Exam 2 study guide, worked

01

Lagrange Multipliers

Lesson 17 · §15.8

Don't know this yet? §s158 in the full guide →

From the official study guide

Method of Lagrange Multipliers

To find the extreme values of $f(x,y)$ subject to a constraint $g(x,y) = k$ , solve the system:

\nabla f = \lambda\,\nabla g \qquad \text{and} \qquad g(x,y) = k

That is:

$f_x = \lambda\, g_x$
$f_y = \lambda\, g_y$
$g(x,y) = k$

Three Variables

To optimize $f(x,y,z)$ subject to $g(x,y,z) = k$ :

\nabla f = \lambda\,\nabla g \qquad \text{and} \qquad g(x,y,z) = k

Note: Evaluate $f$ at all solution points and compare to identify absolute max/min.

Practice

5 problems

1

Find the maximum value of $f(x,y) = 3 + 2x - y$ subject to the constraint $x^2 + 2xy + 4y^2 = 7$ .

no calculator, same as the exam
Answer10answer recomputed · sympy
1. The constraint $g(x,y)=x^2+2xy+4y^2=7$ is a closed curve (an ellipse), so $f$ attains a max and a min on it. Use Lagrange: $\nabla f=\lambda\nabla g$ .
2. $\nabla f=\langle 2,\,-1\rangle$ and $\nabla g=\langle 2x+2y,\;2x+8y\rangle$ , so the system is $2=\lambda(2x+2y)$ , $-1=\lambda(2x+8y)$ , $x^2+2xy+4y^2=7$ .
3. Add twice the second equation to the first to kill $x$ : $2+2(-1)=\lambda\big[(2x+2y)+2(2x+8y)\big]$ gives $0=\lambda(6x+18y)$ . Since $\lambda\ne 0$ (else $\nabla f=\vec 0$ , false), $x=-3y$ .
4. Substitute into the constraint: $9y^2-6y^2+4y^2=7y^2=7$ , so $y=\pm 1$ and $(x,y)=(-3,1)$ or $(3,-1)$ .
5. Evaluate: $f(-3,1)=3-6-1=-4$ and $f(3,-1)=3+6+1=10$ .
6. Maximum value $=10$ (attained at $(3,-1)$ ; the minimum is $-4$ ).
Trap $\nabla g$ of the cross term is the usual casualty: $\partial_x(2xy)=2y$ and $\partial_y(2xy)=2x$ , so $\nabla g=\langle 2x+2y,\,2x+8y\rangle$ , not $\langle 2x+2y,\,8y\rangle$ . 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 $(x-1)^2 + y^2 = 1$ .

no calculator, same as the exam
Answer18answer recomputed · sympy
1. Constraint: the circle $g(x,y)=(x-1)^2+y^2=1$ , centre $(1,0)$ , radius $1$ . $f$ is linear, so its extremes sit where the gradient of $f$ is normal to the circle.
2. Lagrange: $\langle 8,-6\rangle=\lambda\langle 2(x-1),\,2y\rangle$ , i.e. $8=2\lambda(x-1)$ and $-6=2\lambda y$ .
3. So $x-1=\dfrac{4}{\lambda}$ and $y=-\dfrac{3}{\lambda}$ . Plug into the constraint: $\dfrac{16}{\lambda^2}+\dfrac{9}{\lambda^2}=1\Rightarrow \lambda^2=25,\ \lambda=\pm 5$ .
4. $\lambda=5$ : $(x,y)=\left(\tfrac95,-\tfrac35\right)$ , $f=\tfrac{72}{5}+\tfrac{18}{5}=18$ . $\lambda=-5$ : $(x,y)=\left(\tfrac15,\tfrac35\right)$ , $f=\tfrac85-\tfrac{18}{5}=-2$ .
5. Maximum value $=18$ .
6. Sanity check with geometry: $f=8x-6y$ at the centre $(1,0)$ is $8$ , and $|\nabla f|=\sqrt{64+36}=10$ , so on a radius- $1$ circle $f$ ranges over $8\pm 10=\{-2,18\}$ . Same answer.
TrapThe circle is not centred at the origin. Writing the constraint gradient as $\langle 2x,2y\rangle$ (forgetting the shift) is the standard wreck here; it must be $\langle 2(x-1),\,2y\rangle$ . The shortcut $f_{\max}=f(\text{centre})+|\nabla f|\cdot R$ 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 $x^2+y^2=5$ are, respectively:

no calculator, same as the exam
AnswerMaximum 45, Minimum 5answer recomputed · sympy
1. $f(x,y)=(x-2)^2+(y-4)^2$ is the squared distance from $(2,4)$ ; the constraint is the circle of radius $\sqrt5$ about the origin. So this is really a nearest/farthest point problem.
2. Lagrange: $\langle 2(x-2),\,2(y-4)\rangle=\lambda\langle 2x,\,2y\rangle$ , i.e. $x-2=\lambda x$ and $y-4=\lambda y$ .
3. Then $x(1-\lambda)=2$ and $y(1-\lambda)=4$ , so $y=2x$ (dividing; note $\lambda\ne1$ , since $\lambda=1$ would force $2=0$ ).
4. Constraint: $x^2+(2x)^2=5x^2=5\Rightarrow x=\pm1$ , giving $(1,2)$ and $(-1,-2)$ .
5. $f(1,2)=(1-2)^2+(2-4)^2=1+4=5$ . $f(-1,-2)=(-3)^2+(-6)^2=9+36=45$ .
6. Maximum $45$ , minimum $5$ . (Check: $|(2,4)|=\sqrt{20}=2\sqrt5$ , so distances to the circle are $2\sqrt5\mp\sqrt5$ , 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 $\sqrt{45}$ and $\sqrt5$ . Second, the point $(2,4)$ lies *outside* the circle, so both critical points are genuine, one near and one far; the min is not $0$ .
4

Find the maximum of $f(x,y,z) = x+y+z$ subject to $(x-1)^2+y^2+z^2=1$ .

no calculator, same as the exam
Answer $1+\sqrt{3}$ answer recomputed · sympy
1. Constraint: the sphere $g=(x-1)^2+y^2+z^2=1$ of radius $1$ centred at $(1,0,0)$ . $f=x+y+z$ is linear, so Lagrange with $\nabla f=\langle1,1,1\rangle$ .
2. System: $1=2\lambda(x-1)$ , $1=2\lambda y$ , $1=2\lambda z$ . Hence $x-1=y=z=\dfrac{1}{2\lambda}$ .
3. Substitute: $3\left(\dfrac{1}{2\lambda}\right)^2=1\Rightarrow \dfrac{1}{2\lambda}=\pm\dfrac{1}{\sqrt3}$ .
4. Taking the $+$ branch, $x=1+\tfrac{1}{\sqrt3},\ y=z=\tfrac{1}{\sqrt3}$ , so $f=1+\tfrac{3}{\sqrt3}=1+\sqrt3$ . The $-$ branch gives $1-\sqrt3$ .
5. Maximum $=1+\sqrt3$ .
TrapSame shifted-centre trap as before: $\nabla g=\langle 2(x-1),2y,2z\rangle$ . And because the three partials give $x-1=y=z$ (not $x=y=z$ ), the symmetric guess $x=y=z=1/\sqrt3$ is wrong. The geometric check is $f(\text{centre})+|\nabla f|R=1+\sqrt3\cdot1$ .
5

Find the absolute maximum value $M$ and absolute minimum value $m$ of $f(x,y)=x+y$ subject to $x^2-xy+y^2=1$ .

no calculator, same as the exam
Answer $M=2,\;m=-2$ answer recomputed · sympy
1. The constraint $x^2-xy+y^2=1$ is an ellipse (rotated $45^\circ$ ), a closed bounded curve, so a max and a min both exist.
2. Lagrange: $\langle1,1\rangle=\lambda\langle 2x-y,\;2y-x\rangle$ , so $1=\lambda(2x-y)$ and $1=\lambda(2y-x)$ .
3. Setting the two right sides equal: $2x-y=2y-x\Rightarrow 3x=3y\Rightarrow y=x$ .
4. Constraint: $x^2-x^2+x^2=x^2=1$ , so $(1,1)$ and $(-1,-1)$ .
5. $f(1,1)=2$ and $f(-1,-1)=-2$ , so $\boxed{M=2,\ m=-2}$ .
TrapDifferentiating the cross term $-xy$ : $\partial_x=-y$ , $\partial_y=-x$ — it is easy to drop a sign and get $2x+y$ . Also, cancelling $\lambda$ from the two equations is legal here only because $\lambda\ne0$ (if $\lambda=0$ then $1=0$ ), so you should say so rather than divide blindly.

02

Double Integrals over Rectangular Regions

Lesson 18 · §16.1

Don't know this yet? §s161 in the full guide →

From the official study guide

Double Integral over a Rectangle

For $R = [a,b] \times [c,d]$ :

\iint_R f(x,y)\,dA = \int_a^b \int_c^d f(x,y)\,dy\,dx = \int_c^d \int_a^b f(x,y)\,dx\,dy

Fubini's Theorem

If $f$ is continuous on $R = [a,b]\times[c,d]$ , the order of integration may be switched:

\int_a^b\int_c^d f(x,y)\,dy\,dx = \int_c^d\int_a^b f(x,y)\,dx\,dy

Riemann Sum Definition

Partition $[a,b]$ into $m$ subintervals and $[c,d]$ into $n$ subintervals, each of width $\Delta x = \frac{b-a}{m}$ and $\Delta y = \frac{d-c}{n}$ . Choose sample points $(x_{ij}^*, y_{ij}^*)$ in each subrectangle $R_{ij}$ :

\iint_R f(x,y)\,dA = \lim_{m,n\to\infty}\sum_{i=1}^{m}\sum_{j=1}^{n} f(x_{ij}^*,\, y_{ij}^*)\,\Delta A

where $\Delta A = \Delta x\,\Delta y$ is the area of each subrectangle.

Using midpoints: $x_i^* = \frac{x_{i-1}+x_i}{2}$ , $y_j^* = \frac{y_{j-1}+y_j}{2}$
The double integral gives the signed volume between the surface $z = f(x,y)$ and the $xy$ -plane

Practice

5 problems

1

Compute $\displaystyle\int_0^1\int_0^4 \sqrt{xy}\,dx\,dy$ .

no calculator, same as the exam
Answer $\tfrac{32}{9}$ answer recomputed · sympy
1. The integrand factors, $\sqrt{xy}=\sqrt{x}\,\sqrt{y}$ , and the region is the rectangle $[0,4]\times[0,1]$ , so the double integral splits into a product of single integrals.
2. $\displaystyle\int_0^1\!\!\int_0^4\sqrt{x}\sqrt{y}\,dx\,dy=\left(\int_0^4\sqrt{x}\,dx\right)\left(\int_0^1\sqrt{y}\,dy\right)$ .
3. $\displaystyle\int_0^4 x^{1/2}dx=\tfrac23 x^{3/2}\Big|_0^4=\tfrac23\cdot 8=\tfrac{16}{3}$ .
4. $\displaystyle\int_0^1 y^{1/2}dy=\tfrac23$ .
5. Product: $\dfrac{16}{3}\cdot\dfrac23=\dfrac{32}{9}$ .
TrapRead the differentials: $dx\,dy$ 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 $\tfrac23\cdot\tfrac{16}{3}$ 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 $z=e^{x+y}$ .

no calculator, same as the exam
Answer $e^5-e^3-e^2-5$ answer recomputed · sympy
1. The base is the rectangle $R=[0,3]\times[0,2]$ ; the solid sits between the floor $z=1$ and the roof $z=e^{x+y}$ . On $R$ we have $x+y\ge0$ , so $e^{x+y}\ge1$ and the roof really is above the floor everywhere.
2. Volume $=\displaystyle\iint_R\big(e^{x+y}-1\big)\,dA=\int_0^2\!\!\int_0^3\big(e^{x}e^{y}-1\big)\,dx\,dy$ .
3. $\displaystyle\iint_R e^{x}e^{y}\,dA=\left(\int_0^3 e^x dx\right)\left(\int_0^2 e^y dy\right)=(e^3-1)(e^2-1)$ .
4. $\displaystyle\iint_R 1\,dA=\text{area}(R)=3\cdot2=6$ .
5. Volume $=(e^3-1)(e^2-1)-6=e^5-e^3-e^2+1-6=e^5-e^3-e^2-5$ .
TrapForgetting to subtract the floor: integrating $e^{x+y}$ alone gives the volume under the surface down to $z=0$ , not down to $z=1$ . The correction is $-\text{area}(R)=-6$ , and note $(e^3-1)(e^2-1)$ already contributes $+1$ , so the constant is $1-6=-5$ , not $-6$ . Also $e^{x+y}=e^xe^y$ is what lets the integral factor; $e^{x+y}\ne e^x+e^y$ .
3

Find the average value of $f(x,y)=x^2y$ over the rectangle $R$ with vertices $(-1,0)$ , $(-1,5)$ , $(1,5)$ , $(1,0)$ .

no calculator, same as the exam
Answer $\tfrac{5}{6}$ answer recomputed · sympy
1. The four vertices give the rectangle $R=[-1,1]\times[0,5]$ , of area $A=2\cdot5=10$ .
2. Average value $=\dfrac{1}{A}\displaystyle\iint_R f\,dA=\dfrac{1}{10}\int_0^5\!\!\int_{-1}^{1}x^2y\,dx\,dy$ .
3. $\displaystyle\int_{-1}^{1}x^2\,dx=\tfrac{x^3}{3}\Big|_{-1}^{1}=\tfrac23$ and $\displaystyle\int_0^5 y\,dy=\tfrac{25}{2}$ .
4. $\displaystyle\iint_R x^2y\,dA=\tfrac23\cdot\tfrac{25}{2}=\tfrac{25}{3}$ .
5. Average $=\dfrac{25/3}{10}=\dfrac{5}{6}$ .
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: $x^2$ is even, so $\int_{-1}^1 x^2dx=\tfrac23$ , not $0$ — students who remember 'symmetric interval $\Rightarrow$ zero' from odd integrands kill this problem instantly.
4

Evaluate $\displaystyle\int_0^1\int_0^1\int_0^{2-y} e^x\,dz\,dx\,dy$ .

no calculator, same as the exam
Answer $\tfrac{3}{2}(e-1)$ answer recomputed · sympy
1. Innermost is $dz$ and the integrand $e^x$ is free of $z$ , so the first integration is just multiplication by the height: $\displaystyle\int_0^{2-y}e^x\,dz=e^x(2-y)$ .
2. The remaining limits are constants, so $\displaystyle\int_0^1\!\!\int_0^1 e^x(2-y)\,dx\,dy=\left(\int_0^1 e^x dx\right)\left(\int_0^1(2-y)\,dy\right)$ .
3. $\displaystyle\int_0^1 e^x dx=e-1$ .
4. $\displaystyle\int_0^1(2-y)\,dy=2-\tfrac12=\tfrac32$ .
5. Value $=\dfrac32(e-1)$ .
TrapThe order is $dz\,dx\,dy$ , 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 $e^x$ and $(2-y)$ live in different variables.
5

Evaluate $\displaystyle\iint_R \frac{x^2y}{2+x^3}\,dA$ over $R=\{1\le x\le 2,\;0\le y\le 4\}$ .

no calculator, same as the exam
Answer $\tfrac{8}{3}(\ln10-\ln3)$ answer recomputed · sympy
1. $R$ is a rectangle and the integrand splits as $\dfrac{x^2}{2+x^3}\cdot y$ , so the integral is a product of two one-variable integrals.
2. $\displaystyle\iint_R\frac{x^2y}{2+x^3}\,dA=\left(\int_1^2\frac{x^2}{2+x^3}\,dx\right)\left(\int_0^4 y\,dy\right)$ .
3. For the $x$ -integral substitute $u=2+x^3$ , $du=3x^2dx$ , so $x^2dx=\tfrac{du}{3}$ ; as $x:1\to2$ , $u:3\to10$ .
4. $\displaystyle\int_3^{10}\frac{1}{3u}\,du=\tfrac13\big(\ln 10-\ln 3\big)$ .
5. $\displaystyle\int_0^4 y\,dy=8$ .
6. Value $=8\cdot\tfrac13(\ln10-\ln3)=\dfrac{8}{3}\big(\ln10-\ln3\big)=\dfrac83\ln\dfrac{10}{3}$ .
TrapWhen you substitute $u=2+x^3$ you must change the limits to $u:3\to10$ ; leaving $1$ and $2$ gives $\tfrac13\ln\tfrac{2}{1}$ , a completely different number. The other classic is losing the $\tfrac13$ from $du=3x^2\,dx$ .

03

Double Integrals over General Regions

Lesson 19 · §16.2

Don't know this yet? §s162 in the full guide →

From the official study guide

Type I Region (vertical strips)

Region bounded by functions of $x$ : $g_1(x) \le y \le g_2(x)$ , $a \le x \le b$

\iint_D f(x,y)\,dA = \int_a^b\int_{g_1(x)}^{g_2(x)} f(x,y)\,dy\,dx

Type II Region (horizontal strips)

Region bounded by functions of $y$ : $h_1(y) \le x \le h_2(y)$ , $c \le y \le d$

\iint_D f(x,y)\,dA = \int_c^d\int_{h_1(y)}^{h_2(y)} f(x,y)\,dx\,dy

Reversing Order of Integration

Sketch the region, identify bounds for the opposite order, then re-write the integral. This is often necessary when the inner integral has no elementary antiderivative.

Practice

5 problems

1

Evaluate (switching order of integration):
$\displaystyle\int_0^{27}\int_{\sqrt[3]{x}}^{3} 4e^{y^4}\,dy\,dx$

no calculator, same as the exam
Answer $e^{81}-1$ answer recomputed · sympy
1. $e^{y^4}$ has no elementary antiderivative in $y$ , so the inner integral as written is hopeless. That is the signal to switch the order.
2. Read off the region: $0\le x\le 27$ and $\sqrt[3]{x}\le y\le 3$ . The curve $y=\sqrt[3]{x}$ is $x=y^3$ , and at $x=27$ , $y=3$ .
3. Describe the same region with $y$ outer: $y$ runs $0\to3$ , and for each $y$ , $x$ runs from $0$ to $y^3$ .
4. $\displaystyle\int_0^{27}\!\!\int_{\sqrt[3]{x}}^{3}4e^{y^4}\,dy\,dx=\int_0^3\!\!\int_0^{y^3}4e^{y^4}\,dx\,dy=\int_0^3 4y^3e^{y^4}\,dy$ .
5. Substitute $u=y^4$ , $du=4y^3dy$ , limits $u:0\to81$ : $\displaystyle\int_0^{81}e^u\,du=e^{81}-1$ .
6. Value $=e^{81}-1$ .
TrapGetting the new limits backwards. The old inner limit $y\ge\sqrt[3]{x}$ becomes $x\le y^3$ , so the new inner range is $0\le x\le y^3$ — not $y^3\le x\le 27$ . Sketch the region: it is the piece *above* the cube-root curve, i.e. to the *left* of $x=y^3$ . The factor $4$ in the integrand is not decoration; it is exactly the $4y^3$ the substitution needs after the swap.
2

Evaluate $\displaystyle\iint_R \frac{\sin x}{x}\,dA$ 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
Answer $1-\cos(1)$ answer recomputed · sympy
1. Sketch $R$ : bounded below by $y=0$ , above by $y=x$ , on the right by $x=1$ — the triangle with vertices $(0,0),(1,0),(1,1)$ .
2. $\dfrac{\sin x}{x}$ has no elementary antiderivative in $x$ , so we must integrate in $y$ first (as instructed), which makes $x$ a constant.
3. $\displaystyle\iint_R\frac{\sin x}{x}\,dA=\int_0^1\!\!\int_0^{x}\frac{\sin x}{x}\,dy\,dx$ .
4. Inner: $\dfrac{\sin x}{x}\cdot(x-0)=\sin x$ . The offending $x$ in the denominator cancels — that is the whole point of the order.
5. $\displaystyle\int_0^1\sin x\,dx=-\cos x\Big|_0^1=1-\cos(1)$ .
TrapChoosing $dy\,dx$ is not optional here: the other order gives $\int_y^1\frac{\sin x}{x}dx$ , which cannot be evaluated in closed form. Also the inner limits are $0\le y\le x$ — a *variable* upper limit. Writing $\int_0^1$ for the inner $y$ -integral turns the triangle into the unit square and gives $\int_0^1\frac{\sin x}{x}dx$ , 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
Answer $\tfrac{1}{3}$ answer recomputed · sympy
1. 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 $z=2x\ge0$ there, so the volume is the plain double integral.
2. As a type-I region: $0\le x\le 1$ and $0\le y\le 1-x$ .
3. $V=\displaystyle\iint_D 2x\,dA=\int_0^1\!\!\int_0^{1-x}2x\,dy\,dx$ .
4. Inner: $2x(1-x)=2x-2x^2$ .
5. $\displaystyle\int_0^1(2x-2x^2)\,dx=1-\tfrac23=\tfrac13$ .
TrapUsing $\int_0^1\int_0^1$ (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 $\tfrac12$ times anything obvious; you have to actually integrate $2x(1-x)$ .
4

Change the order of integration for $\displaystyle\int_0^2\int_{x^2}^{2x}f(x,y)\,dy\,dx$ .

no calculator, same as the exam
Official answer(original) — $\displaystyle\int_0^4\int_{\sqrt{y}}^{y/2}f\,dx\,dy$ checked · official answer wrong

Correct answer $\displaystyle\int_0^4\!\!\int_{y/2}^{\sqrt{y}}f(x,y)\,dx\,dy$ recomputed · sympy
1. Read the region from the given order $dy\,dx$ : $0\le x\le2$ and $x^2\le y\le 2x$ — the lens between the parabola $y=x^2$ and the line $y=2x$ .
2. Find where the two curves meet: $x^2=2x\Rightarrow x=0$ or $x=2$ , giving the corner points $(0,0)$ and $(2,4)$ . So $y$ ranges over $[0,4]$ overall.
3. Now solve each boundary for $x$ . From $y=2x$ we get $x=\dfrac{y}{2}$ ; from $y=x^2$ (with $x\ge0$ ) we get $x=\sqrt{y}$ .
4. For a fixed $y\in[0,4]$ , which is the left edge? Test $y=1$ : the line gives $x=\tfrac12$ , the parabola gives $x=1$ . So $x$ runs from $\dfrac{y}{2}$ up to $\sqrt{y}$ .
5. Reversed order: $\displaystyle\int_0^4\!\!\int_{y/2}^{\sqrt{y}}f(x,y)\,dx\,dy$ .
6. Confidence check with a test integrand $f=xy$ : the original order gives $\tfrac83$ , and $\int_0^4\int_{y/2}^{\sqrt y}xy\,dx\,dy=\tfrac83$ as well.
TrapInside the lens the parabola is the right boundary and the line is the left one (for $0<x<2$ , $x^2<2x$ means the parabola sits *below* the line, which flips to *right of* when you read horizontally). Writing $\int_{\sqrt y}^{y/2}$ reverses the limits and negates the integral. Also $x=\sqrt y$ , not $x=\pm\sqrt y$ : the region lives in $x\ge0$ .

The official answer is wrongThe official answer prints the inner limits as $\int_{\sqrt y}^{y/2}$ , which is upside down. For $0<y<4$ we have $\tfrac{y}{2}<\sqrt{y}$ (at $y=1$ : $0.5<1$ ), so the lower limit must be $y/2$ and the upper limit $\sqrt y$ . As printed, the integral equals the negative of the original. Numerical check with $f(x,y)=xy$ : the original iterated integral is $\tfrac83$ , the correctly reversed one is $\tfrac83$ , and the printed one is $-\tfrac83$ .
5

Reverse the order and evaluate: $\displaystyle\int_0^{2\sqrt{\ln3}}\!\int_{y/2}^{\sqrt{\ln3}} e^{x^2}\,dx\,dy$ .

no calculator, same as the exam
Answer $2$ answer recomputed · sympy
1. $e^{x^2}$ has no elementary antiderivative, so the order must be reversed. Let $a=\sqrt{\ln 3}$ for short; the limits are $0\le y\le 2a$ and $\dfrac{y}{2}\le x\le a$ .
2. 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.
3. Reading it the other way: $x$ runs $0\to a$ , and for each $x$ , $y$ runs from $0$ to $2x$ .
4. $\displaystyle\int_0^{2a}\!\!\int_{y/2}^{a}e^{x^2}\,dx\,dy=\int_0^{a}\!\!\int_0^{2x}e^{x^2}\,dy\,dx=\int_0^{a}2x\,e^{x^2}\,dx$ .
5. Substitute $u=x^2$ , $du=2x\,dx$ , $u:0\to a^2=\ln3$ : $\displaystyle\int_0^{\ln3}e^u\,du=e^{\ln3}-1=3-1=2$ .
6. Value $=2$ .
TrapThe outer limit $2\sqrt{\ln3}$ and the inner $\sqrt{\ln3}$ 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 $e^{\ln 3}=3$ , so the final answer is the clean $2$ , not $e^{\ln 3}$ left unsimplified.

04

Double Integrals in Polar Coordinates

Lesson 20 · §16.3

Don't know this yet? §s163 in the full guide →

From the official study guide

Polar Coordinate Substitution

Use $x = r\cos\theta$ , $y = r\sin\theta$ , $x^2+y^2=r^2$ :

\iint_D f(x,y)\,dA = \int_\alpha^\beta\int_{r_1(\theta)}^{r_2(\theta)} f(r\cos\theta,\, r\sin\theta)\,r\,dr\,d\theta

Key: Don't forget the Jacobian factor $r$ in $dA = r\,dr\,d\theta$ .

When to Use Polar Coordinates

Region is a disk, annulus, or sector
Integrand contains $x^2 + y^2$
Region has circular symmetry

Practice

5 problems

1

Evaluate $\displaystyle\int_0^3\!\int_0^{\sqrt{9-x^2}}\!\sqrt{x^2+y^2}\,dy\,dx$ by converting to polar.

no calculator, same as the exam
Answer $\tfrac{9\pi}{2}$ answer recomputed · sympy
1. Identify the region: $0\le x\le3$ and $0\le y\le\sqrt{9-x^2}$ , i.e. $x^2+y^2\le9$ with $x,y\ge0$ — the quarter disk of radius $3$ in the first quadrant.
2. Polar: $x=r\cos\theta$ , $y=r\sin\theta$ , $\sqrt{x^2+y^2}=r$ , and $dA=r\,dr\,d\theta$ . The region is $0\le r\le3$ , $0\le\theta\le\dfrac{\pi}{2}$ .
3. $\displaystyle\int_0^3\!\!\int_0^{\sqrt{9-x^2}}\sqrt{x^2+y^2}\,dy\,dx=\int_0^{\pi/2}\!\!\int_0^3 r\cdot r\,dr\,d\theta=\int_0^{\pi/2}\!\!\int_0^3 r^2\,dr\,d\theta$ .
4. Inner: $\dfrac{r^3}{3}\Big|_0^3=9$ .
5. Outer: $9\cdot\dfrac{\pi}{2}=\dfrac{9\pi}{2}$ .
TrapThe integrand becomes $r^2$ , not $r$ : one factor of $r$ is $\sqrt{x^2+y^2}$ and the second is the Jacobian from $dA=r\,dr\,d\theta$ . Dropping the Jacobian gives $\tfrac{9\pi}{4}$ . Second trap: $\theta$ runs only to $\pi/2$ — the limits $x\in[0,3]$ , $y\ge0$ give a quarter disk, not a half disk.
2

Compute $\displaystyle\iint_R e^{-(x^2+y^2)}\,dA$ where $R$ is the disk $x^2+y^2\le4$ .

no calculator, same as the exam
Answer $\pi(1-e^{-4})$ answer recomputed · sympy
1. $R$ is the full disk of radius $2$ and the integrand depends only on $x^2+y^2$ , so polar is the obvious move: $0\le r\le2$ , $0\le\theta\le2\pi$ .
2. $\displaystyle\iint_R e^{-(x^2+y^2)}\,dA=\int_0^{2\pi}\!\!\int_0^{2}e^{-r^2}\,r\,dr\,d\theta$ .
3. Inner: substitute $u=r^2$ , $du=2r\,dr$ , so $\displaystyle\int_0^2 e^{-r^2}r\,dr=\tfrac12\int_0^4 e^{-u}du=\tfrac12\big(1-e^{-4}\big)$ .
4. Outer: multiply by $\displaystyle\int_0^{2\pi}d\theta=2\pi$ .
5. Value $=2\pi\cdot\tfrac12\big(1-e^{-4}\big)=\pi\big(1-e^{-4}\big)$ .
TrapWithout the Jacobian $r$ , $\int e^{-r^2}dr$ 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 $u=r^2$ goes to $4$ , giving $e^{-4}$ ; writing $e^{-2}$ is the standard slip.
3

Rewrite in polar (do not evaluate): $\displaystyle\int_0^1\int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}}(x^2+y^2)^{1/2}\,dx\,dy$ .

no calculator, same as the exam
Answer $\displaystyle\int_0^\pi\!\int_0^1 r^2\,dr\,d\theta$ answer recomputed · sympy
1. Region: $0\le y\le1$ with $-\sqrt{1-y^2}\le x\le\sqrt{1-y^2}$ , i.e. $x^2+y^2\le1$ with $y\ge0$ — the upper half of the unit disk.
2. In polar the half disk is $0\le r\le1$ , $0\le\theta\le\pi$ (the full sweep of the upper half plane).
3. The integrand $(x^2+y^2)^{1/2}=r$ , and $dA=r\,dr\,d\theta$ , so $r\cdot r=r^2$ .
4. $\displaystyle\int_0^1\!\!\int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}}(x^2+y^2)^{1/2}\,dx\,dy=\int_0^{\pi}\!\!\int_0^1 r^2\,dr\,d\theta$ .
5. (Not asked for, but its value is $\dfrac{\pi}{3}$ , and the original Cartesian integral evaluates to $\dfrac{\pi}{3}$ as well.)
Trap $\theta$ ranges over $[0,\pi]$ , not $[0,\pi/2]$ : $x$ takes negative values here, so the region covers the second quadrant too. And the integrand must pick up the Jacobian, $r\cdot r\,dr\,d\theta=r^2dr\,d\theta$ ; writing $\int_0^\pi\int_0^1 r\,dr\,d\theta$ is the most common wrong answer on this exact problem.
4

Evaluate $\displaystyle\int_0^{\sqrt{\pi}}\!\int_0^{\sqrt{\pi-y^2}}\sin(x^2+y^2)\,dx\,dy$ .

no calculator, same as the exam
Answer $\tfrac{\pi}{2}$ answer recomputed · sympy
1. Region: $0\le y\le\sqrt{\pi}$ , $0\le x\le\sqrt{\pi-y^2}$ , i.e. $x^2+y^2\le\pi$ with $x,y\ge0$ — the quarter disk of radius $\sqrt{\pi}$ .
2. $\sin(x^2+y^2)$ has no elementary antiderivative in Cartesian form, so polar is forced: $0\le r\le\sqrt{\pi}$ , $0\le\theta\le\dfrac{\pi}{2}$ .
3. $\displaystyle\int_0^{\sqrt\pi}\!\!\int_0^{\sqrt{\pi-y^2}}\sin(x^2+y^2)\,dx\,dy=\int_0^{\pi/2}\!\!\int_0^{\sqrt{\pi}}\sin(r^2)\,r\,dr\,d\theta$ .
4. Inner: $u=r^2$ , $du=2r\,dr$ , $u:0\to\pi$ , giving $\tfrac12\displaystyle\int_0^{\pi}\sin u\,du=\tfrac12\big(-\cos u\big)\Big|_0^{\pi}=\tfrac12(1+1)=1$ .
5. Outer: $1\cdot\dfrac{\pi}{2}=\dfrac{\pi}{2}$ .
TrapThe radius is $\sqrt{\pi}$ , so $u=r^2$ runs to $\pi$ — not to $\sqrt\pi$ . Writing $\int_0^{\sqrt\pi}\sin u\,du$ 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 $\displaystyle\int_{-3}^{3}\int_{-\sqrt{9-y^2}}^{\sqrt{9-y^2}} e^{x^2+y^2}\,dx\,dy$ .

no calculator, same as the exam
Answer $\pi(e^9-1)$ answer recomputed · sympy
1. Region: $-3\le y\le3$ and $-\sqrt{9-y^2}\le x\le\sqrt{9-y^2}$ — the entire disk $x^2+y^2\le9$ of radius $3$ .
2. Polar: $0\le r\le3$ , $0\le\theta\le2\pi$ , integrand $e^{x^2+y^2}=e^{r^2}$ , and $dA=r\,dr\,d\theta$ .
3. $\displaystyle\int_{-3}^{3}\!\int_{-\sqrt{9-y^2}}^{\sqrt{9-y^2}}e^{x^2+y^2}\,dx\,dy=\int_0^{2\pi}\!\!\int_0^{3}e^{r^2}r\,dr\,d\theta$ .
4. Inner: $u=r^2$ , $du=2r\,dr$ , $u:0\to9$ , so $\tfrac12\displaystyle\int_0^9 e^u\,du=\tfrac12\big(e^9-1\big)$ .
5. Outer: $2\pi\cdot\tfrac12(e^9-1)=\pi\big(e^9-1\big)$ .
TrapBoth limits are symmetric, so $\theta$ sweeps the full $2\pi$ ; using $\pi$ (half disk) or $\pi/2$ halves or quarters the answer. And $u=r^2$ runs to $9$ , giving $e^9$ — writing $e^3$ is the standard misfire. The $r$ from $dA$ is again mandatory.

05

Triple Integrals

Lesson 21 · §16.4

Don't know this yet? §s164 in the full guide →

From the official study guide

Iterated Triple Integral

\iiint_E f(x,y,z)\,dV = \int_a^b\int_{g_1(x)}^{g_2(x)}\int_{u_1(x,y)}^{u_2(x,y)} f(x,y,z)\,dz\,dy\,dx

Applications

Volume: $V = \displaystyle\iiint_E dV$
Average value: $\bar{f} = \dfrac{1}{V}\displaystyle\iiint_E f\,dV$

Note: There are 6 possible orders of integration — choose the order that fits the region and integrand most naturally.

Practice

5 problems

1

Evaluate $\displaystyle\int_0^1\!\int_0^1\!\int_0^{2-y} e^x\,dz\,dx\,dy$ .

no calculator, same as the exam
Answer $\tfrac{3}{2}(e-1)$ answer recomputed · sympy
1. Read the order off the differentials: $dz\,dx\,dy$ . The innermost variable is $z$ , and the integrand $e^x$ does not contain $z$ — so the inner integral is just multiplication by the height.
2. $\displaystyle\int_0^{2-y}e^x\,dz=e^x\,(2-y)$ .
3. Now the $x$ and $y$ limits are both constants, and what is left separates: $\displaystyle\int_0^1\!\int_0^1 e^x(2-y)\,dx\,dy=\left(\int_0^1 e^x\,dx\right)\left(\int_0^1(2-y)\,dy\right)$ .
4. $\displaystyle\int_0^1 e^x\,dx=e-1$ and $\displaystyle\int_0^1(2-y)\,dy=2-\tfrac12=\tfrac32$ .
5. Product: $\tfrac{3}{2}(e-1)$ .
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 $\int_0^1(2-y)\,dy$ 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 $\displaystyle \iiint_E f\,dV=\int_0^a\!\int_0^b\!\int_0^c f\,dz\,dy\,dx$ :

no calculator, same as the exam
Answer $a=2,\;b=4-2x$ answer recomputed · sympy
1. Order $dz\,dy\,dx$ means: $z$ is bounded by surfaces, $y$ by curves in the $xy$ -plane, $x$ by numbers. Build it from the inside out.
2. First octant gives the floor $z=0$ ; the plane gives the ceiling $z=8-4x-2y$ . So $c=8-4x-2y$ .
3. 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 $x\ge0,\ y\ge0$ , is the shadow.
4. For a fixed $x$ , $y$ runs from $0$ up to that line: $b=4-2x$ .
5. $x$ runs until the shadow closes, i.e. until $4-2x=0$ : $a=2$ .
6. $\displaystyle\iiint_E f\,dV=\int_0^2\!\int_0^{4-2x}\!\int_0^{8-4x-2y}f\,dz\,dy\,dx$ .
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 $\iiint_D f\,dV$ as an iterated integral.

no calculator, same as the exam
Official answere.g. $\displaystyle\int_0^1\!\int_x^{2-x}\!\int_0^{2-x-y}f\,dz\,dy\,dx$ , or $\displaystyle\int_0^1\!\int_0^{2-x}\!\int_x^{2-x-z}f\,dy\,dz\,dx$ checked · official answer wrong

Correct answer $\displaystyle\int_0^1\!\int_0^{2-2x}\!\int_x^{2-x-z}f\,dy\,dz\,dx$ recomputed · sympy
1. Four surfaces, so list the inequalities: $x\ge0$ , $z\ge0$ , $y\ge x$ , and $x+y+z\le2$ .
2. Take $z$ innermost: $0\le z\le 2-x-y$ .
3. Shadow in the $xy$ -plane is where that height is non-negative and $y\ge x$ : the triangle $x\le y\le 2-x$ , $x\ge0$ .
4. That triangle closes when $x=2-x$ , i.e. $x=1$ . So $\displaystyle\int_0^1\!\int_x^{2-x}\!\int_0^{2-x-y}f\,dz\,dy\,dx$ .
5. Reordering to $dy\,dz\,dx$ : solve $y\ge x$ and $y\le 2-x-z$ , so $x\le y\le 2-x-z$ .
6. Those two limits cross when $2-x-z=x$ , i.e. at $z=2-2x$ — so $z$ must stop at $2-2x$ , giving $\displaystyle\int_0^1\!\int_0^{2-2x}\!\int_x^{2-x-z}f\,dy\,dz\,dx$ .
7. Sanity check both with $f=1$ : each gives volume $\tfrac23$ .
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 $\tfrac12$ ; the true volume, and the value of the printed first form, is $\tfrac23$ . The $z$ -limit must be $2-2x$ .
4

Let $V$ be bounded by $z=6-x^2-y^2$ and $z=x^2+y^2$ . If $V=\displaystyle\int_{-\sqrt{3}}^{\sqrt{3}}\!\int_a^b\!\int_{x^2+y^2}^{6-x^2-y^2}dz\,dy\,dx$ , what are $a$ and $b$ ?

no calculator, same as the exam
Answer $a=-\sqrt{3-x^2},\;b=\sqrt{3-x^2}$ answer recomputed · sympy
1. The $z$ -limits are already given: the solid sits between the upward paraboloid $z=x^2+y^2$ and the downward one $z=6-x^2-y^2$ .
2. They meet where $x^2+y^2=6-x^2-y^2$ , i.e. $2(x^2+y^2)=6$ , so $x^2+y^2=3$ — a circle of radius $\sqrt3$ .
3. That circle is the shadow's boundary, which is why the outer limits are $\pm\sqrt3$ .
4. Solving $x^2+y^2=3$ for $y$ : $a=-\sqrt{3-x^2}$ , $b=\sqrt{3-x^2}$ .
TrapSetting $x^2+y^2=6$ (from $z=6-x^2-y^2$ 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 $\sqrt3$ , not $\sqrt6$ . The given outer limits $\pm\sqrt3$ are the tell.
5

Evaluate $\displaystyle\int_{-1}^{1}\!\int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}}\!\int_{-1}^{1}(x^2+y^2)^{3/2}\,dz\,dx\,dy$ .

no calculator, same as the exam
Answer $\tfrac{4\pi}{5}$ answer recomputed · sympy
1. Look at the shadow before touching the integrand: $-1\le y\le1$ with $-\sqrt{1-y^2}\le x\le\sqrt{1-y^2}$ is the full unit disk, and $z$ runs over the fixed interval $[-1,1]$ . The solid is a cylinder.
2. $(x^2+y^2)^{3/2}$ is a function of distance from the $z$ -axis alone — cylindrical coordinates, and the $z$ -integral is just a factor of $2$ .
3. $(x^2+y^2)^{3/2}=r^3$ , and $dV=r\,dr\,d\theta\,dz$ , so the integrand becomes $r^3\cdot r=r^4$ .
4. $\displaystyle\int_{-1}^{1}\!\int_0^{2\pi}\!\int_0^1 r^4\,dr\,d\theta\,dz=2\cdot 2\pi\cdot\frac{1}{5}=\frac{4\pi}{5}$ .
TrapConverting $(x^2+y^2)^{3/2}$ to $r^3$ and then integrating $\int_0^1 r^3\,dr=\tfrac14$ — the extra $r$ from $dA=r\,dr\,d\theta$ is dropped and the answer comes out $\pi$ instead of $\tfrac{4\pi}{5}$ . The exponent you integrate is always one higher than the one you converted.

06

Triple Integrals in Cylindrical Coordinates

Lesson 22 · §16.5

Don't know this yet? §s165c in the full guide →

From the official study guide

Cylindrical Coordinates

Relation to rectangular coordinates:

x = r\cos\theta,\quad y = r\sin\theta,\quad z = z

r^2 = x^2+y^2,\quad \tan\theta = \frac{y}{x}

Triple Integral in Cylindrical Coordinates

\iiint_E f(x,y,z)\,dV = \int_\alpha^\beta\int_{r_1}^{r_2}\int_{u_1}^{u_2} f(r\cos\theta,r\sin\theta,z)\,r\,dz\,dr\,d\theta

Key: Don't forget the Jacobian factor $r$ in $dV = r\,dz\,dr\,d\theta$ .

When to Use Cylindrical Coordinates

Region has an axis of symmetry along the $z$ -axis
Integrand or region involves $x^2 + y^2$
Region is bounded by a cylinder $x^2+y^2 = a^2$ , cone, or paraboloid

Practice

5 problems

1

Rewrite $\displaystyle\int_0^{\sqrt{2}}\!\int_0^{\sqrt{4-x^2}}\!\int_{2}^{6}\!e^{-x^2-y^2}\,dz\,dy\,dx$ in cylindrical coordinates.

no calculator, same as the exam
Official answer $\displaystyle\int_2^6\!\int_{\pi/4}^{\pi/2}\!\int_0^2 re^{-r^2}\,dr\,d\theta\,dz$ checked · official answer wrong

Correct answer $\displaystyle\int_2^6\!\int_0^{\pi/4}\!\int_0^{\sqrt2\sec\theta}\!re^{-r^2}\,dr\,d\theta\,dz+\int_2^6\!\int_{\pi/4}^{\pi/2}\!\int_0^2 re^{-r^2}\,dr\,d\theta\,dz$ recomputed · sympy
1. $z$ is already uncoupled: it runs over $[2,6]$ and nothing else mentions it, so it can go anywhere in the order.
2. The integrand $e^{-x^2-y^2}=e^{-r^2}$ — that is the whole reason to convert. In Cartesian it has no elementary antiderivative.
3. $dV=r\,dr\,d\theta\,dz$ , so the integrand becomes $re^{-r^2}$ , which integrates instantly to $-\tfrac12e^{-r^2}$ .
4. Region: $y\le\sqrt{4-x^2}$ is the disk $r\le2$ ; the corner point $(\sqrt2,\sqrt2)$ where $x=\sqrt2$ meets that circle is at $\theta=\frac{\pi}{4}$ .
5. So the wedge $\frac{\pi}{4}\le\theta\le\frac{\pi}{2}$ , $0\le r\le 2$ — which is exactly the set $y\ge x$ inside the disk in the first quadrant.
6. $\displaystyle\int_2^6\!\int_{\pi/4}^{\pi/2}\!\int_0^2 re^{-r^2}\,dr\,d\theta\,dz=4\cdot\frac{\pi}{4}\cdot\frac{1-e^{-4}}{2}=\frac{\pi}{2}\left(1-e^{-4}\right)$ .
TrapConverting the integrand but keeping the Cartesian limits in spirit — reading $0\le x\le\sqrt2$ as ' $\theta$ from $0$ '. The upper $x$ -limit $\sqrt2$ is a vertical line, $r=\sqrt2\sec\theta$ , not a ray; only the presence of the lower boundary $y=x$ turns it into the clean ray $\theta=\frac{\pi}{4}$ . 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, $0\le y\le\sqrt{4-x^2}$ over $0\le x\le\sqrt2$ has area $1+\frac{\pi}{2}$ , while the wedge $\frac{\pi}{4}\le\theta\le\frac{\pi}{2},\ r\le2$ has area $\frac{\pi}{2}$ — the answer is missing the triangular piece below $y=x$ . The printed answer is correct for the inner limit $\int_x^{\sqrt{4-x^2}}$ ; with the printed lower limit $0$ the rewrite needs the two pieces above.
2

Evaluate $\displaystyle\int_0^2\!\int_0^{\sqrt{4-x^2}}\!\int_{\sqrt{x^2+y^2}}^{\sqrt{2(x^2+y^2)}} dz\,dy\,dx$ using cylindrical coordinates.

no calculator, same as the exam
Answer $\tfrac{4\pi}{3}(\sqrt{2}-1)$ answer recomputed · sympy
1. Shadow first: $0\le x\le2$ , $0\le y\le\sqrt{4-x^2}$ is the quarter disk of radius $2$ in the first quadrant, so $0\le\theta\le\frac{\pi}{2}$ , $0\le r\le2$ .
2. The $z$ -limits are both cones: $z=\sqrt{x^2+y^2}=r$ below and $z=\sqrt{2(x^2+y^2)}=\sqrt2\,r$ above.
3. Height of the column at radius $r$ : $\sqrt2\,r-r=(\sqrt2-1)r$ .
4. With $dV=r\,dr\,d\theta\,dz$ : $\displaystyle\int_0^{\pi/2}\!\int_0^2(\sqrt2-1)r\cdot r\,dr\,d\theta=\frac{\pi}{2}(\sqrt2-1)\cdot\frac{8}{3}$ .
5. $=\dfrac{4\pi}{3}\left(\sqrt2-1\right)$ .
Trap $\sqrt{2(x^2+y^2)}$ is $\sqrt2\,r$ , not $2r$ and not $r^2$ — pulling the $2$ out from under the root without the square root gives $\frac{4\pi}{3}$ exactly one unit too large. Second trap: the height $(\sqrt2-1)r$ already carries one power of $r$ , so the integrand is $r^2$ , not $r$ ; forgetting the Jacobian here costs a factor of $\frac{3}{2}\cdot\frac{2}{2}$ and lands on $\pi(\sqrt2-1)$ .
3

Convert to cylindrical and evaluate $\displaystyle\int_0^{10}\!\int_0^{\sqrt{100-x^2}}\!\int_0^{\sqrt{x^2+y^2}}\!\frac{1}{\sqrt{x^2+y^2}}\,dz\,dy\,dx$ .

no calculator, same as the exam
Answer $25\pi$ answer recomputed · sympy
1. The shadow $0\le x\le10$ , $0\le y\le\sqrt{100-x^2}$ is a quarter disk of radius $10$ : $0\le\theta\le\frac{\pi}{2}$ , $0\le r\le10$ .
2. The ceiling $z=\sqrt{x^2+y^2}$ is the cone $z=r$ ; the integrand $\frac{1}{\sqrt{x^2+y^2}}=\frac1r$ .
3. So the full cylindrical integrand is $\frac1r\cdot r=1$ — the Jacobian cancels the singularity. That is the point of the problem.
4. $\displaystyle\int_0^{\pi/2}\!\int_0^{10}\!\int_0^{r}1\,dz\,dr\,d\theta=\int_0^{\pi/2}\!\int_0^{10}r\,dr\,d\theta=\frac{\pi}{2}\cdot 50=25\pi$ .
TrapTwo errors cancel here if you are careless in exactly the wrong way. Dropping the Jacobian leaves $\int\int \frac1r\,\cdot r\,dz$ vs $\int\int\frac1r\,dz$ — the second gives $\int_0^{10}1\,dr=10$ and the answer $5\pi$ . 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 $z=x^2+y^2-9$ and $z=-2(x^2+y^2)$ using cylindrical coordinates.

no calculator, same as the exam
Answer $\tfrac{27\pi}{2}$ answer recomputed · sympy
1. Both surfaces depend only on $r^2=x^2+y^2$ : $z=r^2-9$ (an upward paraboloid dropped to $-9$ ) and $z=-2r^2$ (a downward one from the origin).
2. Which is on top? At $r=0$ : $-9$ versus $0$ , so $z=-2r^2$ is the ceiling and $z=r^2-9$ is the floor.
3. They meet where $r^2-9=-2r^2\Rightarrow 3r^2=9\Rightarrow r=\sqrt3$ . Full revolution, so $0\le\theta\le2\pi$ .
4. Height $=-2r^2-(r^2-9)=9-3r^2$ .
5. $\displaystyle V=\int_0^{2\pi}\!\int_0^{\sqrt3}(9-3r^2)\,r\,dr\,d\theta=2\pi\left[\frac{9r^2}{2}-\frac{3r^4}{4}\right]_0^{\sqrt3}=2\pi\left(\frac{27}{2}-\frac{27}{4}\right)=\frac{27\pi}{2}$ .
TrapAssuming the paraboloid opening upward is the ceiling because it 'goes up'. Over the whole region $r<\sqrt3$ it is the floor, and swapping them makes the volume negative. Test at $r=0$ every time. The other trap is integrating $9-3r^2$ against $dr$ instead of $r\,dr$ , which gives $2\pi(9\sqrt3-3\sqrt3)=12\sqrt3\,\pi$ — a suspiciously irrational volume for a surface of revolution with rational intersection radius.
5

Write the volume of the solid bounded by the cone $z^2=x^2+y^2$ between the planes $z=1$ and $z=2$ as a spherical-coordinates integral.

no calculator, same as the exam
Answer $\displaystyle\int_0^{2\pi}\!\int_0^{\pi/4}\!\int_{\sec\phi}^{2\sec\phi}\rho^2\sin\phi\,d\rho\,d\phi\,d\theta$ answer recomputed · sympy
1. In spherical, the cone $z^2=x^2+y^2$ with $z>0$ is the single surface $\phi=\frac{\pi}{4}$ , and the solid inside it is $0\le\phi\le\frac{\pi}{4}$ .
2. A horizontal plane is not a constant in spherical: $z=\rho\cos\phi$ , so $z=1$ becomes $\rho=\sec\phi$ and $z=2$ becomes $\rho=2\sec\phi$ .
3. Those are the $\rho$ -limits, and they depend on $\phi$ — so $\rho$ must be integrated first.
4. $dV=\rho^2\sin\phi\,d\rho\,d\phi\,d\theta$ , full revolution in $\theta$ :
5. $\displaystyle V=\int_0^{2\pi}\!\int_0^{\pi/4}\!\int_{\sec\phi}^{2\sec\phi}\rho^2\sin\phi\,d\rho\,d\phi\,d\theta$ .
6. Worth confirming in cylindrical, where it is a stack of disks of radius $z$ : $\displaystyle\int_0^{2\pi}\!\int_1^2\!\int_0^{z}r\,dr\,dz\,d\theta=\int_1^2\pi z^2\,dz=\frac{7\pi}{3}$ — and the spherical integral gives the same.
TrapWriting the planes as constant $\rho$ (say $\rho$ from $1$ to $2$ ), which describes the region between two spheres, not two planes — that is a spherical shell wedge, volume $\frac{2\pi}{3}(8-1)(1-\cos\frac\pi4)$ , not $\frac{7\pi}{3}$ . Any flat boundary not through the origin becomes $\rho=\text{(something)}\sec\phi$ and forces $d\rho$ innermost.

07

Triple Integrals in Spherical Coordinates

Lesson 23 · §16.5

Don't know this yet? §s165s in the full guide →

From the official study guide

Spherical Coordinates

x = \rho\sin\phi\cos\theta,\quad y = \rho\sin\phi\sin\theta,\quad z = \rho\cos\phi

\rho^2 = x^2+y^2+z^2, \quad \rho \ge 0,\quad 0 \le \phi \le \pi,\quad 0 \le \theta \le 2\pi

Triple Integral in Spherical Coordinates

\iiint_E f\,dV = \int_0^{2\pi}\int_0^{\phi_{\max}}\int_0^{\rho_{\max}} f\,\rho^2\sin\phi\,d\rho\,d\phi\,d\theta

Key: The Jacobian is $\rho^2\sin\phi$ . Don't forget it!

Useful Identities

Sphere: $\rho = a$
Cone: $\phi = \phi_0$ (half-angle from positive $z$ -axis)
$z = \rho\cos\phi$ , $\;r = \rho\sin\phi$

Practice

5 problems

1

The volume of the region inside $x^2+y^2+z^2=4$ and inside the cone $z=\sqrt{3x^2+3y^2}$ is $\displaystyle\int_0^{2\pi}\!\int_0^a\!\int_0^b c\,d\rho\,d\phi\,d\theta$ . What are $a$ , $b$ , and $c$ ?

no calculator, same as the exam
Answer $a=\pi/6,\;b=2,\;c=\rho^2\sin\phi$ answer recomputed · sympy
1. $c$ is the Jacobian and nothing else: $dV=\rho^2\sin\phi\,d\rho\,d\phi\,d\theta$ , so $c=\rho^2\sin\phi$ .
2. Sphere $x^2+y^2+z^2=4$ is $\rho=2$ , and since $\rho$ is innermost and starts at $0$ , $b=2$ .
3. The cone $z=\sqrt{3x^2+3y^2}=\sqrt3\,\sqrt{x^2+y^2}$ has $\tan\phi=\dfrac{\sqrt{x^2+y^2}}{z}=\dfrac{1}{\sqrt3}$ .
4. So $\phi=\frac{\pi}{6}$ , and 'inside the cone' means $0\le\phi\le\frac{\pi}{6}$ : $a=\frac{\pi}{6}$ .
5. $\displaystyle V=\int_0^{2\pi}\!\int_0^{\pi/6}\!\int_0^{2}\rho^2\sin\phi\,d\rho\,d\phi\,d\theta$ .
Trap $\tan\phi=\frac{r}{z}$ , not $\frac{z}{r}$ — $\phi$ is measured down from the positive $z$ -axis. Taking $\tan\phi=\sqrt3$ gives $\phi=\frac{\pi}{3}$ , which is the complementary wedge and roughly four times the volume. Second trap: writing $c=\rho\sin\phi$ , the *polar* Jacobian with a $\rho$ in it, which is dimensionally wrong for a volume.
2

Write the volume of the solid bounded by the cone $z=\sqrt{x^2+y^2}$ and the sphere $x^2+y^2+z^2=8$ as a spherical-coordinates integral.

no calculator, same as the exam
Answer $\displaystyle\int_0^{2\pi}\!\int_0^{\pi/4}\!\int_0^{\sqrt{8}}\rho^2\sin\phi\,d\rho\,d\phi\,d\theta$ answer recomputed · sympy
1. Cone $z=\sqrt{x^2+y^2}$ : here $\tan\phi=\frac{r}{z}=1$ , so it is the surface $\phi=\frac{\pi}{4}$ , and the solid inside it is $0\le\phi\le\frac{\pi}{4}$ .
2. Sphere $x^2+y^2+z^2=8$ is $\rho^2=8$ , i.e. $\rho=\sqrt8=2\sqrt2$ .
3. Both boundaries are coordinate surfaces in spherical — every limit is a constant, which is the whole reason to use it here.
4. $\displaystyle V=\int_0^{2\pi}\!\int_0^{\pi/4}\!\int_0^{\sqrt8}\rho^2\sin\phi\,d\rho\,d\phi\,d\theta$ .
5. Value: $2\pi\cdot\left(1-\tfrac{\sqrt2}{2}\right)\cdot\frac{(\sqrt8)^3}{3}=\frac{32\pi}{3}\left(\sqrt2-1\right)$ , which the cylindrical set-up $\int_0^{2\pi}\!\int_0^2\left(\sqrt{8-r^2}-r\right)r\,dr\,d\theta$ confirms.
TrapReading $\rho=8$ off $x^2+y^2+z^2=8$ . The equation gives $\rho^2=8$ , so $\rho=2\sqrt2$ — and cubing the wrong one inflates the volume by a factor of $8^3/8^{3/2}\approx22.6$ . 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.

no calculator, same as the exam
Answer $\displaystyle\int_0^{2\pi}\!\int_0^{\pi/3}\!\int_2^{4\cos\phi}\rho^2\sin\phi\,d\rho\,d\phi\,d\theta$ answer recomputed · sympy
1. Put both spheres in spherical form. Origin-centred radius $2$ : $\rho=2$ .
2. The one centred at $(0,0,2)$ : $x^2+y^2+(z-2)^2=4\Rightarrow \rho^2-4\rho\cos\phi=0\Rightarrow\rho=4\cos\phi$ .
3. 'Inside the shifted sphere, outside the origin one' is exactly $2\le\rho\le4\cos\phi$ — the inner radius is the constant, the outer is the $\phi$ -dependent one.
4. Those limits are only valid where $4\cos\phi\ge2$ , i.e. $\cos\phi\ge\frac12$ , i.e. $\phi\le\frac{\pi}{3}$ . That is the $\phi$ -limit.
5. $\displaystyle V=\int_0^{2\pi}\!\int_0^{\pi/3}\!\int_2^{4\cos\phi}\rho^2\sin\phi\,d\rho\,d\phi\,d\theta$ .
6. Check it independently: the lens where two radius- $2$ spheres with centres $2$ apart overlap has volume $\frac{\pi(4R+d)(2R-d)^2}{12}=\frac{10\pi}{3}$ , so the answer is $\frac{32\pi}{3}-\frac{10\pi}{3}=\frac{22\pi}{3}$ — and the integral above gives $\frac{22\pi}{3}$ .
TrapTaking $\phi$ up to $\frac{\pi}{2}$ because the shifted sphere 'sits above the $xy$ -plane'. Past $\phi=\frac{\pi}{3}$ the outer limit $4\cos\phi$ drops below the inner limit $2$ , so the integral starts subtracting. The upper $\phi$ is always where the two $\rho$ -limits meet: solve $4\cos\phi=2$ .
4

Converting $\displaystyle\int_{-1}^{1}\!\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}\!\int_{\sqrt{3(x^2+y^2)}}^{\sqrt{4-x^2-y^2}} f\,dz\,dy\,dx$ to spherical gives $\displaystyle\int_0^a\!\int_0^b\!\int_0^c f\cdot\rho^2\sin\phi\,d\rho\,d\phi\,d\theta$ (order: $d\rho\,d\phi\,d\theta$ , so $a$ is the upper limit on $\theta$ , $b$ on $\phi$ , $c$ on $\rho$ ). Find $\dfrac{bc}{a}$ .

no calculator, same as the exam
Answer $\tfrac{1}{6}$ ( $a=2\pi,\;b=\tfrac{\pi}{6},\;c=2$ )answer recomputed · sympy
1. The shadow $-1\le x\le1$ , $|y|\le\sqrt{1-x^2}$ is the full unit disk, so $\theta$ makes a full turn: $a=2\pi$ .
2. Ceiling $z=\sqrt{4-x^2-y^2}$ is the sphere $\rho=2$ , so $c=2$ .
3. Floor $z=\sqrt{3(x^2+y^2)}=\sqrt3\,r$ is a cone with $\tan\phi=\frac{r}{z}=\frac{1}{\sqrt3}$ , so $\phi$ runs from $0$ to $\frac{\pi}{6}$ : $b=\frac{\pi}{6}$ .
4. Consistency check on the shadow: the cone meets the sphere where $\rho=2$ , $\phi=\frac\pi6$ , giving $r=2\sin\frac\pi6=1$ — exactly the unit disk in the stem. Good.
5. $\dfrac{bc}{a}=\dfrac{\frac{\pi}{6}\cdot 2}{2\pi}=\dfrac{1}{6}$ .
TrapSetting $b=\frac{\pi}{3}$ from $\tan\phi=\sqrt3$ . $\phi$ is the angle from the $z$ -axis, and a steep cone ( $z$ large compared to $r$ ) means a small $\phi$ . The disk-radius check catches it: $\phi=\frac\pi3$ would give shadow radius $2\sin\frac\pi3=\sqrt3\ne1$ .
5

Find the volume of the solid region enclosed by the surface $\rho=12\cos\phi$ .

no calculator, same as the exam
Answer $288\pi$ answer recomputed · sympy
1. $\rho=12\cos\phi$ is already a sphere in disguise: multiply by $\rho$ to get $\rho^2=12\rho\cos\phi$ , i.e. $x^2+y^2+z^2=12z$ .
2. Complete the square: $x^2+y^2+(z-6)^2=36$ — radius $6$ , centred at $(0,0,6)$ .
3. So the answer must be $\frac43\pi(6)^3=288\pi$ ; now get the same thing from the integral.
4. The surface closes up at $\phi=\frac{\pi}{2}$ (where $\rho=0$ ), so $\phi$ runs $0$ to $\frac{\pi}{2}$ — not to $\pi$ .
5. $\displaystyle V=\int_0^{2\pi}\!\int_0^{\pi/2}\!\int_0^{12\cos\phi}\rho^2\sin\phi\,d\rho\,d\phi\,d\theta=\int_0^{2\pi}\!\int_0^{\pi/2}\frac{1728\cos^3\phi}{3}\sin\phi\,d\phi\,d\theta=2\pi\cdot 576\cdot\frac14=288\pi$ .
TrapLetting $\phi$ run to $\pi$ out of habit. For $\phi>\frac{\pi}{2}$ , $12\cos\phi<0$ and there is no sphere there; the extra range adds another $288\pi$ of nonsense (the $\cos^3\phi\sin\phi$ integrand is odd about $\frac\pi2$ , so it actually cancels to $0$ — an error that can *silently* delete the whole answer). The sphere is fully traced by $\phi\in[0,\frac\pi2]$ .

08

Integrals in Mass Calculations

Lesson 24 · §16.6

Don't know this yet? §s166 in the full guide →

From the official study guide

Mass, Moments, and Center of Mass

For a lamina (2D) with density $\delta(x,y)$ :

m = \iint_D \delta(x,y)\,dA

\bar{x} = \frac{1}{m}\iint_D x\,\delta(x,y)\,dA, \qquad \bar{y} = \frac{1}{m}\iint_D y\,\delta(x,y)\,dA

3D Solid

For a solid $E$ with density $\delta(x,y,z)$ :

m = \iiint_E \delta(x,y,z)\,dV

\bar{x} = \frac{1}{m}\iiint_E x\,\delta\,dV,\quad \bar{y} = \frac{1}{m}\iiint_E y\,\delta\,dV,\quad \bar{z} = \frac{1}{m}\iiint_E z\,\delta\,dV

Moments of Inertia

I_x = \iint_D y^2\,\delta\,dA,\quad I_y = \iint_D x^2\,\delta\,dA,\quad I_0 = \iint_D (x^2+y^2)\,\delta\,dA

Practice

5 problems

1

Find the center of mass of the rectangle $\{(x,y)\mid 0\le x\le 2,\;0\le y\le 4\}$ with density $\rho(x,y)=1+y$ .

no calculator, same as the exam
Answer $\left(1,\;\tfrac{22}{9}\right)$ answer recomputed · sympy
1. Mass first: $\displaystyle m=\int_0^2\!\int_0^4(1+y)\,dy\,dx=\int_0^2\left[y+\tfrac{y^2}{2}\right]_0^4dx=\int_0^2 12\,dx=24$ .
2. $\bar x$ : the density does not depend on $x$ and the region is symmetric in $x$ , so $\bar x=1$ by inspection. Formally $M_y=\int_0^2\!\int_0^4 x(1+y)\,dy\,dx=\left(\int_0^2x\,dx\right)(12)=24$ , and $\bar x=\frac{24}{24}=1$ .
3. $\bar y$ : $\displaystyle M_x=\int_0^2\!\int_0^4 y(1+y)\,dy\,dx=2\int_0^4(y+y^2)\,dy=2\left(8+\tfrac{64}{3}\right)=\tfrac{176}{3}$ .
4. $\bar y=\dfrac{M_x}{m}=\dfrac{176/3}{24}=\dfrac{22}{9}\approx2.44$ .
5. Sanity: $\frac{22}{9}>2$ , and it should be — the density grows with $y$ , so mass piles up toward the top.
6. Centre of mass $\left(1,\tfrac{22}{9}\right)$ .
TrapSwapping the moments. $M_x=\iint y\,\delta\,dA$ gives $\bar y$ , not $\bar x$ — the subscript names the axis the moment is about, so it pairs with the *other* coordinate. Using $M_x$ for $\bar x$ here returns $\frac{22}{9}$ for a rectangle only $2$ wide, which is impossible and is the fastest way to catch it.
2

A cube $0\le x,y,z\le 1$ has density $\delta=x+yz$ . Find the $x$ -coordinate of the center of mass, given $\iiint_{\text{cube}}\delta\,dV=\tfrac{3}{4}$ .

no calculator, same as the exam
Answer $\tfrac{11}{18}$ answer recomputed · sympy
1. $\bar x=\dfrac{M_{yz}}{m}$ with $M_{yz}=\displaystyle\iiint x\,\delta\,dV$ ; the mass $m=\tfrac34$ is handed to you.
2. $\displaystyle M_{yz}=\int_0^1\!\int_0^1\!\int_0^1 x(x+yz)\,dz\,dy\,dx=\int\!\!\int\!\!\int\left(x^2+xyz\right)dV$ .
3. Both pieces separate over the cube: $\displaystyle\iiint x^2\,dV=\tfrac13$ and $\displaystyle\iiint xyz\,dV=\left(\tfrac12\right)^3=\tfrac18$ .
4. $M_{yz}=\tfrac13+\tfrac18=\tfrac{11}{24}$ .
5. $\bar x=\dfrac{11/24}{3/4}=\dfrac{11}{24}\cdot\dfrac43=\dfrac{11}{18}$ .
6. Sanity: $\frac{11}{18}>\frac12$ , correct — the $x$ term in the density weights the far side.
TrapMultiplying only the first term by $x$ : $\int x\cdot x\,dV+\int yz\,dV=\tfrac13+\tfrac14$ gives $\tfrac{7}{12}$ and then $\bar x=\tfrac79$ . 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 $y=x$ , $y=\tfrac{1}{2}x$ , $x=1$ with density $\delta(x,y)=2x$ . (Given: $M_x=\iint y\delta\,dA$ , $M_y=\iint x\delta\,dA$ .)

no calculator, same as the exam
Answer $\tfrac{3}{4}$ answer recomputed · sympy
1. Sketch the wedge: between $y=\tfrac12x$ (lower) and $y=x$ (upper), closed off at $x=1$ . They meet at the origin, so $0\le x\le1$ and $\tfrac{x}{2}\le y\le x$ .
2. Vertical strips are the right choice — each strip has height $x-\tfrac{x}{2}=\tfrac{x}{2}$ .
3. $\displaystyle m=\int_0^1\!\int_{x/2}^{x}2x\,dy\,dx=\int_0^1 2x\cdot\frac{x}{2}\,dx=\int_0^1x^2\,dx=\frac13$ .
4. $\displaystyle M_y=\int_0^1\!\int_{x/2}^{x}x\cdot 2x\,dy\,dx=\int_0^1 2x^2\cdot\frac{x}{2}\,dx=\int_0^1 x^3\,dx=\frac14$ .
5. $\bar x=\dfrac{M_y}{m}=\dfrac{1/4}{1/3}=\dfrac34$ .
6. Sanity: $\tfrac34$ 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 $dx\,dy$ without splitting. Horizontal strips run from $x=y$ to $x=2y$ only while $2y\le1$ ; above $y=\tfrac12$ 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 $\rho(x,y,z)=2z$ .

no calculator, same as the exam
Answer $\tfrac{8}{3}$ answer recomputed · sympy
1. Three of the four corners are on the axes, so use the intercept form of the slanted face: $\dfrac{x}{1}+\dfrac{y}{2}+\dfrac{z}{4}=1$ .
2. Solve for the ceiling: $z=4\left(1-x-\dfrac{y}{2}\right)=4-4x-2y$ .
3. Shadow: set $z=0$ to get $x+\frac{y}{2}=1$ , i.e. $y=2(1-x)$ , with $0\le x\le1$ .
4. $\displaystyle m=\int_0^1\!\int_0^{2(1-x)}\!\int_0^{4-4x-2y}2z\,dz\,dy\,dx$ .
5. Inner: $\left[z^2\right]_0^{4-4x-2y}=(4-4x-2y)^2$ .
6. $\displaystyle\int_0^{2(1-x)}(4-4x-2y)^2dy=\left[-\frac{(4-4x-2y)^3}{6}\right]_0^{2(1-x)}=\frac{64(1-x)^3}{6}$ .
7. $\displaystyle m=\frac{64}{6}\int_0^1(1-x)^3dx=\frac{64}{6}\cdot\frac14=\frac{8}{3}$ .
TrapTreating a variable density as a constant and reaching for $\rho\cdot V$ : the volume is $\frac{1\cdot2\cdot4}{6}=\frac43$ , and $2\bar z\cdot V$ is *not* something you can shortcut without already knowing $\bar z$ . Also: $\rho$ 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 $\rho(x,y)=xy$ occupies the region bounded by $y=x^2$ , $y=1$ , $x=0$ . The mass is $\tfrac{1}{6}$ . Find the $y$ -coordinate of the center of mass.

no calculator, same as the exam
Answer $\tfrac{3}{4}$ answer recomputed · sympy
1. Region: $y=x^2$ below, $y=1$ above, $x=0$ on the left — so $0\le x\le1$ and $x^2\le y\le1$ .
2. The mass is given as $\tfrac16$ ; confirm it cheaply: $\displaystyle\int_0^1\!\int_{x^2}^{1}xy\,dy\,dx=\int_0^1\frac{x(1-x^4)}{2}dx=\frac12\left(\frac12-\frac16\right)=\frac16$ . Good.
3. $\displaystyle M_x=\int_0^1\!\int_{x^2}^{1}y\cdot xy\,dy\,dx=\int_0^1 x\cdot\frac{1-x^6}{3}\,dx$ .
4. $=\dfrac13\left(\dfrac12-\dfrac18\right)=\dfrac13\cdot\dfrac38=\dfrac18$ .
5. $\bar y=\dfrac{M_x}{m}=\dfrac{1/8}{1/6}=\dfrac{6}{8}=\dfrac34$ .
6. Sanity: the region hugs the top (it is wide near $y=1$ , pinched near $y=0$ ) and the density $xy$ favours large $y$ , so $\bar y>\tfrac12$ is right.
TrapTaking the region as $0\le y\le x^2$ — 'bounded by $y=x^2$ ' 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 $\bar y$ to $\tfrac{5}{7}$ -ish territory and, more tellingly, changes the mass away from the given $\tfrac16$ — which is exactly why that value was handed to you.

09

Vector Fields

Lesson 25 · §17.1

Don't know this yet? §s171 in the full guide →

From the official study guide

Definition

A vector field assigns a vector to each point in a region:

\vec{F}(x,y) = P(x,y)\,\vec{i} + Q(x,y)\,\vec{j} = \langle P, Q \rangle

\vec{F}(x,y,z) = P\,\vec{i} + Q\,\vec{j} + R\,\vec{k} = \langle P, Q, R \rangle

Conservative Vector Fields

A vector field $\vec{F}$ is conservative if there exists a scalar function $f$ (the potential function) such that:

\vec{F} = \nabla f \qquad \Longleftrightarrow \qquad P = f_x,\; Q = f_y \; (\text{and }R = f_z)

Test for Conservative Field in 2D

\vec{F} = \langle P, Q \rangle \text{ is conservative} \iff \frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}

Practice

4 problems

1

Consider $\vec{F}=\langle e^{yz}-y\sin(xy),\;zxe^{yz}-x\sin(xy),\;xye^{yz}\rangle$ . Find a potential function $\varphi$ if one exists.

no calculator, same as the exam
Answer $xe^{yz}+\cos(xy)$ answer recomputed · sympy
1. We want $\varphi$ with $\nabla\varphi=\vec F$ , so start with the component that is easiest to antidifferentiate and then patch it up.
2. Integrate the first component in $x$ , holding $y,z$ fixed: $\varphi=\displaystyle\int\big(e^{yz}-y\sin(xy)\big)\,dx = xe^{yz}+\cos(xy)+g(y,z)$ .
3. The sign is the whole game here: $\dfrac{d}{dx}\cos(xy)=-y\sin(xy)$ , so the antiderivative of $-y\sin(xy)$ is $+\cos(xy)$ .
4. Now differentiate that candidate in $y$ : $\varphi_y = xze^{yz}-x\sin(xy)+g_y$ .
5. Compare with the given second component $zxe^{yz}-x\sin(xy)$ : they already match, so $g_y=0$ .
6. Differentiate in $z$ : $\varphi_z = xye^{yz}+g_z$ , and the third component is $xye^{yz}$ , so $g_z=0$ too. Hence $g$ is a constant.
7. $\varphi = xe^{yz}+\cos(xy)$ (any additive constant is also a potential).
TrapWriting $\int -y\sin(xy)\,dx = -\cos(xy)$ . 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 $\varphi_x$ check if you also flip the sign when differentiating.
2

Let $a\ne0$ and $C$ be the curve $x^2-2y^2=a$ . Which $\vec{F}$ is orthogonal to the tangent of $C$ at every point $(x_0,y_0)$ on $C$ ?

no calculator, same as the exam
Answer $\langle 2x,-4y\rangle$ answer recomputed · sympy
1. $C$ is the level curve $g(x,y)=a$ of $g(x,y)=x^2-2y^2$ .
2. Key fact: the gradient of $g$ is normal to every level curve of $g$ , hence orthogonal to the tangent of $C$ at each point.
3. $\nabla g=\langle g_x,g_y\rangle = \langle 2x,\,-4y\rangle$ .
4. So $\vec F=\langle 2x,-4y\rangle$ is orthogonal to $\vec T$ at every $(x_0,y_0)$ on $C$ (and so is any scalar multiple of it).
5. Sanity check by hand: implicit differentiation of $x^2-2y^2=a$ gives $2x-4y\,y'=0$ , so $\vec T\parallel\langle 4y,2x\rangle$ , and $\langle 2x,-4y\rangle\cdot\langle 4y,2x\rangle = 8xy-8xy=0$ .
TrapDropping the minus sign and answering $\langle 2x,4y\rangle$ — that is $\nabla(x^2+2y^2)$ , the normal to an ellipse, not to this hyperbola. The other classic miss is handing back the tangent $\langle 4y,2x\rangle$ 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 $\langle -y,x\rangle$ answer recomputed · sympy
1. Counterclockwise rotation means the field is tangent to circles centered at the origin, pointing in the direction of increasing angle.
2. Parametrize a circle the counterclockwise way: $\vec r(t)=\langle\cos t,\sin t\rangle$ .
3. The direction of travel is $\vec r\,{}'(t)=\langle-\sin t,\cos t\rangle$ .
4. Rewrite that at the point itself, using $x=\cos t$ , $y=\sin t$ : the tangent direction is $\langle -y,\,x\rangle$ .
5. Spot-check the description: above the $x$ -axis take $(0,1)$ , giving $\langle-1,0\rangle$ (points left); below take $(0,-1)$ , giving $\langle1,0\rangle$ (points right). Matches.
6. $\vec F=\langle -y,\,x\rangle$ .
TrapAnswering $\langle y,-x\rangle$ , which is the same picture run clockwise. Always test one concrete point above the $x$ -axis rather than trusting the sign pattern; $\langle y,-x\rangle$ gives $\langle1,0\rangle$ at $(0,1)$ , pointing right, which is the wrong rotation.
4

Suppose $f(x,y,z)=xy^2z^3$ . Find $\operatorname{div}(\nabla f)$ at the point $(2,-1,1)$ .

no calculator, same as the exam
Answer16answer recomputed · sympy
1. $\operatorname{div}(\nabla f)$ is the Laplacian $f_{xx}+f_{yy}+f_{zz}$ — a scalar, not a vector.
2. $\nabla f=\langle y^2z^3,\;2xyz^3,\;3xy^2z^2\rangle$ .
3. Take the matching partial of each component: $f_{xx}=\partial_x(y^2z^3)=0$ , $f_{yy}=\partial_y(2xyz^3)=2xz^3$ , $f_{zz}=\partial_z(3xy^2z^2)=6xy^2z$ .
4. So $\operatorname{div}(\nabla f)=0+2xz^3+6xy^2z$ .
5. 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 $\partial_x$ of the first component, $\partial_y$ of the second, $\partial_z$ of the third, not all three of one component. Second, $(-1)^2=+1$ : plugging $y=-1$ into $6xy^2z$ gives $+12$ , and writing $-12$ turns $16$ into $-8$ .

10

Line Integrals of Functions

Lesson 26 · §17.2

Don't know this yet? §s172 in the full guide →

From the official study guide

Line Integral of a Scalar Function

For curve $C$ parametrized by $\vec{r}(t)$ , $a \le t \le b$ :

\int_C f(x,y,z)\,ds = \int_a^b f(\vec{r}(t))\,|\vec{r}'(t)|\,dt

Interpretations

Arc length: $\displaystyle\int_C ds = \int_a^b |\vec{r}'(t)|\,dt$
Mass of a wire with linear density $\delta$ : $m = \displaystyle\int_C \delta\,ds$

Practice

5 problems

1

Compute $\displaystyle\int_C f(x,y)\,ds$ where $f(x,y)=\sin(2\pi x)+xy$ and $C\colon\vec{r}(t)=\langle t,t\rangle$ , $0\le t\le1$ .

no calculator, same as the exam
Answer $\tfrac{\sqrt{2}}{3}$ answer recomputed · sympy
1. Scalar line integral: $\displaystyle\int_C f\,ds=\int_a^b f(\vec r(t))\,|\vec r\,{}'(t)|\,dt$ .
2. Here $\vec r\,{}'(t)=\langle1,1\rangle$ , so $|\vec r\,{}'(t)|=\sqrt{2}$ and $ds=\sqrt2\,dt$ .
3. On $C$ we have $x=t$ , $y=t$ , so $f(\vec r(t))=\sin(2\pi t)+t^2$ .
4. Set up: $\displaystyle\int_0^1\big(\sin(2\pi t)+t^2\big)\sqrt2\,dt$ .
5. $\displaystyle\int_0^1\sin(2\pi t)\,dt=\Big[-\frac{\cos(2\pi t)}{2\pi}\Big]_0^1=-\frac{1}{2\pi}+\frac{1}{2\pi}=0$ — a full period integrates away.
6. $\displaystyle\int_0^1 t^2\,dt=\frac13$ , so the value is $\sqrt2\cdot\frac13=\frac{\sqrt2}{3}$ .
TrapForgetting the $|\vec r\,{}'|=\sqrt2$ factor and reporting $\tfrac13$ . The parametrization $\langle t,t\rangle$ is not unit speed even though it looks tame — $ds\ne dt$ here.
2

Let $C$ be the helix $\vec{r}(t)=\langle\cos t,\sin t,t\rangle$ , $0\le t\le\pi/2$ . Compute $\displaystyle\int_C xy\,ds$ .

no calculator, same as the exam
Answer $\tfrac{\sqrt{2}}{2}$ answer recomputed · sympy
1. $\vec r\,{}'(t)=\langle-\sin t,\cos t,1\rangle$ , so $|\vec r\,{}'(t)|=\sqrt{\sin^2t+\cos^2t+1}=\sqrt2$ .
2. The helix has constant speed $\sqrt2$ , so $ds=\sqrt2\,dt$ .
3. On $C$ , $xy=\cos t\sin t$ .
4. Set up: $\displaystyle\int_0^{\pi/2}\cos t\,\sin t\,\sqrt2\,dt$ .
5. Substitute $u=\sin t$ : $\displaystyle\sqrt2\int_0^1 u\,du=\sqrt2\cdot\frac12=\frac{\sqrt2}{2}$ .
TrapReading the first two components as a unit circle and concluding $|\vec r\,{}'|=1$ . The $z=t$ component contributes a $1$ under the radical, making the speed $\sqrt2$ ; dropping it halves-by- $\sqrt2$ the answer to $\tfrac12$ .
3

Calculate $\displaystyle\int_C\frac{1}{(x-y)^2}\,ds$ where $C$ is the straight line segment from $(1,0)$ to $(5,3)$ .

no calculator, same as the exam
Answer $\tfrac{5}{2}$ answer recomputed · sympy
1. Parametrize the segment from $(1,0)$ to $(5,3)$ as $\vec r(t)=\langle1,0\rangle+t\langle4,3\rangle=\langle1+4t,\;3t\rangle$ , $0\le t\le1$ .
2. $\vec r\,{}'(t)=\langle4,3\rangle$ , so $|\vec r\,{}'(t)|=\sqrt{16+9}=5$ and $ds=5\,dt$ (the length of the segment, spread over $t\in[0,1]$ ).
3. The integrand simplifies beautifully: $x-y=(1+4t)-3t=1+t$ , so $\dfrac{1}{(x-y)^2}=\dfrac{1}{(1+t)^2}$ .
4. Set up: $\displaystyle\int_0^1\frac{5}{(1+t)^2}\,dt$ .
5. $=5\Big[-\frac{1}{1+t}\Big]_0^1 = 5\Big(-\frac12+1\Big)=\frac52$ .
TrapUsing $dx$ instead of $ds$ . If you parametrize by $x$ with $y=\tfrac34(x-1)$ you must carry $ds=\sqrt{1+(3/4)^2}\,dx=\tfrac54\,dx$ ; dropping that factor gives $2$ instead of $\tfrac52$ . Also note the line never crosses $x=y$ , so the integrand stays finite.
4

Let $C$ be the half-circle $x^2+y^2=4$ with $x\ge0$ . Compute $\displaystyle\int_C x\,ds$ .

no calculator, same as the exam
Answer $8$ answer recomputed · sympy
1. The circle has radius $2$ , so $\vec r(t)=\langle2\cos t,\,2\sin t\rangle$ .
2. Pick the range that gives $x\ge0$ , not $y\ge0$ : $x=2\cos t\ge0$ forces $-\tfrac\pi2\le t\le\tfrac\pi2$ (the right half).
3. $\vec r\,{}'(t)=\langle-2\sin t,2\cos t\rangle$ , so $|\vec r\,{}'(t)|=2$ and $ds=2\,dt$ .
4. Set up: $\displaystyle\int_{-\pi/2}^{\pi/2}(2\cos t)(2)\,dt=4\int_{-\pi/2}^{\pi/2}\cos t\,dt$ .
5. $=4\big[\sin t\big]_{-\pi/2}^{\pi/2}=4(1-(-1))=8$ .
TrapTaking $0\le t\le\pi$ out of habit. That traces the upper half ( $y\ge0$ ), over which $\int x\,ds=0$ by symmetry — a suspiciously clean wrong answer. The condition is $x\ge0$ , so center the interval on $t=0$ .
5

Find the arc length of $\vec{r}(t) = \langle 4\cos t,\, 4\sin t,\, 3t\rangle$ for $0\le t\le 2\pi$ .

no calculator, same as the exam
Answer $10\pi$ answer recomputed · sympy
1. Arc length is $\displaystyle L=\int_a^b|\vec r\,{}'(t)|\,dt$ .
2. $\vec r\,{}'(t)=\langle-4\sin t,\,4\cos t,\,3\rangle$ .
3. $|\vec r\,{}'(t)|=\sqrt{16\sin^2t+16\cos^2t+9}=\sqrt{16+9}=5$ — constant speed, the hallmark of a circular helix.
4. Set up: $\displaystyle L=\int_0^{2\pi}5\,dt=5(2\pi)=10\pi$ .
TrapAdding the radius and the pitch coordinate as if speed were $4+3=7$ , or forgetting the $3$ entirely and getting $8\pi$ (the circumference of the shadow circle). The helix is strictly longer than its projection because of the climb.

11

Line Integrals of Vector Fields

Lesson 27 · §17.2

Don't know this yet? §s172 in the full guide →

From the official study guide

Line Integral of a Vector Field (Work)

The work done by $\vec{F}$ along curve $C$ :

\int_C \vec{F}\cdot d\vec{r} = \int_a^b \vec{F}(\vec{r}(t))\cdot\vec{r}'(t)\,dt = \int_C P\,dx + Q\,dy + R\,dz

Key Properties

Reversing orientation changes the sign: $\displaystyle\int_{-C}\vec{F}\cdot d\vec{r} = -\int_C \vec{F}\cdot d\vec{r}$
Linearity: $\displaystyle\int_C (a\vec{F}+b\vec{G})\cdot d\vec{r} = a\int_C\vec{F}\cdot d\vec{r} + b\int_C\vec{G}\cdot d\vec{r}$

Practice

5 problems

1

Given force field $\vec{F}(x,y)=\langle-y,x\rangle$ , find the work to move an object along the ellipse $\vec{r}(t)=\langle2\cos t,3\sin t\rangle$ from $(2,0)$ to $(0,3)$ .

no calculator, same as the exam
Answer $3\pi$ answer recomputed · sympy
1. Work is $\displaystyle W=\int_C\vec F\cdot d\vec r=\int_a^b\vec F(\vec r(t))\cdot\vec r\,{}'(t)\,dt$ .
2. Find the $t$ -range from the endpoints: $(2,0)$ is $t=0$ and $(0,3)$ is $t=\tfrac\pi2$ , so we travel a quarter of the ellipse.
3. On $C$ : $\vec F(\vec r(t))=\langle-y,x\rangle=\langle-3\sin t,\,2\cos t\rangle$ and $\vec r\,{}'(t)=\langle-2\sin t,\,3\cos t\rangle$ .
4. Dot them: $6\sin^2t+6\cos^2t=6$ — the integrand is constant.
5. Set up: $\displaystyle W=\int_0^{\pi/2}6\,dt=6\cdot\frac{\pi}{2}=3\pi$ .
TrapReaching for Green's theorem. $C$ is an open quarter-arc, not a closed curve, so Green does not apply; and $\vec F=\langle-y,x\rangle$ is not conservative ( $P_y=-1\ne1=Q_x$ ), so the path genuinely matters and FTLI is off the table. You must integrate along this arc.
2

Compute the flux of $\vec{F}(x,y)=\langle x,y\rangle$ across the curve $\vec{r}(t)=\langle3\cos t,2\sin t\rangle$ , $-\pi\le t\le\pi$ . (Flux $=\displaystyle\int_C P\,dy-Q\,dx$ .)

no calculator, same as the exam
Answer $12\pi$ answer recomputed · sympy
1. Flux uses the stated formula $\displaystyle\int_C P\,dy-Q\,dx$ with $P=x$ , $Q=y$ — note $dy$ pairs with $P$ , not with $Q$ .
2. From $\vec r(t)=\langle3\cos t,2\sin t\rangle$ : $dx=-3\sin t\,dt$ and $dy=2\cos t\,dt$ .
3. Substitute: $P\,dy-Q\,dx=(3\cos t)(2\cos t)\,dt-(2\sin t)(-3\sin t)\,dt$ .
4. $=6\cos^2t+6\sin^2t=6$ , constant again.
5. Set up: $\displaystyle\int_{-\pi}^{\pi}6\,dt=6(2\pi)=12\pi$ .
6. Cross-check with the divergence form: $\operatorname{div}\vec F=2$ and the ellipse encloses area $\pi(3)(2)=6\pi$ , so the flux is $2\cdot6\pi=12\pi$ .
TrapComputing circulation $\int P\,dx+Q\,dy$ instead of flux. With this field that swap gives $\int(-6\sin t\cos t+6\sin t\cos t)\,dt=0$ , so the mistake produces a clean $0$ and looks deliberate. Keep the pairing $P\,dy-Q\,dx$ straight.
3

Given force field $\vec{F}(x,y,z)=\langle y,z,x\rangle$ , 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
Answer $\tfrac{26}{2}$ answer recomputed · sympy
1. Parametrize the segment: $\vec r(t)=\langle2t,3t,4t\rangle$ , $0\le t\le1$ , so $\vec r\,{}'(t)=\langle2,3,4\rangle$ .
2. Evaluate the field on the curve, remembering the components are shifted: $\vec F=\langle y,z,x\rangle=\langle3t,\,4t,\,2t\rangle$ .
3. Dot: $\vec F\cdot\vec r\,{}'=6t+12t+8t=26t$ .
4. Set up: $\displaystyle W=\int_0^1 26t\,dt=26\cdot\frac12=\frac{26}{2}=13$ .
TrapSubstituting $\vec F=\langle2t,3t,4t\rangle$ , i.e. reading $\langle y,z,x\rangle$ as $\langle x,y,z\rangle$ . That gives $\int_0^1 29t\,dt$ and the wrong number. The cyclic shift is the entire content of this problem. (Also: $\vec F=\langle y,z,x\rangle$ is not conservative, since $P_y=1$ but $Q_x=0$ , so you must integrate.)
4

Find $\displaystyle\int_C\vec{F}\cdot\vec{T}\,ds$ where $\vec{F}(x,y,z)=\langle ye^z,e^y+xe^z,xye^z\rangle$ on a smooth curve from $(0,0,0)$ to $(-1,1,1)$ .

no calculator, same as the exam
Answer $-1$ answer recomputed · sympy
1. The problem says 'a smooth curve' without naming it, which is the signal that $\vec F$ must be conservative — otherwise the answer would not be well defined.
2. Hunt for $\varphi$ : integrate the first component in $x$ , $\displaystyle\int ye^z\,dx = xye^z+g(y,z)$ .
3. Match the second component: $\varphi_y=xe^z+g_y$ must equal $e^y+xe^z$ , so $g_y=e^y$ and $g=e^y+h(z)$ .
4. Match the third: $\varphi_z=xye^z+h'(z)$ must equal $xye^z$ , so $h$ is constant. Take $\varphi=xye^z+e^y$ .
5. FTLI: $\displaystyle\int_C\vec F\cdot\vec T\,ds=\int_C\vec F\cdot d\vec r=\varphi(-1,1,1)-\varphi(0,0,0)$ .
6. $\varphi(-1,1,1)=(-1)(1)e^1+e^1=-e+e=0$ and $\varphi(0,0,0)=0+e^0=1$ .
7. Value $=0-1=-1$ .
TrapEvaluating $\varphi$ at the start minus the end. FTLI is end minus start, and here that sign error hands you $+1$ . Second trap: forgetting the $e^y$ piece because it does not appear in $\varphi_x$ — it is exactly what makes $\varphi(0,0,0)=1$ rather than $0$ .
5

Evaluate $\displaystyle\int_C\vec{F}\cdot d\vec{r}$ where $\vec{F}(x,y,z)=\langle2xy-yz,\;x^2-xz,\;-xy\rangle$ on a smooth curve from $(2,1,0)$ to $(3,2,-1)$ .

no calculator, same as the exam
Answer20answer recomputed · sympy
1. Again the curve is unspecified, so test for conservativity: with $P=2xy-yz$ , $Q=x^2-xz$ , $R=-xy$ we get $P_y=2x-z=Q_x$ , $P_z=-y=R_x$ , $Q_z=-x=R_y$ . All three match, so $\vec F$ is conservative.
2. Build $\varphi$ : $\displaystyle\int(2xy-yz)\,dx = x^2y-xyz+g(y,z)$ .
3. Check $\varphi_y=x^2-xz+g_y$ against $Q=x^2-xz$ : $g_y=0$ . Check $\varphi_z=-xy+g_z$ against $R=-xy$ : $g_z=0$ . So $\varphi=x^2y-xyz$ .
4. FTLI: value $=\varphi(3,2,-1)-\varphi(2,1,0)$ .
5. $\varphi(3,2,-1)=9(2)-(3)(2)(-1)=18+6=24$ .
6. $\varphi(2,1,0)=4(1)-0=4$ .
7. Value $=24-4=20$ .
TrapSign-slipping on $-xyz$ at the endpoint with $z=-1$ : $-(3)(2)(-1)=+6$ , so $\varphi(3,2,-1)=24$ , not $12$ . Writing $18-6$ gives $8$ and is the single most common wrong answer here.

12

Fundamental Theorem of Line Integrals

Lesson 28 · §17.3

Don't know this yet? §s173 in the full guide →

From the official study guide

Fundamental Theorem of Line Integrals

If $\vec{F} = \nabla f$ (conservative) and $C$ goes from $A$ to $B$ :

\int_C \nabla f \cdot d\vec{r} = f(B) - f(A)

Independence of Path

$\displaystyle\int_C \vec{F}\cdot d\vec{r}$ is path-independent if and only if $\vec{F}$ is conservative.

Closed Curves

\oint_C \vec{F}\cdot d\vec{r} = 0 \iff \vec{F} \text{ is conservative}

Finding the Potential Function

Given $\vec{F} = \langle P, Q \rangle$ is conservative:

Integrate $P$ with respect to $x$ to get $f(x,y) = \int P\,dx + g(y)$
Differentiate with respect to $y$ and set equal to $Q$ to find $g'(y)$ , then integrate

Practice

5 problems

1

Find $\displaystyle\int_C\vec{F}\cdot d\vec{r}$ where $\vec{F}(x,y,z)=\langle2xy-yz,\;x^2-xz,\;-xy\rangle$ on a smooth curve from $(2,1,0)$ to $(3,2,-1)$ .

no calculator, same as the exam
Answer20answer recomputed · sympy
1. The path is only described as 'a smooth curve', so the integral had better be path-independent — first verify $\vec F$ is conservative.
2. With $P=2xy-yz$ , $Q=x^2-xz$ , $R=-xy$ : $P_y=2x-z=Q_x$ , $P_z=-y=R_x$ , $Q_z=-x=R_y$ . Conservative.
3. Find the potential: $\displaystyle\varphi=\int(2xy-yz)\,dx = x^2y-xyz+g(y,z)$ , and both $\varphi_y=x^2-xz$ and $\varphi_z=-xy$ already match, so $g$ is constant.
4. $\varphi(x,y,z)=x^2y-xyz$ .
5. FTLI: $\displaystyle\int_C\vec F\cdot d\vec r=\varphi(3,2,-1)-\varphi(2,1,0)=(18+6)-(4-0)$ .
6. $=24-4=20$ .
TrapTrying to parametrize. No curve was given, so any attempt to build one is wasted work — and if you accidentally pick a path while $\vec F$ 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 $\vec{F}=\langle e^{yz}-y\sin(xy),\;zxe^{yz}-x\sin(xy),\;xye^{yz}\rangle$ , find a potential function.

no calculator, same as the exam
Answer $\varphi=xe^{yz}+\cos(xy)$ answer recomputed · sympy
1. Antidifferentiate the first component in $x$ : $\displaystyle\varphi=\int\big(e^{yz}-y\sin(xy)\big)\,dx$ .
2. Term by term, $\int e^{yz}\,dx = xe^{yz}$ (as $x$ varies, $e^{yz}$ is a constant) and $\int -y\sin(xy)\,dx = \cos(xy)$ .
3. So $\varphi = xe^{yz}+\cos(xy)+g(y,z)$ .
4. Test against the second component: $\varphi_y = xze^{yz}-x\sin(xy)+g_y$ , and we need $zxe^{yz}-x\sin(xy)$ , so $g_y=0$ .
5. Test against the third: $\varphi_z = xye^{yz}+g_z$ , and we need $xye^{yz}$ , so $g_z=0$ .
6. Therefore $\varphi = xe^{yz}+\cos(xy)$ , up to an additive constant.
TrapDifferentiating $e^{yz}$ with respect to $y$ and writing $ze^{yz}$ without the $x$ that the antiderivative introduced. The $x$ in $xe^{yz}$ must be carried through, and it is exactly what makes $\varphi_y=xze^{yz}$ match the printed $zxe^{yz}$ .
3

A potential for $\vec{F}=\langle\sin y,\;x\cos y\rangle$ is:

no calculator, same as the exam
Answer $x\sin y+1$ answer recomputed · sympy
1. First confirm a potential exists: $P=\sin y$ , $Q=x\cos y$ , and $P_y=\cos y=Q_x$ . Conservative on all of $\mathbb R^2$ .
2. Integrate $P$ in $x$ : $\displaystyle\varphi=\int\sin y\,dx = x\sin y+g(y)$ .
3. Differentiate in $y$ : $\varphi_y=x\cos y+g'(y)$ , which must equal $Q=x\cos y$ , so $g'(y)=0$ and $g$ is any constant.
4. The family of potentials is $\varphi=x\sin y+C$ . The printed choice takes $C=1$ : $\varphi = x\sin y+1$ .
TrapRejecting the ' $+1$ ' as a typo. Potentials are only determined up to an additive constant, so $x\sin y$ , $x\sin y+1$ , and $x\sin y-7$ 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 $\varphi=-x\cos y$ , from mistakenly antidifferentiating $\sin y$ in $y$ instead of $x$ .
4

Find $\displaystyle\int_C\vec{F}\cdot\vec{T}\,ds$ where $\vec{F}=\langle ye^z,\;e^y+xe^z,\;xye^z\rangle$ on a smooth curve from $(0,0,0)$ to $(-1,1,1)$ .

no calculator, same as the exam
Answer $-1$ answer recomputed · sympy
1. $\int_C\vec F\cdot\vec T\,ds$ is just $\int_C\vec F\cdot d\vec r$ rewritten, since $d\vec r=\vec T\,ds$ .
2. No specific curve is given, so check conservativity: with $P=ye^z$ , $Q=e^y+xe^z$ , $R=xye^z$ we get $P_y=e^z=Q_x$ , $P_z=ye^z=R_x$ , $Q_z=xe^z=R_y$ . Conservative.
3. Potential: $\displaystyle\int ye^z\,dx = xye^z+g(y,z)$ ; then $\varphi_y=xe^z+g_y = e^y+xe^z$ forces $g_y=e^y$ , and $\varphi_z=xye^z+h'(z)=xye^z$ forces $h'=0$ .
4. $\varphi = xye^z+e^y$ .
5. FTLI: $\varphi(-1,1,1)-\varphi(0,0,0) = \big((-1)(1)e+e\big)-\big(0+1\big)$ .
6. $=0-1=-1$ .
TrapSeeing $\vec T\,ds$ and thinking you must compute $|\vec r\,{}'|$ and a unit tangent. You never do — $\vec T\,ds=d\vec r$ , the speed cancels, and the whole problem collapses to two endpoint evaluations. The second trap is $\varphi(0,0,0)=0$ : the $e^y$ term makes it $e^0=1$ .
5

Evaluate $\displaystyle\int_C\nabla(x^2y-xyz)\cdot d\vec{r}$ where $C$ is any path from $(1,0,0)$ to $(2,1,-1)$ .

no calculator, same as the exam
Answer6answer recomputed · sympy
1. The integrand is already a gradient, $\nabla\varphi$ with $\varphi(x,y,z)=x^2y-xyz$ , so FTLI applies immediately and no conservativity test is needed.
2. $\displaystyle\int_C\nabla\varphi\cdot d\vec r = \varphi(\text{end})-\varphi(\text{start})$ , independent of the path.
3. $\varphi(2,1,-1)=2^2(1)-(2)(1)(-1)=4+2=6$ .
4. $\varphi(1,0,0)=1^2(0)-(1)(0)(0)=0$ .
5. Value $=6-0=6$ .
TrapExpanding $\nabla(x^2y-xyz)=\langle2xy-yz,\,x^2-xz,\,-xy\rangle$ 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 adds $2$ ; writing $4-2=2$ is the common wrong answer.

13

Green's Theorem

Lesson 29 · §17.4

Don't know this yet? §s174 in the full guide →

From the official study guide

Green's Theorem

Let $C$ be a positively oriented (counterclockwise), simple closed curve bounding region $D$ :

\oint_C P\,dx + Q\,dy = \iint_D \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right)dA

Area via Green's Theorem

A = \oint_C x\,dy = -\oint_C y\,dx = \frac{1}{2}\oint_C x\,dy - y\,dx

Note: Green's theorem converts a difficult line integral into a double integral, or vice versa.

Practice

5 problems

1

Use Green's Theorem to evaluate $\displaystyle\oint_C x\,dx+(x^2+y^2)\,dy$ 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
1. Green's Theorem converts a counterclockwise closed line integral into a double integral: $\oint_C P\,dx+Q\,dy=\iint_R\left(\dfrac{\partial Q}{\partial x}-\dfrac{\partial P}{\partial y}\right)dA$ .
2. Read off $P=x$ and $Q=x^2+y^2$ . The orientation is already CCW, so no sign flip is needed.
3. $\dfrac{\partial Q}{\partial x}=2x$ and $\dfrac{\partial P}{\partial y}=0$ , so the integrand is just $2x$ .
4. $R$ is the rectangle $0\le x\le2$ , $0\le y\le3$ — a genuine rectangle, so the limits are constants and the order does not matter.
5. $\displaystyle\iint_R 2x\,dA=\int_0^3\!\!\int_0^2 2x\,dx\,dy=\int_0^3\left[x^2\right]_0^2 dy=\int_0^3 4\,dy=12$ .
TrapTrying to parametrize all four sides by hand. That is four integrals, and the $\int x\,dx$ 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 $P_y=0$ — 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 $\displaystyle\int_C\cos^3(x)\,dx+e^x\,dy$ .

no calculator, same as the exam
Answer $e^2-2e+1$ answer recomputed · sympy
1. $P=\cos^3(x)$ , $Q=e^{x}$ . Neither has an elementary-looking line integral, which is the hint that Green's Theorem is the intended route.
2. $\dfrac{\partial Q}{\partial x}=e^{x}$ , $\dfrac{\partial P}{\partial y}=0$ (there is no $y$ in $\cos^3 x$ at all), so $Q_x-P_y=e^{x}$ .
3. 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 $y=0$ .
4. Integrating in $x$ first avoids splitting the region: for each $y\in[0,1]$ , $x$ runs from the left edge $x=y$ to the right edge $x=2-y$ .
5. $\displaystyle\int_0^1\!\!\int_{y}^{2-y}e^{x}\,dx\,dy=\int_0^1\left(e^{2-y}-e^{y}\right)dy$ .
6. $=\left[-e^{2-y}-e^{y}\right]_0^1=(-e-e)-(-e^2-1)=e^2-2e+1$ .
TrapIntegrating $dy$ first. Then you must split at $x=1$ , because the upper boundary changes formula from $y=x$ to $y=2-x$ there: $\int_0^1\!\int_0^x+\int_1^2\!\int_0^{2-x}$ . Students who write a single $\int_0^2\!\int_0^{x}$ silently integrate over a region that is not the triangle. Going $dx$ first needs no split. (Note $e^2-2e+1=(e-1)^2$ , a useful sanity check that it is positive.)
3

Use Green's Theorem to evaluate $\displaystyle\oint_C y\,dx + 2x\,dy$ where $C$ is the square with vertices $(\pm1,\pm1)$ , counterclockwise.

no calculator, same as the exam
Answer $4$ answer recomputed · sympy
1. $P=y$ , $Q=2x$ , and the square is traversed CCW, so Green's Theorem applies directly.
2. $\dfrac{\partial Q}{\partial x}-\dfrac{\partial P}{\partial y}=2-1=1$ .
3. The integrand collapsed to the constant $1$ , so the double integral is literally the area of $R$ .
4. $R$ is the square $-1\le x\le1$ , $-1\le y\le1$ , whose side length is $2$ , so its area is $4$ .
5. $\displaystyle\oint_C y\,dx+2x\,dy=\iint_R 1\,dA=\text{area}(R)=4$ .
TrapComputing $P_x-Q_y$ instead of $Q_x-P_y$ . That gives $0-0=0$ here, a clean-looking wrong answer. The derivative always hits the *other* variable: $Q$ gets $\partial_x$ , $P$ gets $\partial_y$ . Second trap: the vertices $(\pm1,\pm1)$ describe a square of side $2$ , not side $1$ — its area is $4$ , not $1$ .
4

Use Green's theorem to evaluate $\displaystyle\oint_C(y^3)\,dx-(x^3)\,dy$ where $C$ is the circle $x^2+y^2=1$ , CCW.

no calculator, same as the exam
Answer $\frac{-3\pi}{2}$ answer recomputed · sympy
1. Write the integral as $\oint_C P\,dx+Q\,dy$ with $P=y^3$ and $Q=-x^3$ — the minus sign in front of $x^3\,dy$ belongs to $Q$ .
2. $\dfrac{\partial Q}{\partial x}=-3x^2$ , $\dfrac{\partial P}{\partial y}=3y^2$ , so $Q_x-P_y=-3x^2-3y^2=-3(x^2+y^2)$ .
3. $R$ is the unit disk and the integrand depends only on $x^2+y^2$ , so polar is the obvious move: $x^2+y^2=r^2$ and $dA=r\,dr\,d\theta$ .
4. $\displaystyle\iint_R -3(x^2+y^2)\,dA=\int_0^{2\pi}\!\!\int_0^1 -3r^2\cdot r\,dr\,d\theta=\int_0^{2\pi}\!\!\int_0^1 -3r^3\,dr\,d\theta$ .
5. $=2\pi\cdot\left(-\dfrac{3}{4}\right)=-\dfrac{3\pi}{2}$ .
TrapTwo sign traps stacked. First, forgetting that $Q=-x^3$ , which makes $Q_x=+3x^2$ and hands you $+\frac{3\pi}{2}$ — the right magnitude, wrong sign. Second, forgetting the polar Jacobian: $-3r^2\,dr\,d\theta$ instead of $-3r^3\,dr\,d\theta$ gives $-2\pi$ . 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 $\tfrac{x^2}{a^2}+\tfrac{y^2}{b^2}=1$ .

no calculator, same as the exam
Answer $\pi ab$ answer recomputed · sympy
1. Green's Theorem run backwards: pick $P,Q$ with $Q_x-P_y=1$ , and the double integral becomes the area. The symmetric choice is $P=-\tfrac{y}{2}$ , $Q=\tfrac{x}{2}$ , giving the area formula $A=\tfrac12\oint_C x\,dy-y\,dx$ .
2. Parametrize the ellipse CCW: $x=a\cos t$ , $y=b\sin t$ , $0\le t\le2\pi$ .
3. Then $dx=-a\sin t\,dt$ and $dy=b\cos t\,dt$ .
4. $x\,dy-y\,dx=(a\cos t)(b\cos t)\,dt-(b\sin t)(-a\sin t)\,dt=ab(\cos^2t+\sin^2t)\,dt=ab\,dt$ .
5. $A=\dfrac12\displaystyle\int_0^{2\pi}ab\,dt=\dfrac12\cdot ab\cdot 2\pi=\pi ab$ .
6. Sanity check: setting $a=b=R$ gives $\pi R^2$ , the circle.
TrapDropping the $\tfrac12$ . The area formula is $\tfrac12\oint x\,dy-y\,dx$ — the half is there because *each* of $x\,dy$ and $-y\,dx$ separately already integrates to the full area, and you are averaging them, not adding them. Forgetting it doubles the answer to $2\pi ab$ . The other trap is parametrizing clockwise ( $y=-b\sin t$ ), which returns $-\pi ab$ ; area must come out positive, so a negative result means your orientation was backwards.

14

Curl and Divergence

Lesson 30 · §17.5

Don't know this yet? §s175 in the full guide →

From the official study guide

Curl

For $\vec{F} = \langle P, Q, R \rangle$ :

\text{curl}\,\vec{F} = \nabla \times \vec{F} = \begin{vmatrix}\vec{i} & \vec{j} & \vec{k} \\ \partial_x & \partial_y & \partial_z \\ P & Q & R\end{vmatrix} = \left\langle R_y - Q_z,\, P_z - R_x,\, Q_x - P_y \right\rangle

Divergence

\text{div}\,\vec{F} = \nabla \cdot \vec{F} = \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z}

Key Identities

$\vec{F}$ conservative $\Rightarrow$ $\text{curl}\,\vec{F} = \vec{0}$
$\text{div}(\text{curl}\,\vec{F}) = 0$ always
$\text{curl}(\nabla f) = \vec{0}$ always (gradient fields are irrotational)

Practice

5 problems

1

If $\vec{F}=\langle xy,\sin z,\cos y\rangle$ , find $\operatorname{curl}(\vec{F})$ .

no calculator, same as the exam
Answer $\langle-\sin y-\cos z,\,0,\,-x\rangle$ answer recomputed · sympy
1. $\operatorname{curl}\vec F=\nabla\times\vec F=\left\langle \dfrac{\partial R}{\partial y}-\dfrac{\partial Q}{\partial z},\;\dfrac{\partial P}{\partial z}-\dfrac{\partial R}{\partial x},\;\dfrac{\partial Q}{\partial x}-\dfrac{\partial P}{\partial y}\right\rangle$ for $\vec F=\langle P,Q,R\rangle$ .
2. Here $P=xy$ , $Q=\sin z$ , $R=\cos y$ .
3. First component: $R_y-Q_z=\dfrac{\partial}{\partial y}(\cos y)-\dfrac{\partial}{\partial z}(\sin z)=-\sin y-\cos z$ .
4. Second component: $P_z-R_x=\dfrac{\partial}{\partial z}(xy)-\dfrac{\partial}{\partial x}(\cos y)=0-0=0$ .
5. Third component: $Q_x-P_y=\dfrac{\partial}{\partial x}(\sin z)-\dfrac{\partial}{\partial y}(xy)=0-x=-x$ .
6. $\operatorname{curl}\vec F=\langle-\sin y-\cos z,\,0,\,-x\rangle$ .
TrapThe middle component's sign. Expanding the determinant $\begin{vmatrix}\vec i&\vec j&\vec k\\\partial_x&\partial_y&\partial_z\\P&Q&R\end{vmatrix}$ , the $\vec j$ entry carries a minus from cofactor expansion: it is $-(R_x-P_z)=P_z-R_x$ , not $R_x-P_z$ . Write the middle slot as $P_z-R_x$ 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: $\operatorname{curl}$ of a vector field is a vector, while $\operatorname{div}$ is a scalar — never report a single number here.
2

Let $f(x,y,z)=xy^2z^3$ . Compute $\operatorname{div}(\nabla f)$ at $(2,-1,1)$ .

no calculator, same as the exam
Answer16answer recomputed · sympy
1. $\operatorname{div}(\nabla f)=\nabla^2 f=f_{xx}+f_{yy}+f_{zz}$ , the Laplacian. So take second partials, not first.
2. $f=xy^2z^3$ . Gradient: $\nabla f=\langle y^2z^3,\;2xyz^3,\;3xy^2z^2\rangle$ .
3. $f_{xx}=\dfrac{\partial}{\partial x}(y^2z^3)=0$ — $f$ is linear in $x$ .
4. $f_{yy}=\dfrac{\partial}{\partial y}(2xyz^3)=2xz^3$ .
5. $f_{zz}=\dfrac{\partial}{\partial z}(3xy^2z^2)=6xy^2z$ .
6. $\nabla^2 f=0+2xz^3+6xy^2z$ . 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 $\nabla f$ and then try to differentiate, you are differentiating constants and get $0$ . Differentiate fully, *then* substitute. The other trap is $y^2$ at $y=-1$ : $(-1)^2=+1$ , not $-1$ , so the $6xy^2z$ term is $+12$ ; a dropped square gives $4-12=-8$ .
3

For $\vec{F}=\langle e^{yz}-y\sin(xy),\;zxe^{yz}-x\sin(xy),\;xye^{yz}\rangle$ , verify that $\operatorname{curl}\vec{F}=\vec{0}$ and state what this implies.

no calculator, same as the exam
Answer $\vec{F}$ is conservativeprose answer · not machine-checkable
1. Label $P=e^{yz}-y\sin(xy)$ , $Q=zxe^{yz}-x\sin(xy)$ , $R=xye^{yz}$ , and compute the three components of $\nabla\times\vec F$ .
2. $R_y=\dfrac{\partial}{\partial y}\!\left(xye^{yz}\right)=xe^{yz}+xyze^{yz}$ , and $Q_z=\dfrac{\partial}{\partial z}\!\left(zxe^{yz}-x\sin(xy)\right)=xe^{yz}+xyze^{yz}$ . First component $R_y-Q_z=0$ .
3. $P_z=\dfrac{\partial}{\partial z}\!\left(e^{yz}-y\sin(xy)\right)=ye^{yz}$ , and $R_x=\dfrac{\partial}{\partial x}\!\left(xye^{yz}\right)=ye^{yz}$ . Second component $P_z-R_x=0$ .
4. $Q_x=ze^{yz}-\sin(xy)-xy\cos(xy)$ (product rule on $x\sin(xy)$ ), and $P_y=ze^{yz}-\sin(xy)-xy\cos(xy)$ (product rule on $y\sin(xy)$ ). Third component $Q_x-P_y=0$ .
5. So $\operatorname{curl}\vec F=\vec 0$ everywhere.
6. $\vec F$ is defined and smooth on all of $\mathbb R^3$ , which is simply connected, so $\operatorname{curl}\vec F=\vec0$ implies $\vec F$ is conservative: there is a potential $\varphi$ with $\nabla\varphi=\vec F$ .
7. Indeed $\varphi=xe^{yz}+\cos(xy)$ works: $\varphi_x=e^{yz}-y\sin(xy)$ , $\varphi_y=xze^{yz}-x\sin(xy)$ , $\varphi_z=xye^{yz}$ . Consequence: $\int_C\vec F\cdot d\vec r$ is path-independent and $\oint_C\vec F\cdot d\vec r=0$ on any closed curve.
TrapStating the converse without the domain hypothesis. $\operatorname{curl}\vec F=\vec0$ alone does not make a field conservative — the standard counterexample $\vec F=\left\langle\frac{-y}{x^2+y^2},\frac{x}{x^2+y^2},0\right\rangle$ 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 $\vec F$ is smooth on all of $\mathbb R^3$ . The computational trap is $\partial_y\big(y\sin(xy)\big)$ : it needs the product rule, $\sin(xy)+xy\cos(xy)$ , not just $xy\cos(xy)$ .
4

Determine if $\vec{F}=\langle yz,xz,xy\rangle$ is conservative using $\operatorname{curl}\vec{F}$ . If so, find $\varphi$ .

no calculator, same as the exam
AnswerConservative; $\varphi=xyz+C$ prose answer · not machine-checkable
1. Test with the curl. $P=yz$ , $Q=xz$ , $R=xy$ .
2. $R_y-Q_z=x-x=0$ ; $\;P_z-R_x=y-y=0$ ; $\;Q_x-P_y=z-z=0$ . So $\operatorname{curl}\vec F=\vec 0$ , and since $\vec F$ is a polynomial field on all of $\mathbb R^3$ (simply connected), $\vec F$ is conservative.
3. Now build $\varphi$ with $\nabla\varphi=\vec F$ . Start from $\varphi_x=yz$ and integrate in $x$ , treating $y,z$ as constants: $\varphi=xyz+g(y,z)$ — the 'constant' of integration may depend on the other variables.
4. Differentiate that in $y$ : $\varphi_y=xz+g_y(y,z)$ . Match against $Q=xz$ , so $g_y=0$ and $g=h(z)$ .
5. Differentiate in $z$ : $\varphi_z=xy+h'(z)$ . Match against $R=xy$ , so $h'(z)=0$ and $h$ is a genuine constant.
6. $\varphi=xyz+C$ , and one checks $\nabla(xyz)=\langle yz,xz,xy\rangle=\vec F$ .
TrapWriting the first antiderivative as $\varphi=xyz+C$ with a numeric constant. At that stage the unknown is a whole function $g(y,z)$ , because anything free of $x$ dies under $\partial_x$ . Collapsing it to a number too early means you never actually verify $\varphi_y=Q$ and $\varphi_z=R$ — on a field where the components do *not* fit together, that shortcut produces a 'potential' for a non-conservative field. Also: test with $\operatorname{curl}$ , not $\operatorname{div}$ . Here $\operatorname{div}\vec F=0$ too, but that says the field is incompressible, an entirely different statement.
5

Find $\operatorname{div}\vec{F}$ for $\vec{F}=\langle e^{xy},e^{yz},e^{xz}\rangle$ , and verify the identity $\operatorname{div}(\operatorname{curl}\vec{G})=0$ for any smooth $\vec{G}$ .

no calculator, same as the exam
Answer $\operatorname{div}\vec{F}=ye^{xy}+ze^{yz}+xe^{xz}$ answer recomputed · sympy
1. $\operatorname{div}\vec F=\nabla\cdot\vec F=P_x+Q_y+R_z$ — each component differentiated with respect to its own variable, summed to a scalar.
2. $P=e^{xy}$ , so $P_x=ye^{xy}$ (chain rule: $\partial_x(xy)=y$ ).
3. $Q=e^{yz}$ , so $Q_y=ze^{yz}$ .
4. $R=e^{xz}$ , so $R_z=xe^{xz}$ .
5. $\operatorname{div}\vec F=ye^{xy}+ze^{yz}+xe^{xz}$ .
6. For the identity, let $\vec G=\langle G_1,G_2,G_3\rangle$ . Then $\operatorname{curl}\vec G=\langle \partial_yG_3-\partial_zG_2,\;\partial_zG_1-\partial_xG_3,\;\partial_xG_2-\partial_yG_1\rangle$ .
7. $\operatorname{div}(\operatorname{curl}\vec G)=\left(G_{3yx}-G_{2zx}\right)+\left(G_{1zy}-G_{3xy}\right)+\left(G_{2xz}-G_{1yz}\right)$ .
8. By Clairaut's theorem the mixed partials commute ( $G_{3yx}=G_{3xy}$ , etc.), so the six terms cancel in pairs and $\operatorname{div}(\operatorname{curl}\vec G)=0$ for every $\vec G$ with continuous second partials.
TrapDropping the chain-rule factor. $\partial_x e^{xy}=ye^{xy}$ , not $e^{xy}$ — the answer $e^{xy}+e^{yz}+e^{xz}$ 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 $e^{xz}$ and gets $\partial_z$ , producing $x e^{xz}$ , not $z e^{xz}$ . 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 $\operatorname{div}\vec H\ne0$ can never be written as $\operatorname{curl}$ of anything.

15

Surface Integrals

Lessons 31–33 · §17.6

Don't know this yet? §s176p · §s176a · §s176f in the full guide →

From the official study guide

Surface Integral of a Scalar Function

For surface $z = g(x,y)$ over region $D$ :

\iint_S f(x,y,z)\,dS = \iint_D f(x,y,g(x,y))\sqrt{g_x^2 + g_y^2 + 1}\,dA

Surface Area

A(S) = \iint_S dS = \iint_D \sqrt{g_x^2 + g_y^2 + 1}\,dA

Flux Integral (Surface Integral of a Vector Field)

For $\vec{F} = \langle P, Q, R \rangle$ and oriented surface $S$ with unit normal $\hat{n}$ :

\iint_S \vec{F}\cdot d\vec{S} = \iint_S \vec{F}\cdot\hat{n}\,dS

For $z = g(x,y)$ with upward orientation ( $\hat{n}$ pointing upward):

\iint_S \vec{F}\cdot d\vec{S} = \iint_D \left(-P g_x - Q g_y + R\right)dA

Parametric Surface

For surface $\vec{r}(u,v)$ :

\iint_S f\,dS = \iint_D f(\vec{r}(u,v))\,|\vec{r}_u \times \vec{r}_v|\,dA

\iint_S \vec{F}\cdot d\vec{S} = \iint_D \vec{F}(\vec{r}(u,v))\cdot(\vec{r}_u \times \vec{r}_v)\,dA

Practice

6 problems

1

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
Answer $5\sqrt{30}$ answer recomputed · sympy
1. Solve for $z$ to get the surface as a graph: $z=g(x,y)=10-2x-5y$ .
2. For a graph, $dS=\sqrt{1+g_x^2+g_y^2}\;dA$ . Here $g_x=-2$ , $g_y=-5$ , so $dS=\sqrt{1+4+25}\,dA=\sqrt{30}\,dA$ — a constant, as it must be for a plane.
3. Find the shadow $D$ . The first octant needs $x\ge0$ , $y\ge0$ , and $z\ge0$ , and the last of these is $10-2x-5y\ge0$ , i.e. $2x+5y\le10$ .
4. $D$ is the triangle with legs along the axes: $x$ -intercept $x=5$ , $y$ -intercept $y=2$ . Its area is $\tfrac12(5)(2)=5$ .
5. $\displaystyle\text{Area}=\iint_D\sqrt{30}\,dA=\sqrt{30}\cdot\text{area}(D)=5\sqrt{30}$ .
6. Explicitly: $\displaystyle\int_0^5\!\!\int_0^{(10-2x)/5}\sqrt{30}\,dy\,dx=5\sqrt{30}$ .
TrapReporting $5$ — the area of the shadow — by forgetting the $\sqrt{1+g_x^2+g_y^2}$ factor. A tilted plane is always larger than its projection, so the stretch factor is $\ge1$ and your answer must exceed the shadow's area. The other trap is the shadow's shape: the first octant condition $z\ge0$ is what bounds $D$ , giving the triangle $2x+5y\le10$ with intercepts $x=5$ , $y=2$ . Students who integrate over $0\le x\le5$ , $0\le y\le2$ (the full rectangle) double the answer to $10\sqrt{30}$ .
2

Compute the flux of $\vec{F}(x,y)=\langle x,y\rangle$ across $\vec{r}(t)=\langle3\cos t,2\sin t\rangle$ , $-\pi\le t\le\pi$ . (Flux $=\displaystyle\int_C P\,dy-Q\,dx$ .)

no calculator, same as the exam
Answer $12\pi$ answer recomputed · sympy
1. This is the 2D flux (outward-normal) line integral, and the problem hands you the formula: $\text{Flux}=\displaystyle\int_C P\,dy-Q\,dx$ with $\vec F=\langle P,Q\rangle$ .
2. Along the curve, $P=x=3\cos t$ and $Q=y=2\sin t$ .
3. Differentiate the parametrization: $dx=-3\sin t\,dt$ and $dy=2\cos t\,dt$ .
4. $P\,dy-Q\,dx=(3\cos t)(2\cos t)\,dt-(2\sin t)(-3\sin t)\,dt=6\cos^2t\,dt+6\sin^2t\,dt=6\,dt$ .
5. $\displaystyle\int_{-\pi}^{\pi}6\,dt=6(2\pi)=12\pi$ .
6. Cross-check with the divergence form of Green's Theorem: $\operatorname{div}\vec F=1+1=2$ , and the ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$ encloses area $\pi(3)(2)=6\pi$ , so the flux is $2\cdot6\pi=12\pi$ . ✓
TrapComputing $\int P\,dx+Q\,dy$ — the circulation — instead of $\int P\,dy-Q\,dx$ . For this field that gives $\int(-9\cos t\sin t+4\sin t\cos t)dt=0$ , 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: $-\pi\le t\le\pi$ is a full $2\pi$ sweep traversed counterclockwise, so the outward normal convention holds and the flux is positive — halving it to $6\pi$ means you read the range as a half-loop.
3

Evaluate $\iint_S 1\,dS$ where $S$ is the portion of the paraboloid $z=x^2+y^2$ below $z=4$ .

no calculator, same as the exam
Answer $\dfrac{\pi}{6}(17\sqrt{17}-1)$ answer recomputed · sympy
1. $\iint_S 1\,dS$ is the surface area of $S$ . The surface is the graph $z=g(x,y)=x^2+y^2$ .
2. $g_x=2x$ , $g_y=2y$ , so $dS=\sqrt{1+4x^2+4y^2}\;dA$ .
3. The shadow: $z\le4$ means $x^2+y^2\le4$ , the disk of radius $2$ .
4. Polar, since both the integrand and the region depend only on $x^2+y^2=r^2$ : $dS=\sqrt{1+4r^2}\;r\,dr\,d\theta$ — note both the radical and the Jacobian $r$ .
5. $\displaystyle\iint_S dS=\int_0^{2\pi}\!\!\int_0^2\sqrt{1+4r^2}\;r\,dr\,d\theta$ .
6. Inner integral by $u=1+4r^2$ , $du=8r\,dr$ : $\displaystyle\int_0^2 r\sqrt{1+4r^2}\,dr=\left[\frac{(1+4r^2)^{3/2}}{12}\right]_0^2=\frac{17^{3/2}-1}{12}=\frac{17\sqrt{17}-1}{12}$ .
7. Multiply by $\int_0^{2\pi}d\theta=2\pi$ : $\;\dfrac{2\pi(17\sqrt{17}-1)}{12}=\dfrac{\pi}{6}\left(17\sqrt{17}-1\right)$ .
TrapThe radius of the shadow. $z=4$ on $z=x^2+y^2$ means $r^2=4$ , so $r$ runs to $\mathbf{2}$ , not $4$ — using $r\le4$ gives $\frac{\pi}{6}(65\sqrt{65}-1)$ . The second trap is losing one of the two $r$ 's: $\sqrt{1+4r^2}\,r\,dr\,d\theta$ 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 $\vec{F}=\langle x,y,z\rangle$ upward through the surface $z=4-x^2-y^2$ above the $xy$ -plane.

no calculator, same as the exam
Answer $24\pi$ answer recomputed · sympy
1. For an upward-oriented graph $z=g(x,y)$ , $\displaystyle\iint_S\vec F\cdot d\vec S=\iint_D\left(-P\,g_x-Q\,g_y+R\right)dA$ . Note this formula already has the $\sqrt{1+g_x^2+g_y^2}$ built in — it cancels against the normalization of $\hat n$ , so do not multiply by it again.
2. Here $g=4-x^2-y^2$ , so $g_x=-2x$ and $g_y=-2y$ ; and $\vec F=\langle x,y,z\rangle$ with $z$ evaluated *on the surface*: $R=4-x^2-y^2$ .
3. Integrand: $-x(-2x)-y(-2y)+(4-x^2-y^2)=2x^2+2y^2+4-x^2-y^2=x^2+y^2+4$ .
4. The shadow $D$ is where $z\ge0$ : $4-x^2-y^2\ge0$ , the disk $x^2+y^2\le4$ of radius $2$ .
5. Polar: $\displaystyle\int_0^{2\pi}\!\!\int_0^2\left(r^2+4\right)r\,dr\,d\theta=\int_0^{2\pi}\!\!\int_0^2\left(r^3+4r\right)dr\,d\theta$ .
6. $\displaystyle\int_0^2\left(r^3+4r\right)dr=\left[\frac{r^4}{4}+2r^2\right]_0^2=4+8=12$ , and times $2\pi$ gives $24\pi$ .
TrapLeaving $z$ as the free variable $z$ instead of substituting the surface, $z=4-x^2-y^2$ . The flux integral lives *on* the surface, so every appearance of $z$ in $\vec F$ must be replaced. The second trap is the sign pattern: it is $\langle-g_x,-g_y,1\rangle$ for upward — the two tangential terms get minus signs and the $\vec k$ slot gets $+1$ . If the problem had said downward you would negate the whole thing, giving $-24\pi$ . Third: do not tack on an extra $\sqrt{1+4x^2+4y^2}$ ; that factor is only for $dS$ in a scalar surface integral, never for $d\vec S$ in this form.
5

Find the surface area of the cone $z=\sqrt{x^2+y^2}$ below $z=3$ .

no calculator, same as the exam
Answer $9\pi\sqrt{2}$ answer recomputed · sympy
1. The cone is the graph $z=g(x,y)=\sqrt{x^2+y^2}$ .
2. $g_x=\dfrac{x}{\sqrt{x^2+y^2}}$ and $g_y=\dfrac{y}{\sqrt{x^2+y^2}}$ , so $g_x^2+g_y^2=\dfrac{x^2+y^2}{x^2+y^2}=1$ .
3. Therefore $dS=\sqrt{1+1}\,dA=\sqrt{2}\,dA$ — constant, because a cone has the same slope everywhere.
4. The shadow: $z\le3$ means $\sqrt{x^2+y^2}\le3$ , the disk of radius $3$ , whose area is $9\pi$ .
5. $\displaystyle\iint_S dS=\int_0^{2\pi}\!\!\int_0^3\sqrt2\;r\,dr\,d\theta=\sqrt2\cdot 9\pi=9\pi\sqrt2$ .
TrapWriting $dS=\sqrt{1+x^2+y^2}\,dA$ by squaring the *function* instead of its partials. The radical needs $g_x^2+g_y^2$ , 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 $\sqrt2$ . The other trap: the apex $(0,0,0)$ makes $g_x,g_y$ undefined there, but it is a single point of measure zero and the improper integral converges — the $\sqrt2$ still holds. And do not confuse this with the *lateral* cone formula $\pi r\ell$ : here $r=3$ , $\ell=3\sqrt2$ , giving $9\pi\sqrt2$ , which agrees. ✓
6

Evaluate $\iint_S\vec{F}\cdot d\vec{S}$ where $\vec{F}=\langle y,-x,1\rangle$ and $S$ is the disk $z=1$ , $x^2+y^2\le1$ , oriented upward.

no calculator, same as the exam
Answer $\pi$ answer recomputed · sympy
1. $S$ is a flat horizontal disk at height $z=1$ , oriented upward, so the unit normal is simply $\hat n=\vec k=\langle0,0,1\rangle$ and $d\vec S=\langle0,0,1\rangle\,dA$ (the graph $z=g=1$ has $g_x=g_y=0$ , so $dS=dA$ ).
2. Dot the field with the normal: $\vec F\cdot\hat n=\langle y,-x,1\rangle\cdot\langle0,0,1\rangle=1$ . Only the $\vec k$ -component of $\vec F$ survives.
3. $\displaystyle\iint_S\vec F\cdot d\vec S=\iint_{x^2+y^2\le1}1\,dA=\text{area of the unit disk}=\pi$ .
4. Explicitly in polar: $\displaystyle\int_0^{2\pi}\!\!\int_0^1 1\cdot r\,dr\,d\theta=2\pi\cdot\tfrac12=\pi$ .
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 $\langle0,0,-2\rangle$ , so its *circulation* around the boundary is $-2\pi$ , a completely different number that this problem is baiting you toward.) The second trap is orientation: 'oriented upward' fixes $\hat n=+\vec k$ ; downward would give $-\pi$ . Third, the disk has radius $1$ , so its area is $\pi$ , not $2\pi$ .