 1, 2, 3, 4. Each slice leaves a loop. Draw those loops flat on paper and you've made a topographic map. Nothing was lost.
drag to spin
Slide the glass up and down. Every height gives one loop, and the loops printed on the floor are the topographic map. Loops crowded = steep.
level curves (2 inputs) and level surfaces (3 inputs)f(x,y)=kF(x,y,z)=k
Reading a contour map: loops crowded together = steep. Loops far apart = gentle. Closed rings shrinking inward = a peak or a pit. Contours that cross in an X = a saddle. You will use every one of these again in 15.7.
Algorithm · find and sketch the domain
List what can break: denominators (≠ 0), even roots (≥ 0), logs (> 0), arcsin/arccos (input in [−1,1]).
Write each as an inequality in x and y.
Sketch each boundary curve and shade the region satisfying all of them at once.
Mark boundaries solid or dashed — solid if the inequality is ≥, dashed if strict.
Check yourself
What do the level curves of f(x,y)=x+y look like?
Answer
Parallel straight linesx+y=k, evenly spaced — a constant-slope ramp, not a hill.
On a number line you can only come in from the left or the right. On a plane you can spiral, zig-zag, arrive on any curve. Every road must agree or the limit doesn't exist.
To disprove a limit: find two roads that disagree. To prove one: squeeze it, or switch to polar and show the answer doesn't depend on θ as r → 0.
the two-path test — this disproves, it never provesif (x,y)→(a,b)limf along path 1=lim along path 2,then the limit does not exist
Algorithm · show a limit does NOT exist
Try the axes first. Set y=0, take x→0. Then x=0, take y→0. Cheapest possible test.
Try lines through the origin:y=mx. If the answer still contains m, you're done — different slopes give different limits.
Try matching powers. If the numerator has degree p in one variable and the denominator degree q, try y=xp/q or y=x2. This is what catches the sneaky ones that survive every straight line.
State it cleanly: "along y=0 the limit is A; along y=x it is B; A=B, so the limit does not exist."
Algorithm · show a limit DOES exist
Try direct substitution. If f is built from polynomials, roots, exponentials, sines and cosines and nothing blows up — you're done, it's continuous.
Switch to polar:x=rcosθ, y=rsinθ, and let r→0+. If everything left over is bounded and the answer does not depend on θ, that's the limit.
Or squeeze: find 0≤∣f−L∣≤g(x,y) with g→0. Useful bounds: x2+y2x≤1, ∣sinu∣≤∣u∣, x2≤x2+y2.
Checking a hundred paths proves nothing. Polar or squeeze is the only way to actually close it.
Worked example — polar closing the case
(x,y)→(0,0)limx2+y2x3+y3. Every straight line gives 0, but that isn't a proof.
polar
r2r3(cos3θ+sin3θ)=r(cos3θ+sin3θ)
bound
∣cos3θ+sin3θ∣≤2, so the whole thing is ≤2r
conclude
as r→0 this →0 regardless of θ. The limit is 0.
Check yourself
Along y=0 the limit is 3; along y=x it is 5. What do you conclude?
Answer
The limit does not exist. Two roads disagree, and that is a complete proof.
You checked 100 straight lines and all gave 0. Does the limit exist?
Answer
No — you have proved nothing. Paths can only disprove. To prove, switch to polar and show the answer does not depend on θ, or squeeze it.
STEP 02
The official problems
15 problems
Straight from the department's own study guide, for exactly these lessons — worked, with the answer held back until you commit. Do these first.
Find the arc length of r(t)=⟨3cost,3sint,4t⟩ for 0≤t≤2π.
no calculator, same as the exam
AnswerL=10πanswer recomputed · sympy
r′(t)=⟨−3sint,3cost,4⟩.
∣r′∣=9sin2t+9cos2t+16=9+16=5 — constant, the helix is traversed at constant speed.
L=∫02π5dt=10π.
TrapLosing the Pythagorean identity. 9sin2+9cos2=9; if your integrand still contains t, the collapse was missed.
2
Find the curvature κ(t) of r(t)=⟨2cost,2sint,t⟩.
no calculator, same as the exam
Answerκ=52answer recomputed · sympy
For the helix ⟨acost,asint,bt⟩, κ=a2+b2a — constant.
Here a=2, b=1: κ=4+12=52.
By the cross-product formula: r′×r′′=⟨2sint,−2cost,4⟩, whose length is 4+16=25, and ∣r′∣3=(5)3=55. Ratio =5525=52.
TrapUsing κ=a1 — that is the circle of radius a. Climbing in z straightens the curve, so the helix is always less curved than its shadow.
3
Find the curvature of the plane curve y=x3 at the point (0,0).
no calculator, same as the exam
Answerκ=0answer recomputed · sympy
For a graph, κ(x)=[1+(f′(x))2]3/2∣f′′(x)∣.
f′=3x2, f′′=6x; at x=0 both the numerator and f′ vanish.
κ(0)=13/20=0.
TrapAssuming a curve that visibly bends must have κ>0 somewhere along it. y=x3 is *flat to second order* at the origin — it is an inflection point, and curvature is exactly zero there.
4
Find the arc length of r(t)=⟨3t,4t,0⟩ for 1≤t≤5.
no calculator, same as the exam
AnswerL=20answer recomputed · sympy
r′=⟨3,4,0⟩, a constant vector: this is a straight line.
∣r′∣=5.
L=∫155dt=5⋅4=20.
TrapIntegrating from 0 to 5. The interval is [1,5], which has length 4, not 5.
5
A particle moves along r(t)=⟨5sint,3sint,4cost⟩. Find the distance traveled from t=0 to t=π.
Evaluate (x,y)→(0,0)limx4+y2x3y, or show it does not exist.
no calculator, same as the exam
Official answerlimit does not existchecked · official answer wrong
Correct answer0recomputed · sympy
Try paths first. Along y=0: 0. Along x=0: 0. Along y=x2: x4+x4x3⋅x2=2x→0. Every path gives 0, which is a signal to stop hunting and start proving.
Squeeze. By AM–GM, x4+y2≥2x2∣y∣, so for (x,y)=(0,0):
x4+y2x3y≤2x2∣y∣∣x∣3∣y∣=2∣x∣.
2∣x∣→0, so the limit exists and equals 0.
TrapA finite pile of paths never proves a limit exists — but it also never proves it fails. When every path you try agrees, switch to a bound.
The official answer is wrongThe official answer says the limit does not exist. It does: the AM–GM squeeze above bounds the expression by 2∣x∣. (Compare x4+y2x2y, which genuinely fails along y=x2 — one power of x away.)
2
Evaluate (x,y)→(0,0)limx2+y23x2y, or show it does not exist.
no calculator, same as the exam
Official answerlimit does not existchecked · official answer wrong
Correct answer0recomputed · sympy
Every path gives 0: along y=mx, x2(1+m2)3x2(mx)=1+m23mx→0.
Squeeze, and this one is immediate: x2≤x2+y2, so x2+y2x2≤1 and
x2+y23x2y≤3∣y∣.
3∣y∣→0 as (x,y)→(0,0), so the limit exists and equals 0.
TrapDegree counting is the fast check: the numerator is degree 3, the denominator degree 2. Numerator degree strictly larger ⇒ the quotient is squeezed to 0. Only equal degrees can produce a path-dependent limit.
The official answer is wrongThe official answer says the limit does not exist. It does: it is 0, by the one-line bound x2+y23x2y≤3∣y∣. The classic DNE example is x2+y2xy — numerator and denominator both degree 2.
3
Find the value of k such that (x,y)→(0,0)limx2+y2x4−2k(x2+y2)−y4=6.
no calculator, same as the exam
Answerk=−3answer recomputed · sympy
Split it: x2+y2x4−y4−x2+y22k(x2+y2).
First piece factors: x2+y2(x2−y2)(x2+y2)=x2−y2→0.
Second piece is exactly 2k, for every (x,y)=(0,0).
Limit =−2k. Set −2k=6⇒k=−3.
TrapLosing the sign. The term enters as −2k(x2+y2), so the limit is −2k, and k=3 would give −6.
4
Evaluate (x,y)→(0,0)limx2+y2sin(x2+y2).
no calculator, same as the exam
Answer1answer recomputed · sympy
Substitute u=x2+y2. As (x,y)→(0,0), u→0+.
The expression is usinu.
limu→0usinu=1.
TrapThe substitution is legitimate here precisely because x and y only ever appear in the combination x2+y2. It is not legitimate in general.
5
Evaluate (x,y)→(0,0)limx2+y2x3−y3, or show it does not exist.
no calculator, same as the exam
Answer0answer recomputed · sympy
Polar coordinates: x=rcosθ, y=rsinθ, and (x,y)→(0,0) is exactly r→0+.
r2r3(cos3θ−sin3θ)=r(cos3θ−sin3θ).
∣cos3θ−sin3θ∣≤2, so the whole thing is ≤2r→0uniformly in θ.
The limit is 0.
TrapPolar only settles it when the θ factor is bounded and the r factor goes to zero on its own. If θ survives in the limit, polar has told you it does not exist.
STEP 03
More practice
32 questions
Mine, on those sections only. Untimed, no calculator — the quiz doesn't allow one either.
0/32 answered
114.4 · Arc length
Find the length of the helix r(t)=⟨3cost,3sint,4t⟩ for 0≤t≤2π.
Answer B
r′=⟨−3sint,3cost,4⟩, so
∣r′∣=9sin2t+9cos2t+16=25=5 — constant speed.
L=∫02π5dt=10π. Distance = speed × time, exactly.
TrapForgetting to square the 4, or integrating |r| instead of |r′|.
214.4 · Length when the speed is a multiple of tnew
Find the length of r(t)=⟨9t2,7,12t2−4⟩ for 0≤t≤2.
Answer A
r′(t)=⟨18t,0,24t⟩ — the constant 7 differentiates away, and so does the −4.
∣r′(t)∣=324t2+576t2=900t=30t (positive on [0,2], so no absolute value survives).
L=∫0230tdt=15t202=60.
TrapForgetting the 21 from ∫tdt and answering 60⋅2=120, or quoting 30⋅2=60 by luck while meaning the speed at t=2. A speed growing linearly averages to half its final value.
Find the length of r(t)=⟨6sint−6tcost,6cost+6tsint⟩ for 0≤t≤π.
Answer A
Product rule: x′=6cost−6cost+6tsint=6tsint — the two cos terms cancel. Likewise y′=−6sint+6sint+6tcost=6tcost.
∣r′(t)∣=36t2sin2t+36t2cos2t=6t on t≥0.
L=∫0π6tdt=3t20π=3π2.
TrapNot differentiating 6tcost as a product. The curve is built so the non-t terms cancel and the speed collapses to 6t — if your speed still has a loose sin or cos in it, you dropped a term.
Does r(t)=⟨23cost,21cost,sint⟩ use arc length as its parameter?
Answer A
r′(t)=⟨−23sint,−21sint,cost⟩.
∣r′(t)∣2=43sin2t+41sin2t+cos2t=sin2t+cos2t=1.
The test is exactly ∣r′∣=1 for all t — and it holds, so tis arc length, measured from t=0.
TrapAssuming the two coefficients must combine to something larger than one. They are each squared first, and 43+41=1 — the constants are chosen precisely to make the speed unit.
TrapPutting each component's own coefficient in its denominator (option B). Every component is divided by the same constant 7 — the length of the direction vector, not of one entry.
For r(t)=⟨cost,sint,t⟩, find the binormal B at t=0.
Answer A
r′=⟨−sint,cost,1⟩, ∣r′∣=2, so
T(0)=21⟨0,1,1⟩.
T′=21⟨−cost,−sint,0⟩⇒T′(0)=21⟨−1,0,0⟩,
so N(0)=⟨−1,0,0⟩.
B=T×N=21⟨0,1,1⟩×⟨−1,0,0⟩=⟨0,−21,21⟩.
TrapComputing N × T instead of T × N — that flips every sign.
r′(t)=⟨4,3cost,−3sint⟩ and ∣r′∣=16+9cos2t+9sin2t=25=5.
T=51⟨4,3cost,−3sint⟩=⟨54,53cost,−53sint⟩.
T′=⟨0,−53sint,−53cost⟩, so ∣T′∣=53.
κ=∣r′∣∣T′∣=53/5=253.
TrapWriting the third component of T as cost. dtd(3cost)=−3sint — the derivative of cosine is minus sine, and this single slip is the most common lost mark in the whole section.
TrapUsing the y-coordinate in the formula, or forgetting the 3/2 power and answering 21. The exponent is 3/2, not 1 — it comes from ∣v∣3 with ∣v∣=1+(y′)2.
For f(x,y)=x2+4y2, the level curves f=k with k>0 are
Answer B
x2+4y2=k rearranges to kx2+k/4y2=1 — an ellipse
with semi-axes k and k/2, so it is twice as wide as it is tall.
They would be circles only if the two coefficients matched.
TrapSeeing "sum of squares" and answering circles. The unequal coefficients stretch it.
A level curve is what you get by setting z to a constant z0: y−x2=z0, that is y=x2+z0.
Every level is the same parabola y=x2, translated up by z0.
Unlike a paraboloid's circles, these never close and never shrink — the contour map is a family of identical nested parabolas, evenly spaced, so the surface climbs at a steady rate in the y-direction.
TrapAssuming “parabolic surface, therefore circular contours” by analogy with z=x2+y2. Only a sum of two squares gives circles; y appears here to the first power, and that is what makes the levels parabolas rather than closed loops.
A polynomial in x and y is built from products and sums only.
There is no denominator to vanish, no even root to go negative, and no logarithm to hit zero.
Every point of the plane works: the domain is R2.
TrapReading the odd powers or the sign of the output as a constraint. Products and sums are fine everywhere; only division, even roots and logs ever restrict a domain. The range is a separate question entirely.
An even root needs its radicand non-negative: 50−2x2−2y2≥0.
Divide by 2: 25−x2−y2≥0, that is x2+y2≤25.
A closed disc of radius 5 — closed, because equality is allowed and 0=0 is perfectly well defined.
TrapForgetting to divide by the 2 and answering x2+y2≤50, or making the inequality strict. Strict belongs to logarithms and denominators, not to square roots. And the radius is 25=5, not 25.
cos accepts every real input, so the outer function imposes nothing.
The only restriction lives in the fraction: x−5=0.
Domain: everything except the vertical line x=5.
TrapRestricting the argument to [−1,1] (option D). That is the range of cosine, not its domain — it is the constraint you would impose for arccos, not cos. Note also that the numerator vanishing at y=−3 is harmless: it just makes f=cos0=1.
For f(x,y)=3x+6y−18, which set of intercepts identifies its graph, and what are the domain and range?
Answer A
x-intercept: set y=z=0, so 3x=18 and x=6.
y-intercept: set x=z=0, so 6y=18 and y=3. Note the coefficients and the intercepts swap size: the bigger coefficient gives the smaller intercept.
z-intercept: f(0,0)=−18.
A plane is defined for every (x,y) and, being non-constant and linear, it attains every real value: domain R2, range R.
TrapReading the intercepts straight off the coefficients as (3,0,0) and (0,6,0). You divide by the coefficient, you do not copy it. Option C makes the other classic error: a plane has no lowest point, so the range cannot be bounded below.
Describe the graph of f(x,y)=y2−4, and give its domain and range.
Answer A
x does not appear, so the height never changes as you walk in the x-direction: every cross-section x=k is the same parabola z=y2−4.
That sweeps out a cylinder — in the technical sense of a curve dragged along a line, not a round tube. It is dragged along the axis of the missing variable, here x.
y is unrestricted, so the domain is R2. Since y2≥0, z=y2−4≥−4, and every such value occurs: range z≥−4.
TrapCalling it a paraboloid, or dragging it along the wrong axis (option C). A paraboloid curves in both directions (z=x2+y2−4); this one is flat along x, which is exactly what the missing variable means.
For F(x,y)=36−x2−y2, give the graph, domain and range.
Answer A
Squaring: z2=36−x2−y2, so x2+y2+z2=36 — the sphere of radius 6.
But z=⋅≥0, so only the upper half is the graph. A function has one height per point; a whole sphere fails that.
Domain: 36−x2−y2≥0, the closed disc x2+y2≤36 of radius 6.
Range: the radicand runs over [0,36], so z=⋅ runs over [0,6].
TrapAnswering “sphere”, or letting the range run to 36 instead of 6 (option E) — you must take the square root of the largest radicand. The square root is non-negative by definition, and it is also what makes the domain a disc rather than the whole plane.
A surface has one peak and, beside it along the x-axis, one pit, flattening to zero far away. Which contour plot is it?
Answer A
A local maximum and a local minimum both show up as a nest of closed level curves shrinking toward the extreme point. The picture alone does not distinguish them — only the labelled values do.
The two features are separated along x, so the two nests sit left and right in the xy-plane.
Option C is the same picture rotated: that is two features separated along y. Concentric circles (B) means a single summit; crossing asymptotes (D) is a saddle; parallel lines (E) means one variable is absent.
TrapTwo things at once. First the axis: separation along x on the surface means separation along x on the map. Second, do not expect a pit to look different from a peak — it does not. Read the numbers on the contours, not the shape.
Match each function to its surface: (i) sin(x+y), (ii) x2+y21, (iii) x+y1, (iv) e−x2−y2.
Answer A
Read the singularities and bounds first, not the shape.
(i) sin(x+y) is bounded in [−1,1] and depends only on x+y, so it is constant along every line x+y=c — parallel straight ridges and troughs.
(ii) x2+y21→+∞ as (x,y)→(0,0), is radially symmetric, and is positive everywhere: a spike at a single point.
(iii) x+y1 blows up on the whole liney=−x and changes sign across it — two sheets, +∞ on one side, −∞ on the other.
(iv) e−x2−y2 has maximum 1 at the origin and decays to 0 in every direction — a single smooth bump, never negative.
TrapConfusing (ii) and (iii). Both blow up, but x2+y21 fails at a single point and keeps one sign, while x+y1 fails along an entire line and flips sign. Ask “where is this undefined, and does it change sign there?” and the picture follows.
Consider (x,y)→(0,0)limx4+y2x2y. Which statement is correct?
Answer D
Along y=0: the quotient is 0. Along x=0: also 0. Every straight line y=mx also gives 0.
But along the parabola y=x2:
x4+x4x2⋅x2=2x4x4=21.
Two roads disagree, so the limit does not exist.
TrapTesting only straight lines. Matching the powers (here y ~ x²) is what exposes it.
Consider (x,y)→(0,0)limx2+y2x2−y2. Which is correct?
Answer D
Along y=0: x2x2=1. Along x=0: y2−y2=−1.
Two roads disagree, so the limit does not exist.
In polar the quotient is cos2θ−sin2θ=cos2θ — it depends on θ and never settles as r→0.
TrapConcluding "0/0 so it is 0". The axes alone already settle this one.
Along the x-axis (y=0): the expression is x20=0 for all x=0, so the limit along that path is 0.
Along the line y=x: x2+3x2x2=41 for all x=0.
Two paths, two values ⇒ no limit. In general, along y=mx the value is 1+3m2m, so every slope gives a different answer.
TrapConcluding “the limit is 0” from the axes alone. Checking the two axes is never enough — it is the slanted lines that break this one. Option E is also wrong as stated: if the value depends on direction, the limit does not exist at all.
Along y=mx: x6+m2x2x3⋅mx=x4+m2mx2→m20=0 — every straight line gives 0.
Along the curvey=x3: x6+x6x3⋅x3=21, constant.
So the limit does not exist. Matching the shape of the denominator — here x6 against y2, so y∼x3 — is what tells you which curve to try.
TrapTreating “all lines agree” as proof (option B). It never is. Agreeing on infinitely many paths still leaves infinitely many untried; only a squeeze, polar coordinates or continuity can prove a limit exists.
As (x,y)→(0,0), ∣x∣→0, so the expression is squeezed to 0.
(Polar check: r2r3cosθsin2θ=rcosθsin2θ→0 as r→0, uniformly in θ.)
TrapAssuming that because the previous one failed, this must too. The degrees decide: the numerator here is degree 3 and the denominator degree 2, so in polar form a spare factor of r survives and kills it. When numerator and denominator have equal degree, suspect it fails.
A quotient of polynomials is continuous wherever the denominator is nonzero.
y−x3=0 exactly on the cubic y=x3.
So f is continuous on {(x,y):y=x3} — the whole plane with one curve removed.
TrapRemoving the zeros of the numerator too (option C). On the line x=2y the function equals 0, which is a perfectly good value, not a failure. Only the denominator can break continuity here.
Refuse to turn. Walk due east only and record your height — that's a plain 1-D curve, and its slope is ∂f/∂x. Turn 90° and walk due north for ∂f/∂y.
drag to spin
Move the wall and watch the slice curve change shape. A partial derivative is the slope of that one curve — nothing more exotic than Calc I.
definition — the other variable never movesfx(a,b)=h→0limhf(a+h,b)−f(a,b)fy(a,b)=h→0limhf(a,b+h)−f(a,b)
Algorithm · compute a partial derivative
Cover up the other variables with your hand. Pretend they are numbers like 7.
Differentiate normally in the one remaining variable — product rule, chain rule, all of Calc I still applies.
Uncover. The frozen variables come back exactly as they were.
Then plug in the point, if a specific point was asked for. Never plug in first.
∂x∂(x2y3)=2xy3 — the y3 was a constant, so it just rode along.
second and mixed partialsfxx=∂x2∂2ffxy=(fx)y=∂y∂x∂2ffyx=(fy)x
Clairaut's theorem
If the mixed partials are continuous near a point, fxy=fyx — the order you differentiate in doesn't matter. This is free error-checking: compute both and they must match.
Notation trap
fxy means "x first, then y", but ∂y∂x∂2f means the same thing read right-to-left. Subscripts run left-to-right; Leibniz notation runs inside-out. Clairaut makes them equal anyway, but exam wording depends on you knowing which is which.
Check yourself
Compute ∂x∂(x2y3).
Answer
2xy3. The y3 was a constant, so it rode along untouched.
Is fxy the same as fyx?
Answer
Yes, whenever the mixed partials are continuous (Clairaut). Compute both — it is free error-checking.
Draw a tree. To find how fast z changes when t changes, multiply along each branch from t up to z, then add the branches together. Every chain rule in the course is this one picture.
Why add? Because t wiggling changes both x and y, and each of those independently pushes z. The effects stack.
Case 1 · one independent variabledtdz=∂x∂zdtdx+∂y∂zdtdy
Case 2 · two independent variables∂s∂z=∂x∂z∂s∂x+∂y∂z∂s∂y∂t∂z=∂x∂z∂t∂x+∂y∂z∂t∂y
Algorithm · any chain rule, from the tree
Draw the tree. Output on top. Below it, every variable the output directly depends on. Below those, the variables they depend on. Keep going to the bottom.
Label every edge with the derivative of the upper node with respect to the lower one. Use ∂ when the node has more than one child, d when it has exactly one.
Trace every path from the top down to the variable you're differentiating with respect to.
Multiply along each path, then add the paths.
Number of terms in your answer = number of distinct routes down the tree. Count the routes first, then check your answer has that many terms.
implicit differentiation, from F(x,y)=0 and F(x,y,z)=0dxdy=−FyFx∂x∂z=−FzFx,∂y∂z=−FzFy
Where the minus sign comes from. Staying on the level set means F never changes, so dF=0. Expand: Fxdx+Fydy=0. Solve for dy/dx and the minus falls out. Nothing memorised.
Algorithm · implicit partials without the formula
Differentiate the whole equation with respect to the variable you care about.
Treat the dependent variable as a function — every time you differentiate z, write ∂z/∂x.
Collect all the ∂z/∂x terms on one side.
Solve for ∂z/∂x.
Check yourself
z depends on x and y; both depend on t. How many terms in dz/dt?
Answer
Two — one per route down the tree. Count the routes before you compute, then check your answer has that many terms.
Put the two partials side by side into one arrow. That arrow points straight uphill, it is always perpendicular to the contour lines, and its length is how steep that climb is.
Perpendicular is not a coincidence: walking along a contour keeps your height constant, so the height-change in that direction is zero — and zero change means zero dot product with ∇f.
the gradient∇f=⟨fx,fy⟩or∇f=⟨fx,fy,fz⟩
You don't have to walk straight uphill. Pick any direction u — your slope is the shadow of the gradient onto your direction. The dot product from 13.3, doing real work.
drag to spin
Spin u around. Line it up with ∇f for the steepest climb; turn it 90° and you're walking the contour — dead flat. The green bar is the shadow.
directional derivative — u must be a UNIT vectorDuf=∇f⋅u=∣∇f∣cosθ
Algorithm · directional derivative
Compute ∇f symbolically, then evaluate at the point.
Normalise the direction. If you were given a vector, divide by its length. If you were given an angle θ from the x-axis, use u=⟨cosθ,sinθ⟩. If you were given "toward the point Q", use PQ/∣PQ∣.
Dot them.Duf=∇f⋅u.
Skipping step 2 is the most common lost point in Unit 2. A direction vector of length 3 inflates your answer by exactly 3×.
Question
Answer
Why
Fastest increase
direction ∇f, rate ∣∇f∣
cosθ is maximal at θ=0
Fastest decrease
direction −∇f, rate −∣∇f∣
cosθ=−1
No change at all
any u⊥∇f
cos90°=0 — you're walking the contour
Level set direction
tangent to the contour
same thing as the row above
The gradient is perpendicular to its own level set
Walking along a contour keeps your height constant, so the change in that direction is zero — and zero change means a zero dot product with ∇f. That single fact gives you tangent planes to any implicitly defined surface for free.
tangent plane & normal line to the surface F(x,y,z)=kFx(P)(x−x0)+Fy(P)(y−y0)+Fz(P)(z−z0)=0r(t)=r0+t∇F(P)
Algorithm · tangent plane to any surface
Write it as a level surface.z=f(x,y) becomes F(x,y,z)=f(x,y)−z=0.
Compute ∇F and evaluate at the point. That is the normal vector.
Use the point-normal plane equation from 13.5.
For the normal line, use the same ∇F as the direction vector.
Note step 1 reproduces ⟨fx,fy,−1⟩ from 15.6 — the two methods are the same method.
Check yourself
∇f is perpendicular to what?
Answer
Its own level curve. Walking along a contour keeps height constant, so the change is zero — and zero change means zero dot product.
What is the maximum rate of increase at a point?
Answer
∣∇f∣, the length of the gradient — a number, attained in the direction ∇f.
You walk along a contour line. What is Duf?
Answer
0. Dead flat — that is what a contour means.
STEP 02
The official problems
16 problems
Straight from the department's own study guide, for exactly these lessons — worked, with the answer held back until you commit. Do these first.
Let f(x,y,z)=cos(xy)+ey2z+ln(z3). Compute fyxz(0,1,1).
no calculator, same as the exam
Answerfyxz(0,1,1)=0answer recomputed · sympy
Read the order right: fyxz means differentiate by y, then x, then z.
fy=−xsin(xy)+2yzey2z.
fyx=−sin(xy)−xycos(xy) — the ey2z term has no x and dies here.
fyxz=0: what is left has no z in it at all. Evaluating at (0,1,1) is not even needed.
TrapGrinding through the ln(z3) term. It has no x and no y, so it is annihilated by the very first derivative.
2
If f(x,y)=x3y2+sin(xy), find fxy(0,0).
no calculator, same as the exam
Answerfxy(0,0)=1answer recomputed · sympy
fx=3x2y2+ycos(xy).
fxy=6x2y+cos(xy)−xysin(xy) (product rule on ycos(xy)).
At (0,0): 0+cos0−0=1.
TrapEvaluating at (0,0)before differentiating. f(x,0)=0 for all x, so that route reports 0 — you must differentiate symbolically first, then substitute.
3
If f(x,y)=ex2y, find fx(1,2).
no calculator, same as the exam
Answerfx(1,2)=4e2answer recomputed · sympy
Hold y constant: fx=ex2y⋅2xy.
At (1,2): e2⋅2⋅1⋅2=4e2.
TrapDifferentiating the exponent as if y were a variable too. In fx, y is a constant — that is the entire content of 'partial'.
4
Let f(x,y)=x2ln(y). Find fxy and verify Clairaut's theorem by also computing fyx.
no calculator, same as the exam
Answerfxy=fyx=y2xanswer recomputed · sympy
fx=2xlny, so fxy=y2x.
The other order: fy=yx2, so fyx=y2x.
They agree — as Clairaut guarantees, since both mixed partials are continuous on y>0.
TrapBelieving Clairaut always applies. It needs continuity of the mixed partials; the standard counterexample x2+y2xy(x2−y2) has fxy(0,0)=fyx(0,0).
5
If f(x,y)=arctan(xy), find fx+fy at (1,1).
no calculator, same as the exam
Answerfx+fy=0 at (1,1)answer recomputed · sympy
For f=arctanxy: fx=x2+y2−y and fy=x2+y2x.
At (1,1): fx=−21, fy=21.
fx+fy=0.
TrapDropping the inner derivative ∂x∂xy=−x2y. The 1+(y/x)21 factor then has to be cleared of its fractions to reach x2+y2.
Or substitute first: w=st(s+t)+st(s−t)=2s2t, ws=4st=0 at (1,0).
∂s∂w=0.
TrapMissing that y multiplies two branches (w=xy+yz feeds y into both terms), so ys appears once with coefficient x+z.
4
If sin(xy)=e2z, use implicit differentiation to find ∂x∂z at (1,2π,0).
no calculator, same as the exam
Answer∂x∂z=0answer recomputed · sympy
F=sin(xy)−e2z, so Fx=ycos(xy) and Fz=−2e2z.
At (1,2π,0): Fx=2πcos2π=0 and Fz=−2.
∂x∂z=−−20=0.
Trapcos2π=0 is what kills this one. Check the point before doing any algebra — sometimes it collapses the whole expression.
5
If z=f(x,y) with x=rcosθ and y=rsinθ, express ∂r∂z in terms of fx and fy.
no calculator, same as the exam
Answer∂r∂z=fxcosθ+fysinθprose answer · not machine-checkable
∂r∂z=fx∂r∂x+fy∂r∂y.
∂r∂x=cosθ and ∂r∂y=sinθ (with θ held fixed).
∂r∂z=fxcosθ+fysinθ.
Worth noticing: that is ∇f⋅⟨cosθ,sinθ⟩ — the directional derivative straight outward. Lesson 13 in disguise.
TrapWriting ∂θ∂z by mistake; that one is −fxrsinθ+fyrcosθ, and the factors of r are the tell.
6
If F(x,y,z)=x2+y2+z2−3xyz=0 defines z implicitly, find ∂x∂z at (1,1,1), or explain why it does not exist.
no calculator, same as the exam
Answer∂x∂z(1,1,1)=−1answer recomputed · sympy
F=x2+y2+z2−3xyz, so Fx=2x−3yz and Fz=2z−3xy.
At (1,1,1): Fx=2−3=−1 and Fz=2−3=−1.
Fz=−1=0, so the implicit function theorem applies and z really is a function of (x,y) near this point.
∂x∂z=−−1−1=−1.
TrapSkipping the Fz=0 check. When Fz=0 the surface has a vertical tangent plane there and ∂x∂z does not exist — which is exactly what the 'or explain why not' is fishing for.
Find the directional derivative of f(x,y)=x2y−3y2 at the point (2,1) in the direction of the vector v=⟨3,4⟩.
no calculator, same as the exam
AnswerDu^f(2,1)=54answer recomputed · sympy
∇f=⟨2xy,x2−6y⟩; at (2,1) that is ⟨4,4−6⟩=⟨4,−2⟩.
Normalise the direction: ∣⟨3,4⟩∣=5, so u^=⟨53,54⟩.
Du^f=⟨4,−2⟩⋅⟨53,54⟩=512−8=54.
TrapDotting with ⟨3,4⟩ unnormalised, which gives 4 — five times too big. The directional derivative is only a rate if the direction has length 1.
2
Find the maximum rate of change of f(x,y,z)=xyz at the point (4,1,4), and state the direction in which it occurs.
no calculator, same as the exam
AnswerMax rate of change: 21, in direction ⟨2,4,1⟩answer recomputed · sympy
f=xyz, so fx=yz, fy=2yzxz, fz=2yzxy.
At (4,1,4): yz=2, so ∇f=⟨2,416,44⟩=⟨2,4,1⟩.
Maximum rate =∣∇f∣=4+16+1=21.
It occurs in the direction of ∇f itself, ⟨2,4,1⟩.
TrapGiving the maximum rate as a vector or the direction as a number. The rate is ∣∇f∣; the direction is ∇f (or its unit version).
3
At (1,2), in what direction does f(x,y)=xe−y decrease most rapidly? What is that rate of decrease?
no calculator, same as the exam
AnswerDirection: ⟨−1,1⟩; rate of decrease: 2e−2answer recomputed · sympy
∇f=⟨e−y,−xe−y⟩; at (1,2): ⟨e−2,−e−2⟩.
Steepest decrease is −∇f=⟨−e−2,e−2⟩, i.e. the direction ⟨−1,1⟩.
The rate of decrease is ∣∇f∣=e−22.
TrapReporting the rate as negative. 'Rate of decrease' is already signed by the word; the magnitude 2e−2 is the number wanted.
4
Find the directional derivative of f(x,y,z)=x2+2y2+3z2 at (1,1,1) in the direction of ⟨1,−1,0⟩.
no calculator, same as the exam
AnswerDu^f(1,1,1)=−2answer recomputed · sympy
∇f=⟨2x,4y,6z⟩; at (1,1,1): ⟨2,4,6⟩.
u^=2⟨1,−1,0⟩.
Du^f=22−4+0=2−2=−2.
TrapRationalising carelessly: 2−2=−2, not −22.
5
Suppose ∇f(2,3)=⟨4,−3⟩. Find Duf(2,3) where u is the unit vector in the direction of ⟨3,4⟩.
no calculator, same as the exam
AnswerDu^f(2,3)=0answer recomputed · sympy
u^=⟨53,54⟩.
Du^f=⟨4,−3⟩⋅⟨53,54⟩=512−12=0.
Zero is meaningful: ⟨3,4⟩ is perpendicular to ∇f, so this direction runs along the level curve through (2,3).
TrapTreating 0 as a computational accident. A zero directional derivative always means you are moving tangent to the level curve.
STEP 03
More practice
5 questions
Mine, on those sections only. Untimed, no calculator — the quiz doesn't allow one either.
0/5 answered
115.3 · Mixed partials
For f(x,y)=x2y3+x, find fxy at (1,2).
Answer B
fx=2xy3+1, then fxy=∂y∂(2xy3+1)=6xy2.
At (1,2): 6(1)(4)=24.
Clairaut check the other way round: fy=3x2y2⇒fyx=6xy2 ✓ same answer — free error-checking.
TrapDifferentiating twice in the same variable. f_xy means x first, then y.
Let z=x2y with x=s+t and y=st. Find ∂s∂z at (s,t)=(1,2).
Answer D
∂s∂z=zxxs+zyys=(2xy)(1)+(x2)(t).
At s=1,t=2: x=3,y=2, so =2(3)(2)(1)+9(2)=12+18=30.
Two routes down the tree, so two terms — count the routes before you compute.
TrapDropping one branch. Number of terms = number of paths from z down to s.
Mon Oct 5 · 8:00 pm · Lessons 1–16 · 13.1–15.7 (all of Ch. 13–14, Ch. 15 through max/min)
Midterm 1
Lessons 1–16, one hour, multiple choice, no calculator. Stops in the middle of Chapter 15 — Lagrange is NOT on it.
Worth knowing. Lessons 14–16 (tangent planes, max/min) are never quizzed. They arrive between Quiz 5 and this exam, so this is the only place they are ever tested before the final.
The department publishes a study guide for this exam, and it is the source of truth for coverage. Every one of its practice problems is worked out, with the answer held back until you commit to an attempt, at Exam 1, worked. Start there; use this tab for the sections no quiz ever tested.
Only the sections no quiz ever covers are shown here. Everything from the earlier quizzes is on this exam too — work those tabs as well.
Official problems for these lessons
The department's own questions for exactly these lessons are worked below, before anything of mine.
The Earth is round, but your backyard is flat. Zoom in far enough on any smooth surface and it becomes a plane — and using that plane instead of the real surface is what linear approximation means.
drag to spin
The plane is fixed — it never moves. Zooming makes the surface come to it. That is exactly what linear approximation buys you.
tangent plane to z=f(x,y) at (a,b)z=f(a,b)+fx(a,b)(x−a)+fy(a,b)(y−b)
Read it as a sentence: start at the known height, then add "slope × how far you walked" for each direction. Its normal vector is ⟨fx,fy,−1⟩.
differentials — the same statement, about small changesdz=fxdx+fydyΔz≈dz
Algorithm · estimate a value or an error
Pick a nearby easy point(a,b) where f, fx, fy are trivial to evaluate.
Compute the three numbersf(a,b), fx(a,b), fy(a,b).
Set dx=x−a and dy=y−b — the small steps from easy point to real point.
Add it up:f(x,y)≈f(a,b)+fxdx+fydy.
For maximum error, use ∣dx∣,∣dy∣ at their worst and take absolute values of the coefficients: ∣dz∣≤∣fx∣∣dx∣+∣fy∣∣dy∣.
Differentiable is stronger than "both partials exist." A function can have both partials at a point and still not have a tangent plane there. In practice, if fx and fy are continuous near the point, f is differentiable — that's the test you'll actually use.
Check yourself
What is the normal vector to the surface z=f(x,y)?
Answer
⟨fx,fy,−1⟩. (Or write it as F=f−z=0 and take ∇F — same thing.)
Write the differential dz.
Answer
dz=fxdx+fydy. Nudge x, nudge y, add both effects.
If ∇f=0 you're on level ground. But level ground happens at a summit, at the bottom of a bowl, and at a mountain pass — where it goes up one way and down the other.
Contours that form closed rings around a point mean max or min. Contours that cross in an X mean saddle — the classic tell on an exam.
second derivative testD=fxxfyy−(fxy)2evaluated at the critical point
If
Then
Picture
D>0 and fxx<0
local maximum
both slices bend down — a summit
D>0 and fxx>0
local minimum
both slices bend up — a bowl
D<0
saddle point
slices bend opposite ways — a pass
D=0
test fails
look at the contours or test values directly
Algorithm · classify all critical points
Set both partials to zero:fx=0 and fy=0. Solve the system.
Find every solution. Factor rather than divide — dividing by x silently throws away the critical points with x=0.
Compute fxx, fyy, fxy symbolically, once.
Evaluate D at each point and read the table.
Report the function valuef at each max/min, not just the location.
Step 2 is where solutions get lost. If you ever divide by a variable, handle the "= 0" case separately.
Algorithm · absolute extrema on a closed bounded region
Interior: find critical points strictly inside D, and list their f values.
Boundary: break the boundary into pieces. Parametrize each piece so it becomes a one-variable problem.
Optimise each piece with ordinary Calc I: derivative zero, plus the endpoints of that piece.
Corners: evaluate f at every corner of the region.
Compare the whole list. Biggest value is the absolute max, smallest is the absolute min. No second derivative test needed anywhere.
The extreme value theorem guarantees both exist when the region is closed and bounded — so if your list has no winner, your list is incomplete.
Trap
Forgetting the boundary. On a closed region the absolute maximum very often sits on the edge, where ∇f is not zero at all — so it never shows up as a critical point. Steps 2–4 are not optional.
Check yourself
D<0 at a critical point. What is it?
Answer
A saddle — and fxx is irrelevant here.
You found every interior critical point on a closed region. Are you done?
Answer
No. Check every boundary piece and every corner. On a closed region the maximum usually lives on the edge, where ∇f is not zero at all.
STEP 02
The official problems
69 problems
Straight from the department's own study guide, for exactly these lessons — worked, with the answer held back until you commit. Do these first.
The projection of u onto v is projvu=∣v∣2u⋅vv — a vector *along* v.
u⋅v=(2)(1)+(−1)(2)+(3)(2)=2−2+6=6.
∣v∣2=1+4+4=9.
projvu=96⟨1,2,2⟩=32⟨1,2,2⟩=⟨32,34,34⟩.
TrapDividing by ∣v∣ instead of ∣v∣2. That gives the scalar projection compvu=36=2, which is a number, not a vector. Read which one the question wants.
2
Find the area of the triangle with vertices P(1,0,0), Q(0,2,0), and R(0,0,3).
no calculator, same as the exam
Answer27answer recomputed · sympy
Two edge vectors from P: PQ=⟨−1,2,0⟩ and PR=⟨−1,0,3⟩.
PQ×PR=⟨2⋅3−0⋅0,0⋅(−1)−(−1)⋅3,0−(−2)⟩=⟨6,3,2⟩.
∣⟨6,3,2⟩∣=36+9+4=7 — that is the parallelogram area.
A triangle is half of it: area=27.
TrapForgetting the 21. The cross product's length is the parallelogram, and 7 is sitting right there looking like an answer.
3
If u=⟨3,−1,2⟩ and v=⟨1,4,−1⟩, find u×v.
no calculator, same as the exam
Answer⟨−7,5,13⟩answer recomputed · sympy
u×v=i31j−14k2−1.
i: (−1)(−1)−(2)(4)=1−8=−7.
j: −[(3)(−1)−(2)(1)]=−(−5)=5 — note the sign flip on the middle term.
k: (3)(4)−(−1)(1)=12+1=13. So u×v=⟨−7,5,13⟩.
TrapThe minus on the j component. Dropping it gives ⟨−7,−5,13⟩, and the sanity check catches it instantly: u⋅(u×v) must be 0.
4
For what value of c are the vectors ⟨2,c,−1⟩ and ⟨3,−6,c⟩ orthogonal?
no calculator, same as the exam
Answerc=76answer recomputed · sympy
Orthogonal means the dot product vanishes.
⟨2,c,−1⟩⋅⟨3,−6,c⟩=6−6c−c=6−7c.
6−7c=0⇒c=76.
TrapMissing that c appears in both vectors, so the last term (−1)(c) also carries a c. Treating it as a constant gives c=1.
5
Find the angle between u=⟨1,1,0⟩ and v=⟨0,1,1⟩.
no calculator, same as the exam
Answerθ=3πanswer recomputed · sympy
cosθ=∣u∣∣v∣u⋅v.
u⋅v=0+1+0=1, and ∣u∣=∣v∣=2.
cosθ=21⇒θ=3π.
TrapAnswering 6π: cos−121=3π, not 6π. 6π is where the *sine* is 21.
The line through (1,3,−1) and (3,1,2) intersects the plane x+y+z=6 at what point?
no calculator, same as the exam
Answer(3,1,2)answer recomputed · sympy
Direction: v=⟨3−1,1−3,2−(−1)⟩=⟨2,−2,3⟩.
Line: r(t)=⟨1+2t,3−2t,−1+3t⟩.
Substitute into x+y+z=6: (1+2t)+(3−2t)+(−1+3t)=3+3t=6⇒t=1.
r(1)=⟨3,1,2⟩ — the second given point happens to lie on the plane.
TrapSign errors in the direction vector. Notice the x and y parameters cancel, so the whole equation collapses to 3+3t; if yours does not collapse, recheck v.
2
Find the equation of the plane containing (2,1,−1) and parallel to the vectors ⟨1,0,3⟩ and ⟨2,1,0⟩.
no calculator, same as the exam
Answer−3x+6y+z+1=0answer recomputed · sympy
Two vectors parallel to the plane give the normal through a cross product.
Point-normal form at (2,1,−1): −3(x−2)+6(y−1)+1(z+1)=0.
Expand: −3x+6+6y−6+z+1=0⇒−3x+6y+z+1=0.
Trap'Parallel to the vectors' does not mean the vectors are normals. They lie *in* the plane; their cross product is the normal.
3
Find the line of intersection between the planes x+y=2 and y+z=3.
no calculator, same as the exam
Answerx=2+t,y=−t,z=3+tanswer recomputed · sympy
Two equations, three unknowns — one free parameter, which is the line.
Let y=−t. Then x+y=2 gives x=2+t, and y+z=3 gives z=3+t.
x=2+t,y=−t,z=3+t.
Check the direction against n1×n2=⟨1,1,0⟩×⟨0,1,1⟩=⟨1,−1,1⟩ — same direction, as it must be.
TrapAny parametrisation is correct; yours may look nothing like this one. Verify by plugging back into both plane equations rather than comparing to a printed answer.
4
Find the angle between the planes x+y+z=1 and x−y=0.
no calculator, same as the exam
Answerθ=2π (planes are perpendicular)answer recomputed · sympy
The angle between planes is the angle between normals: n1=⟨1,1,1⟩, n2=⟨1,−1,0⟩.
n1⋅n2=1−1+0=0.
Zero dot product ⇒θ=2π: the planes are perpendicular.
TrapReading coefficients off a plane written as x−y=0 and forgetting the missing z is a coefficient of 0, not of 1.
5
For what value of a do the lines r1(t)=⟨1+t,2t,3+t⟩ and r2(s)=⟨3−s,s+1,2s+a⟩ intersect?
no calculator, same as the exam
Answera=2answer recomputed · sympy
Intersection needs a single point, so use different parameters t and s and solve all three components.
x: 1+t=3−s⇒t=2−s. y: 2t=s+1.
Substitute: 2(2−s)=s+1⇒4−2s=s+1⇒s=1, hence t=1.
z: 3+t=2s+a⇒3+1=2+a⇒a=2.
TrapUsing t for both lines. Two lines can cross at a point they reach at different times; forcing one parameter finds only the cases where they arrive together, and reports skew for lines that do meet.
′restored
Find the distance from the point (1,−2,3) to the plane 2x−y+2z=4.
no calculator, same as the exam
Official answer35checked · official answer wrong
Correct answer2recomputed · sympy
D=a2+b2+c2∣ax0+by0+cz0−d∣ with n=⟨2,−1,2⟩.
Numerator: ∣2(1)−(−2)+2(3)−4∣=∣2+2+6−4∣=6.
Denominator: 4+1+4=3.
D=36=2.
TrapForgetting to move d to the same side. The formula measures ax+by+cz−d, so a plane written as 2x−y+2z=4 contributes −4, not +4.
The official answer is wrongThe commented-out source gives 35. The numerator is ∣2+2+6−4∣=6, not 5, so the distance is 36=2. This may be exactly why the problem was commented out.
Commented out on the official page, so it does not appear there. The skill is still on the syllabus, so it is restored here.
AnswerHyperboloid of two sheets (axis along y)prose answer · not machine-checkable
Write it with the positive term first: 4y2−x2−9z2=1.
One positive square, two negative, equals +1 — that is the two-sheet signature.
Trace at y=0: −x2−9z2=1, impossible. Nothing at the origin, so it cannot be one connected sheet.
Hyperboloid of two sheets, axis along y — the axis is always the variable whose square is positive.
TrapOne sheet vs two sheets is decided by the count of minus signs, not by which variable is isolated. One minus = one sheet (connected); two minuses = two sheets.
2
Identify the surface defined by x2+4x+y2−2y+z2=−4. (Hint: complete the square.)
no calculator, same as the exam
AnswerSphere centered at (−2,1,0) with radius 1answer recomputed · sympy
Complete the square in x and y: x2+4x=(x+2)2−4 and y2−2y=(y−1)2−1.
(x+2)2−4+(y−1)2−1+z2=−4.
(x+2)2+(y−1)2+z2=1.
Sphere, centre (−2,1,0), radius 1.
TrapMoving the completion constants the wrong way. The −4 and −1 come out on the left, so they get added to the right: −4+4+1=1.
3
Identify the surface: z=4x2+9y2.
no calculator, same as the exam
AnswerElliptic paraboloid opening upwardprose answer · not machine-checkable
One variable (z) appears to the first power, the other two are squared with the same sign.
That is the paraboloid family. Both coefficients positive ⇒ opens upward.
Traces z=k>0 are ellipses 4x2+9y2=k; traces x=0 and y=0 are upward parabolas.
Elliptic paraboloid opening upward.
TrapCalling it a cone. A cone has z2, not z; the first power is what makes it a paraboloid.
4
Describe the trace of 4x2+9y2−z2=1 in the plane z=2.
TrapStopping at 4x2+9y2=5. A trace is named by its standard form, and the denominators change once you divide through.
5
Identify the surface: x2+4y2=9z2.
no calculator, same as the exam
AnswerElliptic cone with axis along the z-axisprose answer · not machine-checkable
Rewrite as x2+4y2−9z2=0 — equal to zero, not 1.
Zero is the cone signature. Traces z=k=0 are ellipses that grow linearly with ∣k∣; the trace z=0 is the single point at the origin (the vertex).
Elliptic cone with axis along the z-axis.
TrapReading it as a hyperboloid. Same sign pattern — the right-hand side is the entire difference: =1 one sheet, =−1 two sheets, =0 the cone between them.
TrapUsing t again for the tangent line. The line has its own parameter; reusing t makes you solve on the curve instead of on its tangent.
3
Find the unit tangent vector T(0) for r(t)=⟨cost,sint,t⟩.
no calculator, same as the exam
AnswerT(0)=⟨0,21,21⟩answer recomputed · sympy
r′(t)=⟨−sint,cost,1⟩, so r′(0)=⟨0,1,1⟩.
∣r′(0)∣=0+1+1=2.
T(0)=⟨0,21,21⟩.
TrapSkipping the normalisation. T is a unit vector by definition; r′ alone is not one here.
4
Find the speed of a particle with position r(t)=⟨et,e−t,2t⟩ at t=0.
no calculator, same as the exam
AnswerSpeed =2answer recomputed · sympy
r′(t)=⟨et,−e−t,2⟩; at t=0 this is ⟨1,−1,2⟩.
Speed =1+1+2=4=2.
TrapDifferentiating e−t to e−t. The chain rule puts a minus in front, though here it squares away — the sign matters for direction, not for speed.
5
If r(t)=⟨t,t2,t3⟩, compute r′(t)×r′′(t).
no calculator, same as the exam
Answer⟨6t2,−6t,2⟩answer recomputed · sympy
r′=⟨1,2t,3t2⟩ and r′′=⟨0,2,6t⟩.
i: (2t)(6t)−(3t2)(2)=12t2−6t2=6t2.
j: −[(1)(6t)−(3t2)(0)]=−6t.
k: (1)(2)−(2t)(0)=2. So r′×r′′=⟨6t2,−6t,2⟩.
TrapThis exact vector is the numerator of the curvature formula κ=∣r′∣3∣r′×r′′∣ — recognise it and Lesson 8 gets shorter.
6
A particle travels along r(t)=⟨3cost,4sint,0⟩. Find the speed at t=2π.
no calculator, same as the exam
AnswerSpeed =3answer recomputed · sympy
r′(t)=⟨−3sint,4cost,0⟩.
At t=2π: ⟨−3,0,0⟩.
Speed =3.
TrapAssuming an ellipse is traversed at constant speed. It is not — the speed here is 9sin2t+16cos2t, which runs between 3 and 4. Only the circle has constant speed under this parametrisation.
Find the arc length of r(t)=⟨3cost,3sint,4t⟩ for 0≤t≤2π.
no calculator, same as the exam
AnswerL=10πanswer recomputed · sympy
r′(t)=⟨−3sint,3cost,4⟩.
∣r′∣=9sin2t+9cos2t+16=9+16=5 — constant, the helix is traversed at constant speed.
L=∫02π5dt=10π.
TrapLosing the Pythagorean identity. 9sin2+9cos2=9; if your integrand still contains t, the collapse was missed.
2
Find the curvature κ(t) of r(t)=⟨2cost,2sint,t⟩.
no calculator, same as the exam
Answerκ=52answer recomputed · sympy
For the helix ⟨acost,asint,bt⟩, κ=a2+b2a — constant.
Here a=2, b=1: κ=4+12=52.
By the cross-product formula: r′×r′′=⟨2sint,−2cost,4⟩, whose length is 4+16=25, and ∣r′∣3=(5)3=55. Ratio =5525=52.
TrapUsing κ=a1 — that is the circle of radius a. Climbing in z straightens the curve, so the helix is always less curved than its shadow.
3
Find the curvature of the plane curve y=x3 at the point (0,0).
no calculator, same as the exam
Answerκ=0answer recomputed · sympy
For a graph, κ(x)=[1+(f′(x))2]3/2∣f′′(x)∣.
f′=3x2, f′′=6x; at x=0 both the numerator and f′ vanish.
κ(0)=13/20=0.
TrapAssuming a curve that visibly bends must have κ>0 somewhere along it. y=x3 is *flat to second order* at the origin — it is an inflection point, and curvature is exactly zero there.
4
Find the arc length of r(t)=⟨3t,4t,0⟩ for 1≤t≤5.
no calculator, same as the exam
AnswerL=20answer recomputed · sympy
r′=⟨3,4,0⟩, a constant vector: this is a straight line.
∣r′∣=5.
L=∫155dt=5⋅4=20.
TrapIntegrating from 0 to 5. The interval is [1,5], which has length 4, not 5.
5
A particle moves along r(t)=⟨5sint,3sint,4cost⟩. Find the distance traveled from t=0 to t=π.
Evaluate (x,y)→(0,0)limx4+y2x3y, or show it does not exist.
no calculator, same as the exam
Official answerlimit does not existchecked · official answer wrong
Correct answer0recomputed · sympy
Try paths first. Along y=0: 0. Along x=0: 0. Along y=x2: x4+x4x3⋅x2=2x→0. Every path gives 0, which is a signal to stop hunting and start proving.
Squeeze. By AM–GM, x4+y2≥2x2∣y∣, so for (x,y)=(0,0):
x4+y2x3y≤2x2∣y∣∣x∣3∣y∣=2∣x∣.
2∣x∣→0, so the limit exists and equals 0.
TrapA finite pile of paths never proves a limit exists — but it also never proves it fails. When every path you try agrees, switch to a bound.
The official answer is wrongThe official answer says the limit does not exist. It does: the AM–GM squeeze above bounds the expression by 2∣x∣. (Compare x4+y2x2y, which genuinely fails along y=x2 — one power of x away.)
2
Evaluate (x,y)→(0,0)limx2+y23x2y, or show it does not exist.
no calculator, same as the exam
Official answerlimit does not existchecked · official answer wrong
Correct answer0recomputed · sympy
Every path gives 0: along y=mx, x2(1+m2)3x2(mx)=1+m23mx→0.
Squeeze, and this one is immediate: x2≤x2+y2, so x2+y2x2≤1 and
x2+y23x2y≤3∣y∣.
3∣y∣→0 as (x,y)→(0,0), so the limit exists and equals 0.
TrapDegree counting is the fast check: the numerator is degree 3, the denominator degree 2. Numerator degree strictly larger ⇒ the quotient is squeezed to 0. Only equal degrees can produce a path-dependent limit.
The official answer is wrongThe official answer says the limit does not exist. It does: it is 0, by the one-line bound x2+y23x2y≤3∣y∣. The classic DNE example is x2+y2xy — numerator and denominator both degree 2.
3
Find the value of k such that (x,y)→(0,0)limx2+y2x4−2k(x2+y2)−y4=6.
no calculator, same as the exam
Answerk=−3answer recomputed · sympy
Split it: x2+y2x4−y4−x2+y22k(x2+y2).
First piece factors: x2+y2(x2−y2)(x2+y2)=x2−y2→0.
Second piece is exactly 2k, for every (x,y)=(0,0).
Limit =−2k. Set −2k=6⇒k=−3.
TrapLosing the sign. The term enters as −2k(x2+y2), so the limit is −2k, and k=3 would give −6.
4
Evaluate (x,y)→(0,0)limx2+y2sin(x2+y2).
no calculator, same as the exam
Answer1answer recomputed · sympy
Substitute u=x2+y2. As (x,y)→(0,0), u→0+.
The expression is usinu.
limu→0usinu=1.
TrapThe substitution is legitimate here precisely because x and y only ever appear in the combination x2+y2. It is not legitimate in general.
5
Evaluate (x,y)→(0,0)limx2+y2x3−y3, or show it does not exist.
no calculator, same as the exam
Answer0answer recomputed · sympy
Polar coordinates: x=rcosθ, y=rsinθ, and (x,y)→(0,0) is exactly r→0+.
r2r3(cos3θ−sin3θ)=r(cos3θ−sin3θ).
∣cos3θ−sin3θ∣≤2, so the whole thing is ≤2r→0uniformly in θ.
The limit is 0.
TrapPolar only settles it when the θ factor is bounded and the r factor goes to zero on its own. If θ survives in the limit, polar has told you it does not exist.
Let f(x,y,z)=cos(xy)+ey2z+ln(z3). Compute fyxz(0,1,1).
no calculator, same as the exam
Answerfyxz(0,1,1)=0answer recomputed · sympy
Read the order right: fyxz means differentiate by y, then x, then z.
fy=−xsin(xy)+2yzey2z.
fyx=−sin(xy)−xycos(xy) — the ey2z term has no x and dies here.
fyxz=0: what is left has no z in it at all. Evaluating at (0,1,1) is not even needed.
TrapGrinding through the ln(z3) term. It has no x and no y, so it is annihilated by the very first derivative.
2
If f(x,y)=x3y2+sin(xy), find fxy(0,0).
no calculator, same as the exam
Answerfxy(0,0)=1answer recomputed · sympy
fx=3x2y2+ycos(xy).
fxy=6x2y+cos(xy)−xysin(xy) (product rule on ycos(xy)).
At (0,0): 0+cos0−0=1.
TrapEvaluating at (0,0)before differentiating. f(x,0)=0 for all x, so that route reports 0 — you must differentiate symbolically first, then substitute.
3
If f(x,y)=ex2y, find fx(1,2).
no calculator, same as the exam
Answerfx(1,2)=4e2answer recomputed · sympy
Hold y constant: fx=ex2y⋅2xy.
At (1,2): e2⋅2⋅1⋅2=4e2.
TrapDifferentiating the exponent as if y were a variable too. In fx, y is a constant — that is the entire content of 'partial'.
4
Let f(x,y)=x2ln(y). Find fxy and verify Clairaut's theorem by also computing fyx.
no calculator, same as the exam
Answerfxy=fyx=y2xanswer recomputed · sympy
fx=2xlny, so fxy=y2x.
The other order: fy=yx2, so fyx=y2x.
They agree — as Clairaut guarantees, since both mixed partials are continuous on y>0.
TrapBelieving Clairaut always applies. It needs continuity of the mixed partials; the standard counterexample x2+y2xy(x2−y2) has fxy(0,0)=fyx(0,0).
5
If f(x,y)=arctan(xy), find fx+fy at (1,1).
no calculator, same as the exam
Answerfx+fy=0 at (1,1)answer recomputed · sympy
For f=arctanxy: fx=x2+y2−y and fy=x2+y2x.
At (1,1): fx=−21, fy=21.
fx+fy=0.
TrapDropping the inner derivative ∂x∂xy=−x2y. The 1+(y/x)21 factor then has to be cleared of its fractions to reach x2+y2.
Or substitute first: w=st(s+t)+st(s−t)=2s2t, ws=4st=0 at (1,0).
∂s∂w=0.
TrapMissing that y multiplies two branches (w=xy+yz feeds y into both terms), so ys appears once with coefficient x+z.
4
If sin(xy)=e2z, use implicit differentiation to find ∂x∂z at (1,2π,0).
no calculator, same as the exam
Answer∂x∂z=0answer recomputed · sympy
F=sin(xy)−e2z, so Fx=ycos(xy) and Fz=−2e2z.
At (1,2π,0): Fx=2πcos2π=0 and Fz=−2.
∂x∂z=−−20=0.
Trapcos2π=0 is what kills this one. Check the point before doing any algebra — sometimes it collapses the whole expression.
5
If z=f(x,y) with x=rcosθ and y=rsinθ, express ∂r∂z in terms of fx and fy.
no calculator, same as the exam
Answer∂r∂z=fxcosθ+fysinθprose answer · not machine-checkable
∂r∂z=fx∂r∂x+fy∂r∂y.
∂r∂x=cosθ and ∂r∂y=sinθ (with θ held fixed).
∂r∂z=fxcosθ+fysinθ.
Worth noticing: that is ∇f⋅⟨cosθ,sinθ⟩ — the directional derivative straight outward. Lesson 13 in disguise.
TrapWriting ∂θ∂z by mistake; that one is −fxrsinθ+fyrcosθ, and the factors of r are the tell.
6
If F(x,y,z)=x2+y2+z2−3xyz=0 defines z implicitly, find ∂x∂z at (1,1,1), or explain why it does not exist.
no calculator, same as the exam
Answer∂x∂z(1,1,1)=−1answer recomputed · sympy
F=x2+y2+z2−3xyz, so Fx=2x−3yz and Fz=2z−3xy.
At (1,1,1): Fx=2−3=−1 and Fz=2−3=−1.
Fz=−1=0, so the implicit function theorem applies and z really is a function of (x,y) near this point.
∂x∂z=−−1−1=−1.
TrapSkipping the Fz=0 check. When Fz=0 the surface has a vertical tangent plane there and ∂x∂z does not exist — which is exactly what the 'or explain why not' is fishing for.
TrapDividing 2xy=0 by x. That silently discards the x=0 branch — and here it is a whole critical point.
2
Classify the critical point of f(x,y)=x2+xy+y2+3x−3y+4 using the second derivative test.
no calculator, same as the exam
AnswerLocal minimum at (−3,3)answer recomputed · sympy
fx=2x+y+3=0 and fy=x+2y−3=0.
Solve: y=−3−2x, then x+2(−3−2x)−3=0⇒−3x−9=0⇒x=−3, y=3.
fxx=2, fyy=2, fxy=1, so D=4−1=3>0.
D>0 and fxx=2>0: local minimum at (−3,3).
TrapSquaring fxy is required: D=fxxfyy−(fxy)2. Using fxy unsquared gives 3 here too, by coincidence — the habit will fail you elsewhere.
3
Let M and m be the absolute maximum and minimum values of f(x,y)=x2−2x+y2+4 on the disk x2+y2≤4. Find M+m.
no calculator, same as the exam
AnswerM+m=15answer recomputed · sympy
Interior: fx=2x−2=0⇒x=1, fy=2y=0⇒y=0. Inside the disk, and f(1,0)=1−2+0+4=3.
Boundary x2+y2=4: substitute y2=4−x2, so f=x2−2x+(4−x2)+4=8−2x with x∈[−2,2].
That is linear and decreasing: max 12 at x=−2, min 4 at x=2.
Compare everything: M=12, m=3, so M+m=15.
TrapIgnoring the interior critical point because the boundary already gave a max and a min. The absolute minimum here is *inside*, at (1,0).
4
Find and classify all critical points of f(x,y)=xy−x3−y2.
no calculator, same as the exam
AnswerSaddle point at (0,0); local max at (61,121)answer recomputed · sympy
fx=y−3x2=0 and fy=x−2y=0, so x=2y and y=3x2.
y=3(2y)2=12y2⇒y(1−12y)=0⇒y=0 or y=121.
Critical points: (0,0) and (61,121).
fxx=−6x, fyy=−2, fxy=1, so D=12x−1. At (0,0): D=−1<0, saddle. At (61,121): D=1>0 and fxx=−1<0, local maximum.
TrapD depends on x here, so it must be re-evaluated at each point. A single computed D cannot classify two different critical points.
5
Show that f(x,y)=x2−y2 has a saddle point at (0,0) using the second derivative test.
no calculator, same as the exam
AnswerSaddle point at (0,0)answer recomputed · sympy
fx=2x=0 and fy=−2y=0, so the only critical point is (0,0).
fxx=2, fyy=−2, fxy=0.
D=(2)(−2)−0=−4<0, so (0,0) is a saddle point.
The geometry backs it up: along y=0 the surface is x2 (a valley), along x=0 it is −y2 (a ridge). This is the standard saddle.
TrapD<0 is the one case that is fully decisive with no second test needed. Do not go looking at fxx afterwards; its sign is irrelevant when D<0.
6
Find the absolute maximum and absolute minimum of f(x,y)=4xy−x2−2y2+2 on the region 0≤x≤2, 0≤y≤1.
no calculator, same as the exam
AnswerAbs. max =4 at (2,1); abs. min =−2 at (2,0)answer recomputed · sympy
Interior: fx=4y−2x=0 and fy=4x−4y=0 give x=y and x=2y, so only (0,0) — a corner, not interior. No interior critical point.
Edge y=0: f=2−x2 on x∈[0,2], running from 2 down to −2.
Edge y=1: f=4x−x2 on [0,2], running from 0 up to 4 at x=2.
Edge x=0: f=2−2y2∈[0,2]. Edge x=2: f=−2y2+8y−2, from −2 at y=0 to 4 at y=1.
Absolute max =4 at (2,1); absolute min =−2 at (2,0).
TrapForgetting the four corners. Both extremes here happen to be corner points, which the edge parametrisations only catch at their endpoints.
The official set has a hole here.
Lesson 7 covers projectile motion under gravity, and uniform circular motion — and not one of the official problems above tests it.
It is still examinable, so the questions below are not optional this time — they are the only place those topics get rehearsed.
STEP 03
More practice
3 questions
Mine, on those sections only. Untimed, no calculator — the quiz doesn't allow one either.
0/3 answered
115.6 · Linear approximation
Use the linearization of f(x,y)=x2+y2 at (3,4) to estimate f(3.1,3.9).
Answer B
f(3,4)=5; fx=x2+y2x=53, fy=x2+y2y=54.
Δx=0.1,Δy=−0.1:
L=5+53(0.1)+54(−0.1)=5+0.06−0.08=4.98.
The move is mostly "sideways" along the level circle, so the value barely changes.
TrapSign of Δy. 3.9 is 0.1 below 4, so Δy is negative.
Find the absolute maximum of f(x,y)=xy on the closed disk x2+y2≤1.
Answer B
Interior:∇f=⟨y,x⟩=0 only at (0,0), where f=0.
Boundary: parametrize x=cost,y=sint, so f=costsint=21sin2t, whose maximum is 21.
Compare the whole list: {0,21,−21}. Absolute max =21, at (21,21).
TrapStopping at the interior critical point. On a closed region the max usually lives on the edge.
You must stay on a path. Walk along it and watch the contour lines you cross — as long as you're cutting across them you're still gaining height, so keep going. You've maxed out exactly where the path stops crossing and just touches one.
Touching = the two curves share a tangent line = their normals point the same way. The normals are ∇f and ∇g. So ∇f is a multiple of ∇g. The multiple is λ.
one constraint∇f=λ∇gandg(x,y,z)=k
Touching = sharing a tangent line = normals pointing along the same line. The normals are ∇f and ∇g, so one must be a multiple of the other. That multiple is λ — and you almost never need its value.
Algorithm · Lagrange, one constraint
Name f and g.f is what you're optimising; g=k is the restriction.
Write the system:fx=λgx, fy=λgy, fz=λgz, plus g=k. Four equations, four unknowns.
Eliminate λ first. Solve each equation for λ and set them equal — that removes an unknown immediately.
Watch for zeros. Before dividing by anything, handle the case where it equals zero as its own branch. That's where the missing solutions hide.
Evaluate f at every candidate point and just compare. Largest is the max, smallest is the min.
No second derivative test exists here. Comparison is the whole final step.
Worked example
Maximise f=xy on the circle g=x2+y2=8.
system
y=2λx, x=2λy, x2+y2=8
eliminate
2xy=λ=2yx⇒y2=x2⇒y=±x
constraint
2x2=8⇒x=±2, giving (2,2),(−2,−2),(2,−2),(−2,2)
compare
f=4,4,−4,−4. Max 4, min −4.
zero branch
x=0 would force y=0, which fails the constraint — so nothing was lost in step 2.
two constraints — the curve where two surfaces meet∇f=λ∇g+μ∇h,g=k1,h=k2
With two constraints you're stuck on a curve instead of a surface. ∇g and ∇h span the plane of directions perpendicular to that curve, and at an optimum ∇f has to lie in that plane — otherwise some leftover component would still be pushing you along the curve toward a better value.
When to use Lagrange vs. substitution
If the constraint solves cleanly for one variable, substitute and use 15.7 — it's faster. Reach for Lagrange when the constraint is symmetric, messy, or refuses to be solved for anything (a sphere, an ellipsoid, x2+y2+z2=1).
Check yourself
What is the condition at a constrained optimum?
Answer
∇f=λ∇g — the contour just touches the constraint, so their normals line up.
Do you need the second derivative test to finish?
Answer
No. Evaluate f at every candidate point and compare. Biggest is the max, smallest is the min.
Tile the floor. On each tile, build a column as tall as the surface above it. Volume of one column = height × tile area. Add up all the columns.
drag to spin
Each column is height × tile area. Turn the tiles up and the staircase melts into the smooth solid. That limit is the double integral.
definition and Fubini's theorem∬RfdA=m,n→∞limi=1∑mj=1∑nf(xij∗,yij∗)ΔA=∫ab∫cdfdydx=∫cd∫abfdxdy
Over a rectangle the order never matters. And if f(x,y)=g(x)h(y) splits, the double integral splits too: ∬=(∫abg)(∫cdh). Look for that before doing any work.
f=1 gives the plain area of R. That's the sanity check for every setup you write.
Check yourself
What does ∬R1dA give you?
Answer
The plain area of R. Use it to sanity-check any setup.
Over a rectangle, does the order of integration matter?
Answer
No (Fubini). And if f(x,y)=g(x)h(y) the whole thing factors into two single integrals — look for that first.
This is where the points get lost, and it's mechanical. Fire an arrow across the region. Where it enters and exits gives the inner limits. How far the arrow itself slides gives the outer limits — always plain numbers.
Same region, same answer, different work. Swap the order when the inner integral is impossible or the region needs splitting into pieces — that is the whole point of the "reverse the order" problems.
Type I · vertical arrows, y between two curves∬RfdA=∫ab∫g1(x)g2(x)f(x,y)dydx
Type II · horizontal arrows, x between two curves∬RfdA=∫cd∫h1(y)h2(y)f(x,y)dxdy
Algorithm · set up the limits
Sketch the region. Always. Find where the boundary curves intersect by setting them equal.
Choose a sweep direction — vertical (Type I) or horizontal (Type II). Pick whichever needs fewer pieces.
Inner limits: enter → exit. Which curve does the arrow enter through, which does it leave through? Those are functions of the outer variable.
Outer limits: the arrow's travel range. These must be constants.
Check the shape:∫numbernumber∫functionfunction. A variable in the outer limits means you set it up wrong.
Algorithm · reverse the order of integration
Read the given limits backwards into a sketch. Inner limits are curves; outer limits are the strip the arrow slides through. Draw the region they describe — do not skip to algebra.
Re-sweep the other way. Fire the arrow horizontally instead of vertically (or vice versa).
Solve the boundary curves for the other variable.y=x2 becomes x=y.
Split into pieces if the enter-or-exit curve changes partway across.
Now integrate. The whole point is usually that the new inner integral is doable and the old one wasn't.
Classic tell: an inner integrand like ey2 or sin(x2) that has no elementary antiderivative. That's the problem asking you to flip.
Worked example
∫01∫x1ey2dydx — the inner integral is impossible as written.
sketch
0≤x≤1 and x≤y≤1: the triangle with corners (0,0),(1,1),(0,1).
re-sweep
Horizontally: for each y in [0,1], x runs 0→y.
rewrite
∫01∫0yey2dxdy=∫01yey2dy
finish
=21ey201=2e−1
Check yourself
Your outer limits contain a variable. What went wrong?
Answer
The setup. Outer limits are always constants. Re-sweep the region.
The inner integrand is ey2. What is the problem telling you?
Answer
Flip the order. No elementary antiderivative in that variable is the signal, every time.
STEP 02
The official problems
15 problems
Straight from the department's own study guide, for exactly these lessons — worked, with the answer held back until you commit. Do these first.
Find the maximum value of f(x,y)=3+2x−y subject to the constraint x2+2xy+4y2=7.
no calculator, same as the exam
Answer10answer recomputed · sympy
The constraint g(x,y)=x2+2xy+4y2=7 is a closed curve (an ellipse), so f attains a max and a min on it. Use Lagrange: ∇f=λ∇g.
∇f=⟨2,−1⟩ and ∇g=⟨2x+2y,2x+8y⟩, so the system is 2=λ(2x+2y), −1=λ(2x+8y), x2+2xy+4y2=7.
Add twice the second equation to the first to kill x: 2+2(−1)=λ[(2x+2y)+2(2x+8y)] gives 0=λ(6x+18y). Since λ=0 (else ∇f=0, false), x=−3y.
Substitute into the constraint: 9y2−6y2+4y2=7y2=7, so y=±1 and (x,y)=(−3,1) or (3,−1).
Evaluate: f(−3,1)=3−6−1=−4 and f(3,−1)=3+6+1=10.
Maximum value =10 (attained at (3,−1); the minimum is −4).
Trap∇g of the cross term is the usual casualty: ∂x(2xy)=2y and ∂y(2xy)=2x, so ∇g=⟨2x+2y,2x+8y⟩, not⟨2x+2y,8y⟩. Also, Lagrange hands you *critical values*, plural. The largest of the list is the answer, so you must evaluate f at every solution point, not just the first one you find.
2
Find the maximum value of f(x,y)=8x−6y subject to the constraint (x−1)2+y2=1.
no calculator, same as the exam
Answer18answer recomputed · sympy
Constraint: the circle g(x,y)=(x−1)2+y2=1, centre (1,0), radius 1. f is linear, so its extremes sit where the gradient of f is normal to the circle.
Lagrange: ⟨8,−6⟩=λ⟨2(x−1),2y⟩, i.e. 8=2λ(x−1) and −6=2λy.
So x−1=λ4 and y=−λ3. Plug into the constraint: λ216+λ29=1⇒λ2=25,λ=±5.
Sanity check with geometry: f=8x−6y at the centre (1,0) is 8, and ∣∇f∣=64+36=10, so on a radius-1 circle f ranges over 8±10={−2,18}. Same answer.
TrapThe circle is not centred at the origin. Writing the constraint gradient as ⟨2x,2y⟩ (forgetting the shift) is the standard wreck here; it must be ⟨2(x−1),2y⟩. The shortcut fmax=f(centre)+∣∇f∣⋅R works only because f is linear and the constraint is a circle; do not reach for it on an ellipse.
3
The maximum and minimum values of f(x,y)=(x−2)2+(y−4)2 subject to x2+y2=5 are, respectively:
f(x,y)=(x−2)2+(y−4)2 is the squared distance from (2,4); the constraint is the circle of radius 5 about the origin. So this is really a nearest/farthest point problem.
Lagrange: ⟨2(x−2),2(y−4)⟩=λ⟨2x,2y⟩, i.e. x−2=λx and y−4=λy.
Then x(1−λ)=2 and y(1−λ)=4, so y=2x (dividing; note λ=1, since λ=1 would force 2=0).
Constraint: x2+(2x)2=5x2=5⇒x=±1, giving (1,2) and (−1,−2).
Maximum 45, minimum 5. (Check: ∣(2,4)∣=20=25, so distances to the circle are 25∓5, whose squares are 5 and 45.)
TrapTwo traps. First, f is the squared distance, so do not take a square root at the end and report 45 and 5. Second, the point (2,4) lies *outside* the circle, so both critical points are genuine, one near and one far; the min is not0.
4
Find the maximum of f(x,y,z)=x+y+z subject to (x−1)2+y2+z2=1.
no calculator, same as the exam
Answer1+3answer recomputed · sympy
Constraint: the sphere g=(x−1)2+y2+z2=1 of radius 1 centred at (1,0,0). f=x+y+z is linear, so Lagrange with ∇f=⟨1,1,1⟩.
Taking the + branch, x=1+31,y=z=31, so f=1+33=1+3. The − branch gives 1−3.
Maximum =1+3.
TrapSame shifted-centre trap as before: ∇g=⟨2(x−1),2y,2z⟩. And because the three partials give x−1=y=z (not x=y=z), the symmetric guess x=y=z=1/3 is wrong. The geometric check is f(centre)+∣∇f∣R=1+3⋅1.
5
Find the absolute maximum value M and absolute minimum value m of f(x,y)=x+y subject to x2−xy+y2=1.
no calculator, same as the exam
AnswerM=2,m=−2answer recomputed · sympy
The constraint x2−xy+y2=1 is an ellipse (rotated 45∘), a closed bounded curve, so a max and a min both exist.
Lagrange: ⟨1,1⟩=λ⟨2x−y,2y−x⟩, so 1=λ(2x−y) and 1=λ(2y−x).
Setting the two right sides equal: 2x−y=2y−x⇒3x=3y⇒y=x.
Constraint: x2−x2+x2=x2=1, so (1,1) and (−1,−1).
f(1,1)=2 and f(−1,−1)=−2, so M=2,m=−2.
TrapDifferentiating the cross term −xy: ∂x=−y, ∂y=−x — it is easy to drop a sign and get 2x+y. Also, cancelling λ from the two equations is legal here only because λ=0 (if λ=0 then 1=0), so you should say so rather than divide blindly.
The integrand factors, xy=xy, and the region is the rectangle [0,4]×[0,1], so the double integral splits into a product of single integrals.
∫01∫04xydxdy=(∫04xdx)(∫01ydy).
∫04x1/2dx=32x3/204=32⋅8=316.
∫01y1/2dy=32.
Product: 316⋅32=932.
TrapRead the differentials: dxdy means the inner variable is x (limits 0 to 4) and the outer is y (limits 0 to 1). Pairing x with [0,1] gives 32⋅316 by luck here, but on a non-symmetric integrand the same misreading is fatal. The split into a product is legal only because the integrand factors and the limits are constants.
2
Find the volume enclosed by x=0, x=3, y=0, y=2, z=1, and z=ex+y.
no calculator, same as the exam
Answere5−e3−e2−5answer recomputed · sympy
The base is the rectangle R=[0,3]×[0,2]; the solid sits between the floor z=1 and the roof z=ex+y. On R we have x+y≥0, so ex+y≥1 and the roof really is above the floor everywhere.
Volume =∬R(ex+y−1)dA=∫02∫03(exey−1)dxdy.
∬RexeydA=(∫03exdx)(∫02eydy)=(e3−1)(e2−1).
∬R1dA=area(R)=3⋅2=6.
Volume =(e3−1)(e2−1)−6=e5−e3−e2+1−6=e5−e3−e2−5.
TrapForgetting to subtract the floor: integrating ex+y alone gives the volume under the surface down to z=0, not down to z=1. The correction is −area(R)=−6, and note (e3−1)(e2−1) already contributes +1, so the constant is 1−6=−5, not −6. Also ex+y=exey is what lets the integral factor; ex+y=ex+ey.
3
Find the average value of f(x,y)=x2y over the rectangle R with vertices (−1,0), (−1,5), (1,5), (1,0).
no calculator, same as the exam
Answer65answer recomputed · sympy
The four vertices give the rectangle R=[−1,1]×[0,5], of area A=2⋅5=10.
Average value =A1∬RfdA=101∫05∫−11x2ydxdy.
∫−11x2dx=3x3−11=32 and ∫05ydy=225.
∬Rx2ydA=32⋅225=325.
Average =1025/3=65.
TrapDividing by the wrong thing, or not dividing at all: the answer to an *average value* question is the integral over the area, here 10. Second trap: x2 is even, so ∫−11x2dx=32, not 0 — students who remember 'symmetric interval ⇒ zero' from odd integrands kill this problem instantly.
4
Evaluate ∫01∫01∫02−yexdzdxdy.
no calculator, same as the exam
Answer23(e−1)answer recomputed · sympy
Innermost is dz and the integrand ex is free of z, so the first integration is just multiplication by the height: ∫02−yexdz=ex(2−y).
The remaining limits are constants, so ∫01∫01ex(2−y)dxdy=(∫01exdx)(∫01(2−y)dy).
∫01exdx=e−1.
∫01(2−y)dy=2−21=23.
Value =23(e−1).
TrapThe order is dzdxdy, so the z-limit 2−y belongs to the *innermost* integral and depends on the *outermost* variable — perfectly legal, but you must integrate z first and carry (2−y) outward. Attempting to do dx first, or treating the upper limit as 2−x, both break it. Once z is gone the integral separates only because ex and (2−y) live in different variables.
5
Evaluate ∬R2+x3x2ydA over R={1≤x≤2,0≤y≤4}.
no calculator, same as the exam
Answer38(ln10−ln3)answer recomputed · sympy
R is a rectangle and the integrand splits as 2+x3x2⋅y, so the integral is a product of two one-variable integrals.
∬R2+x3x2ydA=(∫122+x3x2dx)(∫04ydy).
For the x-integral substitute u=2+x3, du=3x2dx, so x2dx=3du; as x:1→2, u:3→10.
∫3103u1du=31(ln10−ln3).
∫04ydy=8.
Value =8⋅31(ln10−ln3)=38(ln10−ln3)=38ln310.
TrapWhen you substitute u=2+x3 you must change the limits to u:3→10; leaving 1 and 2 gives 31ln12, a completely different number. The other classic is losing the 31 from du=3x2dx.
TrapGetting the new limits backwards. The old inner limit y≥3x becomes x≤y3, so the new inner range is 0≤x≤y3 — noty3≤x≤27. Sketch the region: it is the piece *above* the cube-root curve, i.e. to the *left* of x=y3. The factor 4 in the integrand is not decoration; it is exactly the 4y3 the substitution needs after the swap.
2
Evaluate ∬RxsinxdA where R is bounded by the x-axis, y=x, and x=1 (integrate in y first, then x).
no calculator, same as the exam
Answer1−cos(1)answer recomputed · sympy
Sketch R: bounded below by y=0, above by y=x, on the right by x=1 — the triangle with vertices (0,0),(1,0),(1,1).
xsinx has no elementary antiderivative in x, so we must integrate in y first (as instructed), which makes x a constant.
∬RxsinxdA=∫01∫0xxsinxdydx.
Inner: xsinx⋅(x−0)=sinx. The offending x in the denominator cancels — that is the whole point of the order.
∫01sinxdx=−cosx01=1−cos(1).
TrapChoosing dydx is not optional here: the other order gives ∫y1xsinxdx, which cannot be evaluated in closed form. Also the inner limits are 0≤y≤x — a *variable* upper limit. Writing ∫01 for the inner y-integral turns the triangle into the unit square and gives ∫01xsinxdx, which is not elementary.
3
Find the volume of the solid under z=2x and above the region in the first quadrant bounded by the coordinate axes and x+y=1.
no calculator, same as the exam
Answer31answer recomputed · sympy
The base region D is the triangle in the first quadrant cut off by x+y=1: vertices (0,0),(1,0),(0,1). The roof is z=2x≥0 there, so the volume is the plain double integral.
As a type-I region: 0≤x≤1 and 0≤y≤1−x.
V=∬D2xdA=∫01∫01−x2xdydx.
Inner: 2x(1−x)=2x−2x2.
∫01(2x−2x2)dx=1−32=31.
TrapUsing ∫01∫01 (the square) instead of the triangle: the hypotenuse y=1−x must appear as the inner limit. Also note the integrand is 2x, so the answer is not the triangle's area 21 times anything obvious; you have to actually integrate 2x(1−x).
4
Change the order of integration for ∫02∫x22xf(x,y)dydx.
no calculator, same as the exam
Official answer(original) — ∫04∫yy/2fdxdychecked · official answer wrong
Read the region from the given order dydx: 0≤x≤2 and x2≤y≤2x — the lens between the parabola y=x2 and the line y=2x.
Find where the two curves meet: x2=2x⇒x=0 or x=2, giving the corner points (0,0) and (2,4). So y ranges over [0,4] overall.
Now solve each boundary for x. From y=2x we get x=2y; from y=x2 (with x≥0) we get x=y.
For a fixed y∈[0,4], which is the left edge? Test y=1: the line gives x=21, the parabola gives x=1. So x runs from 2yup toy.
Reversed order: ∫04∫y/2yf(x,y)dxdy.
Confidence check with a test integrand f=xy: the original order gives 38, and ∫04∫y/2yxydxdy=38 as well.
TrapInside the lens the parabola is the right boundary and the line is the left one (for 0<x<2, x2<2x means the parabola sits *below* the line, which flips to *right of* when you read horizontally). Writing ∫yy/2 reverses the limits and negates the integral. Also x=y, not x=±y: the region lives in x≥0.
The official answer is wrongThe official answer prints the inner limits as ∫yy/2, which is upside down. For 0<y<4 we have 2y<y (at y=1: 0.5<1), so the lower limit must be y/2 and the upper limit y. As printed, the integral equals the negative of the original. Numerical check with f(x,y)=xy: the original iterated integral is 38, the correctly reversed one is 38, and the printed one is −38.
5
Reverse the order and evaluate: ∫02ln3∫y/2ln3ex2dxdy.
no calculator, same as the exam
Answer2answer recomputed · sympy
ex2 has no elementary antiderivative, so the order must be reversed. Let a=ln3 for short; the limits are 0≤y≤2a and 2y≤x≤a.
Sketch: the region is the triangle with vertices (0,0), (a,0) and (a,2a), bounded by the line x=y/2 (i.e. y=2x), the vertical line x=a, and the x-axis.
Reading it the other way: x runs 0→a, and for each x, y runs from 0 to 2x.
TrapThe outer limit 2ln3 and the inner ln3 are *different* numbers — reversing carelessly and writing y from 0 to x (instead of 2x) loses the factor of 2 that the substitution needs, leaving a non-elementary integral. Also eln3=3, so the final answer is the clean 2, not eln3 left unsimplified.
STEP 03
More practice
4 questions
Mine, on those sections only. Untimed, no calculator — the quiz doesn't allow one either.
0/4 answered
115.8 · Lagrange multipliers
Find the maximum of f(x,y)=x+2y subject to x2+y2=5.
Answer C
∇f=⟨1,2⟩=λ⟨2x,2y⟩⇒y=2x.
Substituting: x2+4x2=5⇒x=±1, giving (1,2) and (−1,−2).
f(1,2)=5, f(−1,−2)=−5. Max =5, min =−5.
Cross-check: the maximum of a⋅r on a circle of radius R is ∣a∣R=5⋅5=5 ✓.
TrapReporting λ, or stopping at one of the two candidate points without comparing.
The integrand splits, so the double integral splits too:
∬RxydA=(∫02xdx)(∫03ydy)=(2)(29)=9.
Look for this factorisation before doing any work — over a rectangle it always applies when f=g(x)h(y).
TrapMultiplying the two upper limits (2 × 3 = 6). The antiderivatives are what multiply.
The region is 0≤y≤1,y≤x≤1, i.e. 0≤x≤1,0≤y≤x2. Sketch it, then re-sweep.
∫01∫0x2ex3dydx=∫01x2ex3dx=[31ex3]01=3e−1.
The x2 that the flip produces is exactly the factor that makes the substitution work.
TrapTrying to antidifferentiate e^(x³) in x. That is the signal to flip.
Region: 0≤x≤1,x≤y≤1 — the triangle above y=x. Re-swept horizontally: 0≤y≤1,0≤x≤y.
∫01∫0ysin(y2)dxdy=∫01ysin(y2)dy=[−21cos(y2)]01=21−cos1.
The y the flip produces is exactly what makes the substitution work.
TrapAttempting ∫sin(y²)dy directly. No elementary antiderivative — that is the signal to flip.
Cut the region into wedges instead of tiles. A wedge is nearly a rectangle: one side is dr long, the other is an arc of length rdθ — because the same angle sweeps a longer arc when you're farther out.
Use polar the moment you see x² + y², a circle, a disk, or an annulus. The messy algebra collapses into r².
the r is a stretch factor, not a rule to memorise∬Rf(x,y)dA=∫αβ∫r1(θ)r2(θ)f(rcosθ,rsinθ)dArdrdθ
Algorithm · convert to polar
Spot the trigger:x2+y2, a circle, a disk, an annulus, a wedge, or limits with a2−x2.
Convert the integrand:x→rcosθ, y→rsinθ, and every x2+y2→r2.
Convert the region.θ sweeps the angular range; for each θ, r runs from the inner boundary to the outer boundary.
Attach the r.dA=rdrdθ. Write it before you do anything else.
Integrate — inner in r, outer in θ.
Useful conversions: x2+y2=a2→r=a. y=x→θ=π/4. x2+y2=2ax→r=2acosθ (a circle through the origin).
The single most expensive mistake in the course
Dropping the r.dA=rdrdθ, not drdθ. It comes from the wedge geometry above and it is also exactly what the Jacobian in 16.7 will produce. Write the r the moment you write drdθ.
Check yourself
What is dA in polar?
Answer
rdrdθ.
Where does that r come from?
Answer
The wedge. One side is dr long; the other is an arc of length rdθ, because the same angle sweeps further when you are farther out.
Same idea, now the tiles are tiny cubes. To get limits: look at the solid's shadow on a coordinate plane, then ask how far the solid runs above each shadow point.
f = 1 gives volume. f = density gives mass. There are six possible orders (dz dy dx, dy dz dx, …) — pick whichever makes the inner surfaces simplest.
the shadow method∭EfdV=∬shadow[∫z1(x,y)z2(x,y)fdz]dA
Algorithm · set up a triple integral
Pick the sweep direction. Usually z: fire an arrow straight up through the solid.
Inner limits — enter and exit surfaces.z runs from the bottom surface z1(x,y) to the top surface z2(x,y). Solve both for z.
Project to get the shadow. Where the top and bottom surfaces meet is the shadow's boundary curve — set z1=z2 and eliminate z.
Set up that 2-D shadow region exactly as in 16.2 (or 16.3 if it's round).
Check the shape: function limits inside, then function limits, then numbers outermost.
Six orders exist (dzdydx, dydzdx, …). If the z-sweep gives an ugly shadow, try sweeping along x or y instead — the shadow lands on a different plane.
f=1
Volume of the solid.
f=ρ
Mass. Then xˉ=m1∭xρdV, and similarly for yˉ,zˉ.
Inertia about the z-axis
Iz=∭(x2+y2)ρdV — squared distance from the axis.
Check yourself
What does ∭E1dV give?
Answer
The volume of E.
How do you get the two outer limits of a triple integral?
Answer
Project. The solid's shadow on a coordinate plane is a 2-D region — set it up exactly as in 16.2.
TrapThe integrand becomes r2, not r: one factor of r is x2+y2 and the second is the Jacobian from dA=rdrdθ. Dropping the Jacobian gives 49π. Second trap: θ runs only to π/2 — the limits x∈[0,3], y≥0 give a quarter disk, not a half disk.
2
Compute ∬Re−(x2+y2)dA where R is the disk x2+y2≤4.
no calculator, same as the exam
Answerπ(1−e−4)answer recomputed · sympy
R is the full disk of radius 2 and the integrand depends only on x2+y2, so polar is the obvious move: 0≤r≤2, 0≤θ≤2π.
∬Re−(x2+y2)dA=∫02π∫02e−r2rdrdθ.
Inner: substitute u=r2, du=2rdr, so ∫02e−r2rdr=21∫04e−udu=21(1−e−4).
Outer: multiply by ∫02πdθ=2π.
Value =2π⋅21(1−e−4)=π(1−e−4).
TrapWithout the Jacobian r, ∫e−r2dr is not elementary at all — the r is exactly what makes the substitution work, so its absence should be an immediate alarm. Also r goes to 2, so u=r2 goes to 4, giving e−4; writing e−2 is the standard slip.
3
Rewrite in polar (do not evaluate): ∫01∫−1−y21−y2(x2+y2)1/2dxdy.
no calculator, same as the exam
Answer∫0π∫01r2drdθanswer recomputed · sympy
Region: 0≤y≤1 with −1−y2≤x≤1−y2, i.e. x2+y2≤1 with y≥0 — the upper half of the unit disk.
In polar the half disk is 0≤r≤1, 0≤θ≤π (the full sweep of the upper half plane).
The integrand (x2+y2)1/2=r, and dA=rdrdθ, so r⋅r=r2.
∫01∫−1−y21−y2(x2+y2)1/2dxdy=∫0π∫01r2drdθ.
(Not asked for, but its value is 3π, and the original Cartesian integral evaluates to 3π as well.)
Trapθ ranges over [0,π], not [0,π/2]: x takes negative values here, so the region covers the second quadrant too. And the integrand must pick up the Jacobian, r⋅rdrdθ=r2drdθ; writing ∫0π∫01rdrdθ is the most common wrong answer on this exact problem.
4
Evaluate ∫0π∫0π−y2sin(x2+y2)dxdy.
no calculator, same as the exam
Answer2πanswer recomputed · sympy
Region: 0≤y≤π, 0≤x≤π−y2, i.e. x2+y2≤π with x,y≥0 — the quarter disk of radius π.
sin(x2+y2) has no elementary antiderivative in Cartesian form, so polar is forced: 0≤r≤π, 0≤θ≤2π.
TrapThe radius is π, so u=r2 runs to π — not to π. Writing ∫0πsinudu is the killer here. As always, the Jacobian r is what makes the u-substitution possible; without it the integral is non-elementary.
5
Compute ∫−33∫−9−y29−y2ex2+y2dxdy.
no calculator, same as the exam
Answerπ(e9−1)answer recomputed · sympy
Region: −3≤y≤3 and −9−y2≤x≤9−y2 — the entire disk x2+y2≤9 of radius 3.
Polar: 0≤r≤3, 0≤θ≤2π, integrand ex2+y2=er2, and dA=rdrdθ.
∫−33∫−9−y29−y2ex2+y2dxdy=∫02π∫03er2rdrdθ.
Inner: u=r2, du=2rdr, u:0→9, so 21∫09eudu=21(e9−1).
Outer: 2π⋅21(e9−1)=π(e9−1).
TrapBoth limits are symmetric, so θ sweeps the full2π; using π (half disk) or π/2 halves or quarters the answer. And u=r2 runs to 9, giving e9 — writing e3 is the standard misfire. The r from dA is again mandatory.
Read the order off the differentials: dzdxdy. The innermost variable is z, and the integrand ex does not contain z — so the inner integral is just multiplication by the height.
∫02−yexdz=ex(2−y).
Now the x and y limits are both constants, and what is left separates: ∫01∫01ex(2−y)dxdy=(∫01exdx)(∫01(2−y)dy).
∫01exdx=e−1 and ∫01(2−y)dy=2−21=23.
Product: 23(e−1).
TrapThe limit 2−y sits on the inner integral while y is the outer variable. That is legal — inner limits may involve any variable integrated later — but it means you cannot do the y-integral first. Someone who starts at the outside gets ∫01(2−y)dy sitting where a z-integral belongs and loses the factor entirely.
2
Let E be the solid in the first octant above the xy-plane and below 4x+2y+z=8. For ∭EfdV=∫0a∫0b∫0cfdzdydx:
no calculator, same as the exam
Answera=2,b=4−2xanswer recomputed · sympy
Order dzdydx means: z is bounded by surfaces, y by curves in the xy-plane, x by numbers. Build it from the inside out.
First octant gives the floor z=0; the plane gives the ceiling z=8−4x−2y. So c=8−4x−2y.
Project the solid onto the xy-plane by setting z=0 in the plane: 4x+2y=8, i.e. y=4−2x. That triangle, together with x≥0,y≥0, is the shadow.
For a fixed x, y runs from 0 up to that line: b=4−2x.
x runs until the shadow closes, i.e. until 4−2x=0: a=2.
∭EfdV=∫02∫04−2x∫08−4x−2yfdzdydx.
TrapUsing the x-intercept of the plane, x=2... by luck that is right here, but the reason matters: a is where the shadow closes, not where the plane meets an axis. Get b by setting z=0, not x=0; setting x=0 gives y=4, which is the wrong constant limit and makes the region a box.
3
Let D be the solid bounded by x=0, z=0, y=x, and x+y+z=2. Write ∭DfdV as an iterated integral.
no calculator, same as the exam
Official answere.g. ∫01∫x2−x∫02−x−yfdzdydx , or ∫01∫02−x∫x2−x−zfdydzdxchecked · official answer wrong
Four surfaces, so list the inequalities: x≥0, z≥0, y≥x, and x+y+z≤2.
Take z innermost: 0≤z≤2−x−y.
Shadow in the xy-plane is where that height is non-negative and y≥x: the triangle x≤y≤2−x, x≥0.
That triangle closes when x=2−x, i.e. x=1. So ∫01∫x2−x∫02−x−yfdzdydx.
Reordering to dydzdx: solve y≥x and y≤2−x−z, so x≤y≤2−x−z.
Those two limits cross when 2−x−z=x, i.e. at z=2−2x — so z must stop at 2−2x, giving ∫01∫02−2x∫x2−x−zfdydzdx.
Sanity check both with f=1: each gives volume 32.
TrapWhen you reorder, the new outer limit is where the inner limits collide, not where some surface hits an axis. Writing z from 0 to 2−x (the plane's z-intercept above y=0) puts z in a range where the inner y-limits are upside-down, 2−x−z<x, and the integral quietly subtracts the phantom slab instead of erroring.
The official answer is wrongThe second offered order is wrong as printed. With z running to 2−x, every z>2−2x has 2−x−z<x, so the inner y-integral runs backwards. With f=1 the printed form evaluates to 21; the true volume, and the value of the printed first form, is 32. The z-limit must be 2−2x.
4
Let V be bounded by z=6−x2−y2 and z=x2+y2. If V=∫−33∫ab∫x2+y26−x2−y2dzdydx, what are a and b?
no calculator, same as the exam
Answera=−3−x2,b=3−x2answer recomputed · sympy
The z-limits are already given: the solid sits between the upward paraboloid z=x2+y2 and the downward one z=6−x2−y2.
They meet where x2+y2=6−x2−y2, i.e. 2(x2+y2)=6, so x2+y2=3 — a circle of radius 3.
That circle is the shadow's boundary, which is why the outer limits are ±3.
Solving x2+y2=3 for y: a=−3−x2, b=3−x2.
TrapSetting x2+y2=6 (from z=6−x2−y2 with z=0) instead of equating the two surfaces. The paraboloids never reach z=0 together; the boundary of the shadow is where the ceiling meets the floor, and that is radius 3, not 6. The given outer limits ±3 are the tell.
5
Evaluate ∫−11∫−1−y21−y2∫−11(x2+y2)3/2dzdxdy.
no calculator, same as the exam
Answer54πanswer recomputed · sympy
Look at the shadow before touching the integrand: −1≤y≤1 with −1−y2≤x≤1−y2 is the full unit disk, and z runs over the fixed interval [−1,1]. The solid is a cylinder.
(x2+y2)3/2 is a function of distance from the z-axis alone — cylindrical coordinates, and the z-integral is just a factor of 2.
(x2+y2)3/2=r3, and dV=rdrdθdz, so the integrand becomes r3⋅r=r4.
∫−11∫02π∫01r4drdθdz=2⋅2π⋅51=54π.
TrapConverting (x2+y2)3/2 to r3 and then integrating ∫01r3dr=41 — the extra r from dA=rdrdθ is dropped and the answer comes out π instead of 54π. The exponent you integrate is always one higher than the one you converted.
TrapConverting the integrand but keeping the Cartesian limits in spirit — reading 0≤x≤2 as 'θ from 0'. The upper x-limit 2 is a vertical line, r=2secθ, not a ray; only the presence of the lower boundary y=x turns it into the clean ray θ=4π. Any time a straight edge is not through the origin, polar limits stop being constants.
The official answer is wrongThe printed Cartesian stem and the printed cylindrical answer describe different regions. As written, 0≤y≤4−x2 over 0≤x≤2 has area 1+2π, while the wedge 4π≤θ≤2π,r≤2 has area 2π — the answer is missing the triangular piece below y=x. The printed answer is correct for the inner limit ∫x4−x2; with the printed lower limit 0 the rewrite needs the two pieces above.
2
Evaluate ∫02∫04−x2∫x2+y22(x2+y2)dzdydx using cylindrical coordinates.
no calculator, same as the exam
Answer34π(2−1)answer recomputed · sympy
Shadow first: 0≤x≤2, 0≤y≤4−x2 is the quarter disk of radius 2 in the first quadrant, so 0≤θ≤2π, 0≤r≤2.
The z-limits are both cones: z=x2+y2=r below and z=2(x2+y2)=2r above.
Height of the column at radius r: 2r−r=(2−1)r.
With dV=rdrdθdz: ∫0π/2∫02(2−1)r⋅rdrdθ=2π(2−1)⋅38.
=34π(2−1).
Trap2(x2+y2) is 2r, not 2r and not r2 — pulling the 2 out from under the root without the square root gives 34π exactly one unit too large. Second trap: the height (2−1)r already carries one power of r, so the integrand is r2, not r; forgetting the Jacobian here costs a factor of 23⋅22 and lands on π(2−1).
3
Convert to cylindrical and evaluate ∫010∫0100−x2∫0x2+y2x2+y21dzdydx.
no calculator, same as the exam
Answer25πanswer recomputed · sympy
The shadow 0≤x≤10, 0≤y≤100−x2 is a quarter disk of radius 10: 0≤θ≤2π, 0≤r≤10.
The ceiling z=x2+y2 is the cone z=r; the integrand x2+y21=r1.
So the full cylindrical integrand is r1⋅r=1 — the Jacobian cancels the singularity. That is the point of the problem.
TrapTwo errors cancel here if you are careless in exactly the wrong way. Dropping the Jacobian leaves ∫∫r1⋅rdz vs ∫∫r1dz — the second gives ∫0101dr=10 and the answer 5π. The integrand being 1 after conversion is a result, not an accident: check that you multiplied by r before you simplified.
4
Find the volume of the solid bounded by z=x2+y2−9 and z=−2(x2+y2) using cylindrical coordinates.
no calculator, same as the exam
Answer227πanswer recomputed · sympy
Both surfaces depend only on r2=x2+y2: z=r2−9 (an upward paraboloid dropped to −9) and z=−2r2 (a downward one from the origin).
Which is on top? At r=0: −9 versus 0, so z=−2r2 is the ceiling and z=r2−9 is the floor.
They meet where r2−9=−2r2⇒3r2=9⇒r=3. Full revolution, so 0≤θ≤2π.
TrapAssuming the paraboloid opening upward is the ceiling because it 'goes up'. Over the whole region r<3 it is the floor, and swapping them makes the volume negative. Test at r=0 every time. The other trap is integrating 9−3r2 against dr instead of rdr, which gives 2π(93−33)=123π — a suspiciously irrational volume for a surface of revolution with rational intersection radius.
5
Write the volume of the solid bounded by the cone z2=x2+y2 between the planes z=1 and z=2 as a spherical-coordinates integral.
In spherical, the cone z2=x2+y2 with z>0 is the single surface ϕ=4π, and the solid inside it is 0≤ϕ≤4π.
A horizontal plane is not a constant in spherical: z=ρcosϕ, so z=1 becomes ρ=secϕ and z=2 becomes ρ=2secϕ.
Those are the ρ-limits, and they depend on ϕ — so ρ must be integrated first.
dV=ρ2sinϕdρdϕdθ, full revolution in θ:
V=∫02π∫0π/4∫secϕ2secϕρ2sinϕdρdϕdθ.
Worth confirming in cylindrical, where it is a stack of disks of radius z: ∫02π∫12∫0zrdrdzdθ=∫12πz2dz=37π — and the spherical integral gives the same.
TrapWriting the planes as constant ρ (say ρ from 1 to 2), which describes the region between two spheres, not two planes — that is a spherical shell wedge, volume 32π(8−1)(1−cos4π), not 37π. Any flat boundary not through the origin becomes ρ=(something)secϕ and forces dρ innermost.
STEP 03
More practice
4 questions
Mine, on those sections only. Untimed, no calculator — the quiz doesn't allow one either.
0/4 answered
116.3 · Polar area
Find the area enclosed by one petal of the rose r=2cos(2θ).
Answer B
One petal runs where cos2θ≥0: −4π≤θ≤4π.
A=21∫−π/4π/4(2cos2θ)2dθ=2∫−π/4π/4cos22θdθ=∫−π/4π/4(1+cos4θ)dθ=2π.
TrapUsing 0 ≤ θ ≤ π/2 — that sweeps a petal and a half. Solve r = 0 for the true limits.
Find the volume of the tetrahedron bounded by x=0,y=0,z=0 and x+y+z=1.
Answer C
Sweep in z first: 0≤z≤1−x−y. The shadow on the xy-plane is the triangle 0≤y≤1−x,0≤x≤1.
V=∫01∫01−x∫01−x−ydzdydx=∫012(1−x)2dx=61.
Note the shape: function limits inside, then function, then numbers outermost.
TrapPutting a variable in the outermost limit. The outer limits are always constants.
Distance from the origin, angle down from the north pole, angle around. Balls and cones. The famous ρ2sinφ is not arbitrary — it's the little box's three edge lengths multiplied together.
drag to spin
Drag φ toward either pole and watch the box go skinny. That shrinking edge is ρ sin φ — the whole reason dV carries a sin φ.
the volume element, derived not memoriseddV=radialdρ×north–southρdφ×east–westρsinφdθ=ρ2sinφdρdφdθ
Why sinφ: the circle of latitude at angle φ has radius ρsinφ, not ρ. Near the poles it shrinks to nothing, so the same dθ sweeps almost no distance — and the volume element shrinks with it.
Algorithm · set up in spherical
Use it when you see x2+y2+z2, a sphere, a ball, a cone, or a region between two spheres.
ρ limits: where the ray from the origin enters and leaves the solid. A sphere of radius a gives ρ=a; a plane z=c gives ρ=c/cosφ=csecφ.
φ limits: the cone angle. φ=0 is straight up, φ=π/2 is the xy-plane, φ=π is straight down. The cone z=x2+y2 is φ=π/4.
θ limits: how far around, usually 0 to 2π.
Attach ρ2sinφ and integrate, innermost ρ outward.
Ranges: ρ≥0, 0≤φ≤π, 0≤θ≤2π. φ never goes past π — that would be going round twice.
The ρ² sin φ is not arbitrary: dρ × (ρ dφ) × (ρ sin φ dθ). Two ρ's and one sin φ fall out on their own. φ ∈ [0, π] (top to bottom), θ ∈ [0, 2π] (all the way around).
Check yourself
What is dV in spherical?
Answer
ρ2sinφdρdφdθ.
What is the range of φ?
Answer
0≤φ≤π — top to bottom. Only θ goes all the way round to 2π.
Why the sinφ?
Answer
The circle of latitude has radius ρsinφ, not ρ. Near the poles it shrinks to nothing, so the box goes skinny.
A metal plate with heavy and light spots. Its mass is density piled over the region. Its centre of mass is where you'd stick a pin so it sits level — the average position, with heavy bits voting harder.
Everything here is "∬ (some weight) × ρ dA". Weight = 1 → mass. Weight = x → moment. Weight = distance² → moment of inertia. Same integral, different sticker.
mass and momentsm=∬RρdAMy=∬RxρdAMx=∬RyρdA
centre of massxˉ=mMy=m1∬RxρdAyˉ=mMx=m1∬RyρdA
moments of inertia — distance is SQUAREDIx=∬Ry2ρdAIy=∬Rx2ρdAI0=∬R(x2+y2)ρdA=Ix+Iy
Naming trap
My uses x, and Mx uses y. The subscript names the axis you are taking the moment about, not the variable in the integrand. Same crossover in Ix, which uses y2 — distance from the x-axis is ∣y∣.
Algorithm · any of these quantities
Set up the region once — limits, and polar if it's round.
Pick the weight from the table: 1 for mass, x or y for a moment, x2+y2 for polar inertia.
Integrate ∬(weight)ρdA. Every quantity in this section is that one integral with a different sticker.
For a centre of mass, divide by m at the very end. Moments are not divided; centres of mass are.
Symmetry shortcut: if the region and the density are both symmetric about a line, the centre of mass lies on that line. That can kill half the work.
Uniform density?ρ is a constant, so it factors out of the top and bottom of xˉ and cancels completely. The answer is then called the centroid and depends only on the shape.
Check yourself
Write xˉ for a lamina.
Answer
xˉ=m1∬RxρdA — the moment, divided by the mass.
Why is moment of inertia weighted by distance squared?
Answer
Because it measures resistance to spinning, and mass far from the axis resists far more than mass near it.
Wind, current, gravity, magnetic pull. Two questions decide everything that follows: does the wind go around, and does the wind go out?
Written down: F(x,y) = ⟨P(x,y), Q(x,y)⟩. The left field is ⟨−y, x⟩. The right field is ⟨x, y⟩.
a vector fieldF(x,y)=P(x,y)i+Q(x,y)jF(x,y,z)=Pi+Qj+Rk
Gradient fields are the special ones: F=∇f for some scalar f. Physically, f is a hill (or a potential) and F points the way water would run. Section 17.3 is entirely about spotting these.
Check yourself
What does divergence measure?
Answer
Net outflow from a tiny box: faucet if positive, drain if negative.
What does curl measure?
Answer
Local spin — whether a paddlewheel turns. Not whether the flow curves: straight flow that is faster on one side still spins it.
STEP 02
The official problems
14 problems
Straight from the department's own study guide, for exactly these lessons — worked, with the answer held back until you commit. Do these first.
The volume of the region inside x2+y2+z2=4 and inside the cone z=3x2+3y2 is ∫02π∫0a∫0bcdρdϕdθ. What are a, b, and c?
no calculator, same as the exam
Answera=π/6,b=2,c=ρ2sinϕanswer recomputed · sympy
c is the Jacobian and nothing else: dV=ρ2sinϕdρdϕdθ, so c=ρ2sinϕ.
Sphere x2+y2+z2=4 is ρ=2, and since ρ is innermost and starts at 0, b=2.
The cone z=3x2+3y2=3x2+y2 has tanϕ=zx2+y2=31.
So ϕ=6π, and 'inside the cone' means 0≤ϕ≤6π: a=6π.
V=∫02π∫0π/6∫02ρ2sinϕdρdϕdθ.
Traptanϕ=zr, not rz — ϕ is measured down from the positive z-axis. Taking tanϕ=3 gives ϕ=3π, which is the complementary wedge and roughly four times the volume. Second trap: writing c=ρsinϕ, the *polar* Jacobian with a ρ in it, which is dimensionally wrong for a volume.
2
Write the volume of the solid bounded by the cone z=x2+y2 and the sphere x2+y2+z2=8 as a spherical-coordinates integral.
Cone z=x2+y2: here tanϕ=zr=1, so it is the surface ϕ=4π, and the solid inside it is 0≤ϕ≤4π.
Sphere x2+y2+z2=8 is ρ2=8, i.e. ρ=8=22.
Both boundaries are coordinate surfaces in spherical — every limit is a constant, which is the whole reason to use it here.
V=∫02π∫0π/4∫08ρ2sinϕdρdϕdθ.
Value: 2π⋅(1−22)⋅3(8)3=332π(2−1), which the cylindrical set-up ∫02π∫02(8−r2−r)rdrdθ confirms.
TrapReading ρ=8 off x2+y2+z2=8. The equation gives ρ2=8, so ρ=22 — and cubing the wrong one inflates the volume by a factor of 83/83/2≈22.6. Also note the cone/sphere meet at z=2, r=2: that intersection is *not* needed in spherical, which is the payoff for converting.
3
Write the volume of the region inside the sphere of radius 2 centered at (0,0,2) and outside the sphere of radius 2 centered at the origin as a spherical-coordinates integral.
Put both spheres in spherical form. Origin-centred radius 2: ρ=2.
The one centred at (0,0,2): x2+y2+(z−2)2=4⇒ρ2−4ρcosϕ=0⇒ρ=4cosϕ.
'Inside the shifted sphere, outside the origin one' is exactly 2≤ρ≤4cosϕ — the inner radius is the constant, the outer is the ϕ-dependent one.
Those limits are only valid where 4cosϕ≥2, i.e. cosϕ≥21, i.e. ϕ≤3π. That is the ϕ-limit.
V=∫02π∫0π/3∫24cosϕρ2sinϕdρdϕdθ.
Check it independently: the lens where two radius-2 spheres with centres 2 apart overlap has volume 12π(4R+d)(2R−d)2=310π, so the answer is 332π−310π=322π — and the integral above gives 322π.
TrapTaking ϕ up to 2π because the shifted sphere 'sits above the xy-plane'. Past ϕ=3π the outer limit 4cosϕ drops below the inner limit 2, so the integral starts subtracting. The upper ϕ is always where the two ρ-limits meet: solve 4cosϕ=2.
4
Converting ∫−11∫−1−x21−x2∫3(x2+y2)4−x2−y2fdzdydx to spherical gives ∫0a∫0b∫0cf⋅ρ2sinϕdρdϕdθ (order: dρdϕdθ, so a is the upper limit on θ, b on ϕ, c on ρ). Find abc.
The shadow −1≤x≤1, ∣y∣≤1−x2 is the full unit disk, so θ makes a full turn: a=2π.
Ceiling z=4−x2−y2 is the sphere ρ=2, so c=2.
Floor z=3(x2+y2)=3r is a cone with tanϕ=zr=31, so ϕ runs from 0 to 6π: b=6π.
Consistency check on the shadow: the cone meets the sphere where ρ=2, ϕ=6π, giving r=2sin6π=1 — exactly the unit disk in the stem. Good.
abc=2π6π⋅2=61.
TrapSetting b=3π from tanϕ=3. ϕ is the angle from the z-axis, and a steep cone (z large compared to r) means a smallϕ. The disk-radius check catches it: ϕ=3π would give shadow radius 2sin3π=3=1.
5
Find the volume of the solid region enclosed by the surface ρ=12cosϕ.
no calculator, same as the exam
Answer288πanswer recomputed · sympy
ρ=12cosϕ is already a sphere in disguise: multiply by ρ to get ρ2=12ρcosϕ, i.e. x2+y2+z2=12z.
Complete the square: x2+y2+(z−6)2=36 — radius 6, centred at (0,0,6).
So the answer must be 34π(6)3=288π; now get the same thing from the integral.
The surface closes up at ϕ=2π (where ρ=0), so ϕ runs 0 to 2π — not to π.
TrapLetting ϕ run to π out of habit. For ϕ>2π, 12cosϕ<0 and there is no sphere there; the extra range adds another 288π of nonsense (the cos3ϕsinϕ integrand is odd about 2π, so it actually cancels to 0 — an error that can *silently* delete the whole answer). The sphere is fully traced by ϕ∈[0,2π].
Find the center of mass of the rectangle {(x,y)∣0≤x≤2,0≤y≤4} with density ρ(x,y)=1+y.
no calculator, same as the exam
Answer(1,922)answer recomputed · sympy
Mass first: m=∫02∫04(1+y)dydx=∫02[y+2y2]04dx=∫0212dx=24.
xˉ: the density does not depend on x and the region is symmetric in x, so xˉ=1 by inspection. Formally My=∫02∫04x(1+y)dydx=(∫02xdx)(12)=24, and xˉ=2424=1.
Sanity: 922>2, and it should be — the density grows with y, so mass piles up toward the top.
Centre of mass (1,922).
TrapSwapping the moments. Mx=∬yδdA gives yˉ, not xˉ — the subscript names the axis the moment is about, so it pairs with the *other* coordinate. Using Mx for xˉ here returns 922 for a rectangle only 2 wide, which is impossible and is the fastest way to catch it.
2
A cube 0≤x,y,z≤1 has density δ=x+yz. Find the x-coordinate of the center of mass, given ∭cubeδdV=43.
no calculator, same as the exam
Answer1811answer recomputed · sympy
xˉ=mMyz with Myz=∭xδdV; the mass m=43 is handed to you.
Myz=∫01∫01∫01x(x+yz)dzdydx=∫∫∫(x2+xyz)dV.
Both pieces separate over the cube: ∭x2dV=31 and ∭xyzdV=(21)3=81.
Myz=31+81=2411.
xˉ=3/411/24=2411⋅34=1811.
Sanity: 1811>21, correct — the x term in the density weights the far side.
TrapMultiplying only the first term by x: ∫x⋅xdV+∫yzdV=31+41 gives 127 and then xˉ=97. The moment weights the entire density, so distribute the x across x+yz before integrating.
3
Find the x-coordinate of the center of mass of the plate bounded by y=x, y=21x, x=1 with density δ(x,y)=2x. (Given: Mx=∬yδdA, My=∬xδdA.)
no calculator, same as the exam
Answer43answer recomputed · sympy
Sketch the wedge: between y=21x (lower) and y=x (upper), closed off at x=1. They meet at the origin, so 0≤x≤1 and 2x≤y≤x.
Vertical strips are the right choice — each strip has height x−2x=2x.
Sanity: 43 is past the midpoint of [0,1], as it must be — the strips get both taller and denser as x grows.
TrapIntegrating in the order dxdy without splitting. Horizontal strips run from x=y to x=2y only while 2y≤1; above y=21 the right edge is the line x=1 instead, so that order needs two integrals. The printed answer's order avoids that entirely — pick strips that are described by one pair of curves.
4
Calculate the mass of the tetrahedron with corners (0,0,0), (1,0,0), (0,2,0), (0,0,4) with density ρ(x,y,z)=2z.
no calculator, same as the exam
Answer38answer recomputed · sympy
Three of the four corners are on the axes, so use the intercept form of the slanted face: 1x+2y+4z=1.
Solve for the ceiling: z=4(1−x−2y)=4−4x−2y.
Shadow: set z=0 to get x+2y=1, i.e. y=2(1−x), with 0≤x≤1.
TrapTreating a variable density as a constant and reaching for ρ⋅V: the volume is 61⋅2⋅4=34, and 2zˉ⋅V is *not* something you can shortcut without already knowing zˉ. Also: ρ here is a density, not the spherical radius — the symbol collides with Lesson 23 and the units tell you which.
5
A lamina with density ρ(x,y)=xy occupies the region bounded by y=x2, y=1, x=0. The mass is 61. Find the y-coordinate of the center of mass.
no calculator, same as the exam
Answer43answer recomputed · sympy
Region: y=x2 below, y=1 above, x=0 on the left — so 0≤x≤1 and x2≤y≤1.
The mass is given as 61; confirm it cheaply: ∫01∫x21xydydx=∫012x(1−x4)dx=21(21−61)=61. Good.
Mx=∫01∫x21y⋅xydydx=∫01x⋅31−x6dx.
=31(21−81)=31⋅83=81.
yˉ=mMx=1/61/8=86=43.
Sanity: the region hugs the top (it is wide near y=1, pinched near y=0) and the density xy favours large y, so yˉ>21 is right.
TrapTaking the region as 0≤y≤x2 — 'bounded by y=x2' does not say which side. The third boundary x=0 is the tell: the region above the parabola touches the y-axis, the region below it does not need x=0 to close. Getting this backwards flips yˉ to 75-ish territory and, more tellingly, changes the mass away from the given 61 — which is exactly why that value was handed to you.
Consider F=⟨eyz−ysin(xy),zxeyz−xsin(xy),xyeyz⟩. Find a potential function φ if one exists.
no calculator, same as the exam
Answerxeyz+cos(xy)answer recomputed · sympy
We want φ with ∇φ=F, so start with the component that is easiest to antidifferentiate and then patch it up.
Integrate the first component in x, holding y,z fixed: φ=∫(eyz−ysin(xy))dx=xeyz+cos(xy)+g(y,z).
The sign is the whole game here: dxdcos(xy)=−ysin(xy), so the antiderivative of −ysin(xy) is +cos(xy).
Now differentiate that candidate in y: φy=xzeyz−xsin(xy)+gy.
Compare with the given second component zxeyz−xsin(xy): they already match, so gy=0.
Differentiate in z: φz=xyeyz+gz, and the third component is xyeyz, so gz=0 too. Hence g is a constant.
φ=xeyz+cos(xy) (any additive constant is also a potential).
TrapWriting ∫−ysin(xy)dx=−cos(xy). The −y out front is exactly the inner derivative, so the two minus signs cancel and you get +cos(xy). Getting this backwards still passes a careless φx check if you also flip the sign when differentiating.
2
Let a=0 and C be the curve x2−2y2=a. Which F is orthogonal to the tangent of C at every point (x0,y0) on C?
no calculator, same as the exam
Answer⟨2x,−4y⟩answer recomputed · sympy
C is the level curve g(x,y)=a of g(x,y)=x2−2y2.
Key fact: the gradient of g is normal to every level curve of g, hence orthogonal to the tangent of C at each point.
∇g=⟨gx,gy⟩=⟨2x,−4y⟩.
So F=⟨2x,−4y⟩ is orthogonal to T at every (x0,y0) on C (and so is any scalar multiple of it).
Sanity check by hand: implicit differentiation of x2−2y2=a gives 2x−4yy′=0, so T∥⟨4y,2x⟩, and ⟨2x,−4y⟩⋅⟨4y,2x⟩=8xy−8xy=0.
TrapDropping the minus sign and answering ⟨2x,4y⟩ — that is ∇(x2+2y2), the normal to an ellipse, not to this hyperbola. The other classic miss is handing back the tangent ⟨4y,2x⟩ instead of the normal.
3
Which vector field corresponds to the field with vectors rotating counterclockwise (pointing left above the x-axis, right below it)?
no calculator, same as the exam
Answer⟨−y,x⟩answer recomputed · sympy
Counterclockwise rotation means the field is tangent to circles centered at the origin, pointing in the direction of increasing angle.
Parametrize a circle the counterclockwise way: r(t)=⟨cost,sint⟩.
The direction of travel is r′(t)=⟨−sint,cost⟩.
Rewrite that at the point itself, using x=cost, y=sint: the tangent direction is ⟨−y,x⟩.
Spot-check the description: above the x-axis take (0,1), giving ⟨−1,0⟩ (points left); below take (0,−1), giving ⟨1,0⟩ (points right). Matches.
F=⟨−y,x⟩.
TrapAnswering ⟨y,−x⟩, which is the same picture run clockwise. Always test one concrete point above the x-axis rather than trusting the sign pattern; ⟨y,−x⟩ gives ⟨1,0⟩ at (0,1), pointing right, which is the wrong rotation.
4
Suppose f(x,y,z)=xy2z3. Find div(∇f) at the point (2,−1,1).
no calculator, same as the exam
Answer16answer recomputed · sympy
div(∇f) is the Laplacianfxx+fyy+fzz — a scalar, not a vector.
∇f=⟨y2z3,2xyz3,3xy2z2⟩.
Take the matching partial of each component: fxx=∂x(y2z3)=0, fyy=∂y(2xyz3)=2xz3, fzz=∂z(3xy2z2)=6xy2z.
So div(∇f)=0+2xz3+6xy2z.
At (2,−1,1): 2(2)(1)3+6(2)(−1)2(1)=4+12=16.
TrapTwo traps. First, differentiating the wrong slot — you must take ∂x of the first component, ∂y of the second, ∂z of the third, not all three of one component. Second, (−1)2=+1: plugging y=−1 into 6xy2z gives +12, and writing −12 turns 16 into −8.
STEP 03
More practice
4 questions
Mine, on those sections only. Untimed, no calculator — the quiz doesn't allow one either.
0/4 answered
116.5 · Spherical triple integral
Evaluate ∭Ex2+y2+z2dV where E is the unit ball.
Answer B
In spherical the integrand is just ρ, and dV=ρ2sinφdρdφdθ:
∫02π∫0π∫01ρ⋅ρ2sinφdρdφdθ=∫ρ341⋅∫sinφ2⋅∫dθ2π=π.
TrapLetting φ run to 2π. φ ∈ [0, π]; only θ goes all the way round.
A lamina of constant density occupies the triangle 0≤x≤1,0≤y≤x. Find xˉ.
Answer C
Mass (density 1): m=∫01∫0xdydx=∫01xdx=21.
Moment: ∫01∫0xxdydx=∫01x2dx=31.
xˉ=1/21/3=32.
Sensible: the triangle carries more area at large x, so the balance point sits right of centre.
TrapForgetting to divide by the mass — that leaves the moment, 1/3, option B.
Find the polar moment of inertia I0=∬D(x2+y2)dA for the unit disk with constant density 1.
Answer B
In polar the integrand is r2 and dA=rdrdθ, so you integrate r3:
I0=∫02π∫01r2⋅rdrdθ=2π[4r4]01=2π.
Because distance is squared, mass far from the axis counts far more than mass near it.
TrapDropping the Jacobian r, which turns r³ into r² and gives 2π/3.
divF=∂x(x)+∂y(−y)=1−1=0.
2-D curl =Qx−Py=∂x(−y)−∂y(x)=0−0=0.
So the field neither creates nor spins: it stretches along x exactly as fast as it
compresses along y. Zero curl on all of R2 (simply connected) also makes it
conservative, with potential f=21x2−21y2.
TrapAssuming arrows that "spread out" must have positive divergence — here the two axes cancel.
Walk a curve and add something up as you go. If the something is a height you get a curtain's area. If it's a push you get work.
Slice the curtain into thin strips: each is height f × width ds. And ds = |r′(t)| dt, exactly the arc-length piece from Chapter 2.
scalar line integral — with respect to arc length∫Cfds=∫abf(r(t))r′(t)dt
Now the vector version. Wind straight at your back: full help. Wind from the side: does nothing. Wind in your face: fights you. Only the part of the push along your walk counts.
Dot product again, at every point along the curve. F·dr keeps only the along-the-path share of the push.
work — three notations for one integral∫CF⋅dr=∫abF(r(t))⋅r′(t)dt=∫CPdx+Qdy+Rdz
Algorithm · evaluate any line integral
Parametrize C. Line segment from A to B: r(t)=(1−t)A+tB, 0≤t≤1. Circle of radius a: ⟨acost,asint⟩. Graph y=g(x): ⟨t,g(t)⟩.
Compute r′(t). For a scalar integral you also need ∣r′(t)∣.
Substitute the parametrization into F or f so everything is in terms of t alone.
Dot (vector version) or multiply by ∣r′∣ (scalar version).
Integrate over the parameter range. If C is made of several pieces, do each separately and add.
Orientation trap
∫−CF⋅dr=−∫CF⋅dr — reversing the walk flips the sign of work. But ∫−Cfds=∫Cfds — a curtain has the same area no matter which way you walk along its base. Vector integrals care about direction; scalar ones don't.
Check yourself
What does ∫Cfds give when f=1?
Answer
The length of the curve.
You walk C backwards. What happens to ∫CF⋅dr?
Answer
The sign flips. But ∫Cfds does not change — a curtain has the same area either way.
Climb from the car park to the summit. Take the switchbacks or take the ridge — you gained exactly the same height. Some fields are just "downhill from a hill", and for those the route is irrelevant.
Equivalent statements, all meaning the same thing: F = ∇f · the work is path-independent · every closed loop gives zero · curl F = 0 (on a region with no holes).
fundamental theorem for line integrals∫C∇f⋅dr=f(r(b))−f(r(a))=f(end)−f(start)
This is the FTC from Calc I wearing a coat: integrate a derivative, get boundary values. The whole path collapses to two evaluations.
Four ways of saying the same thing
F is conservative ⟺ F=∇f for some potential f ⟺ ∫CF⋅dr is path-independent ⟺ ∮CF⋅dr=0 around every closed loop. On a simply connected region these are also equivalent to curlF=0.
the test2-D: ∂y∂P=∂x∂Q3-D: curlF=∇×F=0
Algorithm · find the potential function f
Test first. Check Py=Qx (2-D) or curlF=0 (3-D). If it fails, stop — there is no f.
Integrate P with respect to x:f=∫Pdx+g(y,z). The constant of integration is a function of the other variables, not a number.
Differentiate your f with respect to y and set it equal to Q. Solve for gy.
Integrate gy in y, picking up a new constant h(z).
Repeat with z and R to pin down h(z). Add a genuine constant C at the end.
Verify:∇f should give back F exactly.
Then use it: ∫CF⋅dr=f(end)−f(start), and the parametrization never gets written down.
Worked example
F=⟨2xy,x2+2y⟩, from (0,0) to (1,3).
test
Py=2x, Qx=2x ✓ conservative
step 2
f=∫2xydx=x2y+g(y)
step 3
fy=x2+g′(y)=x2+2y⇒g′(y)=2y⇒g=y2
potential
f=x2y+y2
evaluate
f(1,3)−f(0,0)=(3+9)−0=12 — no parametrization needed.
Trap
Curl zero is not enough on a region with a hole. The classic F=⟨x2+y2−y,x2+y2x⟩ passes Py=Qx everywhere it is defined, yet a loop around the origin gives 2π, not 0. The region must be simply connected — no punctures.
Check yourself
What is the 2-D test for a conservative field?
Answer
∂y∂P=∂x∂Q — and the region must have no holes.
F is conservative. What is ∮CF⋅dr around any closed loop?
Answer
0. You end where you started, so the height change is zero.
STEP 02
The official problems
15 problems
Straight from the department's own study guide, for exactly these lessons — worked, with the answer held back until you commit. Do these first.
Compute ∫Cf(x,y)ds where f(x,y)=sin(2πx)+xy and C:r(t)=⟨t,t⟩, 0≤t≤1.
no calculator, same as the exam
Answer32answer recomputed · sympy
Scalar line integral: ∫Cfds=∫abf(r(t))∣r′(t)∣dt.
Here r′(t)=⟨1,1⟩, so ∣r′(t)∣=2 and ds=2dt.
On C we have x=t, y=t, so f(r(t))=sin(2πt)+t2.
Set up: ∫01(sin(2πt)+t2)2dt.
∫01sin(2πt)dt=[−2πcos(2πt)]01=−2π1+2π1=0 — a full period integrates away.
∫01t2dt=31, so the value is 2⋅31=32.
TrapForgetting the ∣r′∣=2 factor and reporting 31. The parametrization ⟨t,t⟩ is not unit speed even though it looks tame — ds=dt here.
2
Let C be the helix r(t)=⟨cost,sint,t⟩, 0≤t≤π/2. Compute ∫Cxyds.
no calculator, same as the exam
Answer22answer recomputed · sympy
r′(t)=⟨−sint,cost,1⟩, so ∣r′(t)∣=sin2t+cos2t+1=2.
The helix has constant speed2, so ds=2dt.
On C, xy=costsint.
Set up: ∫0π/2costsint2dt.
Substitute u=sint: 2∫01udu=2⋅21=22.
TrapReading the first two components as a unit circle and concluding ∣r′∣=1. The z=t component contributes a 1 under the radical, making the speed 2; dropping it halves-by-2 the answer to 21.
3
Calculate ∫C(x−y)21ds where C is the straight line segment from (1,0) to (5,3).
no calculator, same as the exam
Answer25answer recomputed · sympy
Parametrize the segment from (1,0) to (5,3) as r(t)=⟨1,0⟩+t⟨4,3⟩=⟨1+4t,3t⟩, 0≤t≤1.
r′(t)=⟨4,3⟩, so ∣r′(t)∣=16+9=5 and ds=5dt (the length of the segment, spread over t∈[0,1]).
The integrand simplifies beautifully: x−y=(1+4t)−3t=1+t, so (x−y)21=(1+t)21.
Set up: ∫01(1+t)25dt.
=5[−1+t1]01=5(−21+1)=25.
TrapUsing dx instead of ds. If you parametrize by x with y=43(x−1) you must carry ds=1+(3/4)2dx=45dx; dropping that factor gives 2 instead of 25. Also note the line never crosses x=y, so the integrand stays finite.
4
Let C be the half-circle x2+y2=4 with x≥0. Compute ∫Cxds.
no calculator, same as the exam
Answer8answer recomputed · sympy
The circle has radius 2, so r(t)=⟨2cost,2sint⟩.
Pick the range that gives x≥0, not y≥0: x=2cost≥0 forces −2π≤t≤2π (the right half).
r′(t)=⟨−2sint,2cost⟩, so ∣r′(t)∣=2 and ds=2dt.
Set up: ∫−π/2π/2(2cost)(2)dt=4∫−π/2π/2costdt.
=4[sint]−π/2π/2=4(1−(−1))=8.
TrapTaking 0≤t≤π out of habit. That traces the upper half (y≥0), over which ∫xds=0 by symmetry — a suspiciously clean wrong answer. The condition is x≥0, so center the interval on t=0.
5
Find the arc length of r(t)=⟨4cost,4sint,3t⟩ for 0≤t≤2π.
no calculator, same as the exam
Answer10πanswer recomputed · sympy
Arc length is L=∫ab∣r′(t)∣dt.
r′(t)=⟨−4sint,4cost,3⟩.
∣r′(t)∣=16sin2t+16cos2t+9=16+9=5 — constant speed, the hallmark of a circular helix.
Set up: L=∫02π5dt=5(2π)=10π.
TrapAdding the radius and the pitch coordinate as if speed were 4+3=7, or forgetting the 3 entirely and getting 8π (the circumference of the shadow circle). The helix is strictly longer than its projection because of the climb.
Given force field F(x,y)=⟨−y,x⟩, find the work to move an object along the ellipse r(t)=⟨2cost,3sint⟩ from (2,0) to (0,3).
no calculator, same as the exam
Answer3πanswer recomputed · sympy
Work is W=∫CF⋅dr=∫abF(r(t))⋅r′(t)dt.
Find the t-range from the endpoints: (2,0) is t=0 and (0,3) is t=2π, so we travel a quarter of the ellipse.
On C: F(r(t))=⟨−y,x⟩=⟨−3sint,2cost⟩ and r′(t)=⟨−2sint,3cost⟩.
Dot them: 6sin2t+6cos2t=6 — the integrand is constant.
Set up: W=∫0π/26dt=6⋅2π=3π.
TrapReaching for Green's theorem. C is an open quarter-arc, not a closed curve, so Green does not apply; and F=⟨−y,x⟩ is not conservative (Py=−1=1=Qx), so the path genuinely matters and FTLI is off the table. You must integrate along this arc.
2
Compute the flux of F(x,y)=⟨x,y⟩ across the curve r(t)=⟨3cost,2sint⟩, −π≤t≤π. (Flux =∫CPdy−Qdx.)
no calculator, same as the exam
Answer12πanswer recomputed · sympy
Flux uses the stated formula ∫CPdy−Qdx with P=x, Q=y — note dy pairs with P, not with Q.
From r(t)=⟨3cost,2sint⟩: dx=−3sintdt and dy=2costdt.
Cross-check with the divergence form: divF=2 and the ellipse encloses area π(3)(2)=6π, so the flux is 2⋅6π=12π.
TrapComputing circulation ∫Pdx+Qdy instead of flux. With this field that swap gives ∫(−6sintcost+6sintcost)dt=0, so the mistake produces a clean 0 and looks deliberate. Keep the pairing Pdy−Qdx straight.
3
Given force field F(x,y,z)=⟨y,z,x⟩, find the work to move an object along the straight line from (0,0,0) to (2,3,4).
no calculator, same as the exam
Answer226answer recomputed · sympy
Parametrize the segment: r(t)=⟨2t,3t,4t⟩, 0≤t≤1, so r′(t)=⟨2,3,4⟩.
Evaluate the field on the curve, remembering the components are shifted: F=⟨y,z,x⟩=⟨3t,4t,2t⟩.
Dot: F⋅r′=6t+12t+8t=26t.
Set up: W=∫0126tdt=26⋅21=226=13.
TrapSubstituting F=⟨2t,3t,4t⟩, i.e. reading ⟨y,z,x⟩ as ⟨x,y,z⟩. That gives ∫0129tdt and the wrong number. The cyclic shift is the entire content of this problem. (Also: F=⟨y,z,x⟩ is not conservative, since Py=1 but Qx=0, so you must integrate.)
4
Find ∫CF⋅Tds where F(x,y,z)=⟨yez,ey+xez,xyez⟩ on a smooth curve from (0,0,0) to (−1,1,1).
no calculator, same as the exam
Answer−1answer recomputed · sympy
The problem says 'a smooth curve' without naming it, which is the signal that F must be conservative — otherwise the answer would not be well defined.
Hunt for φ: integrate the first component in x, ∫yezdx=xyez+g(y,z).
Match the second component: φy=xez+gy must equal ey+xez, so gy=ey and g=ey+h(z).
Match the third: φz=xyez+h′(z) must equal xyez, so h is constant. Take φ=xyez+ey.
FTLI: ∫CF⋅Tds=∫CF⋅dr=φ(−1,1,1)−φ(0,0,0).
φ(−1,1,1)=(−1)(1)e1+e1=−e+e=0 and φ(0,0,0)=0+e0=1.
Value =0−1=−1.
TrapEvaluating φ at the start minus the end. FTLI is end minus start, and here that sign error hands you +1. Second trap: forgetting the ey piece because it does not appear in φx — it is exactly what makes φ(0,0,0)=1 rather than 0.
5
Evaluate ∫CF⋅dr where F(x,y,z)=⟨2xy−yz,x2−xz,−xy⟩ on a smooth curve from (2,1,0) to (3,2,−1).
no calculator, same as the exam
Answer20answer recomputed · sympy
Again the curve is unspecified, so test for conservativity: with P=2xy−yz, Q=x2−xz, R=−xy we get Py=2x−z=Qx, Pz=−y=Rx, Qz=−x=Ry. All three match, so F is conservative.
Build φ: ∫(2xy−yz)dx=x2y−xyz+g(y,z).
Check φy=x2−xz+gy against Q=x2−xz: gy=0. Check φz=−xy+gz against R=−xy: gz=0. So φ=x2y−xyz.
FTLI: value =φ(3,2,−1)−φ(2,1,0).
φ(3,2,−1)=9(2)−(3)(2)(−1)=18+6=24.
φ(2,1,0)=4(1)−0=4.
Value =24−4=20.
TrapSign-slipping on −xyz at the endpoint with z=−1: −(3)(2)(−1)=+6, so φ(3,2,−1)=24, not 12. Writing 18−6 gives 8 and is the single most common wrong answer here.
Find ∫CF⋅dr where F(x,y,z)=⟨2xy−yz,x2−xz,−xy⟩ on a smooth curve from (2,1,0) to (3,2,−1).
no calculator, same as the exam
Answer20answer recomputed · sympy
The path is only described as 'a smooth curve', so the integral had better be path-independent — first verify F is conservative.
With P=2xy−yz, Q=x2−xz, R=−xy: Py=2x−z=Qx, Pz=−y=Rx, Qz=−x=Ry. Conservative.
Find the potential: φ=∫(2xy−yz)dx=x2y−xyz+g(y,z), and both φy=x2−xz and φz=−xy already match, so g is constant.
φ(x,y,z)=x2y−xyz.
FTLI: ∫CF⋅dr=φ(3,2,−1)−φ(2,1,0)=(18+6)−(4−0).
=24−4=20.
TrapTrying to parametrize. No curve was given, so any attempt to build one is wasted work — and if you accidentally pick a path while F were not conservative, the answer would be meaningless. Confirm the mixed partials first, then use endpoints only. Watch −xyz at z=−1, which contributes +6.
2
For F=⟨eyz−ysin(xy),zxeyz−xsin(xy),xyeyz⟩, find a potential function.
no calculator, same as the exam
Answerφ=xeyz+cos(xy)answer recomputed · sympy
Antidifferentiate the first component in x: φ=∫(eyz−ysin(xy))dx.
Term by term, ∫eyzdx=xeyz (as x varies, eyz is a constant) and ∫−ysin(xy)dx=cos(xy).
So φ=xeyz+cos(xy)+g(y,z).
Test against the second component: φy=xzeyz−xsin(xy)+gy, and we need zxeyz−xsin(xy), so gy=0.
Test against the third: φz=xyeyz+gz, and we need xyeyz, so gz=0.
Therefore φ=xeyz+cos(xy), up to an additive constant.
TrapDifferentiating eyz with respect to y and writing zeyz without the x that the antiderivative introduced. The x in xeyz must be carried through, and it is exactly what makes φy=xzeyz match the printed zxeyz.
3
A potential for F=⟨siny,xcosy⟩ is:
no calculator, same as the exam
Answerxsiny+1answer recomputed · sympy
First confirm a potential exists: P=siny, Q=xcosy, and Py=cosy=Qx. Conservative on all of R2.
Integrate P in x: φ=∫sinydx=xsiny+g(y).
Differentiate in y: φy=xcosy+g′(y), which must equal Q=xcosy, so g′(y)=0 and g is any constant.
The family of potentials is φ=xsiny+C. The printed choice takes C=1: φ=xsiny+1.
TrapRejecting the '+1' as a typo. Potentials are only determined up to an additive constant, so xsiny, xsiny+1, and xsiny−7 are all correct; on a multiple-choice list the constant is there precisely to see whether you know that. The real error to avoid is φ=−xcosy, from mistakenly antidifferentiating siny in y instead of x.
4
Find ∫CF⋅Tds where F=⟨yez,ey+xez,xyez⟩ on a smooth curve from (0,0,0) to (−1,1,1).
no calculator, same as the exam
Answer−1answer recomputed · sympy
∫CF⋅Tds is just ∫CF⋅dr rewritten, since dr=Tds.
No specific curve is given, so check conservativity: with P=yez, Q=ey+xez, R=xyez we get Py=ez=Qx, Pz=yez=Rx, Qz=xez=Ry. Conservative.
Potential: ∫yezdx=xyez+g(y,z); then φy=xez+gy=ey+xez forces gy=ey, and φz=xyez+h′(z)=xyez forces h′=0.
φ=xyez+ey.
FTLI: φ(−1,1,1)−φ(0,0,0)=((−1)(1)e+e)−(0+1).
=0−1=−1.
TrapSeeing Tds and thinking you must compute ∣r′∣ and a unit tangent. You never do — Tds=dr, the speed cancels, and the whole problem collapses to two endpoint evaluations. The second trap is φ(0,0,0)=0: the ey term makes it e0=1.
5
Evaluate ∫C∇(x2y−xyz)⋅dr where C is any path from (1,0,0) to (2,1,−1).
no calculator, same as the exam
Answer6answer recomputed · sympy
The integrand is already a gradient, ∇φ with φ(x,y,z)=x2y−xyz, so FTLI applies immediately and no conservativity test is needed.
∫C∇φ⋅dr=φ(end)−φ(start), independent of the path.
φ(2,1,−1)=22(1)−(2)(1)(−1)=4+2=6.
φ(1,0,0)=12(0)−(1)(0)(0)=0.
Value =6−0=6.
TrapExpanding ∇(x2y−xyz)=⟨2xy−yz,x2−xz,−xy⟩ and then grinding out a line integral along some invented path. All that work is discarded by FTLI. Within the evaluation, the sign trap is −xyz at z=−1, which adds2; writing 4−2=2 is the common wrong answer.
STEP 03
More practice
3 questions
Mine, on those sections only. Untimed, no calculator — the quiz doesn't allow one either.
0/3 answered
117.2 · Line integral · work
Evaluate ∫Cydx+xdy+zdz along the straight segment from (0,1,1) to (2,3,5).
Answer D
r(t)=⟨2t,1+2t,1+4t⟩,0≤t≤1, so dx=2dt,dy=2dt,dz=4dt.
∫01[(1+2t)(2)+(2t)(2)+(1+4t)(4)]dt=∫01(2+4t+4t+4+16t)dt=∫01(6+24t)dt=6+12=18.
TrapMismatching which component multiplies which differential.
F=⟨y+z,x+z,x+y⟩. Find ∫CF⋅dr along any path from (0,0,0) to (1,2,3).
Answer C
∇×F=⟨1−1,1−1,1−1⟩=0 on all of R3, so F is conservative.
Build the potential: f=∫(y+z)dx=xy+xz+g(y,z); matching fy=x+z gives gy=z⇒g=yz.
So f=xy+yz+xz, and ∫CF⋅dr=f(1,2,3)−f(0,0,0)=2+6+3=11.
TrapParametrizing a path. Test for conservative first — it collapses to two evaluations.
Wed Nov 18 · 8:00 pm · Lessons 17–31 · 15.8–17.6 (Lagrange through surface integrals)
Midterm 2
Lessons 17–31, one hour, multiple choice, no calculator. Starts at Lagrange.
Worth knowing. Green's theorem (17.4) and curl / divergence (17.5) are never quizzed — Lesson 29 and 30 fall between Quiz 9 and this exam. The ground rules say Lessons 17–31; the calendar teaches through Lesson 33 before the exam. Assume 17–31 and know 32–33.
The department publishes a study guide for this exam, and it is the source of truth for coverage. Every one of its practice problems is worked out, with the answer held back until you commit to an attempt, at Exam 2, worked. Start there; use this tab for the sections no quiz ever tested.
Only the sections no quiz ever covers are shown here. Everything from the earlier quizzes is on this exam too — work those tabs as well.
Official problems for these lessons
The department's own questions for exactly these lessons are worked below, before anything of mine.
Chop the region into cells and spin each one counterclockwise. Along any wall between two cells, one cell pushes up and its neighbour pushes down — dead even. The only walls with nobody on the other side are on the outside.
Each cell's spin is (Qₓ − Pᵧ) dA — the 2-D curl. Add them all up and you're left holding a single trip around the boundary.
circulation form∮CPdx+Qdy=∬D(∂x∂Q−∂y∂P)dA
flux (divergence) form — same theorem, rotated 90°∮CF⋅nds=∬D(∂x∂P+∂y∂Q)dA=∬DdivFdA
Algorithm · apply Green's theorem
Check the hypotheses.C must be closed, simple (no self-crossings), positively oriented — counterclockwise, region on your left — and F smooth on all of D.
Identify P and Q from ∮Pdx+Qdy.
Compute Qx−Py and simplify. This is usually far smaller than either piece.
Set up ∬D over the enclosed region — polar if it's round.
If C runs clockwise, apply the theorem anyway and negate the result.
Also runs backwards: a horrible double integral can become an easy loop integral. Read the problem for which direction is cheaper.
area for free — choose P and Q so that Qx−Py=1A=∮Cxdy=−∮Cydx=21∮Cxdy−ydx
Algorithm · regions with a hole
Orient both boundaries positively: outer curve counterclockwise, inner curve clockwise — so the region always stays on your left.
Apply Green's theorem to the whole boundaryC=Couter+Cinner.
If Qx−Py=0 on the region between them, the two loop integrals are equal in magnitude — so you can replace a nasty outer curve with a tiny convenient circle around the hole.
That last trick is how the 2π in the 17.3 trap gets computed.
Check yourself
Which way must C be oriented for Green's theorem?
Answer
Counterclockwise, region on your left. If it runs clockwise, apply the theorem anyway and negate.
Give a line integral that returns the enclosed area.
Curl: drop a paddlewheel in and see if it spins. Not "is the water going in a circle" — "is one side pushed harder than the other".
Curl is a vector in 3D: it points along the axis the paddlewheel spins about, by the right-hand rule, and its length is how fast. In 2D only the z-component survives: Qₓ − Pᵧ.
curl — a vector, computed as a determinantcurlF=∇×F=i∂x∂Pj∂y∂Qk∂z∂R
The direction of curl is the axis the paddlewheel spins about (right-hand rule); its length is how fast. In 2-D only the k-component survives, and it is exactly the Qx−Py from Green's theorem.
Divergence: draw a tiny box. Count what flows in, count what flows out. More out than in means something inside is making it.
Divergence is a number, not a vector — there's no direction to "how much is being created here."
divergence — a scalardivF=∇⋅F=∂x∂P+∂y∂Q+∂z∂R
Two identities worth memorising outright
div(curlF)=0 — a spin never creates anything. curl(∇f)=0 — a hill never makes a whirlpool. Together they explain why a conservative field always has zero curl, and why a curl field always has zero flux out of a closed surface.
Type trap
curl of a vector field is a vector. Divergence of a vector field is a number. So div(curlF) makes sense and curl(divF) does not — you can't cross a scalar. Checking the type of your answer catches half of all algebra errors here.
Check yourself
What is div(curlF)?
Answer
0, always. A spin never creates anything.
What is curl(∇f)?
Answer
0, always. A hill never makes a whirlpool.
Is divF a scalar or a vector?
Answer
A scalar. So curl(divF) is not even defined — checking types catches half of all errors here.
STEP 02
The official problems
75 problems
Straight from the department's own study guide, for exactly these lessons — worked, with the answer held back until you commit. Do these first.
Find the maximum value of f(x,y)=3+2x−y subject to the constraint x2+2xy+4y2=7.
no calculator, same as the exam
Answer10answer recomputed · sympy
The constraint g(x,y)=x2+2xy+4y2=7 is a closed curve (an ellipse), so f attains a max and a min on it. Use Lagrange: ∇f=λ∇g.
∇f=⟨2,−1⟩ and ∇g=⟨2x+2y,2x+8y⟩, so the system is 2=λ(2x+2y), −1=λ(2x+8y), x2+2xy+4y2=7.
Add twice the second equation to the first to kill x: 2+2(−1)=λ[(2x+2y)+2(2x+8y)] gives 0=λ(6x+18y). Since λ=0 (else ∇f=0, false), x=−3y.
Substitute into the constraint: 9y2−6y2+4y2=7y2=7, so y=±1 and (x,y)=(−3,1) or (3,−1).
Evaluate: f(−3,1)=3−6−1=−4 and f(3,−1)=3+6+1=10.
Maximum value =10 (attained at (3,−1); the minimum is −4).
Trap∇g of the cross term is the usual casualty: ∂x(2xy)=2y and ∂y(2xy)=2x, so ∇g=⟨2x+2y,2x+8y⟩, not⟨2x+2y,8y⟩. Also, Lagrange hands you *critical values*, plural. The largest of the list is the answer, so you must evaluate f at every solution point, not just the first one you find.
2
Find the maximum value of f(x,y)=8x−6y subject to the constraint (x−1)2+y2=1.
no calculator, same as the exam
Answer18answer recomputed · sympy
Constraint: the circle g(x,y)=(x−1)2+y2=1, centre (1,0), radius 1. f is linear, so its extremes sit where the gradient of f is normal to the circle.
Lagrange: ⟨8,−6⟩=λ⟨2(x−1),2y⟩, i.e. 8=2λ(x−1) and −6=2λy.
So x−1=λ4 and y=−λ3. Plug into the constraint: λ216+λ29=1⇒λ2=25,λ=±5.
Sanity check with geometry: f=8x−6y at the centre (1,0) is 8, and ∣∇f∣=64+36=10, so on a radius-1 circle f ranges over 8±10={−2,18}. Same answer.
TrapThe circle is not centred at the origin. Writing the constraint gradient as ⟨2x,2y⟩ (forgetting the shift) is the standard wreck here; it must be ⟨2(x−1),2y⟩. The shortcut fmax=f(centre)+∣∇f∣⋅R works only because f is linear and the constraint is a circle; do not reach for it on an ellipse.
3
The maximum and minimum values of f(x,y)=(x−2)2+(y−4)2 subject to x2+y2=5 are, respectively:
f(x,y)=(x−2)2+(y−4)2 is the squared distance from (2,4); the constraint is the circle of radius 5 about the origin. So this is really a nearest/farthest point problem.
Lagrange: ⟨2(x−2),2(y−4)⟩=λ⟨2x,2y⟩, i.e. x−2=λx and y−4=λy.
Then x(1−λ)=2 and y(1−λ)=4, so y=2x (dividing; note λ=1, since λ=1 would force 2=0).
Constraint: x2+(2x)2=5x2=5⇒x=±1, giving (1,2) and (−1,−2).
Maximum 45, minimum 5. (Check: ∣(2,4)∣=20=25, so distances to the circle are 25∓5, whose squares are 5 and 45.)
TrapTwo traps. First, f is the squared distance, so do not take a square root at the end and report 45 and 5. Second, the point (2,4) lies *outside* the circle, so both critical points are genuine, one near and one far; the min is not0.
4
Find the maximum of f(x,y,z)=x+y+z subject to (x−1)2+y2+z2=1.
no calculator, same as the exam
Answer1+3answer recomputed · sympy
Constraint: the sphere g=(x−1)2+y2+z2=1 of radius 1 centred at (1,0,0). f=x+y+z is linear, so Lagrange with ∇f=⟨1,1,1⟩.
Taking the + branch, x=1+31,y=z=31, so f=1+33=1+3. The − branch gives 1−3.
Maximum =1+3.
TrapSame shifted-centre trap as before: ∇g=⟨2(x−1),2y,2z⟩. And because the three partials give x−1=y=z (not x=y=z), the symmetric guess x=y=z=1/3 is wrong. The geometric check is f(centre)+∣∇f∣R=1+3⋅1.
5
Find the absolute maximum value M and absolute minimum value m of f(x,y)=x+y subject to x2−xy+y2=1.
no calculator, same as the exam
AnswerM=2,m=−2answer recomputed · sympy
The constraint x2−xy+y2=1 is an ellipse (rotated 45∘), a closed bounded curve, so a max and a min both exist.
Lagrange: ⟨1,1⟩=λ⟨2x−y,2y−x⟩, so 1=λ(2x−y) and 1=λ(2y−x).
Setting the two right sides equal: 2x−y=2y−x⇒3x=3y⇒y=x.
Constraint: x2−x2+x2=x2=1, so (1,1) and (−1,−1).
f(1,1)=2 and f(−1,−1)=−2, so M=2,m=−2.
TrapDifferentiating the cross term −xy: ∂x=−y, ∂y=−x — it is easy to drop a sign and get 2x+y. Also, cancelling λ from the two equations is legal here only because λ=0 (if λ=0 then 1=0), so you should say so rather than divide blindly.
The integrand factors, xy=xy, and the region is the rectangle [0,4]×[0,1], so the double integral splits into a product of single integrals.
∫01∫04xydxdy=(∫04xdx)(∫01ydy).
∫04x1/2dx=32x3/204=32⋅8=316.
∫01y1/2dy=32.
Product: 316⋅32=932.
TrapRead the differentials: dxdy means the inner variable is x (limits 0 to 4) and the outer is y (limits 0 to 1). Pairing x with [0,1] gives 32⋅316 by luck here, but on a non-symmetric integrand the same misreading is fatal. The split into a product is legal only because the integrand factors and the limits are constants.
2
Find the volume enclosed by x=0, x=3, y=0, y=2, z=1, and z=ex+y.
no calculator, same as the exam
Answere5−e3−e2−5answer recomputed · sympy
The base is the rectangle R=[0,3]×[0,2]; the solid sits between the floor z=1 and the roof z=ex+y. On R we have x+y≥0, so ex+y≥1 and the roof really is above the floor everywhere.
Volume =∬R(ex+y−1)dA=∫02∫03(exey−1)dxdy.
∬RexeydA=(∫03exdx)(∫02eydy)=(e3−1)(e2−1).
∬R1dA=area(R)=3⋅2=6.
Volume =(e3−1)(e2−1)−6=e5−e3−e2+1−6=e5−e3−e2−5.
TrapForgetting to subtract the floor: integrating ex+y alone gives the volume under the surface down to z=0, not down to z=1. The correction is −area(R)=−6, and note (e3−1)(e2−1) already contributes +1, so the constant is 1−6=−5, not −6. Also ex+y=exey is what lets the integral factor; ex+y=ex+ey.
3
Find the average value of f(x,y)=x2y over the rectangle R with vertices (−1,0), (−1,5), (1,5), (1,0).
no calculator, same as the exam
Answer65answer recomputed · sympy
The four vertices give the rectangle R=[−1,1]×[0,5], of area A=2⋅5=10.
Average value =A1∬RfdA=101∫05∫−11x2ydxdy.
∫−11x2dx=3x3−11=32 and ∫05ydy=225.
∬Rx2ydA=32⋅225=325.
Average =1025/3=65.
TrapDividing by the wrong thing, or not dividing at all: the answer to an *average value* question is the integral over the area, here 10. Second trap: x2 is even, so ∫−11x2dx=32, not 0 — students who remember 'symmetric interval ⇒ zero' from odd integrands kill this problem instantly.
4
Evaluate ∫01∫01∫02−yexdzdxdy.
no calculator, same as the exam
Answer23(e−1)answer recomputed · sympy
Innermost is dz and the integrand ex is free of z, so the first integration is just multiplication by the height: ∫02−yexdz=ex(2−y).
The remaining limits are constants, so ∫01∫01ex(2−y)dxdy=(∫01exdx)(∫01(2−y)dy).
∫01exdx=e−1.
∫01(2−y)dy=2−21=23.
Value =23(e−1).
TrapThe order is dzdxdy, so the z-limit 2−y belongs to the *innermost* integral and depends on the *outermost* variable — perfectly legal, but you must integrate z first and carry (2−y) outward. Attempting to do dx first, or treating the upper limit as 2−x, both break it. Once z is gone the integral separates only because ex and (2−y) live in different variables.
5
Evaluate ∬R2+x3x2ydA over R={1≤x≤2,0≤y≤4}.
no calculator, same as the exam
Answer38(ln10−ln3)answer recomputed · sympy
R is a rectangle and the integrand splits as 2+x3x2⋅y, so the integral is a product of two one-variable integrals.
∬R2+x3x2ydA=(∫122+x3x2dx)(∫04ydy).
For the x-integral substitute u=2+x3, du=3x2dx, so x2dx=3du; as x:1→2, u:3→10.
∫3103u1du=31(ln10−ln3).
∫04ydy=8.
Value =8⋅31(ln10−ln3)=38(ln10−ln3)=38ln310.
TrapWhen you substitute u=2+x3 you must change the limits to u:3→10; leaving 1 and 2 gives 31ln12, a completely different number. The other classic is losing the 31 from du=3x2dx.
TrapGetting the new limits backwards. The old inner limit y≥3x becomes x≤y3, so the new inner range is 0≤x≤y3 — noty3≤x≤27. Sketch the region: it is the piece *above* the cube-root curve, i.e. to the *left* of x=y3. The factor 4 in the integrand is not decoration; it is exactly the 4y3 the substitution needs after the swap.
2
Evaluate ∬RxsinxdA where R is bounded by the x-axis, y=x, and x=1 (integrate in y first, then x).
no calculator, same as the exam
Answer1−cos(1)answer recomputed · sympy
Sketch R: bounded below by y=0, above by y=x, on the right by x=1 — the triangle with vertices (0,0),(1,0),(1,1).
xsinx has no elementary antiderivative in x, so we must integrate in y first (as instructed), which makes x a constant.
∬RxsinxdA=∫01∫0xxsinxdydx.
Inner: xsinx⋅(x−0)=sinx. The offending x in the denominator cancels — that is the whole point of the order.
∫01sinxdx=−cosx01=1−cos(1).
TrapChoosing dydx is not optional here: the other order gives ∫y1xsinxdx, which cannot be evaluated in closed form. Also the inner limits are 0≤y≤x — a *variable* upper limit. Writing ∫01 for the inner y-integral turns the triangle into the unit square and gives ∫01xsinxdx, which is not elementary.
3
Find the volume of the solid under z=2x and above the region in the first quadrant bounded by the coordinate axes and x+y=1.
no calculator, same as the exam
Answer31answer recomputed · sympy
The base region D is the triangle in the first quadrant cut off by x+y=1: vertices (0,0),(1,0),(0,1). The roof is z=2x≥0 there, so the volume is the plain double integral.
As a type-I region: 0≤x≤1 and 0≤y≤1−x.
V=∬D2xdA=∫01∫01−x2xdydx.
Inner: 2x(1−x)=2x−2x2.
∫01(2x−2x2)dx=1−32=31.
TrapUsing ∫01∫01 (the square) instead of the triangle: the hypotenuse y=1−x must appear as the inner limit. Also note the integrand is 2x, so the answer is not the triangle's area 21 times anything obvious; you have to actually integrate 2x(1−x).
4
Change the order of integration for ∫02∫x22xf(x,y)dydx.
no calculator, same as the exam
Official answer(original) — ∫04∫yy/2fdxdychecked · official answer wrong
Read the region from the given order dydx: 0≤x≤2 and x2≤y≤2x — the lens between the parabola y=x2 and the line y=2x.
Find where the two curves meet: x2=2x⇒x=0 or x=2, giving the corner points (0,0) and (2,4). So y ranges over [0,4] overall.
Now solve each boundary for x. From y=2x we get x=2y; from y=x2 (with x≥0) we get x=y.
For a fixed y∈[0,4], which is the left edge? Test y=1: the line gives x=21, the parabola gives x=1. So x runs from 2yup toy.
Reversed order: ∫04∫y/2yf(x,y)dxdy.
Confidence check with a test integrand f=xy: the original order gives 38, and ∫04∫y/2yxydxdy=38 as well.
TrapInside the lens the parabola is the right boundary and the line is the left one (for 0<x<2, x2<2x means the parabola sits *below* the line, which flips to *right of* when you read horizontally). Writing ∫yy/2 reverses the limits and negates the integral. Also x=y, not x=±y: the region lives in x≥0.
The official answer is wrongThe official answer prints the inner limits as ∫yy/2, which is upside down. For 0<y<4 we have 2y<y (at y=1: 0.5<1), so the lower limit must be y/2 and the upper limit y. As printed, the integral equals the negative of the original. Numerical check with f(x,y)=xy: the original iterated integral is 38, the correctly reversed one is 38, and the printed one is −38.
5
Reverse the order and evaluate: ∫02ln3∫y/2ln3ex2dxdy.
no calculator, same as the exam
Answer2answer recomputed · sympy
ex2 has no elementary antiderivative, so the order must be reversed. Let a=ln3 for short; the limits are 0≤y≤2a and 2y≤x≤a.
Sketch: the region is the triangle with vertices (0,0), (a,0) and (a,2a), bounded by the line x=y/2 (i.e. y=2x), the vertical line x=a, and the x-axis.
Reading it the other way: x runs 0→a, and for each x, y runs from 0 to 2x.
TrapThe outer limit 2ln3 and the inner ln3 are *different* numbers — reversing carelessly and writing y from 0 to x (instead of 2x) loses the factor of 2 that the substitution needs, leaving a non-elementary integral. Also eln3=3, so the final answer is the clean 2, not eln3 left unsimplified.
TrapThe integrand becomes r2, not r: one factor of r is x2+y2 and the second is the Jacobian from dA=rdrdθ. Dropping the Jacobian gives 49π. Second trap: θ runs only to π/2 — the limits x∈[0,3], y≥0 give a quarter disk, not a half disk.
2
Compute ∬Re−(x2+y2)dA where R is the disk x2+y2≤4.
no calculator, same as the exam
Answerπ(1−e−4)answer recomputed · sympy
R is the full disk of radius 2 and the integrand depends only on x2+y2, so polar is the obvious move: 0≤r≤2, 0≤θ≤2π.
∬Re−(x2+y2)dA=∫02π∫02e−r2rdrdθ.
Inner: substitute u=r2, du=2rdr, so ∫02e−r2rdr=21∫04e−udu=21(1−e−4).
Outer: multiply by ∫02πdθ=2π.
Value =2π⋅21(1−e−4)=π(1−e−4).
TrapWithout the Jacobian r, ∫e−r2dr is not elementary at all — the r is exactly what makes the substitution work, so its absence should be an immediate alarm. Also r goes to 2, so u=r2 goes to 4, giving e−4; writing e−2 is the standard slip.
3
Rewrite in polar (do not evaluate): ∫01∫−1−y21−y2(x2+y2)1/2dxdy.
no calculator, same as the exam
Answer∫0π∫01r2drdθanswer recomputed · sympy
Region: 0≤y≤1 with −1−y2≤x≤1−y2, i.e. x2+y2≤1 with y≥0 — the upper half of the unit disk.
In polar the half disk is 0≤r≤1, 0≤θ≤π (the full sweep of the upper half plane).
The integrand (x2+y2)1/2=r, and dA=rdrdθ, so r⋅r=r2.
∫01∫−1−y21−y2(x2+y2)1/2dxdy=∫0π∫01r2drdθ.
(Not asked for, but its value is 3π, and the original Cartesian integral evaluates to 3π as well.)
Trapθ ranges over [0,π], not [0,π/2]: x takes negative values here, so the region covers the second quadrant too. And the integrand must pick up the Jacobian, r⋅rdrdθ=r2drdθ; writing ∫0π∫01rdrdθ is the most common wrong answer on this exact problem.
4
Evaluate ∫0π∫0π−y2sin(x2+y2)dxdy.
no calculator, same as the exam
Answer2πanswer recomputed · sympy
Region: 0≤y≤π, 0≤x≤π−y2, i.e. x2+y2≤π with x,y≥0 — the quarter disk of radius π.
sin(x2+y2) has no elementary antiderivative in Cartesian form, so polar is forced: 0≤r≤π, 0≤θ≤2π.
TrapThe radius is π, so u=r2 runs to π — not to π. Writing ∫0πsinudu is the killer here. As always, the Jacobian r is what makes the u-substitution possible; without it the integral is non-elementary.
5
Compute ∫−33∫−9−y29−y2ex2+y2dxdy.
no calculator, same as the exam
Answerπ(e9−1)answer recomputed · sympy
Region: −3≤y≤3 and −9−y2≤x≤9−y2 — the entire disk x2+y2≤9 of radius 3.
Polar: 0≤r≤3, 0≤θ≤2π, integrand ex2+y2=er2, and dA=rdrdθ.
∫−33∫−9−y29−y2ex2+y2dxdy=∫02π∫03er2rdrdθ.
Inner: u=r2, du=2rdr, u:0→9, so 21∫09eudu=21(e9−1).
Outer: 2π⋅21(e9−1)=π(e9−1).
TrapBoth limits are symmetric, so θ sweeps the full2π; using π (half disk) or π/2 halves or quarters the answer. And u=r2 runs to 9, giving e9 — writing e3 is the standard misfire. The r from dA is again mandatory.
Read the order off the differentials: dzdxdy. The innermost variable is z, and the integrand ex does not contain z — so the inner integral is just multiplication by the height.
∫02−yexdz=ex(2−y).
Now the x and y limits are both constants, and what is left separates: ∫01∫01ex(2−y)dxdy=(∫01exdx)(∫01(2−y)dy).
∫01exdx=e−1 and ∫01(2−y)dy=2−21=23.
Product: 23(e−1).
TrapThe limit 2−y sits on the inner integral while y is the outer variable. That is legal — inner limits may involve any variable integrated later — but it means you cannot do the y-integral first. Someone who starts at the outside gets ∫01(2−y)dy sitting where a z-integral belongs and loses the factor entirely.
2
Let E be the solid in the first octant above the xy-plane and below 4x+2y+z=8. For ∭EfdV=∫0a∫0b∫0cfdzdydx:
no calculator, same as the exam
Answera=2,b=4−2xanswer recomputed · sympy
Order dzdydx means: z is bounded by surfaces, y by curves in the xy-plane, x by numbers. Build it from the inside out.
First octant gives the floor z=0; the plane gives the ceiling z=8−4x−2y. So c=8−4x−2y.
Project the solid onto the xy-plane by setting z=0 in the plane: 4x+2y=8, i.e. y=4−2x. That triangle, together with x≥0,y≥0, is the shadow.
For a fixed x, y runs from 0 up to that line: b=4−2x.
x runs until the shadow closes, i.e. until 4−2x=0: a=2.
∭EfdV=∫02∫04−2x∫08−4x−2yfdzdydx.
TrapUsing the x-intercept of the plane, x=2... by luck that is right here, but the reason matters: a is where the shadow closes, not where the plane meets an axis. Get b by setting z=0, not x=0; setting x=0 gives y=4, which is the wrong constant limit and makes the region a box.
3
Let D be the solid bounded by x=0, z=0, y=x, and x+y+z=2. Write ∭DfdV as an iterated integral.
no calculator, same as the exam
Official answere.g. ∫01∫x2−x∫02−x−yfdzdydx , or ∫01∫02−x∫x2−x−zfdydzdxchecked · official answer wrong
Four surfaces, so list the inequalities: x≥0, z≥0, y≥x, and x+y+z≤2.
Take z innermost: 0≤z≤2−x−y.
Shadow in the xy-plane is where that height is non-negative and y≥x: the triangle x≤y≤2−x, x≥0.
That triangle closes when x=2−x, i.e. x=1. So ∫01∫x2−x∫02−x−yfdzdydx.
Reordering to dydzdx: solve y≥x and y≤2−x−z, so x≤y≤2−x−z.
Those two limits cross when 2−x−z=x, i.e. at z=2−2x — so z must stop at 2−2x, giving ∫01∫02−2x∫x2−x−zfdydzdx.
Sanity check both with f=1: each gives volume 32.
TrapWhen you reorder, the new outer limit is where the inner limits collide, not where some surface hits an axis. Writing z from 0 to 2−x (the plane's z-intercept above y=0) puts z in a range where the inner y-limits are upside-down, 2−x−z<x, and the integral quietly subtracts the phantom slab instead of erroring.
The official answer is wrongThe second offered order is wrong as printed. With z running to 2−x, every z>2−2x has 2−x−z<x, so the inner y-integral runs backwards. With f=1 the printed form evaluates to 21; the true volume, and the value of the printed first form, is 32. The z-limit must be 2−2x.
4
Let V be bounded by z=6−x2−y2 and z=x2+y2. If V=∫−33∫ab∫x2+y26−x2−y2dzdydx, what are a and b?
no calculator, same as the exam
Answera=−3−x2,b=3−x2answer recomputed · sympy
The z-limits are already given: the solid sits between the upward paraboloid z=x2+y2 and the downward one z=6−x2−y2.
They meet where x2+y2=6−x2−y2, i.e. 2(x2+y2)=6, so x2+y2=3 — a circle of radius 3.
That circle is the shadow's boundary, which is why the outer limits are ±3.
Solving x2+y2=3 for y: a=−3−x2, b=3−x2.
TrapSetting x2+y2=6 (from z=6−x2−y2 with z=0) instead of equating the two surfaces. The paraboloids never reach z=0 together; the boundary of the shadow is where the ceiling meets the floor, and that is radius 3, not 6. The given outer limits ±3 are the tell.
5
Evaluate ∫−11∫−1−y21−y2∫−11(x2+y2)3/2dzdxdy.
no calculator, same as the exam
Answer54πanswer recomputed · sympy
Look at the shadow before touching the integrand: −1≤y≤1 with −1−y2≤x≤1−y2 is the full unit disk, and z runs over the fixed interval [−1,1]. The solid is a cylinder.
(x2+y2)3/2 is a function of distance from the z-axis alone — cylindrical coordinates, and the z-integral is just a factor of 2.
(x2+y2)3/2=r3, and dV=rdrdθdz, so the integrand becomes r3⋅r=r4.
∫−11∫02π∫01r4drdθdz=2⋅2π⋅51=54π.
TrapConverting (x2+y2)3/2 to r3 and then integrating ∫01r3dr=41 — the extra r from dA=rdrdθ is dropped and the answer comes out π instead of 54π. The exponent you integrate is always one higher than the one you converted.
TrapConverting the integrand but keeping the Cartesian limits in spirit — reading 0≤x≤2 as 'θ from 0'. The upper x-limit 2 is a vertical line, r=2secθ, not a ray; only the presence of the lower boundary y=x turns it into the clean ray θ=4π. Any time a straight edge is not through the origin, polar limits stop being constants.
The official answer is wrongThe printed Cartesian stem and the printed cylindrical answer describe different regions. As written, 0≤y≤4−x2 over 0≤x≤2 has area 1+2π, while the wedge 4π≤θ≤2π,r≤2 has area 2π — the answer is missing the triangular piece below y=x. The printed answer is correct for the inner limit ∫x4−x2; with the printed lower limit 0 the rewrite needs the two pieces above.
2
Evaluate ∫02∫04−x2∫x2+y22(x2+y2)dzdydx using cylindrical coordinates.
no calculator, same as the exam
Answer34π(2−1)answer recomputed · sympy
Shadow first: 0≤x≤2, 0≤y≤4−x2 is the quarter disk of radius 2 in the first quadrant, so 0≤θ≤2π, 0≤r≤2.
The z-limits are both cones: z=x2+y2=r below and z=2(x2+y2)=2r above.
Height of the column at radius r: 2r−r=(2−1)r.
With dV=rdrdθdz: ∫0π/2∫02(2−1)r⋅rdrdθ=2π(2−1)⋅38.
=34π(2−1).
Trap2(x2+y2) is 2r, not 2r and not r2 — pulling the 2 out from under the root without the square root gives 34π exactly one unit too large. Second trap: the height (2−1)r already carries one power of r, so the integrand is r2, not r; forgetting the Jacobian here costs a factor of 23⋅22 and lands on π(2−1).
3
Convert to cylindrical and evaluate ∫010∫0100−x2∫0x2+y2x2+y21dzdydx.
no calculator, same as the exam
Answer25πanswer recomputed · sympy
The shadow 0≤x≤10, 0≤y≤100−x2 is a quarter disk of radius 10: 0≤θ≤2π, 0≤r≤10.
The ceiling z=x2+y2 is the cone z=r; the integrand x2+y21=r1.
So the full cylindrical integrand is r1⋅r=1 — the Jacobian cancels the singularity. That is the point of the problem.
TrapTwo errors cancel here if you are careless in exactly the wrong way. Dropping the Jacobian leaves ∫∫r1⋅rdz vs ∫∫r1dz — the second gives ∫0101dr=10 and the answer 5π. The integrand being 1 after conversion is a result, not an accident: check that you multiplied by r before you simplified.
4
Find the volume of the solid bounded by z=x2+y2−9 and z=−2(x2+y2) using cylindrical coordinates.
no calculator, same as the exam
Answer227πanswer recomputed · sympy
Both surfaces depend only on r2=x2+y2: z=r2−9 (an upward paraboloid dropped to −9) and z=−2r2 (a downward one from the origin).
Which is on top? At r=0: −9 versus 0, so z=−2r2 is the ceiling and z=r2−9 is the floor.
They meet where r2−9=−2r2⇒3r2=9⇒r=3. Full revolution, so 0≤θ≤2π.
TrapAssuming the paraboloid opening upward is the ceiling because it 'goes up'. Over the whole region r<3 it is the floor, and swapping them makes the volume negative. Test at r=0 every time. The other trap is integrating 9−3r2 against dr instead of rdr, which gives 2π(93−33)=123π — a suspiciously irrational volume for a surface of revolution with rational intersection radius.
5
Write the volume of the solid bounded by the cone z2=x2+y2 between the planes z=1 and z=2 as a spherical-coordinates integral.
In spherical, the cone z2=x2+y2 with z>0 is the single surface ϕ=4π, and the solid inside it is 0≤ϕ≤4π.
A horizontal plane is not a constant in spherical: z=ρcosϕ, so z=1 becomes ρ=secϕ and z=2 becomes ρ=2secϕ.
Those are the ρ-limits, and they depend on ϕ — so ρ must be integrated first.
dV=ρ2sinϕdρdϕdθ, full revolution in θ:
V=∫02π∫0π/4∫secϕ2secϕρ2sinϕdρdϕdθ.
Worth confirming in cylindrical, where it is a stack of disks of radius z: ∫02π∫12∫0zrdrdzdθ=∫12πz2dz=37π — and the spherical integral gives the same.
TrapWriting the planes as constant ρ (say ρ from 1 to 2), which describes the region between two spheres, not two planes — that is a spherical shell wedge, volume 32π(8−1)(1−cos4π), not 37π. Any flat boundary not through the origin becomes ρ=(something)secϕ and forces dρ innermost.
The volume of the region inside x2+y2+z2=4 and inside the cone z=3x2+3y2 is ∫02π∫0a∫0bcdρdϕdθ. What are a, b, and c?
no calculator, same as the exam
Answera=π/6,b=2,c=ρ2sinϕanswer recomputed · sympy
c is the Jacobian and nothing else: dV=ρ2sinϕdρdϕdθ, so c=ρ2sinϕ.
Sphere x2+y2+z2=4 is ρ=2, and since ρ is innermost and starts at 0, b=2.
The cone z=3x2+3y2=3x2+y2 has tanϕ=zx2+y2=31.
So ϕ=6π, and 'inside the cone' means 0≤ϕ≤6π: a=6π.
V=∫02π∫0π/6∫02ρ2sinϕdρdϕdθ.
Traptanϕ=zr, not rz — ϕ is measured down from the positive z-axis. Taking tanϕ=3 gives ϕ=3π, which is the complementary wedge and roughly four times the volume. Second trap: writing c=ρsinϕ, the *polar* Jacobian with a ρ in it, which is dimensionally wrong for a volume.
2
Write the volume of the solid bounded by the cone z=x2+y2 and the sphere x2+y2+z2=8 as a spherical-coordinates integral.
Cone z=x2+y2: here tanϕ=zr=1, so it is the surface ϕ=4π, and the solid inside it is 0≤ϕ≤4π.
Sphere x2+y2+z2=8 is ρ2=8, i.e. ρ=8=22.
Both boundaries are coordinate surfaces in spherical — every limit is a constant, which is the whole reason to use it here.
V=∫02π∫0π/4∫08ρ2sinϕdρdϕdθ.
Value: 2π⋅(1−22)⋅3(8)3=332π(2−1), which the cylindrical set-up ∫02π∫02(8−r2−r)rdrdθ confirms.
TrapReading ρ=8 off x2+y2+z2=8. The equation gives ρ2=8, so ρ=22 — and cubing the wrong one inflates the volume by a factor of 83/83/2≈22.6. Also note the cone/sphere meet at z=2, r=2: that intersection is *not* needed in spherical, which is the payoff for converting.
3
Write the volume of the region inside the sphere of radius 2 centered at (0,0,2) and outside the sphere of radius 2 centered at the origin as a spherical-coordinates integral.
Put both spheres in spherical form. Origin-centred radius 2: ρ=2.
The one centred at (0,0,2): x2+y2+(z−2)2=4⇒ρ2−4ρcosϕ=0⇒ρ=4cosϕ.
'Inside the shifted sphere, outside the origin one' is exactly 2≤ρ≤4cosϕ — the inner radius is the constant, the outer is the ϕ-dependent one.
Those limits are only valid where 4cosϕ≥2, i.e. cosϕ≥21, i.e. ϕ≤3π. That is the ϕ-limit.
V=∫02π∫0π/3∫24cosϕρ2sinϕdρdϕdθ.
Check it independently: the lens where two radius-2 spheres with centres 2 apart overlap has volume 12π(4R+d)(2R−d)2=310π, so the answer is 332π−310π=322π — and the integral above gives 322π.
TrapTaking ϕ up to 2π because the shifted sphere 'sits above the xy-plane'. Past ϕ=3π the outer limit 4cosϕ drops below the inner limit 2, so the integral starts subtracting. The upper ϕ is always where the two ρ-limits meet: solve 4cosϕ=2.
4
Converting ∫−11∫−1−x21−x2∫3(x2+y2)4−x2−y2fdzdydx to spherical gives ∫0a∫0b∫0cf⋅ρ2sinϕdρdϕdθ (order: dρdϕdθ, so a is the upper limit on θ, b on ϕ, c on ρ). Find abc.
The shadow −1≤x≤1, ∣y∣≤1−x2 is the full unit disk, so θ makes a full turn: a=2π.
Ceiling z=4−x2−y2 is the sphere ρ=2, so c=2.
Floor z=3(x2+y2)=3r is a cone with tanϕ=zr=31, so ϕ runs from 0 to 6π: b=6π.
Consistency check on the shadow: the cone meets the sphere where ρ=2, ϕ=6π, giving r=2sin6π=1 — exactly the unit disk in the stem. Good.
abc=2π6π⋅2=61.
TrapSetting b=3π from tanϕ=3. ϕ is the angle from the z-axis, and a steep cone (z large compared to r) means a smallϕ. The disk-radius check catches it: ϕ=3π would give shadow radius 2sin3π=3=1.
5
Find the volume of the solid region enclosed by the surface ρ=12cosϕ.
no calculator, same as the exam
Answer288πanswer recomputed · sympy
ρ=12cosϕ is already a sphere in disguise: multiply by ρ to get ρ2=12ρcosϕ, i.e. x2+y2+z2=12z.
Complete the square: x2+y2+(z−6)2=36 — radius 6, centred at (0,0,6).
So the answer must be 34π(6)3=288π; now get the same thing from the integral.
The surface closes up at ϕ=2π (where ρ=0), so ϕ runs 0 to 2π — not to π.
TrapLetting ϕ run to π out of habit. For ϕ>2π, 12cosϕ<0 and there is no sphere there; the extra range adds another 288π of nonsense (the cos3ϕsinϕ integrand is odd about 2π, so it actually cancels to 0 — an error that can *silently* delete the whole answer). The sphere is fully traced by ϕ∈[0,2π].
Find the center of mass of the rectangle {(x,y)∣0≤x≤2,0≤y≤4} with density ρ(x,y)=1+y.
no calculator, same as the exam
Answer(1,922)answer recomputed · sympy
Mass first: m=∫02∫04(1+y)dydx=∫02[y+2y2]04dx=∫0212dx=24.
xˉ: the density does not depend on x and the region is symmetric in x, so xˉ=1 by inspection. Formally My=∫02∫04x(1+y)dydx=(∫02xdx)(12)=24, and xˉ=2424=1.
Sanity: 922>2, and it should be — the density grows with y, so mass piles up toward the top.
Centre of mass (1,922).
TrapSwapping the moments. Mx=∬yδdA gives yˉ, not xˉ — the subscript names the axis the moment is about, so it pairs with the *other* coordinate. Using Mx for xˉ here returns 922 for a rectangle only 2 wide, which is impossible and is the fastest way to catch it.
2
A cube 0≤x,y,z≤1 has density δ=x+yz. Find the x-coordinate of the center of mass, given ∭cubeδdV=43.
no calculator, same as the exam
Answer1811answer recomputed · sympy
xˉ=mMyz with Myz=∭xδdV; the mass m=43 is handed to you.
Myz=∫01∫01∫01x(x+yz)dzdydx=∫∫∫(x2+xyz)dV.
Both pieces separate over the cube: ∭x2dV=31 and ∭xyzdV=(21)3=81.
Myz=31+81=2411.
xˉ=3/411/24=2411⋅34=1811.
Sanity: 1811>21, correct — the x term in the density weights the far side.
TrapMultiplying only the first term by x: ∫x⋅xdV+∫yzdV=31+41 gives 127 and then xˉ=97. The moment weights the entire density, so distribute the x across x+yz before integrating.
3
Find the x-coordinate of the center of mass of the plate bounded by y=x, y=21x, x=1 with density δ(x,y)=2x. (Given: Mx=∬yδdA, My=∬xδdA.)
no calculator, same as the exam
Answer43answer recomputed · sympy
Sketch the wedge: between y=21x (lower) and y=x (upper), closed off at x=1. They meet at the origin, so 0≤x≤1 and 2x≤y≤x.
Vertical strips are the right choice — each strip has height x−2x=2x.
Sanity: 43 is past the midpoint of [0,1], as it must be — the strips get both taller and denser as x grows.
TrapIntegrating in the order dxdy without splitting. Horizontal strips run from x=y to x=2y only while 2y≤1; above y=21 the right edge is the line x=1 instead, so that order needs two integrals. The printed answer's order avoids that entirely — pick strips that are described by one pair of curves.
4
Calculate the mass of the tetrahedron with corners (0,0,0), (1,0,0), (0,2,0), (0,0,4) with density ρ(x,y,z)=2z.
no calculator, same as the exam
Answer38answer recomputed · sympy
Three of the four corners are on the axes, so use the intercept form of the slanted face: 1x+2y+4z=1.
Solve for the ceiling: z=4(1−x−2y)=4−4x−2y.
Shadow: set z=0 to get x+2y=1, i.e. y=2(1−x), with 0≤x≤1.
TrapTreating a variable density as a constant and reaching for ρ⋅V: the volume is 61⋅2⋅4=34, and 2zˉ⋅V is *not* something you can shortcut without already knowing zˉ. Also: ρ here is a density, not the spherical radius — the symbol collides with Lesson 23 and the units tell you which.
5
A lamina with density ρ(x,y)=xy occupies the region bounded by y=x2, y=1, x=0. The mass is 61. Find the y-coordinate of the center of mass.
no calculator, same as the exam
Answer43answer recomputed · sympy
Region: y=x2 below, y=1 above, x=0 on the left — so 0≤x≤1 and x2≤y≤1.
The mass is given as 61; confirm it cheaply: ∫01∫x21xydydx=∫012x(1−x4)dx=21(21−61)=61. Good.
Mx=∫01∫x21y⋅xydydx=∫01x⋅31−x6dx.
=31(21−81)=31⋅83=81.
yˉ=mMx=1/61/8=86=43.
Sanity: the region hugs the top (it is wide near y=1, pinched near y=0) and the density xy favours large y, so yˉ>21 is right.
TrapTaking the region as 0≤y≤x2 — 'bounded by y=x2' does not say which side. The third boundary x=0 is the tell: the region above the parabola touches the y-axis, the region below it does not need x=0 to close. Getting this backwards flips yˉ to 75-ish territory and, more tellingly, changes the mass away from the given 61 — which is exactly why that value was handed to you.
Consider F=⟨eyz−ysin(xy),zxeyz−xsin(xy),xyeyz⟩. Find a potential function φ if one exists.
no calculator, same as the exam
Answerxeyz+cos(xy)answer recomputed · sympy
We want φ with ∇φ=F, so start with the component that is easiest to antidifferentiate and then patch it up.
Integrate the first component in x, holding y,z fixed: φ=∫(eyz−ysin(xy))dx=xeyz+cos(xy)+g(y,z).
The sign is the whole game here: dxdcos(xy)=−ysin(xy), so the antiderivative of −ysin(xy) is +cos(xy).
Now differentiate that candidate in y: φy=xzeyz−xsin(xy)+gy.
Compare with the given second component zxeyz−xsin(xy): they already match, so gy=0.
Differentiate in z: φz=xyeyz+gz, and the third component is xyeyz, so gz=0 too. Hence g is a constant.
φ=xeyz+cos(xy) (any additive constant is also a potential).
TrapWriting ∫−ysin(xy)dx=−cos(xy). The −y out front is exactly the inner derivative, so the two minus signs cancel and you get +cos(xy). Getting this backwards still passes a careless φx check if you also flip the sign when differentiating.
2
Let a=0 and C be the curve x2−2y2=a. Which F is orthogonal to the tangent of C at every point (x0,y0) on C?
no calculator, same as the exam
Answer⟨2x,−4y⟩answer recomputed · sympy
C is the level curve g(x,y)=a of g(x,y)=x2−2y2.
Key fact: the gradient of g is normal to every level curve of g, hence orthogonal to the tangent of C at each point.
∇g=⟨gx,gy⟩=⟨2x,−4y⟩.
So F=⟨2x,−4y⟩ is orthogonal to T at every (x0,y0) on C (and so is any scalar multiple of it).
Sanity check by hand: implicit differentiation of x2−2y2=a gives 2x−4yy′=0, so T∥⟨4y,2x⟩, and ⟨2x,−4y⟩⋅⟨4y,2x⟩=8xy−8xy=0.
TrapDropping the minus sign and answering ⟨2x,4y⟩ — that is ∇(x2+2y2), the normal to an ellipse, not to this hyperbola. The other classic miss is handing back the tangent ⟨4y,2x⟩ instead of the normal.
3
Which vector field corresponds to the field with vectors rotating counterclockwise (pointing left above the x-axis, right below it)?
no calculator, same as the exam
Answer⟨−y,x⟩answer recomputed · sympy
Counterclockwise rotation means the field is tangent to circles centered at the origin, pointing in the direction of increasing angle.
Parametrize a circle the counterclockwise way: r(t)=⟨cost,sint⟩.
The direction of travel is r′(t)=⟨−sint,cost⟩.
Rewrite that at the point itself, using x=cost, y=sint: the tangent direction is ⟨−y,x⟩.
Spot-check the description: above the x-axis take (0,1), giving ⟨−1,0⟩ (points left); below take (0,−1), giving ⟨1,0⟩ (points right). Matches.
F=⟨−y,x⟩.
TrapAnswering ⟨y,−x⟩, which is the same picture run clockwise. Always test one concrete point above the x-axis rather than trusting the sign pattern; ⟨y,−x⟩ gives ⟨1,0⟩ at (0,1), pointing right, which is the wrong rotation.
4
Suppose f(x,y,z)=xy2z3. Find div(∇f) at the point (2,−1,1).
no calculator, same as the exam
Answer16answer recomputed · sympy
div(∇f) is the Laplacianfxx+fyy+fzz — a scalar, not a vector.
∇f=⟨y2z3,2xyz3,3xy2z2⟩.
Take the matching partial of each component: fxx=∂x(y2z3)=0, fyy=∂y(2xyz3)=2xz3, fzz=∂z(3xy2z2)=6xy2z.
So div(∇f)=0+2xz3+6xy2z.
At (2,−1,1): 2(2)(1)3+6(2)(−1)2(1)=4+12=16.
TrapTwo traps. First, differentiating the wrong slot — you must take ∂x of the first component, ∂y of the second, ∂z of the third, not all three of one component. Second, (−1)2=+1: plugging y=−1 into 6xy2z gives +12, and writing −12 turns 16 into −8.
Compute ∫Cf(x,y)ds where f(x,y)=sin(2πx)+xy and C:r(t)=⟨t,t⟩, 0≤t≤1.
no calculator, same as the exam
Answer32answer recomputed · sympy
Scalar line integral: ∫Cfds=∫abf(r(t))∣r′(t)∣dt.
Here r′(t)=⟨1,1⟩, so ∣r′(t)∣=2 and ds=2dt.
On C we have x=t, y=t, so f(r(t))=sin(2πt)+t2.
Set up: ∫01(sin(2πt)+t2)2dt.
∫01sin(2πt)dt=[−2πcos(2πt)]01=−2π1+2π1=0 — a full period integrates away.
∫01t2dt=31, so the value is 2⋅31=32.
TrapForgetting the ∣r′∣=2 factor and reporting 31. The parametrization ⟨t,t⟩ is not unit speed even though it looks tame — ds=dt here.
2
Let C be the helix r(t)=⟨cost,sint,t⟩, 0≤t≤π/2. Compute ∫Cxyds.
no calculator, same as the exam
Answer22answer recomputed · sympy
r′(t)=⟨−sint,cost,1⟩, so ∣r′(t)∣=sin2t+cos2t+1=2.
The helix has constant speed2, so ds=2dt.
On C, xy=costsint.
Set up: ∫0π/2costsint2dt.
Substitute u=sint: 2∫01udu=2⋅21=22.
TrapReading the first two components as a unit circle and concluding ∣r′∣=1. The z=t component contributes a 1 under the radical, making the speed 2; dropping it halves-by-2 the answer to 21.
3
Calculate ∫C(x−y)21ds where C is the straight line segment from (1,0) to (5,3).
no calculator, same as the exam
Answer25answer recomputed · sympy
Parametrize the segment from (1,0) to (5,3) as r(t)=⟨1,0⟩+t⟨4,3⟩=⟨1+4t,3t⟩, 0≤t≤1.
r′(t)=⟨4,3⟩, so ∣r′(t)∣=16+9=5 and ds=5dt (the length of the segment, spread over t∈[0,1]).
The integrand simplifies beautifully: x−y=(1+4t)−3t=1+t, so (x−y)21=(1+t)21.
Set up: ∫01(1+t)25dt.
=5[−1+t1]01=5(−21+1)=25.
TrapUsing dx instead of ds. If you parametrize by x with y=43(x−1) you must carry ds=1+(3/4)2dx=45dx; dropping that factor gives 2 instead of 25. Also note the line never crosses x=y, so the integrand stays finite.
4
Let C be the half-circle x2+y2=4 with x≥0. Compute ∫Cxds.
no calculator, same as the exam
Answer8answer recomputed · sympy
The circle has radius 2, so r(t)=⟨2cost,2sint⟩.
Pick the range that gives x≥0, not y≥0: x=2cost≥0 forces −2π≤t≤2π (the right half).
r′(t)=⟨−2sint,2cost⟩, so ∣r′(t)∣=2 and ds=2dt.
Set up: ∫−π/2π/2(2cost)(2)dt=4∫−π/2π/2costdt.
=4[sint]−π/2π/2=4(1−(−1))=8.
TrapTaking 0≤t≤π out of habit. That traces the upper half (y≥0), over which ∫xds=0 by symmetry — a suspiciously clean wrong answer. The condition is x≥0, so center the interval on t=0.
5
Find the arc length of r(t)=⟨4cost,4sint,3t⟩ for 0≤t≤2π.
no calculator, same as the exam
Answer10πanswer recomputed · sympy
Arc length is L=∫ab∣r′(t)∣dt.
r′(t)=⟨−4sint,4cost,3⟩.
∣r′(t)∣=16sin2t+16cos2t+9=16+9=5 — constant speed, the hallmark of a circular helix.
Set up: L=∫02π5dt=5(2π)=10π.
TrapAdding the radius and the pitch coordinate as if speed were 4+3=7, or forgetting the 3 entirely and getting 8π (the circumference of the shadow circle). The helix is strictly longer than its projection because of the climb.
Given force field F(x,y)=⟨−y,x⟩, find the work to move an object along the ellipse r(t)=⟨2cost,3sint⟩ from (2,0) to (0,3).
no calculator, same as the exam
Answer3πanswer recomputed · sympy
Work is W=∫CF⋅dr=∫abF(r(t))⋅r′(t)dt.
Find the t-range from the endpoints: (2,0) is t=0 and (0,3) is t=2π, so we travel a quarter of the ellipse.
On C: F(r(t))=⟨−y,x⟩=⟨−3sint,2cost⟩ and r′(t)=⟨−2sint,3cost⟩.
Dot them: 6sin2t+6cos2t=6 — the integrand is constant.
Set up: W=∫0π/26dt=6⋅2π=3π.
TrapReaching for Green's theorem. C is an open quarter-arc, not a closed curve, so Green does not apply; and F=⟨−y,x⟩ is not conservative (Py=−1=1=Qx), so the path genuinely matters and FTLI is off the table. You must integrate along this arc.
2
Compute the flux of F(x,y)=⟨x,y⟩ across the curve r(t)=⟨3cost,2sint⟩, −π≤t≤π. (Flux =∫CPdy−Qdx.)
no calculator, same as the exam
Answer12πanswer recomputed · sympy
Flux uses the stated formula ∫CPdy−Qdx with P=x, Q=y — note dy pairs with P, not with Q.
From r(t)=⟨3cost,2sint⟩: dx=−3sintdt and dy=2costdt.
Cross-check with the divergence form: divF=2 and the ellipse encloses area π(3)(2)=6π, so the flux is 2⋅6π=12π.
TrapComputing circulation ∫Pdx+Qdy instead of flux. With this field that swap gives ∫(−6sintcost+6sintcost)dt=0, so the mistake produces a clean 0 and looks deliberate. Keep the pairing Pdy−Qdx straight.
3
Given force field F(x,y,z)=⟨y,z,x⟩, find the work to move an object along the straight line from (0,0,0) to (2,3,4).
no calculator, same as the exam
Answer226answer recomputed · sympy
Parametrize the segment: r(t)=⟨2t,3t,4t⟩, 0≤t≤1, so r′(t)=⟨2,3,4⟩.
Evaluate the field on the curve, remembering the components are shifted: F=⟨y,z,x⟩=⟨3t,4t,2t⟩.
Dot: F⋅r′=6t+12t+8t=26t.
Set up: W=∫0126tdt=26⋅21=226=13.
TrapSubstituting F=⟨2t,3t,4t⟩, i.e. reading ⟨y,z,x⟩ as ⟨x,y,z⟩. That gives ∫0129tdt and the wrong number. The cyclic shift is the entire content of this problem. (Also: F=⟨y,z,x⟩ is not conservative, since Py=1 but Qx=0, so you must integrate.)
4
Find ∫CF⋅Tds where F(x,y,z)=⟨yez,ey+xez,xyez⟩ on a smooth curve from (0,0,0) to (−1,1,1).
no calculator, same as the exam
Answer−1answer recomputed · sympy
The problem says 'a smooth curve' without naming it, which is the signal that F must be conservative — otherwise the answer would not be well defined.
Hunt for φ: integrate the first component in x, ∫yezdx=xyez+g(y,z).
Match the second component: φy=xez+gy must equal ey+xez, so gy=ey and g=ey+h(z).
Match the third: φz=xyez+h′(z) must equal xyez, so h is constant. Take φ=xyez+ey.
FTLI: ∫CF⋅Tds=∫CF⋅dr=φ(−1,1,1)−φ(0,0,0).
φ(−1,1,1)=(−1)(1)e1+e1=−e+e=0 and φ(0,0,0)=0+e0=1.
Value =0−1=−1.
TrapEvaluating φ at the start minus the end. FTLI is end minus start, and here that sign error hands you +1. Second trap: forgetting the ey piece because it does not appear in φx — it is exactly what makes φ(0,0,0)=1 rather than 0.
5
Evaluate ∫CF⋅dr where F(x,y,z)=⟨2xy−yz,x2−xz,−xy⟩ on a smooth curve from (2,1,0) to (3,2,−1).
no calculator, same as the exam
Answer20answer recomputed · sympy
Again the curve is unspecified, so test for conservativity: with P=2xy−yz, Q=x2−xz, R=−xy we get Py=2x−z=Qx, Pz=−y=Rx, Qz=−x=Ry. All three match, so F is conservative.
Build φ: ∫(2xy−yz)dx=x2y−xyz+g(y,z).
Check φy=x2−xz+gy against Q=x2−xz: gy=0. Check φz=−xy+gz against R=−xy: gz=0. So φ=x2y−xyz.
FTLI: value =φ(3,2,−1)−φ(2,1,0).
φ(3,2,−1)=9(2)−(3)(2)(−1)=18+6=24.
φ(2,1,0)=4(1)−0=4.
Value =24−4=20.
TrapSign-slipping on −xyz at the endpoint with z=−1: −(3)(2)(−1)=+6, so φ(3,2,−1)=24, not 12. Writing 18−6 gives 8 and is the single most common wrong answer here.
Find ∫CF⋅dr where F(x,y,z)=⟨2xy−yz,x2−xz,−xy⟩ on a smooth curve from (2,1,0) to (3,2,−1).
no calculator, same as the exam
Answer20answer recomputed · sympy
The path is only described as 'a smooth curve', so the integral had better be path-independent — first verify F is conservative.
With P=2xy−yz, Q=x2−xz, R=−xy: Py=2x−z=Qx, Pz=−y=Rx, Qz=−x=Ry. Conservative.
Find the potential: φ=∫(2xy−yz)dx=x2y−xyz+g(y,z), and both φy=x2−xz and φz=−xy already match, so g is constant.
φ(x,y,z)=x2y−xyz.
FTLI: ∫CF⋅dr=φ(3,2,−1)−φ(2,1,0)=(18+6)−(4−0).
=24−4=20.
TrapTrying to parametrize. No curve was given, so any attempt to build one is wasted work — and if you accidentally pick a path while F were not conservative, the answer would be meaningless. Confirm the mixed partials first, then use endpoints only. Watch −xyz at z=−1, which contributes +6.
2
For F=⟨eyz−ysin(xy),zxeyz−xsin(xy),xyeyz⟩, find a potential function.
no calculator, same as the exam
Answerφ=xeyz+cos(xy)answer recomputed · sympy
Antidifferentiate the first component in x: φ=∫(eyz−ysin(xy))dx.
Term by term, ∫eyzdx=xeyz (as x varies, eyz is a constant) and ∫−ysin(xy)dx=cos(xy).
So φ=xeyz+cos(xy)+g(y,z).
Test against the second component: φy=xzeyz−xsin(xy)+gy, and we need zxeyz−xsin(xy), so gy=0.
Test against the third: φz=xyeyz+gz, and we need xyeyz, so gz=0.
Therefore φ=xeyz+cos(xy), up to an additive constant.
TrapDifferentiating eyz with respect to y and writing zeyz without the x that the antiderivative introduced. The x in xeyz must be carried through, and it is exactly what makes φy=xzeyz match the printed zxeyz.
3
A potential for F=⟨siny,xcosy⟩ is:
no calculator, same as the exam
Answerxsiny+1answer recomputed · sympy
First confirm a potential exists: P=siny, Q=xcosy, and Py=cosy=Qx. Conservative on all of R2.
Integrate P in x: φ=∫sinydx=xsiny+g(y).
Differentiate in y: φy=xcosy+g′(y), which must equal Q=xcosy, so g′(y)=0 and g is any constant.
The family of potentials is φ=xsiny+C. The printed choice takes C=1: φ=xsiny+1.
TrapRejecting the '+1' as a typo. Potentials are only determined up to an additive constant, so xsiny, xsiny+1, and xsiny−7 are all correct; on a multiple-choice list the constant is there precisely to see whether you know that. The real error to avoid is φ=−xcosy, from mistakenly antidifferentiating siny in y instead of x.
4
Find ∫CF⋅Tds where F=⟨yez,ey+xez,xyez⟩ on a smooth curve from (0,0,0) to (−1,1,1).
no calculator, same as the exam
Answer−1answer recomputed · sympy
∫CF⋅Tds is just ∫CF⋅dr rewritten, since dr=Tds.
No specific curve is given, so check conservativity: with P=yez, Q=ey+xez, R=xyez we get Py=ez=Qx, Pz=yez=Rx, Qz=xez=Ry. Conservative.
Potential: ∫yezdx=xyez+g(y,z); then φy=xez+gy=ey+xez forces gy=ey, and φz=xyez+h′(z)=xyez forces h′=0.
φ=xyez+ey.
FTLI: φ(−1,1,1)−φ(0,0,0)=((−1)(1)e+e)−(0+1).
=0−1=−1.
TrapSeeing Tds and thinking you must compute ∣r′∣ and a unit tangent. You never do — Tds=dr, the speed cancels, and the whole problem collapses to two endpoint evaluations. The second trap is φ(0,0,0)=0: the ey term makes it e0=1.
5
Evaluate ∫C∇(x2y−xyz)⋅dr where C is any path from (1,0,0) to (2,1,−1).
no calculator, same as the exam
Answer6answer recomputed · sympy
The integrand is already a gradient, ∇φ with φ(x,y,z)=x2y−xyz, so FTLI applies immediately and no conservativity test is needed.
∫C∇φ⋅dr=φ(end)−φ(start), independent of the path.
φ(2,1,−1)=22(1)−(2)(1)(−1)=4+2=6.
φ(1,0,0)=12(0)−(1)(0)(0)=0.
Value =6−0=6.
TrapExpanding ∇(x2y−xyz)=⟨2xy−yz,x2−xz,−xy⟩ and then grinding out a line integral along some invented path. All that work is discarded by FTLI. Within the evaluation, the sign trap is −xyz at z=−1, which adds2; writing 4−2=2 is the common wrong answer.
Use Green's Theorem to evaluate ∮Cxdx+(x2+y2)dy where C is the boundary of the rectangle with vertices (0,0), (2,0), (2,3), (0,3), counterclockwise.
no calculator, same as the exam
Answer12answer recomputed · sympy
Green's Theorem converts a counterclockwise closed line integral into a double integral: ∮CPdx+Qdy=∬R(∂x∂Q−∂y∂P)dA.
Read off P=x and Q=x2+y2. The orientation is already CCW, so no sign flip is needed.
∂x∂Q=2x and ∂y∂P=0, so the integrand is just 2x.
R is the rectangle 0≤x≤2, 0≤y≤3 — a genuine rectangle, so the limits are constants and the order does not matter.
∬R2xdA=∫03∫022xdxdy=∫03[x2]02dy=∫034dy=12.
TrapTrying to parametrize all four sides by hand. That is four integrals, and the ∫xdx pieces on the top and bottom edges cancel in a way that is easy to botch — one sign slip on the *return* edges (which run right-to-left and top-to-bottom) and you get 0 or 24. Green's theorem exists precisely to make the orientation bookkeeping automatic; use it. Also note P=x contributes nothing, since Py=0 — the whole answer comes from Q.
2
Let R be the triangle with vertices (0,0), (2,0), (1,1) and C its CCW boundary. Compute ∫Ccos3(x)dx+exdy.
no calculator, same as the exam
Answere2−2e+1answer recomputed · sympy
P=cos3(x), Q=ex. Neither has an elementary-looking line integral, which is the hint that Green's Theorem is the intended route.
∂x∂Q=ex, ∂y∂P=0 (there is no y in cos3x at all), so Qx−Py=ex.
Describe the triangle. The two slanted sides are y=x (from (0,0) to (1,1)) and y=2−x (from (1,1) to (2,0)), with the base y=0.
Integrating in x first avoids splitting the region: for each y∈[0,1], x runs from the left edge x=y to the right edge x=2−y.
∫01∫y2−yexdxdy=∫01(e2−y−ey)dy.
=[−e2−y−ey]01=(−e−e)−(−e2−1)=e2−2e+1.
TrapIntegrating dy first. Then you must split at x=1, because the upper boundary changes formula from y=x to y=2−x there: ∫01∫0x+∫12∫02−x. Students who write a single ∫02∫0x silently integrate over a region that is not the triangle. Going dx first needs no split. (Note e2−2e+1=(e−1)2, a useful sanity check that it is positive.)
3
Use Green's Theorem to evaluate ∮Cydx+2xdy where C is the square with vertices (±1,±1), counterclockwise.
no calculator, same as the exam
Answer4answer recomputed · sympy
P=y, Q=2x, and the square is traversed CCW, so Green's Theorem applies directly.
∂x∂Q−∂y∂P=2−1=1.
The integrand collapsed to the constant 1, so the double integral is literally the area of R.
R is the square −1≤x≤1, −1≤y≤1, whose side length is 2, so its area is 4.
∮Cydx+2xdy=∬R1dA=area(R)=4.
TrapComputing Px−Qy instead of Qx−Py. That gives 0−0=0 here, a clean-looking wrong answer. The derivative always hits the *other* variable: Q gets ∂x, P gets ∂y. Second trap: the vertices (±1,±1) describe a square of side 2, not side 1 — its area is 4, not 1.
4
Use Green's theorem to evaluate ∮C(y3)dx−(x3)dy where C is the circle x2+y2=1, CCW.
no calculator, same as the exam
Answer2−3πanswer recomputed · sympy
Write the integral as ∮CPdx+Qdy with P=y3 and Q=−x3 — the minus sign in front of x3dy belongs to Q.
∂x∂Q=−3x2, ∂y∂P=3y2, so Qx−Py=−3x2−3y2=−3(x2+y2).
R is the unit disk and the integrand depends only on x2+y2, so polar is the obvious move: x2+y2=r2 and dA=rdrdθ.
TrapTwo sign traps stacked. First, forgetting that Q=−x3, which makes Qx=+3x2 and hands you +23π — the right magnitude, wrong sign. Second, forgetting the polar Jacobian: −3r2drdθ instead of −3r3drdθ gives −2π. The answer is negative, and it *should* be: the field circulates clockwise while C runs counterclockwise.
5
Use Green's theorem to find the area of the ellipse a2x2+b2y2=1.
no calculator, same as the exam
Answerπabanswer recomputed · sympy
Green's Theorem run backwards: pick P,Q with Qx−Py=1, and the double integral becomes the area. The symmetric choice is P=−2y, Q=2x, giving the area formula A=21∮Cxdy−ydx.
Parametrize the ellipse CCW: x=acost, y=bsint, 0≤t≤2π.
Sanity check: setting a=b=R gives πR2, the circle.
TrapDropping the 21. The area formula is 21∮xdy−ydx — the half is there because *each* of xdy and −ydx separately already integrates to the full area, and you are averaging them, not adding them. Forgetting it doubles the answer to 2πab. The other trap is parametrizing clockwise (y=−bsint), which returns −πab; area must come out positive, so a negative result means your orientation was backwards.
curlF=∇×F=⟨∂y∂R−∂z∂Q,∂z∂P−∂x∂R,∂x∂Q−∂y∂P⟩ for F=⟨P,Q,R⟩.
Here P=xy, Q=sinz, R=cosy.
First component: Ry−Qz=∂y∂(cosy)−∂z∂(sinz)=−siny−cosz.
Second component: Pz−Rx=∂z∂(xy)−∂x∂(cosy)=0−0=0.
Third component: Qx−Py=∂x∂(sinz)−∂y∂(xy)=0−x=−x.
curlF=⟨−siny−cosz,0,−x⟩.
TrapThe middle component's sign. Expanding the determinant i∂xPj∂yQk∂zR, the j entry carries a minus from cofactor expansion: it is −(Rx−Pz)=Pz−Rx, notRx−Pz. Write the middle slot as Pz−Rx and the sign takes care of itself. (Here it is 0 either way, which is exactly why the habit must be built on a problem where it does not bite.) Second trap: curl of a vector field is a vector, while div is a scalar — never report a single number here.
2
Let f(x,y,z)=xy2z3. Compute div(∇f) at (2,−1,1).
no calculator, same as the exam
Answer16answer recomputed · sympy
div(∇f)=∇2f=fxx+fyy+fzz, the Laplacian. So take second partials, not first.
f=xy2z3. Gradient: ∇f=⟨y2z3,2xyz3,3xy2z2⟩.
fxx=∂x∂(y2z3)=0 — f is linear in x.
fyy=∂y∂(2xyz3)=2xz3.
fzz=∂z∂(3xy2z2)=6xy2z.
∇2f=0+2xz3+6xy2z. At (2,−1,1): 2(2)(1)3+6(2)(−1)2(1)=4+12=16.
TrapEvaluating at the point too early. If you plug (2,−1,1) into ∇f and then try to differentiate, you are differentiating constants and get 0. Differentiate fully, *then* substitute. The other trap is y2 at y=−1: (−1)2=+1, not −1, so the 6xy2z term is +12; a dropped square gives 4−12=−8.
3
For F=⟨eyz−ysin(xy),zxeyz−xsin(xy),xyeyz⟩, verify that curlF=0 and state what this implies.
no calculator, same as the exam
AnswerF is conservativeprose answer · not machine-checkable
Label P=eyz−ysin(xy), Q=zxeyz−xsin(xy), R=xyeyz, and compute the three components of ∇×F.
Ry=∂y∂(xyeyz)=xeyz+xyzeyz, and Qz=∂z∂(zxeyz−xsin(xy))=xeyz+xyzeyz. First component Ry−Qz=0.
Pz=∂z∂(eyz−ysin(xy))=yeyz, and Rx=∂x∂(xyeyz)=yeyz. Second component Pz−Rx=0.
Qx=zeyz−sin(xy)−xycos(xy) (product rule on xsin(xy)), and Py=zeyz−sin(xy)−xycos(xy) (product rule on ysin(xy)). Third component Qx−Py=0.
So curlF=0 everywhere.
F is defined and smooth on all of R3, which is simply connected, so curlF=0impliesF is conservative: there is a potential φ with ∇φ=F.
Indeed φ=xeyz+cos(xy) works: φx=eyz−ysin(xy), φy=xzeyz−xsin(xy), φz=xyeyz. Consequence: ∫CF⋅dr is path-independent and ∮CF⋅dr=0 on any closed curve.
TrapStating the converse without the domain hypothesis. curlF=0 alone does not make a field conservative — the standard counterexample F=⟨x2+y2−y,x2+y2x,0⟩ has zero curl but a nonzero loop integral, because its domain has the z-axis punched out. You get the conclusion here only because this F is smooth on all of R3. The computational trap is ∂y(ysin(xy)): it needs the product rule, sin(xy)+xycos(xy), not just xycos(xy).
4
Determine if F=⟨yz,xz,xy⟩ is conservative using curlF. If so, find φ.
no calculator, same as the exam
AnswerConservative; φ=xyz+Cprose answer · not machine-checkable
Test with the curl. P=yz, Q=xz, R=xy.
Ry−Qz=x−x=0; Pz−Rx=y−y=0; Qx−Py=z−z=0. So curlF=0, and since F is a polynomial field on all of R3 (simply connected), Fis conservative.
Now build φ with ∇φ=F. Start from φx=yz and integrate in x, treating y,z as constants: φ=xyz+g(y,z) — the 'constant' of integration may depend on the other variables.
Differentiate that in y: φy=xz+gy(y,z). Match against Q=xz, so gy=0 and g=h(z).
Differentiate in z: φz=xy+h′(z). Match against R=xy, so h′(z)=0 and h is a genuine constant.
φ=xyz+C, and one checks ∇(xyz)=⟨yz,xz,xy⟩=F.
TrapWriting the first antiderivative as φ=xyz+C with a numeric constant. At that stage the unknown is a whole function g(y,z), because anything free of x dies under ∂x. Collapsing it to a number too early means you never actually verify φy=Q and φz=R — on a field where the components do *not* fit together, that shortcut produces a 'potential' for a non-conservative field. Also: test with curl, not div. Here divF=0 too, but that says the field is incompressible, an entirely different statement.
5
Find divF for F=⟨exy,eyz,exz⟩, and verify the identity div(curlG)=0 for any smooth G.
By Clairaut's theorem the mixed partials commute (G3yx=G3xy, etc.), so the six terms cancel in pairs and div(curlG)=0 for every G with continuous second partials.
TrapDropping the chain-rule factor. ∂xexy=yexy, not exy — the answer exy+eyz+exz is the single most common version of this and it is wrong in all three terms. Watch which variable owns which exponent: the third component is exz and gets ∂z, producing xexz, not zexz. For the identity, the cancellation needs continuous second partials (Clairaut); it is a theorem, not algebra, and the practical payoff is that a field with divH=0 can never be written as curl of anything.
Find the area of the part of the plane 2x+5y+z=10 that lies in the first octant.
no calculator, same as the exam
Answer530answer recomputed · sympy
Solve for z to get the surface as a graph: z=g(x,y)=10−2x−5y.
For a graph, dS=1+gx2+gy2dA. Here gx=−2, gy=−5, so dS=1+4+25dA=30dA — a constant, as it must be for a plane.
Find the shadow D. The first octant needs x≥0, y≥0, and z≥0, and the last of these is 10−2x−5y≥0, i.e. 2x+5y≤10.
D is the triangle with legs along the axes: x-intercept x=5, y-intercept y=2. Its area is 21(5)(2)=5.
Area=∬D30dA=30⋅area(D)=530.
Explicitly: ∫05∫0(10−2x)/530dydx=530.
TrapReporting 5 — the area of the shadow — by forgetting the 1+gx2+gy2 factor. A tilted plane is always larger than its projection, so the stretch factor is ≥1 and your answer must exceed the shadow's area. The other trap is the shadow's shape: the first octant condition z≥0 is what bounds D, giving the triangle 2x+5y≤10 with intercepts x=5, y=2. Students who integrate over 0≤x≤5, 0≤y≤2 (the full rectangle) double the answer to 1030.
2
Compute the flux of F(x,y)=⟨x,y⟩ across r(t)=⟨3cost,2sint⟩, −π≤t≤π. (Flux =∫CPdy−Qdx.)
no calculator, same as the exam
Answer12πanswer recomputed · sympy
This is the 2D flux (outward-normal) line integral, and the problem hands you the formula: Flux=∫CPdy−Qdx with F=⟨P,Q⟩.
Along the curve, P=x=3cost and Q=y=2sint.
Differentiate the parametrization: dx=−3sintdt and dy=2costdt.
Cross-check with the divergence form of Green's Theorem: divF=1+1=2, and the ellipse 9x2+4y2=1 encloses area π(3)(2)=6π, so the flux is 2⋅6π=12π. ✓
TrapComputing ∫Pdx+Qdy — the circulation — instead of ∫Pdy−Qdx. For this field that gives ∫(−9costsint+4sintcost)dt=0, a plausible-looking but wrong answer. Flux swaps the differentials *and* inserts a minus sign; circulation does not. The second trap is the parameter range: −π≤t≤π is a full 2π sweep traversed counterclockwise, so the outward normal convention holds and the flux is positive — halving it to 6π means you read the range as a half-loop.
3
Evaluate ∬S1dS where S is the portion of the paraboloid z=x2+y2 below z=4.
no calculator, same as the exam
Answer6π(1717−1)answer recomputed · sympy
∬S1dS is the surface area of S. The surface is the graph z=g(x,y)=x2+y2.
gx=2x, gy=2y, so dS=1+4x2+4y2dA.
The shadow: z≤4 means x2+y2≤4, the disk of radius 2.
Polar, since both the integrand and the region depend only on x2+y2=r2: dS=1+4r2rdrdθ — note both the radical and the Jacobian r.
∬SdS=∫02π∫021+4r2rdrdθ.
Inner integral by u=1+4r2, du=8rdr: ∫02r1+4r2dr=[12(1+4r2)3/2]02=12173/2−1=121717−1.
Multiply by ∫02πdθ=2π: 122π(1717−1)=6π(1717−1).
TrapThe radius of the shadow. z=4 on z=x2+y2 means r2=4, so r runs to 2, not 4 — using r≤4 gives 6π(6565−1). The second trap is losing one of the two r's: 1+4r2rdrdθ carries the surface radical *and* the polar Jacobian, and dropping the Jacobian leaves an integral with no elementary u-substitution, which should tip you off that something is missing.
4
Find the flux of F=⟨x,y,z⟩ upward through the surface z=4−x2−y2 above the xy-plane.
no calculator, same as the exam
Answer24πanswer recomputed · sympy
For an upward-oriented graph z=g(x,y), ∬SF⋅dS=∬D(−Pgx−Qgy+R)dA. Note this formula already has the 1+gx2+gy2 built in — it cancels against the normalization of n^, so do not multiply by it again.
Here g=4−x2−y2, so gx=−2x and gy=−2y; and F=⟨x,y,z⟩ with z evaluated *on the surface*: R=4−x2−y2.
The shadow D is where z≥0: 4−x2−y2≥0, the disk x2+y2≤4 of radius 2.
Polar: ∫02π∫02(r2+4)rdrdθ=∫02π∫02(r3+4r)drdθ.
∫02(r3+4r)dr=[4r4+2r2]02=4+8=12, and times 2π gives 24π.
TrapLeaving z as the free variable z instead of substituting the surface, z=4−x2−y2. The flux integral lives *on* the surface, so every appearance of z in F must be replaced. The second trap is the sign pattern: it is ⟨−gx,−gy,1⟩ for upward — the two tangential terms get minus signs and the k slot gets +1. If the problem had said downward you would negate the whole thing, giving −24π. Third: do not tack on an extra 1+4x2+4y2; that factor is only for dS in a scalar surface integral, never for dS in this form.
5
Find the surface area of the cone z=x2+y2 below z=3.
no calculator, same as the exam
Answer9π2answer recomputed · sympy
The cone is the graph z=g(x,y)=x2+y2.
gx=x2+y2x and gy=x2+y2y, so gx2+gy2=x2+y2x2+y2=1.
Therefore dS=1+1dA=2dA — constant, because a cone has the same slope everywhere.
The shadow: z≤3 means x2+y2≤3, the disk of radius 3, whose area is 9π.
∬SdS=∫02π∫032rdrdθ=2⋅9π=9π2.
TrapWriting dS=1+x2+y2dA by squaring the *function* instead of its partials. The radical needs gx2+gy2, and for this cone those partials are the components of a unit vector, so their squares sum to exactly 1 and the factor is the clean 2. The other trap: the apex (0,0,0) makes gx,gy undefined there, but it is a single point of measure zero and the improper integral converges — the 2 still holds. And do not confuse this with the *lateral* cone formula πrℓ: here r=3, ℓ=32, giving 9π2, which agrees. ✓
6
Evaluate ∬SF⋅dS where F=⟨y,−x,1⟩ and S is the disk z=1, x2+y2≤1, oriented upward.
no calculator, same as the exam
Answerπanswer recomputed · sympy
S is a flat horizontal disk at height z=1, oriented upward, so the unit normal is simply n^=k=⟨0,0,1⟩ and dS=⟨0,0,1⟩dA (the graph z=g=1 has gx=gy=0, so dS=dA).
Dot the field with the normal: F⋅n^=⟨y,−x,1⟩⋅⟨0,0,1⟩=1. Only the k-component of F survives.
∬SF⋅dS=∬x2+y2≤11dA=area of the unit disk=π.
Explicitly in polar: ∫02π∫011⋅rdrdθ=2π⋅21=π.
TrapTrying to integrate the y and −x components. They are entirely tangential to a horizontal disk and contribute nothing to flux — this field swirls *within* the plane z=1 and never crosses it. (Its curl is ⟨0,0,−2⟩, so its *circulation* around the boundary is −2π, a completely different number that this problem is baiting you toward.) The second trap is orientation: 'oriented upward' fixes n^=+k; downward would give −π. Third, the disk has radius 1, so its area is π, not 2π.
STEP 03
More practice
4 questions
Mine, on those sections only. Untimed, no calculator — the quiz doesn't allow one either.
0/4 answered
117.4 · Green's theorem
C is the circle x2+y2=9 traversed counterclockwise. Find ∮C(y3)dx+(x3+3xy2)dy.
Answer E
P=y3,Q=x3+3xy2.
Py=3y2 and Qx=3x2+3y2, so the scalar curl is Qx−Py=3x2.
Green: ∮C=∬D3x2dA. In polar, x=rcosθ:
∫02π∫033r2cos2θrdrdθ=3⋅434∫02πcos2θdθ=4243⋅π=4243π.
TrapForgetting the r in r dr dθ, or dropping the −P_y term.
F has continuous second partials. Which expression is guaranteed to be 0 (the zero scalar or zero vector)?
Answer B
div(curlF)=0 always — a spin creates nothing.
(The companion identity is curl(∇f)=0.)
Option A is not even defined: divF is a scalar and you cannot take
the curl of a scalar. div(∇f)=∇2f is the Laplacian, generally nonzero.
TrapType errors. curl eats a vector and returns a vector; div eats a vector and returns a scalar.
A surface is a rubber sheet with its own two coordinates. Move u a little and you slide one way; move v and you slide the other. Those two slides span a tiny tile — and 13.4 already told you its area.
The cross product does double duty: its direction is the surface normal, its length is the tile's area. Both of the things a surface integral needs.
tangent vectors, normal, and area elementru=∂u∂r,rv=∂v∂r,ndS=(ru×rv)dudv,dS=∣ru×rv∣dudv
The cross product does double duty: its direction is the surface normal, its length is the tile's area. Both things a surface integral needs, from one computation.
Surface
Parametrization
∣ru×rv∣
Graph z=f(x,y)
⟨x,y,f(x,y)⟩
1+fx2+fy2
Sphere radius a
⟨asinφcosθ,asinφsinθ,acosφ⟩
a2sinφ
Cylinder radius a
⟨acosθ,asinθ,z⟩
a
Cone z=x2+y2
⟨rcosθ,rsinθ,r⟩
r2
Surface of revolution
⟨x,f(x)cosθ,f(x)sinθ⟩
f1+(f′)2
Row 1 and row 2 are worth recognising instantly — they reproduce the surface-area formula from 17.6 and the ρ2sinφ from 16.5.
Check yourself
What does (ru×rv)dudv equal?
Answer
ndS — both the direction and the area, from one computation. Never normalise it and then also multiply by dS.
Briggs files surface area under Surface Integrals (17.6) — which is why it sits here rather than back in Chapter 16.
A patch of hillside is always bigger than the flat patch beneath it. How much bigger depends on the tilt — and the tilt is exactly what the partial derivatives measure.
Same idea as arc length's √(1 + (dy/dx)²), one dimension up. Steeper means more surface hiding above the same footprint.
area of the graph of z=f(x,y) over the region RA(S)=∬R1+(fx)2+(fy)2dA
Same shape as arc length, which was ∫1+(dy/dx)2dx — one dimension up. Flat ground gives fx=fy=0, the root becomes 1, and surface area collapses to plain area. That's your check.
Algorithm · surface area of a graph
Solve for z so you have z=f(x,y).
Compute fx and fy, then 1+fx2+fy2 and simplify before integrating — this is usually where a clean root appears.
Find R: the shadow of the surface on the xy-plane.
Integrate over R, going polar if the shadow is round (it usually is).
For a surface given parametrically instead, jump to 17.6: A=∬∣ru×rv∣dA.
Check yourself
What is the stretch factor for the graph z=f(x,y)?
Answer
1+fx2+fy2.
On flat ground, what does that factor become?
Answer
1, so surface area collapses to plain area. That is the check on every surface-area setup.
Hold a hoop in the wind. Facing it head-on you catch all of it. Turn it edge-on and you catch nothing — even though the wind hasn't changed. Only the part perpendicular to the surface gets through.
drag to spin
Tilt to 90° and the wind hasn't changed at all — but nothing gets through. Only the component along n crosses the surface.
scalar surface integral — mass of a sheet∬SfdS=∬Df(r(u,v))ru×rvdudv
flux of a vector field∬SF⋅dS=∬SF⋅ndS=∬DF⋅(ru×rv)dudv
Algorithm · compute a flux integral
Parametrize S — use the table in 17.6 whenever it fits.
Compute ru×rv. Do not normalise it and do not compute dS separately; the un-normalised cross product already carries both.
Check orientation. Does the cross product point the way the problem wants (outward, upward)? If not, negate it.
Substitute F at the parametrized points, so everything is in u and v.
Dot and integrate over the parameter region D.
For a graph z=g(x,y) oriented upward, this collapses to ∬D(−Pgx−Qgy+R)dA.
Orientation trap
Swapping u and v flips the normal, and flipping the normal negates your flux. Always sanity-check by evaluating the normal at one concrete point: for a sphere with outward orientation it should point away from the origin.
Check yourself
You turn the net edge-on to the wind. How much gets through?
Answer
Nothing.F⊥n makes F⋅n=0, even though the wind never changed.
Should you normalise ru×rv before dotting with F?
Answer
No. The raw cross product already carries dS inside it.
STEP 02
The official problems
6 problems
Straight from the department's own study guide, for exactly these lessons — worked, with the answer held back until you commit. Do these first.
Find the area of the part of the plane 2x+5y+z=10 that lies in the first octant.
no calculator, same as the exam
Answer530answer recomputed · sympy
Solve for z to get the surface as a graph: z=g(x,y)=10−2x−5y.
For a graph, dS=1+gx2+gy2dA. Here gx=−2, gy=−5, so dS=1+4+25dA=30dA — a constant, as it must be for a plane.
Find the shadow D. The first octant needs x≥0, y≥0, and z≥0, and the last of these is 10−2x−5y≥0, i.e. 2x+5y≤10.
D is the triangle with legs along the axes: x-intercept x=5, y-intercept y=2. Its area is 21(5)(2)=5.
Area=∬D30dA=30⋅area(D)=530.
Explicitly: ∫05∫0(10−2x)/530dydx=530.
TrapReporting 5 — the area of the shadow — by forgetting the 1+gx2+gy2 factor. A tilted plane is always larger than its projection, so the stretch factor is ≥1 and your answer must exceed the shadow's area. The other trap is the shadow's shape: the first octant condition z≥0 is what bounds D, giving the triangle 2x+5y≤10 with intercepts x=5, y=2. Students who integrate over 0≤x≤5, 0≤y≤2 (the full rectangle) double the answer to 1030.
2
Compute the flux of F(x,y)=⟨x,y⟩ across r(t)=⟨3cost,2sint⟩, −π≤t≤π. (Flux =∫CPdy−Qdx.)
no calculator, same as the exam
Answer12πanswer recomputed · sympy
This is the 2D flux (outward-normal) line integral, and the problem hands you the formula: Flux=∫CPdy−Qdx with F=⟨P,Q⟩.
Along the curve, P=x=3cost and Q=y=2sint.
Differentiate the parametrization: dx=−3sintdt and dy=2costdt.
Cross-check with the divergence form of Green's Theorem: divF=1+1=2, and the ellipse 9x2+4y2=1 encloses area π(3)(2)=6π, so the flux is 2⋅6π=12π. ✓
TrapComputing ∫Pdx+Qdy — the circulation — instead of ∫Pdy−Qdx. For this field that gives ∫(−9costsint+4sintcost)dt=0, a plausible-looking but wrong answer. Flux swaps the differentials *and* inserts a minus sign; circulation does not. The second trap is the parameter range: −π≤t≤π is a full 2π sweep traversed counterclockwise, so the outward normal convention holds and the flux is positive — halving it to 6π means you read the range as a half-loop.
3
Evaluate ∬S1dS where S is the portion of the paraboloid z=x2+y2 below z=4.
no calculator, same as the exam
Answer6π(1717−1)answer recomputed · sympy
∬S1dS is the surface area of S. The surface is the graph z=g(x,y)=x2+y2.
gx=2x, gy=2y, so dS=1+4x2+4y2dA.
The shadow: z≤4 means x2+y2≤4, the disk of radius 2.
Polar, since both the integrand and the region depend only on x2+y2=r2: dS=1+4r2rdrdθ — note both the radical and the Jacobian r.
∬SdS=∫02π∫021+4r2rdrdθ.
Inner integral by u=1+4r2, du=8rdr: ∫02r1+4r2dr=[12(1+4r2)3/2]02=12173/2−1=121717−1.
Multiply by ∫02πdθ=2π: 122π(1717−1)=6π(1717−1).
TrapThe radius of the shadow. z=4 on z=x2+y2 means r2=4, so r runs to 2, not 4 — using r≤4 gives 6π(6565−1). The second trap is losing one of the two r's: 1+4r2rdrdθ carries the surface radical *and* the polar Jacobian, and dropping the Jacobian leaves an integral with no elementary u-substitution, which should tip you off that something is missing.
4
Find the flux of F=⟨x,y,z⟩ upward through the surface z=4−x2−y2 above the xy-plane.
no calculator, same as the exam
Answer24πanswer recomputed · sympy
For an upward-oriented graph z=g(x,y), ∬SF⋅dS=∬D(−Pgx−Qgy+R)dA. Note this formula already has the 1+gx2+gy2 built in — it cancels against the normalization of n^, so do not multiply by it again.
Here g=4−x2−y2, so gx=−2x and gy=−2y; and F=⟨x,y,z⟩ with z evaluated *on the surface*: R=4−x2−y2.
The shadow D is where z≥0: 4−x2−y2≥0, the disk x2+y2≤4 of radius 2.
Polar: ∫02π∫02(r2+4)rdrdθ=∫02π∫02(r3+4r)drdθ.
∫02(r3+4r)dr=[4r4+2r2]02=4+8=12, and times 2π gives 24π.
TrapLeaving z as the free variable z instead of substituting the surface, z=4−x2−y2. The flux integral lives *on* the surface, so every appearance of z in F must be replaced. The second trap is the sign pattern: it is ⟨−gx,−gy,1⟩ for upward — the two tangential terms get minus signs and the k slot gets +1. If the problem had said downward you would negate the whole thing, giving −24π. Third: do not tack on an extra 1+4x2+4y2; that factor is only for dS in a scalar surface integral, never for dS in this form.
5
Find the surface area of the cone z=x2+y2 below z=3.
no calculator, same as the exam
Answer9π2answer recomputed · sympy
The cone is the graph z=g(x,y)=x2+y2.
gx=x2+y2x and gy=x2+y2y, so gx2+gy2=x2+y2x2+y2=1.
Therefore dS=1+1dA=2dA — constant, because a cone has the same slope everywhere.
The shadow: z≤3 means x2+y2≤3, the disk of radius 3, whose area is 9π.
∬SdS=∫02π∫032rdrdθ=2⋅9π=9π2.
TrapWriting dS=1+x2+y2dA by squaring the *function* instead of its partials. The radical needs gx2+gy2, and for this cone those partials are the components of a unit vector, so their squares sum to exactly 1 and the factor is the clean 2. The other trap: the apex (0,0,0) makes gx,gy undefined there, but it is a single point of measure zero and the improper integral converges — the 2 still holds. And do not confuse this with the *lateral* cone formula πrℓ: here r=3, ℓ=32, giving 9π2, which agrees. ✓
6
Evaluate ∬SF⋅dS where F=⟨y,−x,1⟩ and S is the disk z=1, x2+y2≤1, oriented upward.
no calculator, same as the exam
Answerπanswer recomputed · sympy
S is a flat horizontal disk at height z=1, oriented upward, so the unit normal is simply n^=k=⟨0,0,1⟩ and dS=⟨0,0,1⟩dA (the graph z=g=1 has gx=gy=0, so dS=dA).
Dot the field with the normal: F⋅n^=⟨y,−x,1⟩⋅⟨0,0,1⟩=1. Only the k-component of F survives.
∬SF⋅dS=∬x2+y2≤11dA=area of the unit disk=π.
Explicitly in polar: ∫02π∫011⋅rdrdθ=2π⋅21=π.
TrapTrying to integrate the y and −x components. They are entirely tangential to a horizontal disk and contribute nothing to flux — this field swirls *within* the plane z=1 and never crosses it. (Its curl is ⟨0,0,−2⟩, so its *circulation* around the boundary is −2π, a completely different number that this problem is baiting you toward.) The second trap is orientation: 'oriented upward' fixes n^=+k; downward would give −π. Third, the disk has radius 1, so its area is π, not 2π.
STEP 03
More practice
4 questions
Mine, on those sections only. Untimed, no calculator — the quiz doesn't allow one either.
0/4 answered
117.6 · Surface area
Find the area of the part of the paraboloid z=x2+y2 lying below z=1.
Answer A
zx=2x,zy=2y⇒1+4x2+4y2=1+4r2; the shadow is the unit disk.
A=∫02π∫011+4r2rdrdθ=2π⋅121[(1+4r2)3/2]01=6π(55−1).
TrapDropping the r from dA, which kills the clean u-substitution.
S is the part of the plane z=2x+2y lying above the unit disk x2+y2≤1. Find ∬S1dS (the area of S).
Answer C
zx=2,zy=2⇒dS=1+4+4dA=3dA.
The shadow is the unit disk, of area π, so ∬SdS=3π.
The stretch factor is constant here because the surface is a plane — the tilt never changes.
TrapForgetting the √(1+zₓ²+z_y²) factor and reporting the shadow’s area, π.
Two hours, all course content. Stokes and the divergence theorem appear here and nowhere else.
Worth knowing. Lessons 34–37 are covered in the last two weeks and are on no quiz and no midterm. Every point they are worth is on the final.
The department publishes a study guide for this exam, and it is the source of truth for coverage. Every one of its practice problems is worked out, with the answer held back until you commit to an attempt, at the Final guide, worked. Start there; use this tab for the sections no quiz ever tested.
Only the sections no quiz ever covers are shown here. Everything from the earlier quizzes is on this exam too — work those tabs as well.
Official problems for these lessons
The department's own questions for exactly these lessons are worked below, before anything of mine.
Dip a bent wire loop in soap. The film can be flat or bulge out like a bubble. Both have the same rim, so both have the same total spin through them. Pick the lazy one.
drag to spin
Morph the soap film however you like. The rim never moves, so the answer never changes — Green's theorem is this picture lying flat.
circulation around the rim = curl through any surface it bounds∮CF⋅dr=∬S(curlF)⋅dS
Algorithm · use Stokes' theorem
Decide which side is easier. Nasty curve, simple curl → go right. Nasty surface, simple curve → go left.
Choose the friendliest surface with that rim. A flat disk or the plane region inside C almost always beats the surface you were handed.
Compute curlF as a determinant.
Match orientations by the right-hand rule: curl the fingers of your right hand along C, and your thumb must point the way n points. Mismatch = wrong sign.
Evaluate the surface integral using 17.6.
Green's theorem is exactly this with the surface lying flat in the xy-plane and n=k.
Check yourself
You are told S is 'any smooth surface with boundary C'. Does the choice matter?
Answer
No. Only the rim matters — pick the laziest surface, usually the flat disk.
When is Green's theorem just Stokes' theorem?
Answer
When the surface lies flat in the xy-plane with n=k. Same theorem, tilted.
Fill a solid with tiny cubes. Whatever leaks out the right face of one cube leaks into the left face of the next. Every internal face cancels with its partner. The only faces with nothing behind them are on the outer skin.
Each cube's surplus is div F · dV. Add them all up and you're left with the flux through the outside surface. Exactly Green's proof, in 3D.
flux out of a closed surface = divergence inside∬SF⋅dS=∭EdivFdV
Algorithm · use the divergence theorem
Confirm S is closed — a sealed bag with no openings — and oriented outward.
Compute divF=Px+Qy+Rz. This is usually startlingly simple compared to F itself.
Set up ∭E over the enclosed solid, in whichever coordinates match it (16.4–16.5).
Integrate. Often divF is a constant, and the answer is just that constant times the volume.
Algorithm · when the surface is NOT closed
Cap it. Add a simple surface S2 (usually a flat disk) to seal the opening.
Apply the theorem to the sealed union:∬S∪S2=∭EdivFdV.
Compute the cap's flux separately — it's flat, so it's easy.
Subtract:∬S=∭EdivFdV−∬S2.
Watch the cap's orientation: as part of the sealed bag it must point outward, which for a bottom cap means downward.
This capping trick is the single most-tested idea on the final.
Check yourself
What two conditions must S satisfy?
Answer
It must be closed (a sealed bag) and oriented outward.
The surface is open. Now what?
Answer
Cap it, apply the theorem to the sealed union, compute the cap's flux separately, then subtract it. The cap must point outward too.
STEP 02
The official problems
154 problems
Straight from the department's own study guide, for exactly these lessons — worked, with the answer held back until you commit. Do these first.
The projection of u onto v is projvu=∣v∣2u⋅vv — a vector *along* v.
u⋅v=(2)(1)+(−1)(2)+(3)(2)=2−2+6=6.
∣v∣2=1+4+4=9.
projvu=96⟨1,2,2⟩=32⟨1,2,2⟩=⟨32,34,34⟩.
TrapDividing by ∣v∣ instead of ∣v∣2. That gives the scalar projection compvu=36=2, which is a number, not a vector. Read which one the question wants.
2
Find the area of the triangle with vertices P(1,0,0), Q(0,2,0), and R(0,0,3).
no calculator, same as the exam
Answer27answer recomputed · sympy
Two edge vectors from P: PQ=⟨−1,2,0⟩ and PR=⟨−1,0,3⟩.
PQ×PR=⟨2⋅3−0⋅0,0⋅(−1)−(−1)⋅3,0−(−2)⟩=⟨6,3,2⟩.
∣⟨6,3,2⟩∣=36+9+4=7 — that is the parallelogram area.
A triangle is half of it: area=27.
TrapForgetting the 21. The cross product's length is the parallelogram, and 7 is sitting right there looking like an answer.
3
If u=⟨3,−1,2⟩ and v=⟨1,4,−1⟩, find u×v.
no calculator, same as the exam
Answer⟨−7,5,13⟩answer recomputed · sympy
u×v=i31j−14k2−1.
i: (−1)(−1)−(2)(4)=1−8=−7.
j: −[(3)(−1)−(2)(1)]=−(−5)=5 — note the sign flip on the middle term.
k: (3)(4)−(−1)(1)=12+1=13. So u×v=⟨−7,5,13⟩.
TrapThe minus on the j component. Dropping it gives ⟨−7,−5,13⟩, and the sanity check catches it instantly: u⋅(u×v) must be 0.
4
For what value of c are the vectors ⟨2,c,−1⟩ and ⟨3,−6,c⟩ orthogonal?
no calculator, same as the exam
Answerc=76answer recomputed · sympy
Orthogonal means the dot product vanishes.
⟨2,c,−1⟩⋅⟨3,−6,c⟩=6−6c−c=6−7c.
6−7c=0⇒c=76.
TrapMissing that c appears in both vectors, so the last term (−1)(c) also carries a c. Treating it as a constant gives c=1.
5
Find the angle between u=⟨1,1,0⟩ and v=⟨0,1,1⟩.
no calculator, same as the exam
Answerθ=3πanswer recomputed · sympy
cosθ=∣u∣∣v∣u⋅v.
u⋅v=0+1+0=1, and ∣u∣=∣v∣=2.
cosθ=21⇒θ=3π.
TrapAnswering 6π: cos−121=3π, not 6π. 6π is where the *sine* is 21.
The line through (1,3,−1) and (3,1,2) intersects the plane x+y+z=6 at what point?
no calculator, same as the exam
Answer(3,1,2)answer recomputed · sympy
Direction: v=⟨3−1,1−3,2−(−1)⟩=⟨2,−2,3⟩.
Line: r(t)=⟨1+2t,3−2t,−1+3t⟩.
Substitute into x+y+z=6: (1+2t)+(3−2t)+(−1+3t)=3+3t=6⇒t=1.
r(1)=⟨3,1,2⟩ — the second given point happens to lie on the plane.
TrapSign errors in the direction vector. Notice the x and y parameters cancel, so the whole equation collapses to 3+3t; if yours does not collapse, recheck v.
2
Find the equation of the plane containing (2,1,−1) and parallel to the vectors ⟨1,0,3⟩ and ⟨2,1,0⟩.
no calculator, same as the exam
Answer−3x+6y+z+1=0answer recomputed · sympy
Two vectors parallel to the plane give the normal through a cross product.
Point-normal form at (2,1,−1): −3(x−2)+6(y−1)+1(z+1)=0.
Expand: −3x+6+6y−6+z+1=0⇒−3x+6y+z+1=0.
Trap'Parallel to the vectors' does not mean the vectors are normals. They lie *in* the plane; their cross product is the normal.
3
Find the line of intersection between the planes x+y=2 and y+z=3.
no calculator, same as the exam
Answerx=2+t,y=−t,z=3+tanswer recomputed · sympy
Two equations, three unknowns — one free parameter, which is the line.
Let y=−t. Then x+y=2 gives x=2+t, and y+z=3 gives z=3+t.
x=2+t,y=−t,z=3+t.
Check the direction against n1×n2=⟨1,1,0⟩×⟨0,1,1⟩=⟨1,−1,1⟩ — same direction, as it must be.
TrapAny parametrisation is correct; yours may look nothing like this one. Verify by plugging back into both plane equations rather than comparing to a printed answer.
4
Find the angle between the planes x+y+z=1 and x−y=0.
no calculator, same as the exam
Answerθ=2π (planes are perpendicular)answer recomputed · sympy
The angle between planes is the angle between normals: n1=⟨1,1,1⟩, n2=⟨1,−1,0⟩.
n1⋅n2=1−1+0=0.
Zero dot product ⇒θ=2π: the planes are perpendicular.
TrapReading coefficients off a plane written as x−y=0 and forgetting the missing z is a coefficient of 0, not of 1.
5
For what value of a do the lines r1(t)=⟨1+t,2t,3+t⟩ and r2(s)=⟨3−s,s+1,2s+a⟩ intersect?
no calculator, same as the exam
Answera=2answer recomputed · sympy
Intersection needs a single point, so use different parameters t and s and solve all three components.
x: 1+t=3−s⇒t=2−s. y: 2t=s+1.
Substitute: 2(2−s)=s+1⇒4−2s=s+1⇒s=1, hence t=1.
z: 3+t=2s+a⇒3+1=2+a⇒a=2.
TrapUsing t for both lines. Two lines can cross at a point they reach at different times; forcing one parameter finds only the cases where they arrive together, and reports skew for lines that do meet.
′restored
Find the distance from the point (1,−2,3) to the plane 2x−y+2z=4.
no calculator, same as the exam
Official answer35checked · official answer wrong
Correct answer2recomputed · sympy
D=a2+b2+c2∣ax0+by0+cz0−d∣ with n=⟨2,−1,2⟩.
Numerator: ∣2(1)−(−2)+2(3)−4∣=∣2+2+6−4∣=6.
Denominator: 4+1+4=3.
D=36=2.
TrapForgetting to move d to the same side. The formula measures ax+by+cz−d, so a plane written as 2x−y+2z=4 contributes −4, not +4.
The official answer is wrongThe commented-out source gives 35. The numerator is ∣2+2+6−4∣=6, not 5, so the distance is 36=2. This may be exactly why the problem was commented out.
Commented out on the official page, so it does not appear there. The skill is still on the syllabus, so it is restored here.
AnswerHyperboloid of two sheets (axis along y)prose answer · not machine-checkable
Write it with the positive term first: 4y2−x2−9z2=1.
One positive square, two negative, equals +1 — that is the two-sheet signature.
Trace at y=0: −x2−9z2=1, impossible. Nothing at the origin, so it cannot be one connected sheet.
Hyperboloid of two sheets, axis along y — the axis is always the variable whose square is positive.
TrapOne sheet vs two sheets is decided by the count of minus signs, not by which variable is isolated. One minus = one sheet (connected); two minuses = two sheets.
2
Identify the surface defined by x2+4x+y2−2y+z2=−4. (Hint: complete the square.)
no calculator, same as the exam
AnswerSphere centered at (−2,1,0) with radius 1answer recomputed · sympy
Complete the square in x and y: x2+4x=(x+2)2−4 and y2−2y=(y−1)2−1.
(x+2)2−4+(y−1)2−1+z2=−4.
(x+2)2+(y−1)2+z2=1.
Sphere, centre (−2,1,0), radius 1.
TrapMoving the completion constants the wrong way. The −4 and −1 come out on the left, so they get added to the right: −4+4+1=1.
3
Identify the surface: z=4x2+9y2.
no calculator, same as the exam
AnswerElliptic paraboloid opening upwardprose answer · not machine-checkable
One variable (z) appears to the first power, the other two are squared with the same sign.
That is the paraboloid family. Both coefficients positive ⇒ opens upward.
Traces z=k>0 are ellipses 4x2+9y2=k; traces x=0 and y=0 are upward parabolas.
Elliptic paraboloid opening upward.
TrapCalling it a cone. A cone has z2, not z; the first power is what makes it a paraboloid.
4
Describe the trace of 4x2+9y2−z2=1 in the plane z=2.
TrapStopping at 4x2+9y2=5. A trace is named by its standard form, and the denominators change once you divide through.
5
Identify the surface: x2+4y2=9z2.
no calculator, same as the exam
AnswerElliptic cone with axis along the z-axisprose answer · not machine-checkable
Rewrite as x2+4y2−9z2=0 — equal to zero, not 1.
Zero is the cone signature. Traces z=k=0 are ellipses that grow linearly with ∣k∣; the trace z=0 is the single point at the origin (the vertex).
Elliptic cone with axis along the z-axis.
TrapReading it as a hyperboloid. Same sign pattern — the right-hand side is the entire difference: =1 one sheet, =−1 two sheets, =0 the cone between them.
TrapUsing t again for the tangent line. The line has its own parameter; reusing t makes you solve on the curve instead of on its tangent.
3
Find the unit tangent vector T(0) for r(t)=⟨cost,sint,t⟩.
no calculator, same as the exam
AnswerT(0)=⟨0,21,21⟩answer recomputed · sympy
r′(t)=⟨−sint,cost,1⟩, so r′(0)=⟨0,1,1⟩.
∣r′(0)∣=0+1+1=2.
T(0)=⟨0,21,21⟩.
TrapSkipping the normalisation. T is a unit vector by definition; r′ alone is not one here.
4
Find the speed of a particle with position r(t)=⟨et,e−t,2t⟩ at t=0.
no calculator, same as the exam
AnswerSpeed =2answer recomputed · sympy
r′(t)=⟨et,−e−t,2⟩; at t=0 this is ⟨1,−1,2⟩.
Speed =1+1+2=4=2.
TrapDifferentiating e−t to e−t. The chain rule puts a minus in front, though here it squares away — the sign matters for direction, not for speed.
5
If r(t)=⟨t,t2,t3⟩, compute r′(t)×r′′(t).
no calculator, same as the exam
Answer⟨6t2,−6t,2⟩answer recomputed · sympy
r′=⟨1,2t,3t2⟩ and r′′=⟨0,2,6t⟩.
i: (2t)(6t)−(3t2)(2)=12t2−6t2=6t2.
j: −[(1)(6t)−(3t2)(0)]=−6t.
k: (1)(2)−(2t)(0)=2. So r′×r′′=⟨6t2,−6t,2⟩.
TrapThis exact vector is the numerator of the curvature formula κ=∣r′∣3∣r′×r′′∣ — recognise it and Lesson 8 gets shorter.
6
A particle travels along r(t)=⟨3cost,4sint,0⟩. Find the speed at t=2π.
no calculator, same as the exam
AnswerSpeed =3answer recomputed · sympy
r′(t)=⟨−3sint,4cost,0⟩.
At t=2π: ⟨−3,0,0⟩.
Speed =3.
TrapAssuming an ellipse is traversed at constant speed. It is not — the speed here is 9sin2t+16cos2t, which runs between 3 and 4. Only the circle has constant speed under this parametrisation.
Find the arc length of r(t)=⟨3cost,3sint,4t⟩ for 0≤t≤2π.
no calculator, same as the exam
AnswerL=10πanswer recomputed · sympy
r′(t)=⟨−3sint,3cost,4⟩.
∣r′∣=9sin2t+9cos2t+16=9+16=5 — constant, the helix is traversed at constant speed.
L=∫02π5dt=10π.
TrapLosing the Pythagorean identity. 9sin2+9cos2=9; if your integrand still contains t, the collapse was missed.
2
Find the curvature κ(t) of r(t)=⟨2cost,2sint,t⟩.
no calculator, same as the exam
Answerκ=52answer recomputed · sympy
For the helix ⟨acost,asint,bt⟩, κ=a2+b2a — constant.
Here a=2, b=1: κ=4+12=52.
By the cross-product formula: r′×r′′=⟨2sint,−2cost,4⟩, whose length is 4+16=25, and ∣r′∣3=(5)3=55. Ratio =5525=52.
TrapUsing κ=a1 — that is the circle of radius a. Climbing in z straightens the curve, so the helix is always less curved than its shadow.
3
Find the curvature of the plane curve y=x3 at the point (0,0).
no calculator, same as the exam
Answerκ=0answer recomputed · sympy
For a graph, κ(x)=[1+(f′(x))2]3/2∣f′′(x)∣.
f′=3x2, f′′=6x; at x=0 both the numerator and f′ vanish.
κ(0)=13/20=0.
TrapAssuming a curve that visibly bends must have κ>0 somewhere along it. y=x3 is *flat to second order* at the origin — it is an inflection point, and curvature is exactly zero there.
4
Find the arc length of r(t)=⟨3t,4t,0⟩ for 1≤t≤5.
no calculator, same as the exam
AnswerL=20answer recomputed · sympy
r′=⟨3,4,0⟩, a constant vector: this is a straight line.
∣r′∣=5.
L=∫155dt=5⋅4=20.
TrapIntegrating from 0 to 5. The interval is [1,5], which has length 4, not 5.
5
A particle moves along r(t)=⟨5sint,3sint,4cost⟩. Find the distance traveled from t=0 to t=π.
Evaluate (x,y)→(0,0)limx4+y2x3y, or show it does not exist.
no calculator, same as the exam
Official answerlimit does not existchecked · official answer wrong
Correct answer0recomputed · sympy
Try paths first. Along y=0: 0. Along x=0: 0. Along y=x2: x4+x4x3⋅x2=2x→0. Every path gives 0, which is a signal to stop hunting and start proving.
Squeeze. By AM–GM, x4+y2≥2x2∣y∣, so for (x,y)=(0,0):
x4+y2x3y≤2x2∣y∣∣x∣3∣y∣=2∣x∣.
2∣x∣→0, so the limit exists and equals 0.
TrapA finite pile of paths never proves a limit exists — but it also never proves it fails. When every path you try agrees, switch to a bound.
The official answer is wrongThe official answer says the limit does not exist. It does: the AM–GM squeeze above bounds the expression by 2∣x∣. (Compare x4+y2x2y, which genuinely fails along y=x2 — one power of x away.)
2
Evaluate (x,y)→(0,0)limx2+y23x2y, or show it does not exist.
no calculator, same as the exam
Official answerlimit does not existchecked · official answer wrong
Correct answer0recomputed · sympy
Every path gives 0: along y=mx, x2(1+m2)3x2(mx)=1+m23mx→0.
Squeeze, and this one is immediate: x2≤x2+y2, so x2+y2x2≤1 and
x2+y23x2y≤3∣y∣.
3∣y∣→0 as (x,y)→(0,0), so the limit exists and equals 0.
TrapDegree counting is the fast check: the numerator is degree 3, the denominator degree 2. Numerator degree strictly larger ⇒ the quotient is squeezed to 0. Only equal degrees can produce a path-dependent limit.
The official answer is wrongThe official answer says the limit does not exist. It does: it is 0, by the one-line bound x2+y23x2y≤3∣y∣. The classic DNE example is x2+y2xy — numerator and denominator both degree 2.
3
Find the value of k such that (x,y)→(0,0)limx2+y2x4−2k(x2+y2)−y4=6.
no calculator, same as the exam
Answerk=−3answer recomputed · sympy
Split it: x2+y2x4−y4−x2+y22k(x2+y2).
First piece factors: x2+y2(x2−y2)(x2+y2)=x2−y2→0.
Second piece is exactly 2k, for every (x,y)=(0,0).
Limit =−2k. Set −2k=6⇒k=−3.
TrapLosing the sign. The term enters as −2k(x2+y2), so the limit is −2k, and k=3 would give −6.
4
Evaluate (x,y)→(0,0)limx2+y2sin(x2+y2).
no calculator, same as the exam
Answer1answer recomputed · sympy
Substitute u=x2+y2. As (x,y)→(0,0), u→0+.
The expression is usinu.
limu→0usinu=1.
TrapThe substitution is legitimate here precisely because x and y only ever appear in the combination x2+y2. It is not legitimate in general.
5
Evaluate (x,y)→(0,0)limx2+y2x3−y3, or show it does not exist.
no calculator, same as the exam
Answer0answer recomputed · sympy
Polar coordinates: x=rcosθ, y=rsinθ, and (x,y)→(0,0) is exactly r→0+.
r2r3(cos3θ−sin3θ)=r(cos3θ−sin3θ).
∣cos3θ−sin3θ∣≤2, so the whole thing is ≤2r→0uniformly in θ.
The limit is 0.
TrapPolar only settles it when the θ factor is bounded and the r factor goes to zero on its own. If θ survives in the limit, polar has told you it does not exist.
Let f(x,y,z)=cos(xy)+ey2z+ln(z3). Compute fyxz(0,1,1).
no calculator, same as the exam
Answerfyxz(0,1,1)=0answer recomputed · sympy
Read the order right: fyxz means differentiate by y, then x, then z.
fy=−xsin(xy)+2yzey2z.
fyx=−sin(xy)−xycos(xy) — the ey2z term has no x and dies here.
fyxz=0: what is left has no z in it at all. Evaluating at (0,1,1) is not even needed.
TrapGrinding through the ln(z3) term. It has no x and no y, so it is annihilated by the very first derivative.
2
If f(x,y)=x3y2+sin(xy), find fxy(0,0).
no calculator, same as the exam
Answerfxy(0,0)=1answer recomputed · sympy
fx=3x2y2+ycos(xy).
fxy=6x2y+cos(xy)−xysin(xy) (product rule on ycos(xy)).
At (0,0): 0+cos0−0=1.
TrapEvaluating at (0,0)before differentiating. f(x,0)=0 for all x, so that route reports 0 — you must differentiate symbolically first, then substitute.
3
If f(x,y)=ex2y, find fx(1,2).
no calculator, same as the exam
Answerfx(1,2)=4e2answer recomputed · sympy
Hold y constant: fx=ex2y⋅2xy.
At (1,2): e2⋅2⋅1⋅2=4e2.
TrapDifferentiating the exponent as if y were a variable too. In fx, y is a constant — that is the entire content of 'partial'.
4
Let f(x,y)=x2ln(y). Find fxy and verify Clairaut's theorem by also computing fyx.
no calculator, same as the exam
Answerfxy=fyx=y2xanswer recomputed · sympy
fx=2xlny, so fxy=y2x.
The other order: fy=yx2, so fyx=y2x.
They agree — as Clairaut guarantees, since both mixed partials are continuous on y>0.
TrapBelieving Clairaut always applies. It needs continuity of the mixed partials; the standard counterexample x2+y2xy(x2−y2) has fxy(0,0)=fyx(0,0).
5
If f(x,y)=arctan(xy), find fx+fy at (1,1).
no calculator, same as the exam
Answerfx+fy=0 at (1,1)answer recomputed · sympy
For f=arctanxy: fx=x2+y2−y and fy=x2+y2x.
At (1,1): fx=−21, fy=21.
fx+fy=0.
TrapDropping the inner derivative ∂x∂xy=−x2y. The 1+(y/x)21 factor then has to be cleared of its fractions to reach x2+y2.
Or substitute first: w=st(s+t)+st(s−t)=2s2t, ws=4st=0 at (1,0).
∂s∂w=0.
TrapMissing that y multiplies two branches (w=xy+yz feeds y into both terms), so ys appears once with coefficient x+z.
4
If sin(xy)=e2z, use implicit differentiation to find ∂x∂z at (1,2π,0).
no calculator, same as the exam
Answer∂x∂z=0answer recomputed · sympy
F=sin(xy)−e2z, so Fx=ycos(xy) and Fz=−2e2z.
At (1,2π,0): Fx=2πcos2π=0 and Fz=−2.
∂x∂z=−−20=0.
Trapcos2π=0 is what kills this one. Check the point before doing any algebra — sometimes it collapses the whole expression.
5
If z=f(x,y) with x=rcosθ and y=rsinθ, express ∂r∂z in terms of fx and fy.
no calculator, same as the exam
Answer∂r∂z=fxcosθ+fysinθprose answer · not machine-checkable
∂r∂z=fx∂r∂x+fy∂r∂y.
∂r∂x=cosθ and ∂r∂y=sinθ (with θ held fixed).
∂r∂z=fxcosθ+fysinθ.
Worth noticing: that is ∇f⋅⟨cosθ,sinθ⟩ — the directional derivative straight outward. Lesson 13 in disguise.
TrapWriting ∂θ∂z by mistake; that one is −fxrsinθ+fyrcosθ, and the factors of r are the tell.
6
If F(x,y,z)=x2+y2+z2−3xyz=0 defines z implicitly, find ∂x∂z at (1,1,1), or explain why it does not exist.
no calculator, same as the exam
Answer∂x∂z(1,1,1)=−1answer recomputed · sympy
F=x2+y2+z2−3xyz, so Fx=2x−3yz and Fz=2z−3xy.
At (1,1,1): Fx=2−3=−1 and Fz=2−3=−1.
Fz=−1=0, so the implicit function theorem applies and z really is a function of (x,y) near this point.
∂x∂z=−−1−1=−1.
TrapSkipping the Fz=0 check. When Fz=0 the surface has a vertical tangent plane there and ∂x∂z does not exist — which is exactly what the 'or explain why not' is fishing for.
Find the maximum value of f(x,y)=3+2x−y subject to the constraint x2+2xy+4y2=7.
no calculator, same as the exam
Answer10answer recomputed · sympy
The constraint g(x,y)=x2+2xy+4y2=7 is a closed curve (an ellipse), so f attains a max and a min on it. Use Lagrange: ∇f=λ∇g.
∇f=⟨2,−1⟩ and ∇g=⟨2x+2y,2x+8y⟩, so the system is 2=λ(2x+2y), −1=λ(2x+8y), x2+2xy+4y2=7.
Add twice the second equation to the first to kill x: 2+2(−1)=λ[(2x+2y)+2(2x+8y)] gives 0=λ(6x+18y). Since λ=0 (else ∇f=0, false), x=−3y.
Substitute into the constraint: 9y2−6y2+4y2=7y2=7, so y=±1 and (x,y)=(−3,1) or (3,−1).
Evaluate: f(−3,1)=3−6−1=−4 and f(3,−1)=3+6+1=10.
Maximum value =10 (attained at (3,−1); the minimum is −4).
Trap∇g of the cross term is the usual casualty: ∂x(2xy)=2y and ∂y(2xy)=2x, so ∇g=⟨2x+2y,2x+8y⟩, not⟨2x+2y,8y⟩. Also, Lagrange hands you *critical values*, plural. The largest of the list is the answer, so you must evaluate f at every solution point, not just the first one you find.
2
Find the maximum value of f(x,y)=8x−6y subject to the constraint (x−1)2+y2=1.
no calculator, same as the exam
Answer18answer recomputed · sympy
Constraint: the circle g(x,y)=(x−1)2+y2=1, centre (1,0), radius 1. f is linear, so its extremes sit where the gradient of f is normal to the circle.
Lagrange: ⟨8,−6⟩=λ⟨2(x−1),2y⟩, i.e. 8=2λ(x−1) and −6=2λy.
So x−1=λ4 and y=−λ3. Plug into the constraint: λ216+λ29=1⇒λ2=25,λ=±5.
Sanity check with geometry: f=8x−6y at the centre (1,0) is 8, and ∣∇f∣=64+36=10, so on a radius-1 circle f ranges over 8±10={−2,18}. Same answer.
TrapThe circle is not centred at the origin. Writing the constraint gradient as ⟨2x,2y⟩ (forgetting the shift) is the standard wreck here; it must be ⟨2(x−1),2y⟩. The shortcut fmax=f(centre)+∣∇f∣⋅R works only because f is linear and the constraint is a circle; do not reach for it on an ellipse.
3
The maximum and minimum values of f(x,y)=(x−2)2+(y−4)2 subject to x2+y2=5 are, respectively:
f(x,y)=(x−2)2+(y−4)2 is the squared distance from (2,4); the constraint is the circle of radius 5 about the origin. So this is really a nearest/farthest point problem.
Lagrange: ⟨2(x−2),2(y−4)⟩=λ⟨2x,2y⟩, i.e. x−2=λx and y−4=λy.
Then x(1−λ)=2 and y(1−λ)=4, so y=2x (dividing; note λ=1, since λ=1 would force 2=0).
Constraint: x2+(2x)2=5x2=5⇒x=±1, giving (1,2) and (−1,−2).
Maximum 45, minimum 5. (Check: ∣(2,4)∣=20=25, so distances to the circle are 25∓5, whose squares are 5 and 45.)
TrapTwo traps. First, f is the squared distance, so do not take a square root at the end and report 45 and 5. Second, the point (2,4) lies *outside* the circle, so both critical points are genuine, one near and one far; the min is not0.
4
Find the maximum of f(x,y,z)=x+y+z subject to (x−1)2+y2+z2=1.
no calculator, same as the exam
Answer1+3answer recomputed · sympy
Constraint: the sphere g=(x−1)2+y2+z2=1 of radius 1 centred at (1,0,0). f=x+y+z is linear, so Lagrange with ∇f=⟨1,1,1⟩.
Taking the + branch, x=1+31,y=z=31, so f=1+33=1+3. The − branch gives 1−3.
Maximum =1+3.
TrapSame shifted-centre trap as before: ∇g=⟨2(x−1),2y,2z⟩. And because the three partials give x−1=y=z (not x=y=z), the symmetric guess x=y=z=1/3 is wrong. The geometric check is f(centre)+∣∇f∣R=1+3⋅1.
5
Find the absolute maximum value M and absolute minimum value m of f(x,y)=x+y subject to x2−xy+y2=1.
no calculator, same as the exam
AnswerM=2,m=−2answer recomputed · sympy
The constraint x2−xy+y2=1 is an ellipse (rotated 45∘), a closed bounded curve, so a max and a min both exist.
Lagrange: ⟨1,1⟩=λ⟨2x−y,2y−x⟩, so 1=λ(2x−y) and 1=λ(2y−x).
Setting the two right sides equal: 2x−y=2y−x⇒3x=3y⇒y=x.
Constraint: x2−x2+x2=x2=1, so (1,1) and (−1,−1).
f(1,1)=2 and f(−1,−1)=−2, so M=2,m=−2.
TrapDifferentiating the cross term −xy: ∂x=−y, ∂y=−x — it is easy to drop a sign and get 2x+y. Also, cancelling λ from the two equations is legal here only because λ=0 (if λ=0 then 1=0), so you should say so rather than divide blindly.
The integrand factors, xy=xy, and the region is the rectangle [0,4]×[0,1], so the double integral splits into a product of single integrals.
∫01∫04xydxdy=(∫04xdx)(∫01ydy).
∫04x1/2dx=32x3/204=32⋅8=316.
∫01y1/2dy=32.
Product: 316⋅32=932.
TrapRead the differentials: dxdy means the inner variable is x (limits 0 to 4) and the outer is y (limits 0 to 1). Pairing x with [0,1] gives 32⋅316 by luck here, but on a non-symmetric integrand the same misreading is fatal. The split into a product is legal only because the integrand factors and the limits are constants.
2
Find the volume enclosed by x=0, x=3, y=0, y=2, z=1, and z=ex+y.
no calculator, same as the exam
Answere5−e3−e2−5answer recomputed · sympy
The base is the rectangle R=[0,3]×[0,2]; the solid sits between the floor z=1 and the roof z=ex+y. On R we have x+y≥0, so ex+y≥1 and the roof really is above the floor everywhere.
Volume =∬R(ex+y−1)dA=∫02∫03(exey−1)dxdy.
∬RexeydA=(∫03exdx)(∫02eydy)=(e3−1)(e2−1).
∬R1dA=area(R)=3⋅2=6.
Volume =(e3−1)(e2−1)−6=e5−e3−e2+1−6=e5−e3−e2−5.
TrapForgetting to subtract the floor: integrating ex+y alone gives the volume under the surface down to z=0, not down to z=1. The correction is −area(R)=−6, and note (e3−1)(e2−1) already contributes +1, so the constant is 1−6=−5, not −6. Also ex+y=exey is what lets the integral factor; ex+y=ex+ey.
3
Find the average value of f(x,y)=x2y over the rectangle R with vertices (−1,0), (−1,5), (1,5), (1,0).
no calculator, same as the exam
Answer65answer recomputed · sympy
The four vertices give the rectangle R=[−1,1]×[0,5], of area A=2⋅5=10.
Average value =A1∬RfdA=101∫05∫−11x2ydxdy.
∫−11x2dx=3x3−11=32 and ∫05ydy=225.
∬Rx2ydA=32⋅225=325.
Average =1025/3=65.
TrapDividing by the wrong thing, or not dividing at all: the answer to an *average value* question is the integral over the area, here 10. Second trap: x2 is even, so ∫−11x2dx=32, not 0 — students who remember 'symmetric interval ⇒ zero' from odd integrands kill this problem instantly.
4
Evaluate ∫01∫01∫02−yexdzdxdy.
no calculator, same as the exam
Answer23(e−1)answer recomputed · sympy
Innermost is dz and the integrand ex is free of z, so the first integration is just multiplication by the height: ∫02−yexdz=ex(2−y).
The remaining limits are constants, so ∫01∫01ex(2−y)dxdy=(∫01exdx)(∫01(2−y)dy).
∫01exdx=e−1.
∫01(2−y)dy=2−21=23.
Value =23(e−1).
TrapThe order is dzdxdy, so the z-limit 2−y belongs to the *innermost* integral and depends on the *outermost* variable — perfectly legal, but you must integrate z first and carry (2−y) outward. Attempting to do dx first, or treating the upper limit as 2−x, both break it. Once z is gone the integral separates only because ex and (2−y) live in different variables.
5
Evaluate ∬R2+x3x2ydA over R={1≤x≤2,0≤y≤4}.
no calculator, same as the exam
Answer38(ln10−ln3)answer recomputed · sympy
R is a rectangle and the integrand splits as 2+x3x2⋅y, so the integral is a product of two one-variable integrals.
∬R2+x3x2ydA=(∫122+x3x2dx)(∫04ydy).
For the x-integral substitute u=2+x3, du=3x2dx, so x2dx=3du; as x:1→2, u:3→10.
∫3103u1du=31(ln10−ln3).
∫04ydy=8.
Value =8⋅31(ln10−ln3)=38(ln10−ln3)=38ln310.
TrapWhen you substitute u=2+x3 you must change the limits to u:3→10; leaving 1 and 2 gives 31ln12, a completely different number. The other classic is losing the 31 from du=3x2dx.
TrapGetting the new limits backwards. The old inner limit y≥3x becomes x≤y3, so the new inner range is 0≤x≤y3 — noty3≤x≤27. Sketch the region: it is the piece *above* the cube-root curve, i.e. to the *left* of x=y3. The factor 4 in the integrand is not decoration; it is exactly the 4y3 the substitution needs after the swap.
2
Evaluate ∬RxsinxdA where R is bounded by the x-axis, y=x, and x=1 (integrate in y first, then x).
no calculator, same as the exam
Answer1−cos(1)answer recomputed · sympy
Sketch R: bounded below by y=0, above by y=x, on the right by x=1 — the triangle with vertices (0,0),(1,0),(1,1).
xsinx has no elementary antiderivative in x, so we must integrate in y first (as instructed), which makes x a constant.
∬RxsinxdA=∫01∫0xxsinxdydx.
Inner: xsinx⋅(x−0)=sinx. The offending x in the denominator cancels — that is the whole point of the order.
∫01sinxdx=−cosx01=1−cos(1).
TrapChoosing dydx is not optional here: the other order gives ∫y1xsinxdx, which cannot be evaluated in closed form. Also the inner limits are 0≤y≤x — a *variable* upper limit. Writing ∫01 for the inner y-integral turns the triangle into the unit square and gives ∫01xsinxdx, which is not elementary.
3
Find the volume of the solid under z=2x and above the region in the first quadrant bounded by the coordinate axes and x+y=1.
no calculator, same as the exam
Answer31answer recomputed · sympy
The base region D is the triangle in the first quadrant cut off by x+y=1: vertices (0,0),(1,0),(0,1). The roof is z=2x≥0 there, so the volume is the plain double integral.
As a type-I region: 0≤x≤1 and 0≤y≤1−x.
V=∬D2xdA=∫01∫01−x2xdydx.
Inner: 2x(1−x)=2x−2x2.
∫01(2x−2x2)dx=1−32=31.
TrapUsing ∫01∫01 (the square) instead of the triangle: the hypotenuse y=1−x must appear as the inner limit. Also note the integrand is 2x, so the answer is not the triangle's area 21 times anything obvious; you have to actually integrate 2x(1−x).
4
Change the order of integration for ∫02∫x22xf(x,y)dydx.
no calculator, same as the exam
Official answer(original) — ∫04∫yy/2fdxdychecked · official answer wrong
Read the region from the given order dydx: 0≤x≤2 and x2≤y≤2x — the lens between the parabola y=x2 and the line y=2x.
Find where the two curves meet: x2=2x⇒x=0 or x=2, giving the corner points (0,0) and (2,4). So y ranges over [0,4] overall.
Now solve each boundary for x. From y=2x we get x=2y; from y=x2 (with x≥0) we get x=y.
For a fixed y∈[0,4], which is the left edge? Test y=1: the line gives x=21, the parabola gives x=1. So x runs from 2yup toy.
Reversed order: ∫04∫y/2yf(x,y)dxdy.
Confidence check with a test integrand f=xy: the original order gives 38, and ∫04∫y/2yxydxdy=38 as well.
TrapInside the lens the parabola is the right boundary and the line is the left one (for 0<x<2, x2<2x means the parabola sits *below* the line, which flips to *right of* when you read horizontally). Writing ∫yy/2 reverses the limits and negates the integral. Also x=y, not x=±y: the region lives in x≥0.
The official answer is wrongThe official answer prints the inner limits as ∫yy/2, which is upside down. For 0<y<4 we have 2y<y (at y=1: 0.5<1), so the lower limit must be y/2 and the upper limit y. As printed, the integral equals the negative of the original. Numerical check with f(x,y)=xy: the original iterated integral is 38, the correctly reversed one is 38, and the printed one is −38.
5
Reverse the order and evaluate: ∫02ln3∫y/2ln3ex2dxdy.
no calculator, same as the exam
Answer2answer recomputed · sympy
ex2 has no elementary antiderivative, so the order must be reversed. Let a=ln3 for short; the limits are 0≤y≤2a and 2y≤x≤a.
Sketch: the region is the triangle with vertices (0,0), (a,0) and (a,2a), bounded by the line x=y/2 (i.e. y=2x), the vertical line x=a, and the x-axis.
Reading it the other way: x runs 0→a, and for each x, y runs from 0 to 2x.
TrapThe outer limit 2ln3 and the inner ln3 are *different* numbers — reversing carelessly and writing y from 0 to x (instead of 2x) loses the factor of 2 that the substitution needs, leaving a non-elementary integral. Also eln3=3, so the final answer is the clean 2, not eln3 left unsimplified.
TrapThe integrand becomes r2, not r: one factor of r is x2+y2 and the second is the Jacobian from dA=rdrdθ. Dropping the Jacobian gives 49π. Second trap: θ runs only to π/2 — the limits x∈[0,3], y≥0 give a quarter disk, not a half disk.
2
Compute ∬Re−(x2+y2)dA where R is the disk x2+y2≤4.
no calculator, same as the exam
Answerπ(1−e−4)answer recomputed · sympy
R is the full disk of radius 2 and the integrand depends only on x2+y2, so polar is the obvious move: 0≤r≤2, 0≤θ≤2π.
∬Re−(x2+y2)dA=∫02π∫02e−r2rdrdθ.
Inner: substitute u=r2, du=2rdr, so ∫02e−r2rdr=21∫04e−udu=21(1−e−4).
Outer: multiply by ∫02πdθ=2π.
Value =2π⋅21(1−e−4)=π(1−e−4).
TrapWithout the Jacobian r, ∫e−r2dr is not elementary at all — the r is exactly what makes the substitution work, so its absence should be an immediate alarm. Also r goes to 2, so u=r2 goes to 4, giving e−4; writing e−2 is the standard slip.
3
Rewrite in polar (do not evaluate): ∫01∫−1−y21−y2(x2+y2)1/2dxdy.
no calculator, same as the exam
Answer∫0π∫01r2drdθanswer recomputed · sympy
Region: 0≤y≤1 with −1−y2≤x≤1−y2, i.e. x2+y2≤1 with y≥0 — the upper half of the unit disk.
In polar the half disk is 0≤r≤1, 0≤θ≤π (the full sweep of the upper half plane).
The integrand (x2+y2)1/2=r, and dA=rdrdθ, so r⋅r=r2.
∫01∫−1−y21−y2(x2+y2)1/2dxdy=∫0π∫01r2drdθ.
(Not asked for, but its value is 3π, and the original Cartesian integral evaluates to 3π as well.)
Trapθ ranges over [0,π], not [0,π/2]: x takes negative values here, so the region covers the second quadrant too. And the integrand must pick up the Jacobian, r⋅rdrdθ=r2drdθ; writing ∫0π∫01rdrdθ is the most common wrong answer on this exact problem.
4
Evaluate ∫0π∫0π−y2sin(x2+y2)dxdy.
no calculator, same as the exam
Answer2πanswer recomputed · sympy
Region: 0≤y≤π, 0≤x≤π−y2, i.e. x2+y2≤π with x,y≥0 — the quarter disk of radius π.
sin(x2+y2) has no elementary antiderivative in Cartesian form, so polar is forced: 0≤r≤π, 0≤θ≤2π.
TrapThe radius is π, so u=r2 runs to π — not to π. Writing ∫0πsinudu is the killer here. As always, the Jacobian r is what makes the u-substitution possible; without it the integral is non-elementary.
5
Compute ∫−33∫−9−y29−y2ex2+y2dxdy.
no calculator, same as the exam
Answerπ(e9−1)answer recomputed · sympy
Region: −3≤y≤3 and −9−y2≤x≤9−y2 — the entire disk x2+y2≤9 of radius 3.
Polar: 0≤r≤3, 0≤θ≤2π, integrand ex2+y2=er2, and dA=rdrdθ.
∫−33∫−9−y29−y2ex2+y2dxdy=∫02π∫03er2rdrdθ.
Inner: u=r2, du=2rdr, u:0→9, so 21∫09eudu=21(e9−1).
Outer: 2π⋅21(e9−1)=π(e9−1).
TrapBoth limits are symmetric, so θ sweeps the full2π; using π (half disk) or π/2 halves or quarters the answer. And u=r2 runs to 9, giving e9 — writing e3 is the standard misfire. The r from dA is again mandatory.
Read the order off the differentials: dzdxdy. The innermost variable is z, and the integrand ex does not contain z — so the inner integral is just multiplication by the height.
∫02−yexdz=ex(2−y).
Now the x and y limits are both constants, and what is left separates: ∫01∫01ex(2−y)dxdy=(∫01exdx)(∫01(2−y)dy).
∫01exdx=e−1 and ∫01(2−y)dy=2−21=23.
Product: 23(e−1).
TrapThe limit 2−y sits on the inner integral while y is the outer variable. That is legal — inner limits may involve any variable integrated later — but it means you cannot do the y-integral first. Someone who starts at the outside gets ∫01(2−y)dy sitting where a z-integral belongs and loses the factor entirely.
2
Let E be the solid in the first octant above the xy-plane and below 4x+2y+z=8. For ∭EfdV=∫0a∫0b∫0cfdzdydx:
no calculator, same as the exam
Answera=2,b=4−2xanswer recomputed · sympy
Order dzdydx means: z is bounded by surfaces, y by curves in the xy-plane, x by numbers. Build it from the inside out.
First octant gives the floor z=0; the plane gives the ceiling z=8−4x−2y. So c=8−4x−2y.
Project the solid onto the xy-plane by setting z=0 in the plane: 4x+2y=8, i.e. y=4−2x. That triangle, together with x≥0,y≥0, is the shadow.
For a fixed x, y runs from 0 up to that line: b=4−2x.
x runs until the shadow closes, i.e. until 4−2x=0: a=2.
∭EfdV=∫02∫04−2x∫08−4x−2yfdzdydx.
TrapUsing the x-intercept of the plane, x=2... by luck that is right here, but the reason matters: a is where the shadow closes, not where the plane meets an axis. Get b by setting z=0, not x=0; setting x=0 gives y=4, which is the wrong constant limit and makes the region a box.
3
Let D be the solid bounded by x=0, z=0, y=x, and x+y+z=2. Write ∭DfdV as an iterated integral.
no calculator, same as the exam
Official answere.g. ∫01∫x2−x∫02−x−yfdzdydx , or ∫01∫02−x∫x2−x−zfdydzdxchecked · official answer wrong
Four surfaces, so list the inequalities: x≥0, z≥0, y≥x, and x+y+z≤2.
Take z innermost: 0≤z≤2−x−y.
Shadow in the xy-plane is where that height is non-negative and y≥x: the triangle x≤y≤2−x, x≥0.
That triangle closes when x=2−x, i.e. x=1. So ∫01∫x2−x∫02−x−yfdzdydx.
Reordering to dydzdx: solve y≥x and y≤2−x−z, so x≤y≤2−x−z.
Those two limits cross when 2−x−z=x, i.e. at z=2−2x — so z must stop at 2−2x, giving ∫01∫02−2x∫x2−x−zfdydzdx.
Sanity check both with f=1: each gives volume 32.
TrapWhen you reorder, the new outer limit is where the inner limits collide, not where some surface hits an axis. Writing z from 0 to 2−x (the plane's z-intercept above y=0) puts z in a range where the inner y-limits are upside-down, 2−x−z<x, and the integral quietly subtracts the phantom slab instead of erroring.
The official answer is wrongThe second offered order is wrong as printed. With z running to 2−x, every z>2−2x has 2−x−z<x, so the inner y-integral runs backwards. With f=1 the printed form evaluates to 21; the true volume, and the value of the printed first form, is 32. The z-limit must be 2−2x.
4
Let V be bounded by z=6−x2−y2 and z=x2+y2. If V=∫−33∫ab∫x2+y26−x2−y2dzdydx, what are a and b?
no calculator, same as the exam
Answera=−3−x2,b=3−x2answer recomputed · sympy
The z-limits are already given: the solid sits between the upward paraboloid z=x2+y2 and the downward one z=6−x2−y2.
They meet where x2+y2=6−x2−y2, i.e. 2(x2+y2)=6, so x2+y2=3 — a circle of radius 3.
That circle is the shadow's boundary, which is why the outer limits are ±3.
Solving x2+y2=3 for y: a=−3−x2, b=3−x2.
TrapSetting x2+y2=6 (from z=6−x2−y2 with z=0) instead of equating the two surfaces. The paraboloids never reach z=0 together; the boundary of the shadow is where the ceiling meets the floor, and that is radius 3, not 6. The given outer limits ±3 are the tell.
5
Evaluate ∫−11∫−1−y21−y2∫−11(x2+y2)3/2dzdxdy.
no calculator, same as the exam
Answer54πanswer recomputed · sympy
Look at the shadow before touching the integrand: −1≤y≤1 with −1−y2≤x≤1−y2 is the full unit disk, and z runs over the fixed interval [−1,1]. The solid is a cylinder.
(x2+y2)3/2 is a function of distance from the z-axis alone — cylindrical coordinates, and the z-integral is just a factor of 2.
(x2+y2)3/2=r3, and dV=rdrdθdz, so the integrand becomes r3⋅r=r4.
∫−11∫02π∫01r4drdθdz=2⋅2π⋅51=54π.
TrapConverting (x2+y2)3/2 to r3 and then integrating ∫01r3dr=41 — the extra r from dA=rdrdθ is dropped and the answer comes out π instead of 54π. The exponent you integrate is always one higher than the one you converted.
TrapConverting the integrand but keeping the Cartesian limits in spirit — reading 0≤x≤2 as 'θ from 0'. The upper x-limit 2 is a vertical line, r=2secθ, not a ray; only the presence of the lower boundary y=x turns it into the clean ray θ=4π. Any time a straight edge is not through the origin, polar limits stop being constants.
The official answer is wrongThe printed Cartesian stem and the printed cylindrical answer describe different regions. As written, 0≤y≤4−x2 over 0≤x≤2 has area 1+2π, while the wedge 4π≤θ≤2π,r≤2 has area 2π — the answer is missing the triangular piece below y=x. The printed answer is correct for the inner limit ∫x4−x2; with the printed lower limit 0 the rewrite needs the two pieces above.
2
Evaluate ∫02∫04−x2∫x2+y22(x2+y2)dzdydx using cylindrical coordinates.
no calculator, same as the exam
Answer34π(2−1)answer recomputed · sympy
Shadow first: 0≤x≤2, 0≤y≤4−x2 is the quarter disk of radius 2 in the first quadrant, so 0≤θ≤2π, 0≤r≤2.
The z-limits are both cones: z=x2+y2=r below and z=2(x2+y2)=2r above.
Height of the column at radius r: 2r−r=(2−1)r.
With dV=rdrdθdz: ∫0π/2∫02(2−1)r⋅rdrdθ=2π(2−1)⋅38.
=34π(2−1).
Trap2(x2+y2) is 2r, not 2r and not r2 — pulling the 2 out from under the root without the square root gives 34π exactly one unit too large. Second trap: the height (2−1)r already carries one power of r, so the integrand is r2, not r; forgetting the Jacobian here costs a factor of 23⋅22 and lands on π(2−1).
3
Convert to cylindrical and evaluate ∫010∫0100−x2∫0x2+y2x2+y21dzdydx.
no calculator, same as the exam
Answer25πanswer recomputed · sympy
The shadow 0≤x≤10, 0≤y≤100−x2 is a quarter disk of radius 10: 0≤θ≤2π, 0≤r≤10.
The ceiling z=x2+y2 is the cone z=r; the integrand x2+y21=r1.
So the full cylindrical integrand is r1⋅r=1 — the Jacobian cancels the singularity. That is the point of the problem.
TrapTwo errors cancel here if you are careless in exactly the wrong way. Dropping the Jacobian leaves ∫∫r1⋅rdz vs ∫∫r1dz — the second gives ∫0101dr=10 and the answer 5π. The integrand being 1 after conversion is a result, not an accident: check that you multiplied by r before you simplified.
4
Find the volume of the solid bounded by z=x2+y2−9 and z=−2(x2+y2) using cylindrical coordinates.
no calculator, same as the exam
Answer227πanswer recomputed · sympy
Both surfaces depend only on r2=x2+y2: z=r2−9 (an upward paraboloid dropped to −9) and z=−2r2 (a downward one from the origin).
Which is on top? At r=0: −9 versus 0, so z=−2r2 is the ceiling and z=r2−9 is the floor.
They meet where r2−9=−2r2⇒3r2=9⇒r=3. Full revolution, so 0≤θ≤2π.
TrapAssuming the paraboloid opening upward is the ceiling because it 'goes up'. Over the whole region r<3 it is the floor, and swapping them makes the volume negative. Test at r=0 every time. The other trap is integrating 9−3r2 against dr instead of rdr, which gives 2π(93−33)=123π — a suspiciously irrational volume for a surface of revolution with rational intersection radius.
5
Write the volume of the solid bounded by the cone z2=x2+y2 between the planes z=1 and z=2 as a spherical-coordinates integral.
In spherical, the cone z2=x2+y2 with z>0 is the single surface ϕ=4π, and the solid inside it is 0≤ϕ≤4π.
A horizontal plane is not a constant in spherical: z=ρcosϕ, so z=1 becomes ρ=secϕ and z=2 becomes ρ=2secϕ.
Those are the ρ-limits, and they depend on ϕ — so ρ must be integrated first.
dV=ρ2sinϕdρdϕdθ, full revolution in θ:
V=∫02π∫0π/4∫secϕ2secϕρ2sinϕdρdϕdθ.
Worth confirming in cylindrical, where it is a stack of disks of radius z: ∫02π∫12∫0zrdrdzdθ=∫12πz2dz=37π — and the spherical integral gives the same.
TrapWriting the planes as constant ρ (say ρ from 1 to 2), which describes the region between two spheres, not two planes — that is a spherical shell wedge, volume 32π(8−1)(1−cos4π), not 37π. Any flat boundary not through the origin becomes ρ=(something)secϕ and forces dρ innermost.
The volume of the region inside x2+y2+z2=4 and inside the cone z=3x2+3y2 is ∫02π∫0a∫0bcdρdϕdθ. What are a, b, and c?
no calculator, same as the exam
Answera=π/6,b=2,c=ρ2sinϕanswer recomputed · sympy
c is the Jacobian and nothing else: dV=ρ2sinϕdρdϕdθ, so c=ρ2sinϕ.
Sphere x2+y2+z2=4 is ρ=2, and since ρ is innermost and starts at 0, b=2.
The cone z=3x2+3y2=3x2+y2 has tanϕ=zx2+y2=31.
So ϕ=6π, and 'inside the cone' means 0≤ϕ≤6π: a=6π.
V=∫02π∫0π/6∫02ρ2sinϕdρdϕdθ.
Traptanϕ=zr, not rz — ϕ is measured down from the positive z-axis. Taking tanϕ=3 gives ϕ=3π, which is the complementary wedge and roughly four times the volume. Second trap: writing c=ρsinϕ, the *polar* Jacobian with a ρ in it, which is dimensionally wrong for a volume.
2
Write the volume of the solid bounded by the cone z=x2+y2 and the sphere x2+y2+z2=8 as a spherical-coordinates integral.
Cone z=x2+y2: here tanϕ=zr=1, so it is the surface ϕ=4π, and the solid inside it is 0≤ϕ≤4π.
Sphere x2+y2+z2=8 is ρ2=8, i.e. ρ=8=22.
Both boundaries are coordinate surfaces in spherical — every limit is a constant, which is the whole reason to use it here.
V=∫02π∫0π/4∫08ρ2sinϕdρdϕdθ.
Value: 2π⋅(1−22)⋅3(8)3=332π(2−1), which the cylindrical set-up ∫02π∫02(8−r2−r)rdrdθ confirms.
TrapReading ρ=8 off x2+y2+z2=8. The equation gives ρ2=8, so ρ=22 — and cubing the wrong one inflates the volume by a factor of 83/83/2≈22.6. Also note the cone/sphere meet at z=2, r=2: that intersection is *not* needed in spherical, which is the payoff for converting.
3
Write the volume of the region inside the sphere of radius 2 centered at (0,0,2) and outside the sphere of radius 2 centered at the origin as a spherical-coordinates integral.
Put both spheres in spherical form. Origin-centred radius 2: ρ=2.
The one centred at (0,0,2): x2+y2+(z−2)2=4⇒ρ2−4ρcosϕ=0⇒ρ=4cosϕ.
'Inside the shifted sphere, outside the origin one' is exactly 2≤ρ≤4cosϕ — the inner radius is the constant, the outer is the ϕ-dependent one.
Those limits are only valid where 4cosϕ≥2, i.e. cosϕ≥21, i.e. ϕ≤3π. That is the ϕ-limit.
V=∫02π∫0π/3∫24cosϕρ2sinϕdρdϕdθ.
Check it independently: the lens where two radius-2 spheres with centres 2 apart overlap has volume 12π(4R+d)(2R−d)2=310π, so the answer is 332π−310π=322π — and the integral above gives 322π.
TrapTaking ϕ up to 2π because the shifted sphere 'sits above the xy-plane'. Past ϕ=3π the outer limit 4cosϕ drops below the inner limit 2, so the integral starts subtracting. The upper ϕ is always where the two ρ-limits meet: solve 4cosϕ=2.
4
Converting ∫−11∫−1−x21−x2∫3(x2+y2)4−x2−y2fdzdydx to spherical gives ∫0a∫0b∫0cf⋅ρ2sinϕdρdϕdθ (order: dρdϕdθ, so a is the upper limit on θ, b on ϕ, c on ρ). Find abc.
The shadow −1≤x≤1, ∣y∣≤1−x2 is the full unit disk, so θ makes a full turn: a=2π.
Ceiling z=4−x2−y2 is the sphere ρ=2, so c=2.
Floor z=3(x2+y2)=3r is a cone with tanϕ=zr=31, so ϕ runs from 0 to 6π: b=6π.
Consistency check on the shadow: the cone meets the sphere where ρ=2, ϕ=6π, giving r=2sin6π=1 — exactly the unit disk in the stem. Good.
abc=2π6π⋅2=61.
TrapSetting b=3π from tanϕ=3. ϕ is the angle from the z-axis, and a steep cone (z large compared to r) means a smallϕ. The disk-radius check catches it: ϕ=3π would give shadow radius 2sin3π=3=1.
5
Find the volume of the solid region enclosed by the surface ρ=12cosϕ.
no calculator, same as the exam
Answer288πanswer recomputed · sympy
ρ=12cosϕ is already a sphere in disguise: multiply by ρ to get ρ2=12ρcosϕ, i.e. x2+y2+z2=12z.
Complete the square: x2+y2+(z−6)2=36 — radius 6, centred at (0,0,6).
So the answer must be 34π(6)3=288π; now get the same thing from the integral.
The surface closes up at ϕ=2π (where ρ=0), so ϕ runs 0 to 2π — not to π.
TrapLetting ϕ run to π out of habit. For ϕ>2π, 12cosϕ<0 and there is no sphere there; the extra range adds another 288π of nonsense (the cos3ϕsinϕ integrand is odd about 2π, so it actually cancels to 0 — an error that can *silently* delete the whole answer). The sphere is fully traced by ϕ∈[0,2π].
Find the center of mass of the rectangle {(x,y)∣0≤x≤2,0≤y≤4} with density ρ(x,y)=1+y.
no calculator, same as the exam
Answer(1,922)answer recomputed · sympy
Mass first: m=∫02∫04(1+y)dydx=∫02[y+2y2]04dx=∫0212dx=24.
xˉ: the density does not depend on x and the region is symmetric in x, so xˉ=1 by inspection. Formally My=∫02∫04x(1+y)dydx=(∫02xdx)(12)=24, and xˉ=2424=1.
Sanity: 922>2, and it should be — the density grows with y, so mass piles up toward the top.
Centre of mass (1,922).
TrapSwapping the moments. Mx=∬yδdA gives yˉ, not xˉ — the subscript names the axis the moment is about, so it pairs with the *other* coordinate. Using Mx for xˉ here returns 922 for a rectangle only 2 wide, which is impossible and is the fastest way to catch it.
2
A cube 0≤x,y,z≤1 has density δ=x+yz. Find the x-coordinate of the center of mass, given ∭cubeδdV=43.
no calculator, same as the exam
Answer1811answer recomputed · sympy
xˉ=mMyz with Myz=∭xδdV; the mass m=43 is handed to you.
Myz=∫01∫01∫01x(x+yz)dzdydx=∫∫∫(x2+xyz)dV.
Both pieces separate over the cube: ∭x2dV=31 and ∭xyzdV=(21)3=81.
Myz=31+81=2411.
xˉ=3/411/24=2411⋅34=1811.
Sanity: 1811>21, correct — the x term in the density weights the far side.
TrapMultiplying only the first term by x: ∫x⋅xdV+∫yzdV=31+41 gives 127 and then xˉ=97. The moment weights the entire density, so distribute the x across x+yz before integrating.
3
Find the x-coordinate of the center of mass of the plate bounded by y=x, y=21x, x=1 with density δ(x,y)=2x. (Given: Mx=∬yδdA, My=∬xδdA.)
no calculator, same as the exam
Answer43answer recomputed · sympy
Sketch the wedge: between y=21x (lower) and y=x (upper), closed off at x=1. They meet at the origin, so 0≤x≤1 and 2x≤y≤x.
Vertical strips are the right choice — each strip has height x−2x=2x.
Sanity: 43 is past the midpoint of [0,1], as it must be — the strips get both taller and denser as x grows.
TrapIntegrating in the order dxdy without splitting. Horizontal strips run from x=y to x=2y only while 2y≤1; above y=21 the right edge is the line x=1 instead, so that order needs two integrals. The printed answer's order avoids that entirely — pick strips that are described by one pair of curves.
4
Calculate the mass of the tetrahedron with corners (0,0,0), (1,0,0), (0,2,0), (0,0,4) with density ρ(x,y,z)=2z.
no calculator, same as the exam
Answer38answer recomputed · sympy
Three of the four corners are on the axes, so use the intercept form of the slanted face: 1x+2y+4z=1.
Solve for the ceiling: z=4(1−x−2y)=4−4x−2y.
Shadow: set z=0 to get x+2y=1, i.e. y=2(1−x), with 0≤x≤1.
TrapTreating a variable density as a constant and reaching for ρ⋅V: the volume is 61⋅2⋅4=34, and 2zˉ⋅V is *not* something you can shortcut without already knowing zˉ. Also: ρ here is a density, not the spherical radius — the symbol collides with Lesson 23 and the units tell you which.
5
A lamina with density ρ(x,y)=xy occupies the region bounded by y=x2, y=1, x=0. The mass is 61. Find the y-coordinate of the center of mass.
no calculator, same as the exam
Answer43answer recomputed · sympy
Region: y=x2 below, y=1 above, x=0 on the left — so 0≤x≤1 and x2≤y≤1.
The mass is given as 61; confirm it cheaply: ∫01∫x21xydydx=∫012x(1−x4)dx=21(21−61)=61. Good.
Mx=∫01∫x21y⋅xydydx=∫01x⋅31−x6dx.
=31(21−81)=31⋅83=81.
yˉ=mMx=1/61/8=86=43.
Sanity: the region hugs the top (it is wide near y=1, pinched near y=0) and the density xy favours large y, so yˉ>21 is right.
TrapTaking the region as 0≤y≤x2 — 'bounded by y=x2' does not say which side. The third boundary x=0 is the tell: the region above the parabola touches the y-axis, the region below it does not need x=0 to close. Getting this backwards flips yˉ to 75-ish territory and, more tellingly, changes the mass away from the given 61 — which is exactly why that value was handed to you.
Consider F=⟨eyz−ysin(xy),zxeyz−xsin(xy),xyeyz⟩. Find a potential function φ if one exists.
no calculator, same as the exam
Answerxeyz+cos(xy)answer recomputed · sympy
We want φ with ∇φ=F, so start with the component that is easiest to antidifferentiate and then patch it up.
Integrate the first component in x, holding y,z fixed: φ=∫(eyz−ysin(xy))dx=xeyz+cos(xy)+g(y,z).
The sign is the whole game here: dxdcos(xy)=−ysin(xy), so the antiderivative of −ysin(xy) is +cos(xy).
Now differentiate that candidate in y: φy=xzeyz−xsin(xy)+gy.
Compare with the given second component zxeyz−xsin(xy): they already match, so gy=0.
Differentiate in z: φz=xyeyz+gz, and the third component is xyeyz, so gz=0 too. Hence g is a constant.
φ=xeyz+cos(xy) (any additive constant is also a potential).
TrapWriting ∫−ysin(xy)dx=−cos(xy). The −y out front is exactly the inner derivative, so the two minus signs cancel and you get +cos(xy). Getting this backwards still passes a careless φx check if you also flip the sign when differentiating.
2
Let a=0 and C be the curve x2−2y2=a. Which F is orthogonal to the tangent of C at every point (x0,y0) on C?
no calculator, same as the exam
Answer⟨2x,−4y⟩answer recomputed · sympy
C is the level curve g(x,y)=a of g(x,y)=x2−2y2.
Key fact: the gradient of g is normal to every level curve of g, hence orthogonal to the tangent of C at each point.
∇g=⟨gx,gy⟩=⟨2x,−4y⟩.
So F=⟨2x,−4y⟩ is orthogonal to T at every (x0,y0) on C (and so is any scalar multiple of it).
Sanity check by hand: implicit differentiation of x2−2y2=a gives 2x−4yy′=0, so T∥⟨4y,2x⟩, and ⟨2x,−4y⟩⋅⟨4y,2x⟩=8xy−8xy=0.
TrapDropping the minus sign and answering ⟨2x,4y⟩ — that is ∇(x2+2y2), the normal to an ellipse, not to this hyperbola. The other classic miss is handing back the tangent ⟨4y,2x⟩ instead of the normal.
3
Which vector field corresponds to the field with vectors rotating counterclockwise (pointing left above the x-axis, right below it)?
no calculator, same as the exam
Answer⟨−y,x⟩answer recomputed · sympy
Counterclockwise rotation means the field is tangent to circles centered at the origin, pointing in the direction of increasing angle.
Parametrize a circle the counterclockwise way: r(t)=⟨cost,sint⟩.
The direction of travel is r′(t)=⟨−sint,cost⟩.
Rewrite that at the point itself, using x=cost, y=sint: the tangent direction is ⟨−y,x⟩.
Spot-check the description: above the x-axis take (0,1), giving ⟨−1,0⟩ (points left); below take (0,−1), giving ⟨1,0⟩ (points right). Matches.
F=⟨−y,x⟩.
TrapAnswering ⟨y,−x⟩, which is the same picture run clockwise. Always test one concrete point above the x-axis rather than trusting the sign pattern; ⟨y,−x⟩ gives ⟨1,0⟩ at (0,1), pointing right, which is the wrong rotation.
4
Suppose f(x,y,z)=xy2z3. Find div(∇f) at the point (2,−1,1).
no calculator, same as the exam
Answer16answer recomputed · sympy
div(∇f) is the Laplacianfxx+fyy+fzz — a scalar, not a vector.
∇f=⟨y2z3,2xyz3,3xy2z2⟩.
Take the matching partial of each component: fxx=∂x(y2z3)=0, fyy=∂y(2xyz3)=2xz3, fzz=∂z(3xy2z2)=6xy2z.
So div(∇f)=0+2xz3+6xy2z.
At (2,−1,1): 2(2)(1)3+6(2)(−1)2(1)=4+12=16.
TrapTwo traps. First, differentiating the wrong slot — you must take ∂x of the first component, ∂y of the second, ∂z of the third, not all three of one component. Second, (−1)2=+1: plugging y=−1 into 6xy2z gives +12, and writing −12 turns 16 into −8.
Compute ∫Cf(x,y)ds where f(x,y)=sin(2πx)+xy and C:r(t)=⟨t,t⟩, 0≤t≤1.
no calculator, same as the exam
Answer32answer recomputed · sympy
Scalar line integral: ∫Cfds=∫abf(r(t))∣r′(t)∣dt.
Here r′(t)=⟨1,1⟩, so ∣r′(t)∣=2 and ds=2dt.
On C we have x=t, y=t, so f(r(t))=sin(2πt)+t2.
Set up: ∫01(sin(2πt)+t2)2dt.
∫01sin(2πt)dt=[−2πcos(2πt)]01=−2π1+2π1=0 — a full period integrates away.
∫01t2dt=31, so the value is 2⋅31=32.
TrapForgetting the ∣r′∣=2 factor and reporting 31. The parametrization ⟨t,t⟩ is not unit speed even though it looks tame — ds=dt here.
2
Let C be the helix r(t)=⟨cost,sint,t⟩, 0≤t≤π/2. Compute ∫Cxyds.
no calculator, same as the exam
Answer22answer recomputed · sympy
r′(t)=⟨−sint,cost,1⟩, so ∣r′(t)∣=sin2t+cos2t+1=2.
The helix has constant speed2, so ds=2dt.
On C, xy=costsint.
Set up: ∫0π/2costsint2dt.
Substitute u=sint: 2∫01udu=2⋅21=22.
TrapReading the first two components as a unit circle and concluding ∣r′∣=1. The z=t component contributes a 1 under the radical, making the speed 2; dropping it halves-by-2 the answer to 21.
3
Calculate ∫C(x−y)21ds where C is the straight line segment from (1,0) to (5,3).
no calculator, same as the exam
Answer25answer recomputed · sympy
Parametrize the segment from (1,0) to (5,3) as r(t)=⟨1,0⟩+t⟨4,3⟩=⟨1+4t,3t⟩, 0≤t≤1.
r′(t)=⟨4,3⟩, so ∣r′(t)∣=16+9=5 and ds=5dt (the length of the segment, spread over t∈[0,1]).
The integrand simplifies beautifully: x−y=(1+4t)−3t=1+t, so (x−y)21=(1+t)21.
Set up: ∫01(1+t)25dt.
=5[−1+t1]01=5(−21+1)=25.
TrapUsing dx instead of ds. If you parametrize by x with y=43(x−1) you must carry ds=1+(3/4)2dx=45dx; dropping that factor gives 2 instead of 25. Also note the line never crosses x=y, so the integrand stays finite.
4
Let C be the half-circle x2+y2=4 with x≥0. Compute ∫Cxds.
no calculator, same as the exam
Answer8answer recomputed · sympy
The circle has radius 2, so r(t)=⟨2cost,2sint⟩.
Pick the range that gives x≥0, not y≥0: x=2cost≥0 forces −2π≤t≤2π (the right half).
r′(t)=⟨−2sint,2cost⟩, so ∣r′(t)∣=2 and ds=2dt.
Set up: ∫−π/2π/2(2cost)(2)dt=4∫−π/2π/2costdt.
=4[sint]−π/2π/2=4(1−(−1))=8.
TrapTaking 0≤t≤π out of habit. That traces the upper half (y≥0), over which ∫xds=0 by symmetry — a suspiciously clean wrong answer. The condition is x≥0, so center the interval on t=0.
5
Find the arc length of r(t)=⟨4cost,4sint,3t⟩ for 0≤t≤2π.
no calculator, same as the exam
Answer10πanswer recomputed · sympy
Arc length is L=∫ab∣r′(t)∣dt.
r′(t)=⟨−4sint,4cost,3⟩.
∣r′(t)∣=16sin2t+16cos2t+9=16+9=5 — constant speed, the hallmark of a circular helix.
Set up: L=∫02π5dt=5(2π)=10π.
TrapAdding the radius and the pitch coordinate as if speed were 4+3=7, or forgetting the 3 entirely and getting 8π (the circumference of the shadow circle). The helix is strictly longer than its projection because of the climb.
Given force field F(x,y)=⟨−y,x⟩, find the work to move an object along the ellipse r(t)=⟨2cost,3sint⟩ from (2,0) to (0,3).
no calculator, same as the exam
Answer3πanswer recomputed · sympy
Work is W=∫CF⋅dr=∫abF(r(t))⋅r′(t)dt.
Find the t-range from the endpoints: (2,0) is t=0 and (0,3) is t=2π, so we travel a quarter of the ellipse.
On C: F(r(t))=⟨−y,x⟩=⟨−3sint,2cost⟩ and r′(t)=⟨−2sint,3cost⟩.
Dot them: 6sin2t+6cos2t=6 — the integrand is constant.
Set up: W=∫0π/26dt=6⋅2π=3π.
TrapReaching for Green's theorem. C is an open quarter-arc, not a closed curve, so Green does not apply; and F=⟨−y,x⟩ is not conservative (Py=−1=1=Qx), so the path genuinely matters and FTLI is off the table. You must integrate along this arc.
2
Compute the flux of F(x,y)=⟨x,y⟩ across the curve r(t)=⟨3cost,2sint⟩, −π≤t≤π. (Flux =∫CPdy−Qdx.)
no calculator, same as the exam
Answer12πanswer recomputed · sympy
Flux uses the stated formula ∫CPdy−Qdx with P=x, Q=y — note dy pairs with P, not with Q.
From r(t)=⟨3cost,2sint⟩: dx=−3sintdt and dy=2costdt.
Cross-check with the divergence form: divF=2 and the ellipse encloses area π(3)(2)=6π, so the flux is 2⋅6π=12π.
TrapComputing circulation ∫Pdx+Qdy instead of flux. With this field that swap gives ∫(−6sintcost+6sintcost)dt=0, so the mistake produces a clean 0 and looks deliberate. Keep the pairing Pdy−Qdx straight.
3
Given force field F(x,y,z)=⟨y,z,x⟩, find the work to move an object along the straight line from (0,0,0) to (2,3,4).
no calculator, same as the exam
Answer226answer recomputed · sympy
Parametrize the segment: r(t)=⟨2t,3t,4t⟩, 0≤t≤1, so r′(t)=⟨2,3,4⟩.
Evaluate the field on the curve, remembering the components are shifted: F=⟨y,z,x⟩=⟨3t,4t,2t⟩.
Dot: F⋅r′=6t+12t+8t=26t.
Set up: W=∫0126tdt=26⋅21=226=13.
TrapSubstituting F=⟨2t,3t,4t⟩, i.e. reading ⟨y,z,x⟩ as ⟨x,y,z⟩. That gives ∫0129tdt and the wrong number. The cyclic shift is the entire content of this problem. (Also: F=⟨y,z,x⟩ is not conservative, since Py=1 but Qx=0, so you must integrate.)
4
Find ∫CF⋅Tds where F(x,y,z)=⟨yez,ey+xez,xyez⟩ on a smooth curve from (0,0,0) to (−1,1,1).
no calculator, same as the exam
Answer−1answer recomputed · sympy
The problem says 'a smooth curve' without naming it, which is the signal that F must be conservative — otherwise the answer would not be well defined.
Hunt for φ: integrate the first component in x, ∫yezdx=xyez+g(y,z).
Match the second component: φy=xez+gy must equal ey+xez, so gy=ey and g=ey+h(z).
Match the third: φz=xyez+h′(z) must equal xyez, so h is constant. Take φ=xyez+ey.
FTLI: ∫CF⋅Tds=∫CF⋅dr=φ(−1,1,1)−φ(0,0,0).
φ(−1,1,1)=(−1)(1)e1+e1=−e+e=0 and φ(0,0,0)=0+e0=1.
Value =0−1=−1.
TrapEvaluating φ at the start minus the end. FTLI is end minus start, and here that sign error hands you +1. Second trap: forgetting the ey piece because it does not appear in φx — it is exactly what makes φ(0,0,0)=1 rather than 0.
5
Evaluate ∫CF⋅dr where F(x,y,z)=⟨2xy−yz,x2−xz,−xy⟩ on a smooth curve from (2,1,0) to (3,2,−1).
no calculator, same as the exam
Answer20answer recomputed · sympy
Again the curve is unspecified, so test for conservativity: with P=2xy−yz, Q=x2−xz, R=−xy we get Py=2x−z=Qx, Pz=−y=Rx, Qz=−x=Ry. All three match, so F is conservative.
Build φ: ∫(2xy−yz)dx=x2y−xyz+g(y,z).
Check φy=x2−xz+gy against Q=x2−xz: gy=0. Check φz=−xy+gz against R=−xy: gz=0. So φ=x2y−xyz.
FTLI: value =φ(3,2,−1)−φ(2,1,0).
φ(3,2,−1)=9(2)−(3)(2)(−1)=18+6=24.
φ(2,1,0)=4(1)−0=4.
Value =24−4=20.
TrapSign-slipping on −xyz at the endpoint with z=−1: −(3)(2)(−1)=+6, so φ(3,2,−1)=24, not 12. Writing 18−6 gives 8 and is the single most common wrong answer here.
Find ∫CF⋅dr where F(x,y,z)=⟨2xy−yz,x2−xz,−xy⟩ on a smooth curve from (2,1,0) to (3,2,−1).
no calculator, same as the exam
Answer20answer recomputed · sympy
The path is only described as 'a smooth curve', so the integral had better be path-independent — first verify F is conservative.
With P=2xy−yz, Q=x2−xz, R=−xy: Py=2x−z=Qx, Pz=−y=Rx, Qz=−x=Ry. Conservative.
Find the potential: φ=∫(2xy−yz)dx=x2y−xyz+g(y,z), and both φy=x2−xz and φz=−xy already match, so g is constant.
φ(x,y,z)=x2y−xyz.
FTLI: ∫CF⋅dr=φ(3,2,−1)−φ(2,1,0)=(18+6)−(4−0).
=24−4=20.
TrapTrying to parametrize. No curve was given, so any attempt to build one is wasted work — and if you accidentally pick a path while F were not conservative, the answer would be meaningless. Confirm the mixed partials first, then use endpoints only. Watch −xyz at z=−1, which contributes +6.
2
For F=⟨eyz−ysin(xy),zxeyz−xsin(xy),xyeyz⟩, find a potential function.
no calculator, same as the exam
Answerφ=xeyz+cos(xy)answer recomputed · sympy
Antidifferentiate the first component in x: φ=∫(eyz−ysin(xy))dx.
Term by term, ∫eyzdx=xeyz (as x varies, eyz is a constant) and ∫−ysin(xy)dx=cos(xy).
So φ=xeyz+cos(xy)+g(y,z).
Test against the second component: φy=xzeyz−xsin(xy)+gy, and we need zxeyz−xsin(xy), so gy=0.
Test against the third: φz=xyeyz+gz, and we need xyeyz, so gz=0.
Therefore φ=xeyz+cos(xy), up to an additive constant.
TrapDifferentiating eyz with respect to y and writing zeyz without the x that the antiderivative introduced. The x in xeyz must be carried through, and it is exactly what makes φy=xzeyz match the printed zxeyz.
3
A potential for F=⟨siny,xcosy⟩ is:
no calculator, same as the exam
Answerxsiny+1answer recomputed · sympy
First confirm a potential exists: P=siny, Q=xcosy, and Py=cosy=Qx. Conservative on all of R2.
Integrate P in x: φ=∫sinydx=xsiny+g(y).
Differentiate in y: φy=xcosy+g′(y), which must equal Q=xcosy, so g′(y)=0 and g is any constant.
The family of potentials is φ=xsiny+C. The printed choice takes C=1: φ=xsiny+1.
TrapRejecting the '+1' as a typo. Potentials are only determined up to an additive constant, so xsiny, xsiny+1, and xsiny−7 are all correct; on a multiple-choice list the constant is there precisely to see whether you know that. The real error to avoid is φ=−xcosy, from mistakenly antidifferentiating siny in y instead of x.
4
Find ∫CF⋅Tds where F=⟨yez,ey+xez,xyez⟩ on a smooth curve from (0,0,0) to (−1,1,1).
no calculator, same as the exam
Answer−1answer recomputed · sympy
∫CF⋅Tds is just ∫CF⋅dr rewritten, since dr=Tds.
No specific curve is given, so check conservativity: with P=yez, Q=ey+xez, R=xyez we get Py=ez=Qx, Pz=yez=Rx, Qz=xez=Ry. Conservative.
Potential: ∫yezdx=xyez+g(y,z); then φy=xez+gy=ey+xez forces gy=ey, and φz=xyez+h′(z)=xyez forces h′=0.
φ=xyez+ey.
FTLI: φ(−1,1,1)−φ(0,0,0)=((−1)(1)e+e)−(0+1).
=0−1=−1.
TrapSeeing Tds and thinking you must compute ∣r′∣ and a unit tangent. You never do — Tds=dr, the speed cancels, and the whole problem collapses to two endpoint evaluations. The second trap is φ(0,0,0)=0: the ey term makes it e0=1.
5
Evaluate ∫C∇(x2y−xyz)⋅dr where C is any path from (1,0,0) to (2,1,−1).
no calculator, same as the exam
Answer6answer recomputed · sympy
The integrand is already a gradient, ∇φ with φ(x,y,z)=x2y−xyz, so FTLI applies immediately and no conservativity test is needed.
∫C∇φ⋅dr=φ(end)−φ(start), independent of the path.
φ(2,1,−1)=22(1)−(2)(1)(−1)=4+2=6.
φ(1,0,0)=12(0)−(1)(0)(0)=0.
Value =6−0=6.
TrapExpanding ∇(x2y−xyz)=⟨2xy−yz,x2−xz,−xy⟩ and then grinding out a line integral along some invented path. All that work is discarded by FTLI. Within the evaluation, the sign trap is −xyz at z=−1, which adds2; writing 4−2=2 is the common wrong answer.
Use Green's Theorem to evaluate ∮Cxdx+(x2+y2)dy where C is the boundary of the rectangle with vertices (0,0), (2,0), (2,3), (0,3), counterclockwise.
no calculator, same as the exam
Answer12answer recomputed · sympy
Green's Theorem converts a counterclockwise closed line integral into a double integral: ∮CPdx+Qdy=∬R(∂x∂Q−∂y∂P)dA.
Read off P=x and Q=x2+y2. The orientation is already CCW, so no sign flip is needed.
∂x∂Q=2x and ∂y∂P=0, so the integrand is just 2x.
R is the rectangle 0≤x≤2, 0≤y≤3 — a genuine rectangle, so the limits are constants and the order does not matter.
∬R2xdA=∫03∫022xdxdy=∫03[x2]02dy=∫034dy=12.
TrapTrying to parametrize all four sides by hand. That is four integrals, and the ∫xdx pieces on the top and bottom edges cancel in a way that is easy to botch — one sign slip on the *return* edges (which run right-to-left and top-to-bottom) and you get 0 or 24. Green's theorem exists precisely to make the orientation bookkeeping automatic; use it. Also note P=x contributes nothing, since Py=0 — the whole answer comes from Q.
2
Let R be the triangle with vertices (0,0), (2,0), (1,1) and C its CCW boundary. Compute ∫Ccos3(x)dx+exdy.
no calculator, same as the exam
Answere2−2e+1answer recomputed · sympy
P=cos3(x), Q=ex. Neither has an elementary-looking line integral, which is the hint that Green's Theorem is the intended route.
∂x∂Q=ex, ∂y∂P=0 (there is no y in cos3x at all), so Qx−Py=ex.
Describe the triangle. The two slanted sides are y=x (from (0,0) to (1,1)) and y=2−x (from (1,1) to (2,0)), with the base y=0.
Integrating in x first avoids splitting the region: for each y∈[0,1], x runs from the left edge x=y to the right edge x=2−y.
∫01∫y2−yexdxdy=∫01(e2−y−ey)dy.
=[−e2−y−ey]01=(−e−e)−(−e2−1)=e2−2e+1.
TrapIntegrating dy first. Then you must split at x=1, because the upper boundary changes formula from y=x to y=2−x there: ∫01∫0x+∫12∫02−x. Students who write a single ∫02∫0x silently integrate over a region that is not the triangle. Going dx first needs no split. (Note e2−2e+1=(e−1)2, a useful sanity check that it is positive.)
3
Use Green's Theorem to evaluate ∮Cydx+2xdy where C is the square with vertices (±1,±1), counterclockwise.
no calculator, same as the exam
Answer4answer recomputed · sympy
P=y, Q=2x, and the square is traversed CCW, so Green's Theorem applies directly.
∂x∂Q−∂y∂P=2−1=1.
The integrand collapsed to the constant 1, so the double integral is literally the area of R.
R is the square −1≤x≤1, −1≤y≤1, whose side length is 2, so its area is 4.
∮Cydx+2xdy=∬R1dA=area(R)=4.
TrapComputing Px−Qy instead of Qx−Py. That gives 0−0=0 here, a clean-looking wrong answer. The derivative always hits the *other* variable: Q gets ∂x, P gets ∂y. Second trap: the vertices (±1,±1) describe a square of side 2, not side 1 — its area is 4, not 1.
4
Use Green's theorem to evaluate ∮C(y3)dx−(x3)dy where C is the circle x2+y2=1, CCW.
no calculator, same as the exam
Answer2−3πanswer recomputed · sympy
Write the integral as ∮CPdx+Qdy with P=y3 and Q=−x3 — the minus sign in front of x3dy belongs to Q.
∂x∂Q=−3x2, ∂y∂P=3y2, so Qx−Py=−3x2−3y2=−3(x2+y2).
R is the unit disk and the integrand depends only on x2+y2, so polar is the obvious move: x2+y2=r2 and dA=rdrdθ.
TrapTwo sign traps stacked. First, forgetting that Q=−x3, which makes Qx=+3x2 and hands you +23π — the right magnitude, wrong sign. Second, forgetting the polar Jacobian: −3r2drdθ instead of −3r3drdθ gives −2π. The answer is negative, and it *should* be: the field circulates clockwise while C runs counterclockwise.
5
Use Green's theorem to find the area of the ellipse a2x2+b2y2=1.
no calculator, same as the exam
Answerπabanswer recomputed · sympy
Green's Theorem run backwards: pick P,Q with Qx−Py=1, and the double integral becomes the area. The symmetric choice is P=−2y, Q=2x, giving the area formula A=21∮Cxdy−ydx.
Parametrize the ellipse CCW: x=acost, y=bsint, 0≤t≤2π.
Sanity check: setting a=b=R gives πR2, the circle.
TrapDropping the 21. The area formula is 21∮xdy−ydx — the half is there because *each* of xdy and −ydx separately already integrates to the full area, and you are averaging them, not adding them. Forgetting it doubles the answer to 2πab. The other trap is parametrizing clockwise (y=−bsint), which returns −πab; area must come out positive, so a negative result means your orientation was backwards.
curlF=∇×F=⟨∂y∂R−∂z∂Q,∂z∂P−∂x∂R,∂x∂Q−∂y∂P⟩ for F=⟨P,Q,R⟩.
Here P=xy, Q=sinz, R=cosy.
First component: Ry−Qz=∂y∂(cosy)−∂z∂(sinz)=−siny−cosz.
Second component: Pz−Rx=∂z∂(xy)−∂x∂(cosy)=0−0=0.
Third component: Qx−Py=∂x∂(sinz)−∂y∂(xy)=0−x=−x.
curlF=⟨−siny−cosz,0,−x⟩.
TrapThe middle component's sign. Expanding the determinant i∂xPj∂yQk∂zR, the j entry carries a minus from cofactor expansion: it is −(Rx−Pz)=Pz−Rx, notRx−Pz. Write the middle slot as Pz−Rx and the sign takes care of itself. (Here it is 0 either way, which is exactly why the habit must be built on a problem where it does not bite.) Second trap: curl of a vector field is a vector, while div is a scalar — never report a single number here.
2
Let f(x,y,z)=xy2z3. Compute div(∇f) at (2,−1,1).
no calculator, same as the exam
Answer16answer recomputed · sympy
div(∇f)=∇2f=fxx+fyy+fzz, the Laplacian. So take second partials, not first.
f=xy2z3. Gradient: ∇f=⟨y2z3,2xyz3,3xy2z2⟩.
fxx=∂x∂(y2z3)=0 — f is linear in x.
fyy=∂y∂(2xyz3)=2xz3.
fzz=∂z∂(3xy2z2)=6xy2z.
∇2f=0+2xz3+6xy2z. At (2,−1,1): 2(2)(1)3+6(2)(−1)2(1)=4+12=16.
TrapEvaluating at the point too early. If you plug (2,−1,1) into ∇f and then try to differentiate, you are differentiating constants and get 0. Differentiate fully, *then* substitute. The other trap is y2 at y=−1: (−1)2=+1, not −1, so the 6xy2z term is +12; a dropped square gives 4−12=−8.
3
For F=⟨eyz−ysin(xy),zxeyz−xsin(xy),xyeyz⟩, verify that curlF=0 and state what this implies.
no calculator, same as the exam
AnswerF is conservativeprose answer · not machine-checkable
Label P=eyz−ysin(xy), Q=zxeyz−xsin(xy), R=xyeyz, and compute the three components of ∇×F.
Ry=∂y∂(xyeyz)=xeyz+xyzeyz, and Qz=∂z∂(zxeyz−xsin(xy))=xeyz+xyzeyz. First component Ry−Qz=0.
Pz=∂z∂(eyz−ysin(xy))=yeyz, and Rx=∂x∂(xyeyz)=yeyz. Second component Pz−Rx=0.
Qx=zeyz−sin(xy)−xycos(xy) (product rule on xsin(xy)), and Py=zeyz−sin(xy)−xycos(xy) (product rule on ysin(xy)). Third component Qx−Py=0.
So curlF=0 everywhere.
F is defined and smooth on all of R3, which is simply connected, so curlF=0impliesF is conservative: there is a potential φ with ∇φ=F.
Indeed φ=xeyz+cos(xy) works: φx=eyz−ysin(xy), φy=xzeyz−xsin(xy), φz=xyeyz. Consequence: ∫CF⋅dr is path-independent and ∮CF⋅dr=0 on any closed curve.
TrapStating the converse without the domain hypothesis. curlF=0 alone does not make a field conservative — the standard counterexample F=⟨x2+y2−y,x2+y2x,0⟩ has zero curl but a nonzero loop integral, because its domain has the z-axis punched out. You get the conclusion here only because this F is smooth on all of R3. The computational trap is ∂y(ysin(xy)): it needs the product rule, sin(xy)+xycos(xy), not just xycos(xy).
4
Determine if F=⟨yz,xz,xy⟩ is conservative using curlF. If so, find φ.
no calculator, same as the exam
AnswerConservative; φ=xyz+Cprose answer · not machine-checkable
Test with the curl. P=yz, Q=xz, R=xy.
Ry−Qz=x−x=0; Pz−Rx=y−y=0; Qx−Py=z−z=0. So curlF=0, and since F is a polynomial field on all of R3 (simply connected), Fis conservative.
Now build φ with ∇φ=F. Start from φx=yz and integrate in x, treating y,z as constants: φ=xyz+g(y,z) — the 'constant' of integration may depend on the other variables.
Differentiate that in y: φy=xz+gy(y,z). Match against Q=xz, so gy=0 and g=h(z).
Differentiate in z: φz=xy+h′(z). Match against R=xy, so h′(z)=0 and h is a genuine constant.
φ=xyz+C, and one checks ∇(xyz)=⟨yz,xz,xy⟩=F.
TrapWriting the first antiderivative as φ=xyz+C with a numeric constant. At that stage the unknown is a whole function g(y,z), because anything free of x dies under ∂x. Collapsing it to a number too early means you never actually verify φy=Q and φz=R — on a field where the components do *not* fit together, that shortcut produces a 'potential' for a non-conservative field. Also: test with curl, not div. Here divF=0 too, but that says the field is incompressible, an entirely different statement.
5
Find divF for F=⟨exy,eyz,exz⟩, and verify the identity div(curlG)=0 for any smooth G.
By Clairaut's theorem the mixed partials commute (G3yx=G3xy, etc.), so the six terms cancel in pairs and div(curlG)=0 for every G with continuous second partials.
TrapDropping the chain-rule factor. ∂xexy=yexy, not exy — the answer exy+eyz+exz is the single most common version of this and it is wrong in all three terms. Watch which variable owns which exponent: the third component is exz and gets ∂z, producing xexz, not zexz. For the identity, the cancellation needs continuous second partials (Clairaut); it is a theorem, not algebra, and the practical payoff is that a field with divH=0 can never be written as curl of anything.
Find the area of the part of the plane 2x+5y+z=10 that lies in the first octant.
no calculator, same as the exam
Answer530answer recomputed · sympy
Solve for z to get the surface as a graph: z=g(x,y)=10−2x−5y.
For a graph, dS=1+gx2+gy2dA. Here gx=−2, gy=−5, so dS=1+4+25dA=30dA — a constant, as it must be for a plane.
Find the shadow D. The first octant needs x≥0, y≥0, and z≥0, and the last of these is 10−2x−5y≥0, i.e. 2x+5y≤10.
D is the triangle with legs along the axes: x-intercept x=5, y-intercept y=2. Its area is 21(5)(2)=5.
Area=∬D30dA=30⋅area(D)=530.
Explicitly: ∫05∫0(10−2x)/530dydx=530.
TrapReporting 5 — the area of the shadow — by forgetting the 1+gx2+gy2 factor. A tilted plane is always larger than its projection, so the stretch factor is ≥1 and your answer must exceed the shadow's area. The other trap is the shadow's shape: the first octant condition z≥0 is what bounds D, giving the triangle 2x+5y≤10 with intercepts x=5, y=2. Students who integrate over 0≤x≤5, 0≤y≤2 (the full rectangle) double the answer to 1030.
2
Compute the flux of F(x,y)=⟨x,y⟩ across r(t)=⟨3cost,2sint⟩, −π≤t≤π. (Flux =∫CPdy−Qdx.)
no calculator, same as the exam
Answer12πanswer recomputed · sympy
This is the 2D flux (outward-normal) line integral, and the problem hands you the formula: Flux=∫CPdy−Qdx with F=⟨P,Q⟩.
Along the curve, P=x=3cost and Q=y=2sint.
Differentiate the parametrization: dx=−3sintdt and dy=2costdt.
Cross-check with the divergence form of Green's Theorem: divF=1+1=2, and the ellipse 9x2+4y2=1 encloses area π(3)(2)=6π, so the flux is 2⋅6π=12π. ✓
TrapComputing ∫Pdx+Qdy — the circulation — instead of ∫Pdy−Qdx. For this field that gives ∫(−9costsint+4sintcost)dt=0, a plausible-looking but wrong answer. Flux swaps the differentials *and* inserts a minus sign; circulation does not. The second trap is the parameter range: −π≤t≤π is a full 2π sweep traversed counterclockwise, so the outward normal convention holds and the flux is positive — halving it to 6π means you read the range as a half-loop.
3
Evaluate ∬S1dS where S is the portion of the paraboloid z=x2+y2 below z=4.
no calculator, same as the exam
Answer6π(1717−1)answer recomputed · sympy
∬S1dS is the surface area of S. The surface is the graph z=g(x,y)=x2+y2.
gx=2x, gy=2y, so dS=1+4x2+4y2dA.
The shadow: z≤4 means x2+y2≤4, the disk of radius 2.
Polar, since both the integrand and the region depend only on x2+y2=r2: dS=1+4r2rdrdθ — note both the radical and the Jacobian r.
∬SdS=∫02π∫021+4r2rdrdθ.
Inner integral by u=1+4r2, du=8rdr: ∫02r1+4r2dr=[12(1+4r2)3/2]02=12173/2−1=121717−1.
Multiply by ∫02πdθ=2π: 122π(1717−1)=6π(1717−1).
TrapThe radius of the shadow. z=4 on z=x2+y2 means r2=4, so r runs to 2, not 4 — using r≤4 gives 6π(6565−1). The second trap is losing one of the two r's: 1+4r2rdrdθ carries the surface radical *and* the polar Jacobian, and dropping the Jacobian leaves an integral with no elementary u-substitution, which should tip you off that something is missing.
4
Find the flux of F=⟨x,y,z⟩ upward through the surface z=4−x2−y2 above the xy-plane.
no calculator, same as the exam
Answer24πanswer recomputed · sympy
For an upward-oriented graph z=g(x,y), ∬SF⋅dS=∬D(−Pgx−Qgy+R)dA. Note this formula already has the 1+gx2+gy2 built in — it cancels against the normalization of n^, so do not multiply by it again.
Here g=4−x2−y2, so gx=−2x and gy=−2y; and F=⟨x,y,z⟩ with z evaluated *on the surface*: R=4−x2−y2.
The shadow D is where z≥0: 4−x2−y2≥0, the disk x2+y2≤4 of radius 2.
Polar: ∫02π∫02(r2+4)rdrdθ=∫02π∫02(r3+4r)drdθ.
∫02(r3+4r)dr=[4r4+2r2]02=4+8=12, and times 2π gives 24π.
TrapLeaving z as the free variable z instead of substituting the surface, z=4−x2−y2. The flux integral lives *on* the surface, so every appearance of z in F must be replaced. The second trap is the sign pattern: it is ⟨−gx,−gy,1⟩ for upward — the two tangential terms get minus signs and the k slot gets +1. If the problem had said downward you would negate the whole thing, giving −24π. Third: do not tack on an extra 1+4x2+4y2; that factor is only for dS in a scalar surface integral, never for dS in this form.
5
Find the surface area of the cone z=x2+y2 below z=3.
no calculator, same as the exam
Answer9π2answer recomputed · sympy
The cone is the graph z=g(x,y)=x2+y2.
gx=x2+y2x and gy=x2+y2y, so gx2+gy2=x2+y2x2+y2=1.
Therefore dS=1+1dA=2dA — constant, because a cone has the same slope everywhere.
The shadow: z≤3 means x2+y2≤3, the disk of radius 3, whose area is 9π.
∬SdS=∫02π∫032rdrdθ=2⋅9π=9π2.
TrapWriting dS=1+x2+y2dA by squaring the *function* instead of its partials. The radical needs gx2+gy2, and for this cone those partials are the components of a unit vector, so their squares sum to exactly 1 and the factor is the clean 2. The other trap: the apex (0,0,0) makes gx,gy undefined there, but it is a single point of measure zero and the improper integral converges — the 2 still holds. And do not confuse this with the *lateral* cone formula πrℓ: here r=3, ℓ=32, giving 9π2, which agrees. ✓
6
Evaluate ∬SF⋅dS where F=⟨y,−x,1⟩ and S is the disk z=1, x2+y2≤1, oriented upward.
no calculator, same as the exam
Answerπanswer recomputed · sympy
S is a flat horizontal disk at height z=1, oriented upward, so the unit normal is simply n^=k=⟨0,0,1⟩ and dS=⟨0,0,1⟩dA (the graph z=g=1 has gx=gy=0, so dS=dA).
Dot the field with the normal: F⋅n^=⟨y,−x,1⟩⋅⟨0,0,1⟩=1. Only the k-component of F survives.
∬SF⋅dS=∬x2+y2≤11dA=area of the unit disk=π.
Explicitly in polar: ∫02π∫011⋅rdrdθ=2π⋅21=π.
TrapTrying to integrate the y and −x components. They are entirely tangential to a horizontal disk and contribute nothing to flux — this field swirls *within* the plane z=1 and never crosses it. (Its curl is ⟨0,0,−2⟩, so its *circulation* around the boundary is −2π, a completely different number that this problem is baiting you toward.) The second trap is orientation: 'oriented upward' fixes n^=+k; downward would give −π. Third, the disk has radius 1, so its area is π, not 2π.
Use Stokes' Theorem to evaluate ∮CF⋅dr where F=⟨y2,x,z2⟩ and C is the boundary of the triangle with vertices (1,0,0), (0,1,0), (0,0,1), oriented counterclockwise when viewed from above.
no calculator, same as the exam
Answer61answer recomputed · sympy
Stokes' Theorem says ∮CF⋅dr=∬Scurl(F)⋅dS for any surface S with ∂S=C. The point of the whole theorem here is that the line integral is three separate segments, each needing its own parametrization, while the surface integral is one flat triangle. Convert.
The three vertices (1,0,0),(0,1,0),(0,0,1) all satisfy x+y+z=1, so take S to be the flat triangle in that plane: z=g(x,y)=1−x−y over the shadow triangle D:x≥0,y≥0,x+y≤1.
For a graph z=g(x,y) with upward normal, dS=⟨−gx,−gy,1⟩dA=⟨1,1,1⟩dA. Counterclockwise viewed from above is exactly the orientation that pairs with the upward normal, so no sign flip.
curl(F)⋅⟨1,1,1⟩=(0)(1)+(0)(1)+(1−2y)(1)=1−2y, and note the z's have vanished, so no substitution of the plane is even needed.
TrapTwo traps stacked. First, people write dS=⟨−gx,−gy,1⟩dA but then also multiply by 1+gx2+gy2=3 — that factor belongs to dS (scalar surface area), not to dS, and including both double-counts. Second, the shadow region is the triangle x+y≤1 in the first quadrant, not the unit square: the inner limit is y:0→1−x, and using 0→1 turns 61 into 0.
2
Let F=⟨−y2,x,z2⟩. Use Stokes' Theorem to evaluate ∮CF⋅dr where C is the circle x2+y2=4 in the plane z=3, traversed counterclockwise when viewed from above.
no calculator, same as the exam
Answer4πanswer recomputed · sympy
C is a circle sitting in a horizontal plane, so the cheapest surface to span it with is the flat disk in that same plane — the normal is constant and the surface integral collapses to a plain double integral.
Take S:x2+y2≤4,z=3. Counterclockwise from above ⇒ upward normal n=⟨0,0,1⟩ and dS=⟨0,0,1⟩dA.
curl(F)⋅n=1+2y, so ∬Scurl(F)⋅dS=∬x2+y2≤4(1+2y)dA.
In polar, ∫02π∫02(1+2rsinθ)rdrdθ. The 2rsinθ piece integrates to 0 because ∫02πsinθdθ=0 (the disk is symmetric about y=0 and y is odd).
What is left is ∬x2+y2≤41dA=area=π(2)2=4π.
TrapThe sign on −y2. Since ∂y∂(−y2)=−2y, the k-component is 1−(−2y)=1+2y, not 1−2y. Either way the y-term dies by symmetry, so this particular problem forgives it — but only by luck, and the same slip elsewhere does not cancel. Also note z=3 is irrelevant to the answer: the curl has no z in it, so the height of the plane never enters.
3
Evaluate ∬Scurl(F)⋅dS where F=⟨yz,xz,xy⟩ and S is the part of the sphere x2+y2+z2=4 with z≥0, oriented with upward normal. (Hint: use Stokes' Theorem and evaluate the line integral on ∂S instead.)
no calculator, same as the exam
Answer0answer recomputed · sympy
Here Stokes is run in the other direction: a curved surface integral is traded for a line integral. ∂S is where the hemisphere meets z=0, namely the circle x2+y2=4 in the xy-plane. Upward normal on S pairs with counterclockwise traversal of that circle seen from above.
Parametrize r(t)=⟨2cost,2sint,0⟩, 0≤t≤2π, so r′(t)=⟨−2sint,2cost,0⟩.
On C we have z=0, so F=⟨yz,xz,xy⟩=⟨0,0,4costsint⟩ — the first two components are killed by z=0.
F⋅r′(t)=0⋅(−2sint)+0⋅(2cost)+4costsint⋅0=0 identically, since r′ has no k-component while F has only a k-component.
∮CF⋅dr=∫02π0dt=0, hence ∬Scurl(F)⋅dS=0.
Sanity check the fast way: curl⟨yz,xz,xy⟩=⟨x−x,y−y,z−z⟩=0, because F=∇(xyz) is conservative. A zero curl makes every such flux zero.
TrapTrying to parametrize the hemisphere and grind out ∬curl(F)⋅dS in spherical coordinates. It works, but it is minutes of work for a field whose curl is 0. Also: the boundary is the equator circle at z=0, not the whole sphere and not the point at the top — a hemisphere with z≥0 has exactly one boundary curve.
4
Use Stokes' Theorem to evaluate ∮CF⋅dr where F=⟨xz,yz,xy⟩ and C is the boundary of the surface z=1−x2−y2 above the xy-plane, traversed counterclockwise when viewed from above.
no calculator, same as the exam
Answer0answer recomputed · sympy
C is where z=1−x2−y2 meets z=0: setting z=0 gives x2+y2=1, the unit circle in the xy-plane. Stokes lets us span it with either the paraboloid cap or the flat disk; both are legal, so pick deliberately.
Cheap route. Use the flat disk S:x2+y2≤1,z=0 with n=⟨0,0,1⟩. Then curl(F)⋅n=0 because the curl has no k-component, and the integral is 0 on sight.
Honest route, to confirm the choice of surface did not smuggle anything in: on the paraboloid, dS=⟨2x,2y,1⟩dA over x2+y2≤1, and curl(F)⋅⟨2x,2y,1⟩=2x(x−y)+2y(x−y)=2(x2−y2).
∫02πcos2θdθ=0, so the value is 0 — same as the flat disk, as Stokes guarantees.
TrapReaching for the paraboloid because the problem named it. Stokes only cares about the boundary curve, so you may replace the cap with the flat disk it bounds and the k-component-free curl kills the integral instantly. The other trap is the boundary itself: 'above the xy-plane' means C is x2+y2=1 at z=0, not a curve on the paraboloid at some other height.
5
Let S be the part of the paraboloid z=9−x2−y2 above z=0 with upward normal, and F=⟨−y3,x3,−z3⟩. Use Stokes' Theorem to find ∬Scurl(F)⋅dS.
Cross-check with Stokes on the boundary circle x2+y2=9, z=0: r(t)=⟨3cost,3sint,0⟩, F=⟨−27sin3t,27cos3t,0⟩, and F⋅r′=81(sin4t+cos4t), whose integral over [0,2π] is 81⋅23π=2243π. Same value.
TrapForgetting the polar Jacobian: the integrand is 3r2 and dA=rdrdθ, so you integrate 3r3. Dropping the extra r gives 2π⋅3⋅9=54π instead of 2243π. Second trap: the shadow radius is 3, not 9 — z=9−x2−y2 hits z=0 at x2+y2=9, i.e. r=3.
Use the Divergence Theorem to evaluate ∬SF⋅dS where F=⟨x3,y3,z3⟩ and S is the sphere x2+y2+z2=1 with outward normal.
no calculator, same as the exam
Answer512πanswer recomputed · sympy
S is closed with outward normal, which is exactly the hypothesis of the Divergence Theorem: ∬SF⋅dS=∭Ediv(F)dV. Going the other way, parametrizing a sphere and dotting with a normal, is far worse.
div(F)=∂x∂(x3)+∂y∂(y3)+∂z∂(z3)=3x2+3y2+3z2=3ρ2 in spherical coordinates.
E is the solid unit ball, so spherical coordinates with 0≤ρ≤1, 0≤ϕ≤π, 0≤θ≤2π and dV=ρ2sinϕdρdϕdθ.
∭E3ρ2dV=∫02π∫0π∫013ρ2⋅ρ2sinϕdρdϕdθ — note the integrand is 3ρ4sinϕ.
The three factors separate: (∫02πdθ)(∫0πsinϕdϕ)(∫013ρ4dρ)=(2π)(2)(53)=512π.
TrapWriting dV=ρ2sinϕ and then integrating 3ρ2 as if it were 3ρ2⋅ρsinϕ or just 3ρ2dρ. The full integrand is 3ρ4sinϕ; dropping either the ρ2 of dV or the sinϕ changes the answer to 2π or 34π-type garbage. Also ∫0πsinϕdϕ=2, not 0 — students who reflexively write 0 for a sine integral lose the whole thing.
2
Use the Divergence Theorem to compute the flux of F=⟨xy,y2,yz⟩ outward through the surface of the unit cube [0,1]3.
no calculator, same as the exam
Answer2answer recomputed · sympy
The cube has six faces, so the direct flux computation is six separate double integrals with six different normals. The Divergence Theorem replaces all of it with one triple integral over [0,1]3.
div(F)=∂x∂(xy)+∂y∂(y2)+∂z∂(yz)=y+2y+y=4y.
∬SF⋅dS=∭[0,1]34ydV=∫01∫01∫014ydxdydz.
The region is a box and the integrand depends only on y, so it factors: (∫01dx)(∫014ydy)(∫01dz)=1⋅2⋅1.
Flux =2.
TrapDifferentiating the wrong variable in each slot. div takes ∂x of the first component, ∂y of the second, ∂z of the third: ∂x(xy)=y and ∂z(yz)=y, so the three terms are y+2y+y=4y, not x+2y+y. A secondary trap is assuming the answer must be 0 by symmetry — y is not symmetric on [0,1], only on an interval centered at the origin.
3
Let S be the closed surface consisting of the hemisphere z=4−x2−y2 and the disk x2+y2≤4 in the xy-plane, both with outward normal. Evaluate ∬SF⋅dS where F=⟨x,y,z⟩.
no calculator, same as the exam
Answer16πanswer recomputed · sympy
The hemisphere alone is not closed, but the problem already glued the flat disk on, so S=∂E where E is the solid half-ball x2+y2+z2≤4,z≥0. Outward normal on both pieces means the Divergence Theorem applies as stated.
div⟨x,y,z⟩=1+1+1=3, a constant — which is why the flux reduces to 3Vol(E).
∬SF⋅dS=∭E3dV=3⋅Vol(E).
E is half of a ball of radius 2: Vol(E)=21⋅34π(2)3=316π. In spherical this is ∫02π∫0π/2∫02ρ2sinϕdρdϕdθ, with ϕ stopping at π/2 because z≥0.
Flux =3⋅316π=16π.
TrapLetting ϕ run to π and getting 32π. For the upper half-ball ϕ∈[0,π/2] — ϕ=π/2 is the equator, ϕ=π is the south pole. The second trap is applying the theorem to the hemisphere alone: it is not closed, so you would need the disk anyway. Here the disk happens to contribute 0, since on z=0 the outward normal is ⟨0,0,−1⟩ and F⋅n=−z=0, but that is a fact you must state, not one you may assume.
4
Use the Divergence Theorem to find the outward flux of F=⟨x2z,y2z,z2⟩ through the boundary of the solid region E={(x,y,z)∣x2+y2≤4,0≤z≤3} (a cylinder).
no calculator, same as the exam
Answer36πanswer recomputed · sympy
The boundary is a can: top disk, bottom disk, and curved side, each with a different normal. One triple integral over the solid beats three surface integrals.
div(F)=∂x∂(x2z)+∂y∂(y2z)+∂z∂(z2)=2xz+2yz+2z.
The solid is a cylinder, so cylindrical coordinates: x=rcosθ, y=rsinθ, 0≤r≤2, 0≤θ≤2π, 0≤z≤3, dV=rdzdrdθ.
The first two terms vanish: ∫02πcosθdθ=∫02πsinθdθ=0, i.e. x and y are odd over a disk centered on the axis.
What survives is ∫02π∫02∫032zrdzdrdθ=(∫032zdz)(∬r≤2dA)=9⋅4π=36π.
TrapJustifying the death of 2xz and 2yz with the wrong symmetry. The z-interval [0,3] is not symmetric about 0, so nothing cancels in z; the cancellation is in θ, because ∫02πcosθdθ=∫02πsinθdθ=0 over the full disk. Beware also that this problem will not punish a dropped Jacobian: ∫02rdr=∫02dr=2, so omitting the r still lands on 36π here. That is a coincidence of the radius being 2, and it will not save you on the next cylinder.
5
Let S be the outward-oriented boundary of the region between the spheres x2+y2+z2=1 and x2+y2+z2=4. Evaluate ∬SF⋅dS where F=⟨x,y,z⟩.
no calculator, same as the exam
Answer28πanswer recomputed · sympy
E is a spherical shell 1≤ρ≤2. Its boundary is two spheres, and 'outward' for the solid means outward on ρ=2 but pointing *toward the origin* on ρ=1. The Divergence Theorem handles that bookkeeping for free.
div⟨x,y,z⟩=3.
∬SF⋅dS=∭E3dV=3⋅Vol(E), and Vol(E)=34π(23−13)=328π.
As a spherical integral: ∫02π∫0π∫123ρ2sinϕdρdϕdθ, where ∫123ρ2dρ=23−13=7, ∫0πsinϕdϕ=2, ∫02πdθ=2π.
Flux =7⋅2⋅2π=28π, matching 3⋅328π.
TrapSetting the inner limit to ρ=0 and getting 32π — the ball of radius 1 is not part of E, it is a hole. Equally common is computing the two spheres separately and adding them with the same sign: on the inner sphere the outward-from-E normal is −ρ^, so its flux is −4π, and 32π+(−4π)=28π. Adding +4π gives 36π.
The official set has a hole here.
Lesson 7 covers projectile motion under gravity, and uniform circular motion — and not one of the official problems above tests it.
It is still examinable, so the questions below are not optional this time — they are the only place those topics get rehearsed.
STEP 03
More practice
7 questions
Mine, on those sections only. Untimed, no calculator — the quiz doesn't allow one either.
0/7 answered
117.7 · Stokes · spatial triangle
C is the triangle with vertices (1,0,0),(0,1,0),(0,0,1), oriented counterclockwise viewed from the point (1,1,1). For F=⟨z,x,y⟩, find ∮CF⋅dr.
Answer B
∇×F=⟨1,1,1⟩. The triangle lies in x+y+z=1, whose unit normal
(pointing toward (1,1,1)) is n=31⟨1,1,1⟩.
(∇×F)⋅n=33=3, a constant.
The triangle's area is 23, so ∮=3⋅23=23.
TrapUsing the un-normalised ⟨1,1,1⟩ as n, or the projected area instead of the true area.
S is any upward-oriented smooth surface whose boundary is the circle x2+y2=4 in the plane z=1, oriented counterclockwise seen from above. For F=⟨−y,x,z2⟩, find ∬S(∇×F)⋅ndS.
Answer C
Stokes turns the surface integral into one lap around the rim, so the surface itself is irrelevant.
Parametrize C:r(t)=⟨2cost,2sint,1⟩, r′=⟨−2sint,2cost,0⟩.
F(r(t))=⟨−2sint,2cost,1⟩.
F⋅r′=4sin2t+4cos2t=4, so ∮=4(2π)=8π.
(Equivalently ∇×F=⟨0,0,2⟩ and the flux through the disk is 2⋅π(2)2=8π.)
TrapTrying to build the unknown surface. Stokes says only the rim matters.
C is the boundary of the first-octant part of the plane x+y+z=1, oriented counterclockwise viewed from above. For F=⟨y,z,x⟩, find ∮CF⋅dr.
Answer B
∇×F=⟨−1,−1,−1⟩. The triangle lies in x+y+z=1 with upward unit normal n=31⟨1,1,1⟩.
(∇×F)⋅n=−33=−3, constant.
Triangle area =23, so ∮=−3⋅23=−23.
TrapLosing the sign. A negative circulation is a real answer, not an error.
Find the outward flux of F=⟨x,y,z⟩ across the sphere of radius 2 centred at the origin.
Answer D
divF=1+1+1=3, a constant, so the triple integral is
3×(volume)=3⋅34π(2)3=32π.
Check directly: on the sphere F⋅n=∣r∣=2, and the area is 4π(2)2=16π, giving 2⋅16π=32π ✓.
TrapUsing radius 1, or forgetting that div is constant so no integration is needed.
Find the outward flux of F=⟨x,y,z⟩ through the closed surface of the solid cylinder x2+y2≤1, 0≤z≤2.
Answer D
divF=3, constant, so the flux is 3×volume.
Volume =π(1)2(2)=2π, giving 6π.
Doing it by faces would mean three separate surface integrals — the theorem replaces all of them.
TrapForgetting the two flat caps if you insist on computing faces directly.
S is the paraboloid z=4−x2−y2 with z≥0, oriented upward — it is open at the bottom. Find the flux of F=⟨x,y,z⟩ through S.
Answer D
Cap it with the disk D:z=0,x2+y2≤4, oriented downward so the union is a sealed outward-oriented bag.
Volume: ∫02π∫02(4−r2)rdrdθ=2π[2r2−4r4]02=8π.
divF=3, so the sealed flux is 3(8π)=24π.
On the cap z=0: n=⟨0,0,−1⟩ and F⋅n=−z=0, so the cap contributes nothing.
Therefore the flux through S alone is 24π−0=24π.
TrapApplying the divergence theorem to S by itself. It is not closed — you must cap it, then subtract the cap.