Question 5 of 8: Closed-Surface Flux Through a Paraboloid Cap via the Divergence Theorem

Question 5: 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 (curved paraboloid cap plus flat disk base) of the solid bounded above by $z=4-x^2-y^2$ and below by $z=0$. $\mathbf F=(yz,\,-2xy,\,3z)$.

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

Approach. $S$ is described as "the surface of the region," i.e. the entire closed boundary of a solid, so apply the divergence theorem, $\iint_S\mathbf F\cdot d\mathbf S=\iiint_V\nabla\cdot\mathbf F\,dV$, instead of separately parametrizing the curved cap and the flat disk.

1. Divergence. $$\nabla\cdot\mathbf F=\frac{\partial(yz)}{\partial x}+\frac{\partial(-2xy)}{\partial y}+\frac{\partial(3z)}{\partial z}=0-2x+3=3-2x.$$
2. Volume of the solid. The paraboloid meets $z=0$ where $x^2+y^2=4$, so the base disk has radius 2. In cylindrical coordinates, $$V=\int_0^{2\pi}\!\!\int_0^2(4-r^2)\,r\,dr\,d\theta=2\pi\Big[2r^2-\tfrac{r^4}{4}\Big]_0^2=2\pi(8-4)=8\pi.$$
3. Integrate the divergence. The solid is symmetric under $x\to-x$, so $\iiint_V(-2x)\,dV=0$ by symmetry, leaving only the constant term: $$\iiint_V(3-2x)\,dV=3V-0=3(8\pi)=24\pi.$$

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