Question 7 of 8: Closed-Surface Flux Through a Paraboloid Cap via the Divergence Theorem

Question 7: 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 opening upward, capped by the plane $z=2$) and vector field $\mathbf F=(xy^2,2xyz,-xz^2)$.

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

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

1. Divergence and enclosed region. $$\operatorname{div}\mathbf F=\frac{\partial(xy^2)}{\partial x}+\frac{\partial(2xyz)}{\partial y}+\frac{\partial(-xz^2)}{\partial z}=y^2+2xz-2xz=y^2.$$ The paraboloid meets $z=2$ where $r^2-2=2\Rightarrow r=2$, so the solid is $0\le r\le2$, $r^2-2\le z\le2$.
2. Inner ($z$) integral. With $y=r\sin\theta$, the integrand $y^2=r^2\sin^2\theta$ doesn't depend on $z$, so integrating over the $z$-range of length $2-(r^2-2)=4-r^2$: $$\int_{r^2-2}^{2}r^2\sin^2\theta\,dz=r^2\sin^2\theta\,(4-r^2).$$
3. Radial and angular integrals. $$\iiint_V y^2\,dV=\int_0^{2\pi}\!\!\int_0^2r^2\sin^2\theta\,(4-r^2)\,r\,dr\,d\theta=\left(\int_0^{2\pi}\sin^2\theta\,d\theta\right)\left(\int_0^2r^3(4-r^2)\,dr\right).$$
4. Evaluate each factor. $\displaystyle\int_0^{2\pi}\sin^2\theta\,d\theta=\pi$. $\displaystyle\int_0^2r^3(4-r^2)\,dr=\int_0^2(4r^3-r^5)\,dr=\left[r^4-\frac{r^6}{6}\right]_0^2=16-\frac{32}{3}=\frac{16}{3}$.
5. Combine. $$\iint_S\mathbf F\cdot\mathbf n\,dA=\pi\cdot\frac{16}{3}=\boxed{\dfrac{16\pi}{3}}\approx16.76$$