Chapter 36 — Quiz

Ten questions covering parametrized surfaces, the area element $dS$, scalar surface integrals, orientation, and flux. Try each before opening the answer. Section references point back to the relevant part of the chapter.

1. A parametrized surface needs how many parameters, and what does its area element $dS$ equal?

  • A) One parameter; $dS = \|\mathbf{r}'(t)\|\,dt$
  • B) Two parameters; $dS = \|\mathbf{r}_u \times \mathbf{r}_v\|\,du\,dv$
  • C) Two parameters; $dS = (\mathbf{r}_u\cdot\mathbf{r}_v)\,du\,dv$
  • D) Three parameters; $dS = du\,dv\,dw$
Answer**B.** A surface is two-dimensional, so it needs two parameters, and the area element is the magnitude of the cross product of the tangent vectors. *(Sections 36.2–36.3.)*

2. For a graph $z = g(x,y)$, the surface area element is:

  • A) $dA$
  • B) $\sqrt{1 + g_x^2 + g_y^2}\,dA$
  • C) $(1 + g_x + g_y)\,dA$
  • D) $g_x g_y\,dA$
Answer**B.** From $\mathbf{r}_x\times\mathbf{r}_y = \langle -g_x,-g_y,1\rangle$, the magnitude is $\sqrt{1+g_x^2+g_y^2}$. Writing $dS=dA$ (option A) drops the tilt factor — the chapter's flagged Common Pitfall. *(Section 36.3.)*

3. The surface area element on a sphere of radius $R$ (spherical angles $\phi,\theta$) is:

  • A) $R\,d\phi\,d\theta$
  • B) $R^2\,d\phi\,d\theta$
  • C) $R^2\sin\phi\,d\phi\,d\theta$
  • D) $\sin\phi\,d\phi\,d\theta$
Answer**C.** $\|\mathbf{r}_\phi\times\mathbf{r}_\theta\| = R^2\sin\phi$. Integrating over the full sphere recovers $4\pi R^2$. *(Section 36.3.)*

4. The scalar surface integral $\iint_S f\,dS$ is computed as which double integral over the parameter domain $D$?

  • A) $\iint_D f(\mathbf{r}(u,v))\,du\,dv$
  • B) $\iint_D f(\mathbf{r}(u,v))\,\|\mathbf{r}_u\times\mathbf{r}_v\|\,du\,dv$
  • C) $\iint_D f(\mathbf{r}(u,v))\,(\mathbf{r}_u\times\mathbf{r}_v)\,du\,dv$
  • D) $\iint_D f(\mathbf{r}(u,v))\,(\mathbf{r}_u\cdot\mathbf{r}_v)\,du\,dv$
Answer**B.** Substitute the parametrization into $f$, multiply by the scalar stretching factor, and integrate over the flat domain $D$. *(Section 36.4.)*

5. Flux is defined as $\iint_S\mathbf{F}\cdot d\mathbf{S} = \iint_S(\mathbf{F}\cdot\hat{\mathbf{n}})\,dS$. The integrand $\mathbf{F}\cdot\hat{\mathbf{n}}$ measures:

  • A) the part of $\mathbf{F}$ tangent to $S$
  • B) the magnitude $\|\mathbf{F}\|$
  • C) the component of $\mathbf{F}$ piercing $S$ (along the normal)
  • D) the area of $S$
Answer**C.** The normal component is the part that actually crosses the surface; the tangential part slides along $S$ and contributes nothing. *(Section 36.6.)*

6. For a graph $z = g(x,y)$ with upward normal and $\mathbf{F} = \langle P,Q,R\rangle$, the flux equals:

  • A) $\iint_D(P g_x + Q g_y + R)\,dA$
  • B) $\iint_D(-P g_x - Q g_y + R)\,dA$
  • C) $\iint_D R\,dA$
  • D) $\iint_D\sqrt{1+g_x^2+g_y^2}\,dA$
Answer**B.** This comes from dotting $\mathbf{F}$ with $\mathbf{r}_x\times\mathbf{r}_y = \langle -g_x,-g_y,1\rangle$; no square root appears because the working flux formula uses the un-normalized normal. *(Section 36.7.)*

7. The outward flux of the inverse-square field $\mathbf{F} = \mathbf{r}/\|\mathbf{r}\|^3$ through a sphere of radius $R$ centered at the origin is:

  • A) $0$
  • B) $4\pi$ (independent of $R$)
  • C) $4\pi R^2$
  • D) $\infty$
Answer**B.** $\mathbf{F}\cdot\hat{\mathbf n} = 1/R^2$, times area $4\pi R^2$ gives $4\pi$; the radius cancels. This is Gauss's law in disguise. *(Section 36.6.)*

8. Gauss's law (integral form) states that the electric flux through any closed surface equals:

  • A) $0$ always
  • B) $Q_{\text{enc}}/\varepsilon_0$
  • C) $\nabla\cdot\mathbf{E}$
  • D) $q/(4\pi\varepsilon_0 r^2)$
Answer**B.** Flux through a closed surface equals the enclosed charge divided by $\varepsilon_0$ — only charges *inside* count. *(Section 36.9.)*

9. A Möbius band is the standard example of a surface that is:

  • A) closed
  • B) non-orientable (no consistent unit normal)
  • C) flat
  • D) not actually a surface
Answer**B.** Sliding a normal once around the loop reverses it, so no global $\hat{\mathbf n}$ exists and flux is undefined there. *(Section 36.5.)*

10. Heat flux follows Fourier's law $\mathbf{q} = -k\nabla T$. The rate of heat energy crossing a surface $S$ is:

  • A) $\iint_S k T\,dS$
  • B) $\iint_S(-k\nabla T)\cdot d\mathbf{S}$
  • C) $\iiint_E\nabla\cdot\mathbf{q}\,dV$
  • D) $\oint_C\mathbf{q}\cdot d\mathbf{r}$
Answer**B.** The heat rate is the flux of $\mathbf{q} = -k\nabla T$ through $S$; the minus sign sends heat from hot to cold. *(Section 36.8.)*

Scoring Guide

  • 9–10 correct: Excellent — you have command of parametrization, the area element, and flux.
  • 7–8 correct: Solid. Revisit the one or two ideas you missed (often the graph flux sign or the $dS$ factor).
  • 5–6 correct: Reread Sections 36.3 (the area element) and 36.6–36.7 (flux), then redo the exercises in Parts B and E.
  • Below 5: Work back through the chapter's worked examples before moving on to Chapter 37.