Closed-Surface Flux Integral via the Divergence Theorem (Paraboloid Cap): 04-BS-1, May 2018, Q7

Question 7: Closed-Surface Flux Integral via the Divergence Theorem (Paraboloid Cap) (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Closed surface $S$: the paraboloid $z=4-x^2-y^2$ capped below by the disk $z=0$, together bounding a solid $V$ (this requires $z\ge0\Rightarrow x^2+y^2\le4$). $\mathbf F=(yz,\,-2xy,\,3z)$.

Find. The total outward flux $\displaystyle\iint_S\mathbf F\cdot d\mathbf S$.

Paraboloid cap $z=4-x^2-y^2$ over $z\ge0$, closed below by the disk $x^2+y^2\le4$ at $z=0$; $S$ is the full closed boundary.

Approach. $S$ is closed (bounds the solid $V$ under the paraboloid, above the plane), so apply the divergence theorem rather than parametrizing the paraboloid and the disk separately.

Compute the divergence. $$\nabla\cdot\mathbf F=\frac{\partial(yz)}{\partial x}+\frac{\partial(-2xy)}{\partial y}+\frac{\partial(3z)}{\partial z}=0-2x+3=3-2x.$$
Set up the volume integral. $V$ is $0\le z\le4-x^2-y^2$ over the disk $x^2+y^2\le4$. By symmetry of the disk about $x=0$, $\iiint_Vx\,dV=0$, so only the constant term survives: $$\iiint_V(3-2x)\,dV=3\iiint_VdV=3\,\mathrm{Vol}(V).$$
Compute the paraboloid-cap volume. In polar coordinates, $$\mathrm{Vol}(V)=\int_0^{2\pi}\!\!\int_0^2(4-r^2)\,r\,dr\,d\theta=2\pi\left[2r^2-\frac{r^4}4\right]_0^2=2\pi(8-4)=8\pi.$$
Total flux. $$\iint_S\mathbf F\cdot d\mathbf S=3(8\pi)=24\pi.$$

$$\iint_S\mathbf F\cdot d\mathbf S=\boxed{24\pi}$$

Quantity	Result
$\nabla\cdot\mathbf F$	$3-2x$
$\mathrm{Vol}(V)$	$8\pi$
Total flux	$24\pi$

Concept & Theory

Topic. Divergence (Gauss's) theorem for a closed paraboloid-cap surface.

Key relations. $\displaystyle\iint_{\partial V}\mathbf F\cdot d\mathbf S=\iiint_V(\nabla\cdot\mathbf F)\,dV$; the volume under a paraboloid $z=a-r^2$ capped at $z=0$ is $V=\pi a^2/2$.

Why this works. $S$ is described as "the surface of the region bounded above/below," i.e. the full closed boundary, so the divergence theorem replaces two separate surface integrals (curved paraboloid + flat disk, each needing its own parametrization and normal) with a single volume integral. The odd-in-$x$ term in the divergence integrates to zero over the disk by symmetry, leaving only a constant times the volume.

Common pitfall. Parametrizing the paraboloid and disk separately and computing two surface integrals instead of recognizing the closed-surface shortcut; forgetting that the region requires $z\ge0$, which caps the disk radius at $r=2$ (where the paraboloid meets $z=0$), not some other radius.

Source. Stewart, Calculus: Early Transcendentals, Ch. 16 (§16.9 The Divergence Theorem); Kreyszig, Advanced Engineering Mathematics, Ch. 10 (§10.7 Gauss's Divergence Theorem).

Question 7 of 8: Closed-Surface Flux Integral via the Divergence Theorem (Paraboloid Cap)

Question 7: Closed-Surface Flux Integral via the Divergence Theorem (Paraboloid Cap) (20 marks)