Question 4 of 8: Closed-Surface Flux Through a Cone via the Divergence Theorem

Question 4: Closed-Surface Flux Through a Cone 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 (slanted cone surface plus flat base disk) of the solid bounded above by $z=4-\sqrt{x^2+y^2}$ and below by $z=0$. $\mathbf F=(4x,\,2x^2,\,-3)$.

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 slanted cone surface and the flat disk base.

1. Divergence. $$\nabla\cdot\mathbf F=\frac{\partial(4x)}{\partial x}+\frac{\partial(2x^2)}{\partial y}+\frac{\partial(-3)}{\partial z}=4+0+0=4.$$
2. Volume of the cone. The cone $z=4-r$ meets $z=0$ at $r=4$ (base radius 4), with apex height 4: $$V=\frac13\pi r_{\text{base}}^2h=\frac13\pi(4)^2(4)=\frac{64\pi}{3}.$$
3. Integrate the (constant) divergence. $$\iiint_V4\,dV=4\cdot\frac{64\pi}{3}=\frac{256\pi}{3}.$$

$$\iint_S\mathbf F\cdot d\mathbf S=\boxed{\dfrac{256\pi}{3}}$$