Closed-Surface Flux Through a Paraboloid Cap via the Divergence Theorem: 04-BS-1, December 2016, Q6

Question 6: Closed-Surface Flux Through a Paraboloid Cap via the Divergence Theorem 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. A closed surface $S$ (paraboloid cap over the plane $z=0$) and vector field $\mathbf F=(xz,-2y,3x)$.

The paraboloid cap ($z=4-r^2$) sits over the disk $r\le2$ in $z=0$; together they bound a closed solid, so the divergence theorem applies to the full outward flux.

Find. $\displaystyle\iint_S\mathbf F\cdot d\mathbf S$.

Approach. Since $S$ is closed, apply the divergence theorem and integrate $\operatorname{div}\mathbf F$ over the enclosed solid using cylindrical coordinates.

Divergence and solid region. $\operatorname{div}\mathbf F=\dfrac{\partial(xz)}{\partial x}+\dfrac{\partial(-2y)}{\partial y}+\dfrac{\partial(3x)}{\partial z}=z-2+0=z-2$. The enclosed solid is $0\le z\le4-r^2$, $0\le r\le2$ (where $z\ge0\Rightarrow r\le2$).
Inner ($z$) integral. With $h=4-r^2$, $$\int_0^h(z-2)\,dz=\frac{h^2}{2}-2h=\frac{h(h-4)}{2}=\frac{(4-r^2)(-r^2)}{2}=\frac{r^4-4r^2}{2}.$$
Radial and angular integrals. $$\iiint_V(z-2)\,dV=\int_0^{2\pi}\!\!\int_0^2\frac{r^4-4r^2}{2}\,r\,dr\,d\theta=2\pi\int_0^2\frac{r^5-4r^3}{2}\,dr=\pi\int_0^2(r^5-4r^3)\,dr.$$
Evaluate. $\displaystyle\int_0^2r^5\,dr=\frac{64}{6}=\frac{32}{3}$, $\displaystyle\int_0^2r^3\,dr=4$, so $$\iint_S\mathbf F\cdot d\mathbf S=\pi\left(\frac{32}{3}-16\right)=\pi\left(\frac{32-48}{3}\right)=\boxed{-\dfrac{16\pi}{3}}\approx-16.76$$

Quantity	Result
$\operatorname{div}\mathbf F$	$z-2$
Enclosed region	$0\le r\le2$, $0\le z\le4-r^2$
Flux $\iint_S\mathbf F\cdot d\mathbf S$	$-\dfrac{16\pi}{3}\approx-16.76$

Concept & Theory

Topic. The divergence theorem applied to a closed surface built from a paraboloid cap and a flat disk.

Key relations. $\displaystyle\iint_S\mathbf F\cdot d\mathbf S=\iiint_V\operatorname{div}\mathbf F\,dV$ for $S$ the closed boundary of a solid $V$, with $\mathbf F\cdot d\mathbf S$ taken in the outward-normal sense — this replaces two separate surface-integral computations (over the curved cap and the flat disk) with a single, usually simpler, volume integral.

Why this works. $S$ here consists of two pieces (paraboloid cap $+$ disk base), so a direct surface integral would require parametrizing and orienting each piece separately and adding; the divergence theorem sidesteps that entirely, and because $\operatorname{div}\mathbf F=z-2$ depends only on $z$, cylindrical coordinates reduce the triple integral to two elementary polynomial integrals in $r$.

Common pitfall. Forgetting the theorem requires a closed surface and attempting to apply it directly to just the curved cap; sign errors in $h(h-4)/2$ when substituting $h=4-r^2$; dropping the extra factor of $r$ from the cylindrical volume element $dV=r\,dr\,d\theta\,dz$.

Source. Kreyszig, Advanced Engineering Mathematics, Ch. 10 (§10.7 Divergence Theorem of Gauss).

Question 6 of 8: Closed-Surface Flux Through a Paraboloid Cap via the Divergence Theorem

Question 6: Closed-Surface Flux Through a Paraboloid Cap via the Divergence Theorem 20 marks