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

Question 8: 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. Closed surface $S$: the full boundary (paraboloid cup plus flat cap) of the solid bounded below by $z=x^2+y^2-2$ and above by $z=2$. $\mathbf F=(xy^2,\,2xyz,\,-xz^2)$.

Find. $\displaystyle\iint_S\mathbf F\cdot\mathbf n\,dA$.

Approach. $S$ is described as the full boundary of an enclosed solid region, so apply the divergence theorem, $\iint_S\mathbf F\cdot\mathbf n\,dA=\iiint_V\nabla\cdot\mathbf F\,dV$, instead of parametrizing the curved paraboloid and flat cap separately.

1. Divergence of $\mathbf F$. $$\nabla\cdot\mathbf F=\frac{\partial(xy^2)}{\partial x}+\frac{\partial(2xyz)}{\partial y}+\frac{\partial(-xz^2)}{\partial z}=y^2+2xz-2xz=y^2.$$ The $2xz$ terms cancel exactly, leaving a purely $y$-dependent divergence.
2. Describe the solid in cylindrical coordinates. The paraboloid $z=r^2-2$ meets the cap $z=2$ where $r^2-2=2\Rightarrow r=2$. So the solid is $0\le r\le2,\ 0\le\theta\le2\pi,\ r^2-2\le z\le2$, with height $4-r^2$ at radius $r$.
3. Set up the volume integral of $y^2=r^2\sin^2\theta$. $$\iiint_Vy^2\,dV=\int_0^{2\pi}\!\!\sin^2\theta\,d\theta\int_0^2 r^2\cdot(4-r^2)\cdot r\,dr=\left(\int_0^{2\pi}\sin^2\theta\,d\theta\right)\left(\int_0^2 r^3(4-r^2)\,dr\right).$$
4. Evaluate each factor. $\displaystyle\int_0^{2\pi}\sin^2\theta\,d\theta=\pi$. $\displaystyle\int_0^2 r^3(4-r^2)\,dr=\int_0^2(4r^3-r^5)\,dr=\Big[r^4-\tfrac{r^6}{6}\Big]_0^2=16-\tfrac{64}{6}=\tfrac{16}{3}$.
5. Combine. $$\iint_S\mathbf F\cdot\mathbf n\,dA=\pi\cdot\frac{16}{3}=\boxed{\dfrac{16\pi}{3}}.$$