Surface integrals and the divergence theorem: dS, flux and Stokes

Worked solutions

Problem 1

1. $r = (2\sin\phi\cos\theta,\ 2\sin\phi\sin\theta,\ 2\cos\phi)$$\phi$ is measured down from the positive $z$-axis, $\theta$ around it.
2. $r_\phi = (2\cos\phi\cos\theta,\ 2\cos\phi\sin\theta,\ -2\sin\phi)$, $r_\theta = (-2\sin\phi\sin\theta,\ 2\sin\phi\cos\theta,\ 0)$Differentiate each coordinate.
3. $r_\phi \times r_\theta = 4\sin\phi\,(\sin\phi\cos\theta,\ \sin\phi\sin\theta,\ \cos\phi)$The components are $4\sin^2\phi\cos\theta$, $4\sin^2\phi\sin\theta$ and $4\sin\phi\cos\phi$. The vector in parentheses is $r/2$, so the cross product points outward.
4. $\|r_\phi \times r_\theta\| = 4\sin\phi$The vector in parentheses is a unit vector, and $\sin\phi \ge 0$ on $[0, \pi]$.
5. Area $= \int_0^{2\pi}\int_0^\pi 4\sin\phi\,d\phi\,d\theta = 4\cdot2\cdot2\pi = 16\pi$It agrees with $4\pi a^2$. Integrating $d\phi\,d\theta$ with no factor gives $2\pi^2 \approx 19.74$ instead of $16\pi \approx 50.27$: the rectangle of angles is not the sphere (Mistake 1).

Problem 2

1. $g_x = 2x$, $g_y = 2y$, so $dS = \sqrt{1 + 4x^2 + 4y^2}\,dA$The graph formula.
2. The shadow is the disk $x^2 + y^2 \le 2$The paraboloid meets $z = 2$ where $r^2 = 2$.
3. Area $= \int_0^{2\pi}\int_0^{\sqrt2}\sqrt{1 + 4r^2}\,r\,dr\,d\theta$Polar coordinates, keeping the $r$ of $dA = r\,dr\,d\theta$.
4. $\int_0^{\sqrt2}\sqrt{1 + 4r^2}\,r\,dr = \tfrac18\int_1^9\sqrt u\,du = \tfrac{1}{12}(27 - 1) = \tfrac{13}{6}$Substitute $u = 1 + 4r^2$, $du = 8r\,dr$.
5. Area $= 2\pi\cdot\tfrac{13}{6} = \tfrac{13\pi}{3} \approx 13.61$More than twice the area $2\pi \approx 6.28$ of the disk underneath, because the surface is steep near the rim. Leaving out the square root gives exactly the disk's area (Mistake 1).

Problem 3

1. $z = 2\cos\phi$ and $dS = 4\sin\phi\,d\phi\,d\theta$, with $0 \le \phi \le \tfrac\pi2$Problem 1 with the upper half only.
2. Mass $= \int_0^{2\pi}\int_0^{\pi/2} 2\cos\phi\cdot4\sin\phi\,d\phi\,d\theta$A scalar surface integral; the direction of the normal plays no part.
3. $\int_0^{\pi/2}\cos\phi\sin\phi\,d\phi = \tfrac12$It is $\tfrac12\sin^2\phi$ from $0$ to $\tfrac\pi2$.
4. Mass $= 8\cdot\tfrac12\cdot2\pi = 8\pi \approx 25.13$The $\theta$-integral gives $2\pi$.
5. Mass $8\pi$; the average height is $8\pi/8\pi = 1$The area of the hemisphere is $2\pi a^2 = 8\pi$. The average height of the shell is $\tfrac a2$, higher than the $\tfrac{3a}{8}$ of the solid half-ball, whose volume crowds near the flat base.

Problem 4

1. $g = 1 - x^2 - y^2$, so $(-g_x, -g_y, 1) = (2x, 2y, 1)$It points up, as required: the $z$-component is $1$.
2. $F \cdot (2x, 2y, 1) = 2x^2 + 2y^2 + (1 - x^2 - y^2) = 1 + r^2$On the surface, $z = 1 - r^2$.
3. The shadow is the unit diskThe cap meets $z = 0$ where $r = 1$.
4. Flux $= \int_0^{2\pi}\int_0^1(1 + r^2)\,r\,dr\,d\theta = 2\pi\big(\tfrac12 + \tfrac14\big)$The $\|r_x \times r_y\|$ cancels against the $1/\|r_x \times r_y\|$ in the unit normal, so no square root appears.
5. Flux $= \tfrac{3\pi}{2} \approx 4.71$A check with the divergence theorem: $\nabla \cdot F = 3$ and the solid under the cap has volume $\tfrac\pi2$, giving $\tfrac{3\pi}{2}$ out of the closed surface; on the flat bottom $F \cdot (0, 0, -1) = -z = 0$, so all of it leaves through the cap. With $(g_x, g_y, 1)$ as the normal the answer is $-\tfrac\pi2$ (Mistake 2).

Problem 5

1. $n = \tfrac12(x, y, z)$, so $F \cdot n = \tfrac{z^2}{2}$The outward unit normal of a sphere is the position vector divided by the radius.
2. $F \cdot n = 2\cos^2\phi$ and $dS = 4\sin\phi\,d\phi\,d\theta$$z = 2\cos\phi$.
3. Flux $= \int_0^{2\pi}\int_0^\pi 8\cos^2\phi\sin\phi\,d\phi\,d\theta$Both hemispheres add: below the equator $z < 0$ and the field points down, which is still outward.
4. $\int_0^\pi\cos^2\phi\sin\phi\,d\phi = \tfrac23$It is $-\tfrac13\cos^3\phi$ from $0$ to $\pi$.
5. Flux $= 8\cdot\tfrac23\cdot2\pi = \tfrac{32\pi}{3} \approx 33.51$This is the volume of the ball, $\tfrac43\pi\cdot2^3$, as the divergence theorem predicts: $\nabla \cdot F = 1$.

Problem 6

1. $\nabla \cdot F = 2x + 2y + 2z$Sum of $P_x$, $Q_y$, $R_z$.
2. $\iiint 2x\,dV = 2\cdot\tfrac12\cdot6 = 6$, $\iiint 2y\,dV = 2\cdot1\cdot6 = 12$, $\iiint 2z\,dV = 2\cdot\tfrac32\cdot6 = 18$The box has volume $6$, and the average of $x$, $y$, $z$ over it is the center $(\tfrac12, 1, \tfrac32)$.
3. Total $= 6 + 12 + 18 = 36$The divergence theorem's answer.
4. On $x = 1$: $F \cdot n = 1$ over area $6$; on $y = 2$: $4$ over area $3$; on $z = 3$: $9$ over area $2$The faces $x = 0$, $y = 0$, $z = 0$ contribute $0$, because there the normal component of $F$ is $-x^2 = 0$, $-y^2 = 0$, $-z^2 = 0$.
5. Flux $= 36$; the faces give $6 + 12 + 18 = 36$Each face's flux matches one term of the triple integral.

Problem 7

1. $\nabla \cdot F = 3x^2 + 3y^2 + 3z^2 = 3\rho^2$In spherical coordinates.
2. $\iiint 3\rho^2\,dV = \int_0^{2\pi}\int_0^\pi\int_0^1 3\rho^4\sin\phi\,d\rho\,d\phi\,d\theta$$dV = \rho^2\sin\phi\,d\rho\,d\phi\,d\theta$.
3. $= 3\cdot\tfrac15\cdot2\cdot2\pi = \tfrac{12\pi}{5}$The limits are numbers, so the integral splits.
4. On the sphere $n = (x, y, z)$, so $F \cdot n = x^4 + y^4 + z^4$, and $\iint x^4\,dS = \tfrac{4\pi}{5}$With $z = \cos\phi$: $\iint z^4\,dS = 2\pi\int_0^\pi\cos^4\phi\sin\phi\,d\phi = 2\pi\cdot\tfrac25$. By symmetry $x^4$ and $y^4$ give the same.
5. Flux $= \tfrac{12\pi}{5} \approx 7.54$ both ways$3\cdot\tfrac{4\pi}{5}$ from the surface.

Problem 8

1. $S$ and the unit disk $D$ in the plane $z = 0$ together bound the half-ball $E$The divergence theorem needs a closed surface; $S$ alone is not one.
2. $\nabla \cdot F = 1 + 1 + 1 = 3$, so $\iiint_E 3\,dV = 3\cdot\tfrac{2\pi}{3} = 2\pi$The half-ball has volume $\tfrac23\pi$.
3. On $D$ the outward normal is $(0, 0, -1)$ and $F \cdot n = -(z + x^2) = -x^2$Outward from $E$ is down on the base.
4. $\iint_D -x^2\,dA = -\int_0^{2\pi}\int_0^1 r^3\cos^2\theta\,dr\,d\theta = -\tfrac\pi4$$\tfrac14\cdot\pi$.
5. Flux through $S$ $= 2\pi - \big(-\tfrac\pi4\big) = \tfrac{9\pi}{4} \approx 7.07$The total out of $E$ is $2\pi$, and the base takes $-\tfrac\pi4$ of it, so the dome carries the rest. Stopping at $2\pi$ forgets the base (Mistake 3).

Problem 9

1. $\nabla \times F = (x - 2z,\ 1 - y,\ 2)$$R_y - Q_z = x - 2z$, $P_z - R_x = 1 - y$, $Q_x - P_y = 1 + 1$.
2. With $n = (x, y, z)$: $(\nabla \times F) \cdot n = x^2 - 2xz + y - y^2 + 2z$The upward normal of the upper hemisphere is outward.
3. $\iint_S x^2\,dS = \iint_S y^2\,dS = \tfrac{2\pi}{3}$, $\iint_S xz\,dS = \iint_S y\,dS = 0$, $\iint_S z\,dS = \pi$Half of $\tfrac{4\pi}{3}$ for each square; the odd terms cancel between $\theta$ and $\theta + \pi$; and $\iint z\,dS$ is Problem 3 with radius $1$.
4. $\iint_S (\nabla \times F) \cdot n\,dS = \tfrac{2\pi}{3} - \tfrac{2\pi}{3} + 2\pi = 2\pi$Add the pieces.
5. Both sides equal $2\pi$Counterclockwise seen from above, $r = (\cos t, \sin t, 0)$, so $F = (-\sin t, \cos t, \sin t\cos t)$ and $F \cdot r' = \sin^2 t + \cos^2 t = 1$, giving $2\pi$. The flat unit disk has the same boundary and gives $\iint 2\,dA = 2\pi$ too. Clockwise gives $-2\pi$ (Mistake 4).

Problem 10

1. $P_x = \dfrac{1}{\rho^3} - \dfrac{3x^2}{\rho^5}$, and similarly for $Q_y$ and $R_z$Here $\rho = \sqrt{x^2 + y^2 + z^2}$, and $\partial\rho/\partial x = x/\rho$.
2. $\nabla \cdot F = \dfrac{3}{\rho^3} - \dfrac{3\rho^2}{\rho^5} = 0$ for $\rho \ne 0$The three $x^2$, $y^2$, $z^2$ terms add up to $\rho^2$.
3. On the sphere of radius $a$: $F \cdot n = \dfrac{1}{a^2}$, so the flux is $\dfrac{1}{a^2}\cdot4\pi a^2 = 4\pi$$F$ points straight out and has length $\tfrac{1}{a^2}$. The radius cancels.
4. Cube: apply the divergence theorem to the region between the cube and a small sphere inside it, where $F$ is smooth: the flux out of the cube equals the flux out of the small sphere, $4\pi$The origin is not in that region, and $\nabla \cdot F = 0$ there. Each face of the cube carries $\tfrac{4\pi}{6} = \tfrac{2\pi}{3}$ by symmetry.
5. Flux $4\pi$ through every sphere about the origin, $4\pi$ through the cube, and $0$ through the sphere centered at $(3, 0, 0)$That sphere does not contain the origin, so the divergence theorem applies on its ball and gives $\iiint 0\,dV = 0$. This is Gauss's law: the flux counts the charge inside, whatever the surface. Using $\nabla \cdot F = 0$ on a ball that contains the origin gives $0$ instead of $4\pi$ (Mistake 5).