Chapter 36 — Exercises
These exercises build the surface-integral toolkit one layer at a time: first parametrize surfaces and find the area element $dS$, then integrate scalars over them, then orient them and push vector fields through. Work as many as you can by hand; the Python in Section 36.12 is for checking, not for replacing, the hand calculation.
Difficulty tiers: ⭐ routine practice · ⭐⭐ standard · ⭐⭐⭐ challenging · ⭐⭐⭐⭐ deep / multi-step.
Throughout, $\mathbf{r}(u,v)$ denotes a parametrized surface, $dS = \|\mathbf{r}_u \times \mathbf{r}_v\|\,du\,dv$ the scalar area element, and $d\mathbf{S} = (\mathbf{r}_u \times \mathbf{r}_v)\,du\,dv$ the oriented vector element.
Part A — Parametrized Surfaces (Section 36.2)
A1. ⭐ Parametrize the cylinder $x^2 + y^2 = R^2$, $0 \le z \le h$, and state the parameter domain.
Answer
$\mathbf{r}(\theta, z) = \langle R\cos\theta,\ R\sin\theta,\ z\rangle$, with $0 \le \theta < 2\pi$, $0 \le z \le h$.A2. ⭐ Parametrize the part of the plane $2x + 3y + z = 6$ lying above the rectangle $0\le x\le 1$, $0\le y\le 2$, using $x$ and $y$ as parameters.
Answer
$\mathbf{r}(x,y) = \langle x,\ y,\ 6 - 2x - 3y\rangle$, $(x,y)\in[0,1]\times[0,2]$.A3. ⭐ Identify the surface $\mathbf{r}(u,v) = \langle 2u,\ 3v,\ u - v\rangle$. Is it a plane? If so, give a normal vector.
Answer
It is a plane through the origin spanned by $\mathbf{r}_u = \langle 2,0,1\rangle$ and $\mathbf{r}_v = \langle 0,3,-1\rangle$. A normal is $\mathbf{r}_u\times\mathbf{r}_v = \langle 0\cdot(-1)-1\cdot 3,\ 1\cdot 0 - 2\cdot(-1),\ 2\cdot 3 - 0\rangle = \langle -3, 2, 6\rangle$.A4. ⭐⭐ Describe the surface $\mathbf{r}(u,v) = \langle u\cos v,\ u\sin v,\ u^2\rangle$, $u\ge 0$, $0\le v < 2\pi$. (Hint: relate $z$ to $x^2+y^2$.)
Answer
$x^2 + y^2 = u^2 = z$, so $z = x^2 + y^2$: a circular **paraboloid** opening upward, vertex at the origin.A5. ⭐⭐ Parametrize the part of the sphere $x^2+y^2+z^2 = 16$ in the first octant ($x,y,z\ge 0$). Give the ranges of $\phi$ and $\theta$.
Answer
$\mathbf{r}(\phi,\theta) = \langle 4\sin\phi\cos\theta,\ 4\sin\phi\sin\theta,\ 4\cos\phi\rangle$ with $0\le\phi\le\pi/2$ (so $z\ge 0$) and $0\le\theta\le\pi/2$ (so $x,y\ge 0$).A6. ⭐⭐ The curve $y = e^x$, $0\le x\le 1$, is revolved about the $x$-axis. Write a parametrization $\mathbf{r}(x,\theta)$ of the surface of revolution (Section 36.11).
Answer
$\mathbf{r}(x,\theta) = \langle x,\ e^x\cos\theta,\ e^x\sin\theta\rangle$, $0\le x\le 1$, $0\le\theta<2\pi$.Part B — The Surface Area Element and Surface Area (Section 36.3)
B1. ⭐ State the area element $dS$ for a graph $z = g(x,y)$, then specialize to $g(x,y) = 2x + 3y$.
Answer
$dS = \sqrt{1 + g_x^2 + g_y^2}\,dA$. For $g = 2x+3y$: $g_x = 2$, $g_y = 3$, so $dS = \sqrt{1 + 4 + 9}\,dA = \sqrt{14}\,dA$ (constant — the surface is a flat tilted plane).B2. ⭐ Compute $\mathbf{r}_\theta\times\mathbf{r}_z$ and $\|\mathbf{r}_\theta\times\mathbf{r}_z\|$ for the cylinder $\mathbf{r}(\theta,z) = \langle R\cos\theta, R\sin\theta, z\rangle$.
Answer
$\mathbf{r}_\theta = \langle -R\sin\theta, R\cos\theta, 0\rangle$, $\mathbf{r}_z = \langle 0,0,1\rangle$. Cross product $= \langle R\cos\theta, R\sin\theta, 0\rangle$, magnitude $R$. So $dS = R\,d\theta\,dz$.B3. ⭐⭐ Find the area of the part of the plane $z = 2x + 2y + 1$ above the unit disk $x^2 + y^2 \le 1$.
Answer
$g_x = g_y = 2$, so $\sqrt{1+4+4} = 3$. Area $= \iint_D 3\,dA = 3\cdot(\pi\cdot 1^2) = 3\pi$.B4. ⭐⭐ Find the area of the part of the paraboloid $z = x^2 + y^2$ below $z = 4$. (Area element $\sqrt{1 + 4x^2 + 4y^2}\,dA$.)
Answer
$D$ is the disk $x^2+y^2\le 4$. In polar, $\int_0^{2\pi}\!\!\int_0^2 \sqrt{1+4\rho^2}\,\rho\,d\rho\,d\theta$. Let $u = 1+4\rho^2$, $du = 8\rho\,d\rho$: inner $= \tfrac18\int_1^{17}\sqrt u\,du = \tfrac{1}{12}(17^{3/2}-1)$. Times $2\pi$: $\dfrac{\pi}{6}(17^{3/2}-1)\approx 36.18$.B5. ⭐⭐ Confirm the lateral area of a cylinder of radius $R$, height $h$, by integrating its $dS$.
Answer
From B2, $dS = R\,d\theta\,dz$. Area $=\int_0^{2\pi}\!\!\int_0^h R\,dz\,d\theta = R\cdot h\cdot 2\pi = 2\pi R h$. ✓B6. ⭐⭐⭐ For the cone $\mathbf{r}(u,v) = \langle u\cos v, u\sin v, u\rangle$, $0\le u\le 1$, $0\le v<2\pi$, compute $\|\mathbf{r}_u\times\mathbf{r}_v\|$ and the lateral area; check against $\pi R\ell$ with $R=1$, $\ell=\sqrt2$.
Answer
$\mathbf{r}_u = \langle\cos v,\sin v,1\rangle$, $\mathbf{r}_v = \langle -u\sin v, u\cos v, 0\rangle$. Cross $= \langle -u\cos v, -u\sin v, u\rangle$, magnitude $u\sqrt2$. Area $= \int_0^{2\pi}\!\!\int_0^1 u\sqrt2\,du\,dv = \sqrt2\cdot\tfrac12\cdot 2\pi = \pi\sqrt2$. The formula gives $\pi\cdot 1\cdot\sqrt2 = \pi\sqrt2$. ✓B7. ⭐⭐⭐ Find the area of $z = \tfrac{2}{3}(x^{3/2} + y^{3/2})$ over $0\le x\le 1$, $0\le y\le 1$. (Engineered so $1 + g_x^2 + g_y^2 = 1 + x + y$.)
Answer
$g_x = x^{1/2}$, $g_y = y^{1/2}$, so the integrand is $\sqrt{1+x+y}$. Area $= \int_0^1\!\!\int_0^1 (1+x+y)^{1/2}\,dx\,dy$. Inner: $\tfrac23(1+x+y)^{3/2}\big|_{x=0}^{1} = \tfrac23[(2+y)^{3/2}-(1+y)^{3/2}]$. Then $\int_0^1$: $\tfrac23\cdot\tfrac25[(2+y)^{5/2}-(1+y)^{5/2}]_0^1 = \tfrac{4}{15}[(3^{5/2}-2^{5/2})-(2^{5/2}-1)] = \tfrac{4}{15}[3^{5/2} - 2\cdot 2^{5/2} + 1]\approx 1.41$.Part C — Scalar Surface Integrals (Section 36.4)
C1. ⭐ Evaluate $\iint_S 1\,dS$ for $S$ the upper hemisphere of radius $2$.
Answer
$\iint_S 1\,dS$ is just the area: half of $4\pi(2)^2 = 16\pi$, so $8\pi$.C2. ⭐⭐ Evaluate $\iint_S z\,dS$ where $S$ is the part of $z = 4 - x - y$ above the triangle with vertices $(0,0),(1,0),(0,1)$.
Answer
$g_x = g_y = -1$, $\sqrt{1+1+1}=\sqrt3$, and $z = 4-x-y$. So $\iint_S z\,dS = \sqrt3\iint_D(4-x-y)\,dA$ over the triangle. $\iint_D 4\,dA = 4\cdot\tfrac12 = 2$; $\iint_D x\,dA = \iint_D y\,dA = \tfrac{1}{6}$ (centroid of a unit-leg right triangle at $1/3$, area $1/2$). So $\sqrt3(2 - \tfrac16 - \tfrac16) = \sqrt3\cdot\tfrac{5}{3} = \dfrac{5\sqrt3}{3}\approx 2.89$.C3. ⭐⭐ A spherical shell of radius $R$ has density $\sigma(\phi,\theta) = \sigma_0\cos^2\phi$. Find its mass.
Answer
$M = \iint_S\sigma\,dS = \int_0^{2\pi}\!\!\int_0^\pi \sigma_0\cos^2\phi\cdot R^2\sin\phi\,d\phi\,d\theta$. Inner: $\int_0^\pi\cos^2\phi\sin\phi\,d\phi = [-\tfrac13\cos^3\phi]_0^\pi = \tfrac23$. So $M = \sigma_0 R^2\cdot\tfrac23\cdot 2\pi = \dfrac{4\pi\sigma_0 R^2}{3}$.C4. ⭐⭐ Find the average value of $f = x^2 + y^2$ over the sphere of radius $R$.
Answer
By symmetry $\iint_S x^2\,dS = \iint_S y^2\,dS = \iint_S z^2\,dS$, and they sum to $\iint_S(x^2+y^2+z^2)\,dS = R^2\cdot 4\pi R^2 = 4\pi R^4$. So $\iint_S(x^2+y^2)\,dS = \tfrac23\cdot 4\pi R^4$. Divide by area $4\pi R^2$: average $= \tfrac23 R^2$.C5. ⭐⭐⭐ Find $\bar z$ for a uniform-density upper hemispherical shell of radius $R$.
Answer
$\bar z = \frac{1}{\text{Area}}\iint_S z\,dS$. With $z = R\cos\phi$, $dS = R^2\sin\phi\,d\phi\,d\theta$, $\phi\in[0,\pi/2]$: $\iint_S z\,dS = \int_0^{2\pi}\!\!\int_0^{\pi/2} R\cos\phi\cdot R^2\sin\phi\,d\phi\,d\theta = 2\pi R^3\cdot\tfrac12 = \pi R^3$. Area $= 2\pi R^2$, so $\bar z = \pi R^3/(2\pi R^2) = R/2$.C6. ⭐⭐⭐ Evaluate $\iint_S(x^2+y^2+z^2)\,dS$ over the cylinder side $x^2+y^2=1$, $0\le z\le 3$.
Answer
On the cylinder $x^2+y^2 = 1$, so integrand $= 1 + z^2$, and $dS = d\theta\,dz$ (since $R=1$). $\int_0^{2\pi}\!\!\int_0^3(1+z^2)\,dz\,d\theta = 2\pi[z+\tfrac{z^3}{3}]_0^3 = 2\pi(3+9) = 24\pi$.Part D — Orientation and the Unit Normal (Section 36.5)
D1. ⭐ Write the upward unit normal $\hat{\mathbf{n}}$ for the graph $z = g(x,y)$.
Answer
$\hat{\mathbf{n}} = \dfrac{\langle -g_x, -g_y, 1\rangle}{\sqrt{1+g_x^2+g_y^2}}$ — positive $z$-component, hence upward.D2. ⭐⭐ For $G(x,y,z) = x^2+y^2-z$, find $\hat{\mathbf{n}} = \nabla G/\|\nabla G\|$ at $(1,0,1)$. Up or down?
Answer
$\nabla G = \langle 2x, 2y, -1\rangle = \langle 2,0,-1\rangle$ at the point; $\|\nabla G\| = \sqrt5$. So $\hat{\mathbf{n}} = \langle 2,0,-1\rangle/\sqrt5$ — negative $z$-component, so it points **downward** (into the paraboloid's bowl).D3. ⭐⭐ Verify that the sphere's $\mathbf{r}_\phi\times\mathbf{r}_\theta = R^2\sin\phi\,\langle\sin\phi\cos\theta,\sin\phi\sin\theta,\cos\phi\rangle$ points outward.
Answer
The bracket is exactly $\mathbf{r}/R$ (the outward radial unit vector), and $R^2\sin\phi\ge 0$ for $\phi\in[0,\pi]$. So the cross product is a positive multiple of the outward radial direction — it points outward, matching the closed-surface convention.D4. ⭐⭐⭐ Explain why a Möbius band has no consistent unit normal and what that means for $\iint_S\mathbf{F}\cdot d\mathbf{S}$ (Section 36.5 Warning).
Answer
Slide a normal once around the band's loop and it returns pointing the opposite way, so no continuous global choice of $\hat{\mathbf n}$ exists — the band is non-orientable. Flux requires a fixed "positive side"; with no consistent normal, "which way is across?" has no answer, so the flux integral is **undefined** on a Möbius band.Part E — Flux of a Vector Field (Sections 36.6–36.7)
E1. ⭐⭐ Upward flux of $\mathbf{F} = \langle 0,0,3\rangle$ through the disk $z=0$, $x^2+y^2\le 4$, oriented upward.
Answer
$\hat{\mathbf{n}} = \langle 0,0,1\rangle$, $\mathbf{F}\cdot\hat{\mathbf n} = 3$, area $= 4\pi$. Flux $= 12\pi$. (Uniform upward field through a horizontal disk: simply field strength times area.)E2. ⭐⭐ Upward flux of $\mathbf{F} = \langle x,y,z\rangle$ through $z = 1-x-y$ over the triangle $x\ge0,y\ge0,x+y\le1$.
Answer
$g_x = g_y = -1$, so flux $= \iint_D(-P g_x - Q g_y + R)\,dA = \iint_D(x + y + (1-x-y))\,dA = \iint_D 1\,dA = \tfrac12$ (area of the triangle).E3. ⭐⭐ Outward flux of $\mathbf{F} = \langle x,y,z\rangle$ through the sphere of radius $R$.
Answer
On the sphere $\hat{\mathbf{n}} = \mathbf{r}/R$, so $\mathbf{F}\cdot\hat{\mathbf n} = \mathbf{r}\cdot\mathbf{r}/R = R^2/R = R$ (constant). Flux $= R\cdot 4\pi R^2 = 4\pi R^3$.E4. ⭐⭐⭐ Outward flux of $\mathbf{F} = \langle y,x,z\rangle$ through the cylinder side $x^2+y^2=1$, $0\le z\le 2$, with $\hat{\mathbf{n}} = \langle\cos\theta,\sin\theta,0\rangle$, $dS = d\theta\,dz$.
Answer
On the wall $x = \cos\theta$, $y=\sin\theta$, so $\mathbf{F}=\langle\sin\theta,\cos\theta,z\rangle$. Then $\mathbf{F}\cdot\hat{\mathbf n} = \sin\theta\cos\theta + \cos\theta\sin\theta = 2\sin\theta\cos\theta = \sin 2\theta$. Flux $= \int_0^2\!\!\int_0^{2\pi}\sin2\theta\,d\theta\,dz = \int_0^2 0\,dz = 0$.E5. ⭐⭐⭐ Upward flux of $\mathbf{F} = \langle -y,x,z\rangle$ through $z = 1-x^2-y^2$, $z\ge0$, oriented upward.
Answer
$g_x = -2x$, $g_y = -2y$. Flux $= \iint_D(-P g_x - Q g_y + R)\,dA = \iint_D(2xy - 2xy + (1-x^2-y^2))\,dA = \iint_D(1-x^2-y^2)\,dA$ over $x^2+y^2\le1$. Polar: $\int_0^{2\pi}\!\!\int_0^1(1-\rho^2)\rho\,d\rho\,d\theta = 2\pi(\tfrac12-\tfrac14) = \tfrac{\pi}{2}$.E6. ⭐⭐⭐ Total outward flux of $\mathbf{F} = \langle x,y,0\rangle$ through the closed cylinder $x^2+y^2=1$, $0\le z\le2$, with caps at $z=0$ and $z=2$.
Answer
**Side:** $\hat{\mathbf n}=\langle\cos\theta,\sin\theta,0\rangle$, $\mathbf{F}\cdot\hat{\mathbf n}=\cos^2\theta+\sin^2\theta=1$, area $2\pi(1)(2)=4\pi$; flux $=4\pi$. **Top cap** ($\hat{\mathbf n}=\langle0,0,1\rangle$): $\mathbf{F}\cdot\hat{\mathbf n}=0$. **Bottom cap** ($\hat{\mathbf n}=\langle0,0,-1\rangle$): $\mathbf{F}\cdot\hat{\mathbf n}=0$. Total $=4\pi$. (Check via Divergence Theorem, Chapter 37: $\nabla\cdot\mathbf F=2$, volume $\pi(1)^2(2)=2\pi$, $2\cdot2\pi=4\pi$. ✓)Part F — Applied Flux (Sections 36.8–36.9), at least two fields
F1. ⭐⭐ (Fluid flow — engineering.) Water flows with $\mathbf{v} = \langle 0,0,2\rangle$ m/s. A horizontal circular drain cover of radius $0.5$ m sits in $z=0$, oriented upward. Find the volume flow rate $\iint_S\mathbf{v}\cdot d\mathbf{S}$.
Answer
$\mathbf{v}\cdot\hat{\mathbf n} = 2$, area $= \pi(0.5)^2 = 0.25\pi$. Flow rate $= 2\cdot 0.25\pi = 0.5\pi\approx 1.57$ m³/s.F2. ⭐⭐ (Electric flux — physics.) A charge $q$ sits at the origin. By Gauss's law, state the flux through (a) a sphere of radius $1$, (b) a sphere of radius $100$, (c) a cube of side $4$ centered on it. Why are they equal?
Answer
All three equal $q/\varepsilon_0$. Gauss's law (Section 36.9) depends only on the *enclosed* charge, not on the shape or size of the closed surface — the inverse-square field is divergence-free away from the charge, so any surface wrapping $q$ catches the same flux.F3. ⭐⭐⭐ (Heat flow — mechanical engineering.) With $T = 100 - x^2 - y^2 - z^2$, $k=2$, the heat flux is $\mathbf{q} = -k\nabla T$. Find the outward heat flux through the sphere of radius $3$ centered at the origin.
Answer
$\nabla T = \langle -2x,-2y,-2z\rangle$, so $\mathbf{q} = -2\langle -2x,-2y,-2z\rangle = \langle 4x,4y,4z\rangle = 4\mathbf{r}$. On the sphere $\hat{\mathbf n} = \mathbf{r}/3$, so $\mathbf{q}\cdot\hat{\mathbf n} = 4\mathbf{r}\cdot\mathbf{r}/3 = 4\cdot9/3 = 12$. Flux $= 12\cdot 4\pi(3)^2 = 12\cdot 36\pi = 432\pi$. (Positive: heat flows outward from the hot center.)F4. ⭐⭐⭐ (Electric flux — physics.) Charges $+4$ and $-1$ (with $\varepsilon_0=1$) sit inside closed surface $S$; a charge $+9$ sits outside. Find $\oiint_S\mathbf{E}\cdot d\mathbf{S}$.
Answer
$Q_{\text{enc}} = 4 + (-1) = 3$, so flux $= 3/\varepsilon_0 = 3$. The outside $+9$ contributes zero net flux: its field enters $S$ on one side and exits the other, cancelling exactly (Section 36.9).Part G — Synthesis and Extension (⭐⭐⭐⭐)
G1. ⭐⭐⭐⭐ (Shape-dependence of flux.) Consider $\mathbf{F} = \langle x,y,0\rangle$. (a) Confirm the closed-cylinder flux of E6 equals $\iiint_E\nabla\cdot\mathbf{F}\,dV$ (preview of the Divergence Theorem, Chapter 37; $\nabla\cdot\mathbf F = 2$). (b) For the inverse-square field $\mathbf{F} = \mathbf{r}/\|\mathbf{r}\|^3$, explain (citing the worked example in Section 36.6) why its outward flux through any closed surface enclosing the origin is $4\pi$, while through a closed surface not enclosing the origin it is $0$.
Answer
(a) $\nabla\cdot\mathbf F = 2$; enclosed volume $= \pi(1)^2(2) = 2\pi$; $\iiint_E 2\,dV = 4\pi$, matching E6's surface total. (b) The Section 36.6 worked example gives flux $4\pi$ through any centered sphere; the field is divergence-free everywhere except the origin, so by the Divergence Theorem any two surfaces enclosing the origin give the same flux $4\pi$. A surface *not* enclosing the origin bounds a region where $\nabla\cdot\mathbf F = 0$ throughout, so its flux is $0$.G2. ⭐⭐⭐⭐ (Re-deriving a surface-of-revolution area.) From $\mathbf{r}(x,\theta) = \langle x, f(x)\cos\theta, f(x)\sin\theta\rangle$ (Section 36.11), compute $\mathbf{r}_x\times\mathbf{r}_\theta$, show $\|\mathbf{r}_x\times\mathbf{r}_\theta\| = |f|\sqrt{1+(f')^2}$, conclude $S = 2\pi\int_a^b|f|\sqrt{1+(f')^2}\,dx$, and apply to $f = \sqrt{R^2-x^2}$ on $[-R,R]$.
Answer
$\mathbf{r}_x = \langle 1, f'\cos\theta, f'\sin\theta\rangle$, $\mathbf{r}_\theta = \langle 0, -f\sin\theta, f\cos\theta\rangle$. Cross product $= \langle f f', -f\cos\theta, -f\sin\theta\rangle$, magnitude $\sqrt{f^2(f')^2 + f^2} = |f|\sqrt{1+(f')^2}$. Integrate $\theta$ over $[0,2\pi]$: $S = 2\pi\int_a^b|f|\sqrt{1+(f')^2}\,dx$. For $f=\sqrt{R^2-x^2}$: $f' = -x/\sqrt{R^2-x^2}$, so $1+(f')^2 = R^2/(R^2-x^2)$ and $|f|\sqrt{1+(f')^2} = \sqrt{R^2-x^2}\cdot\frac{R}{\sqrt{R^2-x^2}} = R$. Then $S = 2\pi\int_{-R}^R R\,dx = 2\pi R\cdot 2R = 4\pi R^2$. ✓G3. ⭐⭐⭐⭐ (Mesh flux — data science / engineering.) A triangulated surface has three triangles:
| Triangle | Centroid $\mathbf{c}_T$ | Unit normal $\hat{\mathbf{n}}_T$ | Area $A_T$ |
|---|---|---|---|
| 1 | $(1,0,0)$ | $(1,0,0)$ | $0.5$ |
| 2 | $(0,1,0)$ | $(0,1,0)$ | $0.5$ |
| 3 | $(0,0,1)$ | $(0,0,1)$ | $1.0$ |
For $\mathbf{F} = \langle x,y,z\rangle$, approximate $\sum_T\mathbf{F}(\mathbf{c}_T)\cdot\hat{\mathbf{n}}_T\,A_T$ (Section 36.13).
Answer
T1: $\mathbf F(1,0,0)\cdot(1,0,0) = 1$, times $0.5 = 0.5$. T2: $\mathbf F(0,1,0)\cdot(0,1,0) = 1$, times $0.5 = 0.5$. T3: $\mathbf F(0,0,1)\cdot(0,0,1) = 1$, times $1.0 = 1.0$. Sum $= 0.5+0.5+1.0 = 2.0$.Tier Summary
| Tier | Count | Exercises |
|---|---|---|
| ⭐ | 7 | A1, A2, A3, B1, B2, C1, D1 |
| ⭐⭐ | 16 | A4, A5, A6, B3, B4, B5, C2, C3, C4, D2, D3, E1, E2, E3, F1, F2 |
| ⭐⭐⭐ | 10 | B6, B7, C5, C6, D4, E4, E5, E6, F3, F4 |
| ⭐⭐⭐⭐ | 3 | G1, G2, G3 |
| Total | 36 |
Worked solutions to selected exercises appear in
appendices/answers-to-selected.md.