Purdue MA 26100 · official Final study guide

Final, worked.

The department's own study guide is the source of truth for what Final covers, and it gives you a bare answer under each problem with no work in between. This is that guide, section for section and problem for problem — with the work filled in, the traps named, and the answer kept shut until you commit to an attempt. All 10 of its problems.

CoverageComprehensive — all course material (Lessons 1–37)

Official exam dateNot published — set by the registrar

Your examFinals week · date TBD

Answers verified10 of 10 recomputed
0 prose, checked by hand

Source of truth: the official Final study guide, read 2026-09-14. Every section heading, formula box and problem statement below is the instructor's, unedited. The worked steps, trap notes and corrections are mine. Every answer below matched what was computed for it. · Reset my progress · Exam 1 → · Exam 2 →

Stokes' Theorem

Lessons 34–35 · §17.7

Don't know this yet? §s177 in the full guide →

From the official study guide

Stokes' Theorem

Let $S$ be an oriented, piecewise-smooth surface bounded by a simple closed curve $C = \partial S$ . Let $\vec{F} = \langle P, Q, R \rangle$ have continuous partial derivatives on an open region containing $S$ . Then:

\oint_C \vec{F}\cdot d\vec{r} = \iint_S \operatorname{curl}(\vec{F})\cdot d\vec{S}

Equivalently, using $d\vec{S} = \hat{n}\,dS$ :

\oint_C \vec{F}\cdot d\vec{r} = \iint_S \operatorname{curl}(\vec{F})\cdot\hat{n}\,dS

Orientation Convention

The boundary curve $C$ is positively oriented relative to $S$ when, walking along $C$ with the surface on your left, $\hat{n}$ points away from your head. Equivalently: $\hat{n}$ and the direction of traversal of $C$ satisfy the right-hand rule.

Curl Formula

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

\operatorname{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} = \langle R_y - Q_z,\; P_z - R_x,\; Q_x - P_y \rangle

Computing the Surface Integral Side

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

\iint_S \operatorname{curl}(\vec{F})\cdot d\vec{S} = \iint_D \left[-(R_y-Q_z)\,g_x - (P_z-R_x)\,g_y + (Q_x-P_y)\right]dA

Key: Stokes' theorem converts a line integral into a surface integral (or vice versa). It generalizes Green's Theorem: when $S$ is flat in the $xy$ -plane, Stokes' theorem reduces exactly to Green's theorem.

Connection to Green's Theorem

Green's Theorem is the special case of Stokes' Theorem where $S$ lies in the $xy$ -plane ( $z=0$ , $\hat{n}=\vec{k}$ ):

\oint_C P\,dx + Q\,dy = \iint_D (Q_x - P_y)\,dA

Consequences

If $\operatorname{curl}(\vec{F}) = \vec{0}$ (i.e., $\vec{F}$ is conservative), then $\oint_C \vec{F}\cdot d\vec{r} = 0$ for every closed curve $C$ .
Two surfaces $S_1$ and $S_2$ with the same oriented boundary $C$ give the same surface integral of $\operatorname{curl}(\vec{F})$ .
Stokes' theorem can simplify computation by replacing a hard surface with an easier one sharing the same boundary.

Practice

5 problems

1

Use Stokes' Theorem to evaluate $\displaystyle\oint_C \vec{F}\cdot d\vec{r}$ where $\vec{F}=\langle y^2, x, z^2\rangle$ and $C$ is the boundary of the triangle with vertices $(1,0,0)$ , $(0,1,0)$ , $(0,0,1)$ , oriented counterclockwise when viewed from above.

no calculator, same as the exam
Answer $\dfrac{1}{6}$ answer recomputed · sympy
1. Stokes' Theorem says $\oint_C \vec F\cdot d\vec r=\iint_S \operatorname{curl}(\vec F)\cdot d\vec S$ for any surface $S$ with $\partial S=C$ . The point of the whole theorem here is that the line integral is three separate segments, each needing its own parametrization, while the surface integral is one flat triangle. Convert.
2. $\operatorname{curl}(\vec F)=\left\langle \dfrac{\partial}{\partial y}(z^2)-\dfrac{\partial}{\partial z}(x),\; \dfrac{\partial}{\partial z}(y^2)-\dfrac{\partial}{\partial x}(z^2),\; \dfrac{\partial}{\partial x}(x)-\dfrac{\partial}{\partial y}(y^2)\right\rangle=\langle 0,0,1-2y\rangle$ .
3. The three vertices $(1,0,0),(0,1,0),(0,0,1)$ all satisfy $x+y+z=1$ , so take $S$ to be the flat triangle in that plane: $z=g(x,y)=1-x-y$ over the shadow triangle $D:\ x\ge0,\ y\ge0,\ x+y\le1$ .
4. For a graph $z=g(x,y)$ with upward normal, $d\vec S=\langle -g_x,-g_y,1\rangle\,dA=\langle 1,1,1\rangle\,dA$ . Counterclockwise viewed from above is exactly the orientation that pairs with the upward normal, so no sign flip.
5. $\operatorname{curl}(\vec F)\cdot\langle 1,1,1\rangle=(0)(1)+(0)(1)+(1-2y)(1)=1-2y$ , and note the $z$ 's have vanished, so no substitution of the plane is even needed.
6. $\displaystyle\iint_D (1-2y)\,dA=\int_0^1\!\!\int_0^{1-x}(1-2y)\,dy\,dx=\int_0^1\Big[y-y^2\Big]_0^{1-x}dx=\int_0^1 \big[(1-x)-(1-x)^2\big]\,dx$ .
7. $\displaystyle=\int_0^1 (x-x^2)\,dx=\tfrac12-\tfrac13=\dfrac{1}{6}$ .
TrapTwo traps stacked. First, people write $d\vec S=\langle -g_x,-g_y,1\rangle\,dA$ but then also multiply by $\sqrt{1+g_x^2+g_y^2}=\sqrt3$ — that factor belongs to $dS$ (scalar surface area), not to $d\vec S$ , and including both double-counts. Second, the shadow region is the triangle $x+y\le 1$ in the first quadrant, not the unit square: the inner limit is $y:0\to 1-x$ , and using $0\to1$ turns $\tfrac16$ into $0$ .
2

Let $\vec{F}=\langle -y^2, x, z^2\rangle$ . Use Stokes' Theorem to evaluate $\displaystyle\oint_C\vec{F}\cdot d\vec{r}$ where $C$ is the circle $x^2+y^2=4$ in the plane $z=3$ , traversed counterclockwise when viewed from above.

no calculator, same as the exam
Answer $4\pi$ answer recomputed · sympy
1. $C$ is a circle sitting in a horizontal plane, so the cheapest surface to span it with is the flat disk in that same plane — the normal is constant and the surface integral collapses to a plain double integral.
2. $\operatorname{curl}(\vec F)=\left\langle \dfrac{\partial}{\partial y}(z^2)-\dfrac{\partial}{\partial z}(x),\; \dfrac{\partial}{\partial z}(-y^2)-\dfrac{\partial}{\partial x}(z^2),\; \dfrac{\partial}{\partial x}(x)-\dfrac{\partial}{\partial y}(-y^2)\right\rangle=\langle 0,0,1+2y\rangle$ .
3. Take $S:\ x^2+y^2\le 4,\ z=3$ . Counterclockwise from above $\Rightarrow$ upward normal $\vec n=\langle0,0,1\rangle$ and $d\vec S=\langle0,0,1\rangle\,dA$ .
4. $\operatorname{curl}(\vec F)\cdot\vec n=1+2y$ , so $\displaystyle\iint_S\operatorname{curl}(\vec F)\cdot d\vec S=\iint_{x^2+y^2\le4}(1+2y)\,dA$ .
5. In polar, $\displaystyle\int_0^{2\pi}\!\!\int_0^2 (1+2r\sin\theta)\,r\,dr\,d\theta$ . The $2r\sin\theta$ piece integrates to $0$ because $\int_0^{2\pi}\sin\theta\,d\theta=0$ (the disk is symmetric about $y=0$ and $y$ is odd).
6. What is left is $\displaystyle\iint_{x^2+y^2\le4}1\,dA=\text{area}=\pi(2)^2=4\pi$ .
TrapThe sign on $-y^2$ . Since $\dfrac{\partial}{\partial y}(-y^2)=-2y$ , the $k$ -component is $1-(-2y)=1+2y$ , not $1-2y$ . Either way the $y$ -term dies by symmetry, so this particular problem forgives it — but only by luck, and the same slip elsewhere does not cancel. Also note $z=3$ is irrelevant to the answer: the curl has no $z$ in it, so the height of the plane never enters.
3

Evaluate $\displaystyle\iint_S \operatorname{curl}(\vec{F})\cdot d\vec{S}$ where $\vec{F}=\langle yz, xz, xy\rangle$ and $S$ is the part of the sphere $x^2+y^2+z^2=4$ with $z\ge0$ , oriented with upward normal. (Hint: use Stokes' Theorem and evaluate the line integral on $\partial S$ instead.)

no calculator, same as the exam
Answer $0$ answer recomputed · sympy
1. Here Stokes is run in the other direction: a curved surface integral is traded for a line integral. $\partial S$ is where the hemisphere meets $z=0$ , namely the circle $x^2+y^2=4$ in the $xy$ -plane. Upward normal on $S$ pairs with counterclockwise traversal of that circle seen from above.
2. Parametrize $\vec r(t)=\langle 2\cos t,\,2\sin t,\,0\rangle$ , $0\le t\le 2\pi$ , so $\vec r\,{}'(t)=\langle -2\sin t,\,2\cos t,\,0\rangle$ .
3. On $C$ we have $z=0$ , so $\vec F=\langle yz,xz,xy\rangle=\langle 0,\,0,\,4\cos t\sin t\rangle$ — the first two components are killed by $z=0$ .
4. $\vec F\cdot\vec r\,{}'(t)=0\cdot(-2\sin t)+0\cdot(2\cos t)+4\cos t\sin t\cdot 0=0$ identically, since $\vec r\,{}'$ has no $k$ -component while $\vec F$ has only a $k$ -component.
5. $\displaystyle\oint_C\vec F\cdot d\vec r=\int_0^{2\pi}0\,dt=0$ , hence $\displaystyle\iint_S\operatorname{curl}(\vec F)\cdot d\vec S=0$ .
6. Sanity check the fast way: $\operatorname{curl}\langle yz,xz,xy\rangle=\langle x-x,\;y-y,\;z-z\rangle=\vec 0$ , because $\vec F=\nabla(xyz)$ is conservative. A zero curl makes every such flux zero.
TrapTrying to parametrize the hemisphere and grind out $\iint\operatorname{curl}(\vec F)\cdot d\vec S$ in spherical coordinates. It works, but it is minutes of work for a field whose curl is $\vec 0$ . Also: the boundary is the equator circle at $z=0$ , not the whole sphere and not the point at the top — a hemisphere with $z\ge0$ has exactly one boundary curve.
4

Use Stokes' Theorem to evaluate $\displaystyle\oint_C \vec{F}\cdot d\vec{r}$ where $\vec{F}=\langle xz, yz, xy\rangle$ and $C$ is the boundary of the surface $z = 1 - x^2 - y^2$ above the $xy$ -plane, traversed counterclockwise when viewed from above.

no calculator, same as the exam
Answer $0$ answer recomputed · sympy
1. $C$ is where $z=1-x^2-y^2$ meets $z=0$ : setting $z=0$ gives $x^2+y^2=1$ , the unit circle in the $xy$ -plane. Stokes lets us span it with either the paraboloid cap or the flat disk; both are legal, so pick deliberately.
2. $\operatorname{curl}(\vec F)=\left\langle \dfrac{\partial}{\partial y}(xy)-\dfrac{\partial}{\partial z}(yz),\; \dfrac{\partial}{\partial z}(xz)-\dfrac{\partial}{\partial x}(xy),\; \dfrac{\partial}{\partial x}(yz)-\dfrac{\partial}{\partial y}(xz)\right\rangle=\langle x-y,\;x-y,\;0\rangle$ .
3. Cheap route. Use the flat disk $S:\ x^2+y^2\le1,\ z=0$ with $\vec n=\langle0,0,1\rangle$ . Then $\operatorname{curl}(\vec F)\cdot\vec n=0$ because the curl has no $k$ -component, and the integral is $0$ on sight.
4. Honest route, to confirm the choice of surface did not smuggle anything in: on the paraboloid, $d\vec S=\langle 2x,2y,1\rangle\,dA$ over $x^2+y^2\le1$ , and $\operatorname{curl}(\vec F)\cdot\langle2x,2y,1\rangle=2x(x-y)+2y(x-y)=2(x^2-y^2)$ .
5. $\displaystyle\iint_{x^2+y^2\le1}2(x^2-y^2)\,dA=\int_0^{2\pi}\!\!\int_0^1 2r^2(\cos^2\theta-\sin^2\theta)\,r\,dr\,d\theta=2\Big(\int_0^1 r^3dr\Big)\Big(\int_0^{2\pi}\cos 2\theta\,d\theta\Big)$ .
6. $\int_0^{2\pi}\cos2\theta\,d\theta=0$ , so the value is $0$ — same as the flat disk, as Stokes guarantees.
TrapReaching for the paraboloid because the problem named it. Stokes only cares about the boundary curve, so you may replace the cap with the flat disk it bounds and the $k$ -component-free curl kills the integral instantly. The other trap is the boundary itself: 'above the $xy$ -plane' means $C$ is $x^2+y^2=1$ at $z=0$ , not a curve on the paraboloid at some other height.
5

Let $S$ be the part of the paraboloid $z=9-x^2-y^2$ above $z=0$ with upward normal, and $\vec{F}=\langle -y^3, x^3, -z^3\rangle$ . Use Stokes' Theorem to find $\displaystyle\iint_S \operatorname{curl}(\vec{F})\cdot d\vec{S}$ .

no calculator, same as the exam
Answer $\dfrac{243\pi}{2}$ answer recomputed · sympy
1. $\operatorname{curl}(\vec F)=\left\langle \dfrac{\partial}{\partial y}(-z^3)-\dfrac{\partial}{\partial z}(x^3),\; \dfrac{\partial}{\partial z}(-y^3)-\dfrac{\partial}{\partial x}(-z^3),\; \dfrac{\partial}{\partial x}(x^3)-\dfrac{\partial}{\partial y}(-y^3)\right\rangle=\langle 0,\,0,\,3x^2+3y^2\rangle$ .
2. Two ways to finish. Direct: $S$ is $z=9-x^2-y^2$ with upward normal, so $d\vec S=\langle 2x,2y,1\rangle\,dA$ over the shadow disk $x^2+y^2\le9$ (set $z=0$ to get radius $3$ ).
3. $\operatorname{curl}(\vec F)\cdot\langle2x,2y,1\rangle=0\cdot 2x+0\cdot 2y+(3x^2+3y^2)(1)=3(x^2+y^2)$ , so the messy $\langle2x,2y\rangle$ part contributes nothing.
4. Polar: $\displaystyle\iint_{x^2+y^2\le9}3(x^2+y^2)\,dA=\int_0^{2\pi}\!\!\int_0^3 3r^2\cdot r\,dr\,d\theta=2\pi\cdot 3\cdot\frac{3^4}{4}=\frac{243\pi}{2}$ .
5. Cross-check with Stokes on the boundary circle $x^2+y^2=9$ , $z=0$ : $\vec r(t)=\langle3\cos t,3\sin t,0\rangle$ , $\vec F=\langle -27\sin^3t,\,27\cos^3 t,\,0\rangle$ , and $\vec F\cdot\vec r\,{}'=81(\sin^4t+\cos^4t)$ , whose integral over $[0,2\pi]$ is $81\cdot\frac{3\pi}{2}=\frac{243\pi}{2}$ . Same value.
TrapForgetting the polar Jacobian: the integrand is $3r^2$ and $dA=r\,dr\,d\theta$ , so you integrate $3r^3$ . Dropping the extra $r$ gives $2\pi\cdot 3\cdot 9=54\pi$ instead of $\tfrac{243\pi}{2}$ . Second trap: the shadow radius is $3$ , not $9$ — $z=9-x^2-y^2$ hits $z=0$ at $x^2+y^2=9$ , i.e. $r=3$ .

Divergence Theorem

Lessons 36–37 · §17.8

Don't know this yet? §s178 in the full guide →

From the official study guide

Divergence Theorem (Gauss's Theorem)

Let $E$ be a simple solid region with outward-oriented boundary surface $S = \partial E$ . Let $\vec{F} = \langle P, Q, R \rangle$ have continuous partial derivatives on an open region containing $E$ . Then:

\iint_S \vec{F}\cdot d\vec{S} = \iiint_E \operatorname{div}(\vec{F})\,dV

Divergence Formula

\operatorname{div}(\vec{F}) = \nabla\cdot\vec{F} = \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z}

Orientation Convention

The boundary surface $S = \partial E$ must be oriented with the outward normal (pointing away from the interior of $E$ ). If the surface is given with inward normal, multiply the result by $-1$ .

Key: The Divergence Theorem converts a flux surface integral into a triple integral over the enclosed volume — often much easier to evaluate. It is especially useful when the surface consists of multiple pieces.

Physical Interpretation

The Divergence Theorem states that the net flux of $\vec{F}$ out of a closed surface equals the total divergence (source/sink density) integrated over the enclosed volume. If $\operatorname{div}(\vec{F}) = 0$ everywhere in $E$ , then the net flux through $\partial E$ is zero.

Strategy: Replacing a Hard Surface

If $S$ is an open surface (not closed), close it by adding a cap $S_{\text{cap}}$ , apply the Divergence Theorem to the closed surface $S \cup S_{\text{cap}}$ , then subtract the cap's contribution:

\iint_S \vec{F}\cdot d\vec{S} = \iiint_E \operatorname{div}(\vec{F})\,dV - \iint_{S_{\text{cap}}} \vec{F}\cdot d\vec{S}

Key Identities

$\operatorname{div}(\operatorname{curl}(\vec{F})) = 0$ for any smooth $\vec{F}$ — so $\iint_S \operatorname{curl}(\vec{F})\cdot d\vec{S} = 0$ for any closed $S$
If $\operatorname{div}(\vec{F}) = 0$ in $E$ , the net outward flux through $\partial E$ is zero
Volume of $E$ : $V = \frac{1}{3}\iint_S \vec{r}\cdot\hat{n}\,dS = \frac{1}{3}\iint_S \langle x,y,z\rangle\cdot d\vec{S}$

Regions with Holes (Two Boundaries)

If $D$ is a solid region with boundary $S_1$ (outer) and $S_2$ (inner), where $S_1$ has normal outward and $S_2$ has normal inward (into $D$ ):

\iiint_D \operatorname{div}(\vec{F})\,dV = \iint_{S_1}\vec{F}\cdot d\vec{S} - \iint_{S_2}\vec{F}\cdot d\vec{S}

Use this when a smaller solid (e.g., a sphere of radius $r_1$ ) is removed from inside a larger one (e.g., a sphere of radius $r_2$ ). The integral over the annular region equals the outer flux minus the inner flux.

Practice

5 problems

1

Use the Divergence Theorem to evaluate $\displaystyle\iint_S \vec{F}\cdot d\vec{S}$ where $\vec{F}=\langle x^3, y^3, z^3\rangle$ and $S$ is the sphere $x^2+y^2+z^2=1$ with outward normal.

no calculator, same as the exam
Answer $\dfrac{12\pi}{5}$ answer recomputed · sympy
1. $S$ is closed with outward normal, which is exactly the hypothesis of the Divergence Theorem: $\displaystyle\iint_S\vec F\cdot d\vec S=\iiint_E \operatorname{div}(\vec F)\,dV$ . Going the other way, parametrizing a sphere and dotting with a normal, is far worse.
2. $\operatorname{div}(\vec F)=\dfrac{\partial}{\partial x}(x^3)+\dfrac{\partial}{\partial y}(y^3)+\dfrac{\partial}{\partial z}(z^3)=3x^2+3y^2+3z^2=3\rho^2$ in spherical coordinates.
3. $E$ is the solid unit ball, so spherical coordinates with $0\le\rho\le1$ , $0\le\phi\le\pi$ , $0\le\theta\le2\pi$ and $dV=\rho^2\sin\phi\,d\rho\,d\phi\,d\theta$ .
4. $\displaystyle\iiint_E 3\rho^2\,dV=\int_0^{2\pi}\!\!\int_0^{\pi}\!\!\int_0^1 3\rho^2\cdot\rho^2\sin\phi\,d\rho\,d\phi\,d\theta$ — note the integrand is $3\rho^4\sin\phi$ .
5. The three factors separate: $\displaystyle\Big(\int_0^{2\pi}d\theta\Big)\Big(\int_0^{\pi}\sin\phi\,d\phi\Big)\Big(\int_0^1 3\rho^4 d\rho\Big)=(2\pi)(2)\left(\frac35\right)=\frac{12\pi}{5}$ .
TrapWriting $dV=\rho^2\sin\phi$ and then integrating $3\rho^2$ as if it were $3\rho^2\cdot\rho\sin\phi$ or just $3\rho^2\,d\rho$ . The full integrand is $3\rho^4\sin\phi$ ; dropping either the $\rho^2$ of $dV$ or the $\sin\phi$ changes the answer to $2\pi$ or $\tfrac{4\pi}{3}$ -type garbage. Also $\int_0^{\pi}\sin\phi\,d\phi=2$ , not $0$ — students who reflexively write $0$ for a sine integral lose the whole thing.
2

Use the Divergence Theorem to compute the flux of $\vec{F}=\langle xy, y^2, yz\rangle$ outward through the surface of the unit cube $[0,1]^3$ .

no calculator, same as the exam
Answer $2$ answer recomputed · sympy
1. The cube has six faces, so the direct flux computation is six separate double integrals with six different normals. The Divergence Theorem replaces all of it with one triple integral over $[0,1]^3$ .
2. $\operatorname{div}(\vec F)=\dfrac{\partial}{\partial x}(xy)+\dfrac{\partial}{\partial y}(y^2)+\dfrac{\partial}{\partial z}(yz)=y+2y+y=4y$ .
3. $\displaystyle\iint_S\vec F\cdot d\vec S=\iiint_{[0,1]^3}4y\,dV=\int_0^1\!\!\int_0^1\!\!\int_0^1 4y\,dx\,dy\,dz$ .
4. The region is a box and the integrand depends only on $y$ , so it factors: $\displaystyle\Big(\int_0^1 dx\Big)\Big(\int_0^1 4y\,dy\Big)\Big(\int_0^1 dz\Big)=1\cdot 2\cdot 1$ .
5. Flux $=2$ .
TrapDifferentiating the wrong variable in each slot. $\operatorname{div}$ takes $\partial_x$ of the first component, $\partial_y$ of the second, $\partial_z$ of the third: $\partial_x(xy)=y$ and $\partial_z(yz)=y$ , so the three terms are $y+2y+y=4y$ , not $x+2y+y$ . A secondary trap is assuming the answer must be $0$ by symmetry — $y$ is not symmetric on $[0,1]$ , only on an interval centered at the origin.
3

Let $S$ be the closed surface consisting of the hemisphere $z=\sqrt{4-x^2-y^2}$ and the disk $x^2+y^2\le4$ in the $xy$ -plane, both with outward normal. Evaluate $\displaystyle\iint_S \vec{F}\cdot d\vec{S}$ where $\vec{F}=\langle x,y,z\rangle$ .

no calculator, same as the exam
Answer $16\pi$ answer recomputed · sympy
1. The hemisphere alone is not closed, but the problem already glued the flat disk on, so $S=\partial E$ where $E$ is the solid half-ball $x^2+y^2+z^2\le4,\ z\ge0$ . Outward normal on both pieces means the Divergence Theorem applies as stated.
2. $\operatorname{div}\langle x,y,z\rangle=1+1+1=3$ , a constant — which is why the flux reduces to $3\,\text{Vol}(E)$ .
3. $\displaystyle\iint_S\vec F\cdot d\vec S=\iiint_E 3\,dV=3\cdot\text{Vol}(E)$ .
4. $E$ is half of a ball of radius $2$ : $\text{Vol}(E)=\tfrac12\cdot\tfrac43\pi(2)^3=\tfrac{16\pi}{3}$ . In spherical this is $\int_0^{2\pi}\!\!\int_0^{\pi/2}\!\!\int_0^2 \rho^2\sin\phi\,d\rho\,d\phi\,d\theta$ , with $\phi$ stopping at $\pi/2$ because $z\ge0$ .
5. Flux $=3\cdot\dfrac{16\pi}{3}=16\pi$ .
TrapLetting $\phi$ run to $\pi$ and getting $32\pi$ . For the upper half-ball $\phi\in[0,\pi/2]$ — $\phi=\pi/2$ is the equator, $\phi=\pi$ is the south pole. The second trap is applying the theorem to the hemisphere alone: it is not closed, so you would need the disk anyway. Here the disk happens to contribute $0$ , since on $z=0$ the outward normal is $\langle0,0,-1\rangle$ and $\vec F\cdot\vec n=-z=0$ , but that is a fact you must state, not one you may assume.
4

Use the Divergence Theorem to find the outward flux of $\vec{F}=\langle x^2z, y^2z, z^2\rangle$ through the boundary of the solid region $E = \{(x,y,z)\mid x^2+y^2\le4,\;0\le z\le3\}$ (a cylinder).

no calculator, same as the exam
Answer $36\pi$ answer recomputed · sympy
1. The boundary is a can: top disk, bottom disk, and curved side, each with a different normal. One triple integral over the solid beats three surface integrals.
2. $\operatorname{div}(\vec F)=\dfrac{\partial}{\partial x}(x^2z)+\dfrac{\partial}{\partial y}(y^2z)+\dfrac{\partial}{\partial z}(z^2)=2xz+2yz+2z$ .
3. The solid is a cylinder, so cylindrical coordinates: $x=r\cos\theta$ , $y=r\sin\theta$ , $0\le r\le2$ , $0\le\theta\le2\pi$ , $0\le z\le3$ , $dV=r\,dz\,dr\,d\theta$ .
4. $\displaystyle\iiint_E(2xz+2yz+2z)\,dV=\int_0^{2\pi}\!\!\int_0^2\!\!\int_0^3\big(2rz\cos\theta+2rz\sin\theta+2z\big)\,r\,dz\,dr\,d\theta$ .
5. The first two terms vanish: $\int_0^{2\pi}\cos\theta\,d\theta=\int_0^{2\pi}\sin\theta\,d\theta=0$ , i.e. $x$ and $y$ are odd over a disk centered on the axis.
6. What survives is $\displaystyle\int_0^{2\pi}\!\!\int_0^2\!\!\int_0^3 2z\,r\,dz\,dr\,d\theta=\Big(\int_0^3 2z\,dz\Big)\Big(\iint_{r\le2}dA\Big)=9\cdot 4\pi=36\pi$ .
TrapJustifying the death of $2xz$ and $2yz$ with the wrong symmetry. The $z$ -interval $[0,3]$ is not symmetric about $0$ , so nothing cancels in $z$ ; the cancellation is in $\theta$ , because $\int_0^{2\pi}\cos\theta\,d\theta=\int_0^{2\pi}\sin\theta\,d\theta=0$ over the full disk. Beware also that this problem will not punish a dropped Jacobian: $\int_0^2 r\,dr=\int_0^2 dr=2$ , so omitting the $r$ still lands on $36\pi$ here. That is a coincidence of the radius being $2$ , and it will not save you on the next cylinder.
5

Let $S$ be the outward-oriented boundary of the region between the spheres $x^2+y^2+z^2=1$ and $x^2+y^2+z^2=4$ . Evaluate $\displaystyle\iint_S \vec{F}\cdot d\vec{S}$ where $\vec{F}=\langle x, y, z\rangle$ .

no calculator, same as the exam
Answer $28\pi$ answer recomputed · sympy
1. $E$ is a spherical shell $1\le\rho\le2$ . Its boundary is two spheres, and 'outward' for the solid means outward on $\rho=2$ but pointing *toward the origin* on $\rho=1$ . The Divergence Theorem handles that bookkeeping for free.
2. $\operatorname{div}\langle x,y,z\rangle=3$ .
3. $\displaystyle\iint_S\vec F\cdot d\vec S=\iiint_E 3\,dV=3\cdot\text{Vol}(E)$ , and $\text{Vol}(E)=\tfrac43\pi(2^3-1^3)=\tfrac{28\pi}{3}$ .
4. As a spherical integral: $\displaystyle\int_0^{2\pi}\!\!\int_0^{\pi}\!\!\int_1^2 3\rho^2\sin\phi\,d\rho\,d\phi\,d\theta$ , where $\int_1^2 3\rho^2 d\rho=2^3-1^3=7$ , $\int_0^{\pi}\sin\phi\,d\phi=2$ , $\int_0^{2\pi}d\theta=2\pi$ .
5. Flux $=7\cdot 2\cdot 2\pi=28\pi$ , matching $3\cdot\dfrac{28\pi}{3}$ .
TrapSetting the inner limit to $\rho=0$ and getting $32\pi$ — the ball of radius $1$ is not part of $E$ , it is a hole. Equally common is computing the two spheres separately and adding them with the same sign: on the inner sphere the outward-from- $E$ normal is $-\hat\rho$ , so its flux is $-4\pi$ , and $32\pi+(-4\pi)=28\pi$ . Adding $+4\pi$ gives $36\pi$ .