Question 6 of 8: Flux Through a Closed Surface (Divergence Theorem)

Question 6: Flux Through a Closed Surface (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. $\mathbf F=(xz,\,-2y,\,3x)$; $S$ = the full closed boundary (paraboloid cap $+$ base disk) of the solid $0\le z\le4-x^2-y^2$.

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

Approach. $S$ is closed, so apply the divergence theorem: $\iint_S\mathbf F\cdot d\mathbf S=\iiint_V\operatorname{div}\mathbf F\,dV$, evaluated in cylindrical coordinates over the paraboloid-capped disk $x^2+y^2\le4$.

1. Divergence. $$\operatorname{div}\mathbf F=\frac{\partial}{\partial x}(xz)+\frac{\partial}{\partial y}(-2y)+\frac{\partial}{\partial z}(3x)=z-2.$$
2. Set up the cylindrical triple integral. With $x=r\cos\theta,\ y=r\sin\theta$, $0\le r\le2$, $0\le z\le4-r^2$: $$\iiint_V(z-2)\,dV=\int_0^{2\pi}\!\!\int_0^2\!\!\int_0^{4-r^2}(z-2)\,r\,dz\,dr\,d\theta.$$
3. Integrate over $z$. $$\int_0^{4-r^2}(z-2)\,dz=\left[\frac{z^2}2-2z\right]_0^{4-r^2}=\frac{(4-r^2)^2}2-2(4-r^2)=\frac{(4-r^2)\bigl[(4-r^2)-4\bigr]}2=-\frac{r^2(4-r^2)}2.$$
4. Integrate over $r$ and $\theta$. $$\iiint_V(z-2)\,dV=2\pi\int_0^2\!\left(-\frac{r^2(4-r^2)}2\right)r\,dr=-\pi\int_0^2\bigl(4r^3-r^5\bigr)\,dr=-\pi\left[r^4-\frac{r^6}6\right]_0^2=-\pi\left(16-\frac{64}6\right)=-\pi\cdot\frac{16}3.$$

$$\iint_S\mathbf F\cdot d\mathbf S=\boxed{-\dfrac{16\pi}3}$$