Chapter 37 — Quiz

Q: State Stokes' theorem, identifying which side is an integral over a *boundary* and which is an integral of a *derivative*.

$\partial S\nabla\times\mathbf{F}$ (curl) over the surface interior. This is the slogan "boundary integral = region integral of a derivative" in 3D. (§37.2)

Q: Compute and for .

. The curl is $$ This is the standard counterclockwise swirl: zero divergence, constant upward curl. (§37.2, §37.3; curl/divergence from Chapter 34)

Q: Verify Stokes' theorem for on the unit disk in the -plane, oriented upward. Give both sides.

Curl-flux: , so . Circulation: parametrize the unit circle as ; the integrand is , so . Both sides equal . ✓ (§37.3, Example 1)

Q: State the Divergence theorem and explain in words what it says physically.

$$ Physically: the net flux out through the skin of a solid equals the total source strength inside. Sum every tiny source and sink in the volume and you get exactly the net amount escaping the boundary — nothing is created or destroyed in transit. The boundary must carry the outward normal. (§37.4)

Q: Use the Divergence theorem to find the outward flux of through the unit sphere.

, so the flux is . A direct surface integral confirms it: on the unit sphere , so the flux is the surface area . ✓ (§37.5, Example 3)

Q: Find the outward flux of through the surface of the unit cube .

. The flux is . By symmetry each term contributes equally: , and likewise for and , for a total of . One volume integral replaced six face integrals. (§37.5, Example 4)

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Chapter 37 — Quiz

10 questions on Stokes' theorem, the Divergence theorem, their shared structure, and Maxwell's equations. Answers and section references in the <details> blocks. Aim to reason each one out before expanding the answer.

Q1. State Stokes' theorem, identifying which side is an integral over a boundary and which is an integral of a derivative.

Answer

$$\oint_{\partial S}\mathbf{F}\cdot d\mathbf{r} = \iint_S(\nabla\times\mathbf{F})\cdot d\mathbf{S}.$$ The left side is a line integral over the *boundary curve* $\partial S$ (circulation); the right side integrates the *derivative* $\nabla\times\mathbf{F}$ (curl) over the surface interior. This is the slogan "boundary integral = region integral of a derivative" in 3D. (§37.2)

Q2. Compute $\nabla\cdot\mathbf{F}$ and $\nabla\times\mathbf{F}$ for $\mathbf{F} = \langle -y,\, x,\, 0\rangle$.

Answer

$\nabla\cdot\mathbf{F} = \partial_x(-y) + \partial_y(x) + \partial_z(0) = 0$. The curl is $$\nabla\times\mathbf{F} = \langle\, 0-0,\ 0-0,\ 1-(-1)\,\rangle = \langle 0,0,2\rangle.$$ This is the standard counterclockwise swirl: zero divergence, constant upward curl. (§37.2, §37.3; curl/divergence from Chapter 34)

Q3. Verify Stokes' theorem for $\mathbf{F} = \langle -y, x, 0\rangle$ on the unit disk in the $xy$-plane, oriented upward. Give both sides.

Answer

Curl-flux: $(\nabla\times\mathbf{F})\cdot\hat{\mathbf{z}} = 2$, so $\iint_D 2\,dA = 2(\pi\cdot 1^2) = 2\pi$. Circulation: parametrize the unit circle as $\langle\cos t,\sin t,0\rangle$; the integrand is $\sin^2 t + \cos^2 t = 1$, so $\int_0^{2\pi}1\,dt = 2\pi$. Both sides equal $2\pi$. ✓ (§37.3, Example 1)

Q4. State the Divergence theorem and explain in words what it says physically.

Answer

$$\oiint_{\partial E}\mathbf{F}\cdot d\mathbf{S} = \iiint_E(\nabla\cdot\mathbf{F})\,dV.$$ Physically: the net flux *out through the skin* of a solid equals the total *source strength inside*. Sum every tiny source and sink in the volume and you get exactly the net amount escaping the boundary — nothing is created or destroyed in transit. The boundary must carry the *outward* normal. (§37.4)

Q5. Use the Divergence theorem to find the outward flux of $\mathbf{F} = \langle x, y, z\rangle$ through the unit sphere.

Answer

$\nabla\cdot\mathbf{F} = 1+1+1 = 3$, so the flux is $\iiint_E 3\,dV = 3\cdot\text{Vol(ball)} = 3\cdot\tfrac{4}{3}\pi = 4\pi$. A direct surface integral confirms it: on the unit sphere $\mathbf{F}\cdot\mathbf{n} = x^2+y^2+z^2 = 1$, so the flux is the surface area $4\pi$. ✓ (§37.5, Example 3)

Q6. Why is the right-hand side of Stokes' theorem the same for every surface sharing a given boundary curve? Name the underlying identity.

Answer

If two surfaces $S_1, S_2$ share a boundary, gluing them forms a *closed* surface bounding a solid $E$. By the Divergence theorem, $\oiint(\nabla\times\mathbf{F})\cdot d\mathbf{S} = \iiint_E\nabla\cdot(\nabla\times\mathbf{F})\,dV = \iiint_E 0\,dV = 0$, since **the divergence of a curl is zero**, $\nabla\cdot(\nabla\times\mathbf{F}) = 0$. So both curl-fluxes equal the same boundary line integral. This is why you may always swap a hard surface for a friendlier one with the same rim. (§37.2, §37.9)

Q7. Find the outward flux of $\mathbf{F} = \langle x^2, y^2, z^2\rangle$ through the surface of the unit cube $[0,1]^3$.

Answer

$\nabla\cdot\mathbf{F} = 2x + 2y + 2z$. The flux is $\iiint_{[0,1]^3}(2x+2y+2z)\,dV$. By symmetry each term contributes equally: $\iiint 2x\,dV = 2\cdot\tfrac12\cdot1\cdot1 = 1$, and likewise for $2y$ and $2z$, for a total of $3$. One volume integral replaced six face integrals. (§37.5, Example 4)

Q8. Which theorem converts the integral form of Faraday's law into its differential form $\nabla\times\mathbf{E} = -\partial_t\mathbf{B}$, and which converts Gauss's law into $\nabla\cdot\mathbf{E} = \rho/\varepsilon_0$?

Answer

**Stokes' theorem** converts Faraday's circulation law: $\oint_{\partial S}\mathbf{E}\cdot d\mathbf{r}$ becomes $\iint_S(\nabla\times\mathbf{E})\cdot d\mathbf{S}$, and matching integrands over every surface gives $\nabla\times\mathbf{E} = -\partial_t\mathbf{B}$. **The Divergence theorem** converts Gauss's flux law: $\oiint\mathbf{E}\cdot d\mathbf{S}$ becomes $\iiint(\nabla\cdot\mathbf{E})\,dV$, and matching integrands over every region gives $\nabla\cdot\mathbf{E} = \rho/\varepsilon_0$. The two theorems are the entire dictionary between the integral and differential columns of Maxwell's equations. (§37.7)

Q9. A common error: a student picks the upward normal $+\hat{\mathbf{z}}$ for a disk but walks the boundary clockwise (viewed from above) and gets $\oint = -2\pi$ while the surface integral gives $+2\pi$. What went wrong?

Answer

The orientations are inconsistent. Stokes' theorem requires the boundary to be traversed so the surface is on your left when your head points along $\mathbf{n}$ — equivalently, right-hand rule: fingers curl the way you walk, thumb along $\mathbf{n}$. With $+\hat{\mathbf{z}}$ the boundary must go *counterclockwise*. Walking clockwise flips exactly one side's sign. The theorem is fine; the bookkeeping wasn't. Always fix the normal first, then let the right hand choose the direction. (§37.3, Common Pitfall)

Q10. The four "big theorems" — FTC, Green's, Stokes', Divergence — are all one slogan. State the slogan, and name the chapter where each appears.

Answer

Slogan: **$\int_{\partial M}(\text{field}) = \int_M(\text{a derivative of the field})$** — the integral of a derivative over a region equals an integral over its boundary. FTC (Chapter 14, $M = [a,b]$, derivative $f'$); Green's theorem (Chapter 35, $M$ a planar region, derivative $Q_x - P_y$); Stokes' theorem (Chapter 37, $M$ a surface, derivative $\nabla\times\mathbf{F}$); Divergence theorem (Chapter 37, $M$ a solid, derivative $\nabla\cdot\mathbf{F}$). Chapter 38 collapses all four into the single equation $\int_{\partial M}\omega = \int_M d\omega$. (§37.6, §37.11)

Scoring Guide

Score	Interpretation
9–10	Mastery. You can state both theorems, verify them by hand, choose the easier side, and explain the Maxwell connection. Proceed to Chapter 38's unification with confidence.
7–8	Solid. Review whichever of orientation (Q9), surface-independence (Q6), or the Maxwell dictionary (Q8) tripped you, then move on.
5–6	Developing. Re-read §37.2–37.5 and rework the verification examples (Q3, Q5, Q7) until both sides land on the same number without effort.
0–4	Revisit the chapter from §37.1. Focus first on computing curl and divergence (Chapter 34) and on the slogan in §37.6 before attempting applications.