Question 5 of 8: Closed-Surface Flux Integral via the Divergence Theorem

Question 5: Closed-Surface Flux Integral 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. Solid region: the quarter of the elliptical cylinder $x^2+4y^2\leq1$ lying in the first quadrant ($x\geq0,y\geq0$), extruded from $z=0$ to $z=4$. $S$ is the entire boundary (closed) surface of that solid: the curved elliptical side, the two flat rectangular cut faces ($x=0$ and $y=0$), and the top/bottom quarter-ellipse caps. $\mathbf F=(y^3,x^3,z^3)$.

Find. The total outward flux $\displaystyle\oiint_S\mathbf F\cdot d\mathbf S$.

Approach. $S$ is a closed surface (bounds a solid region $V$), so apply the divergence theorem, $\oiint_S\mathbf F\cdot d\mathbf S=\iiint_V\nabla\cdot\mathbf F\,dV$, rather than parametrizing four separate pieces.

1. Compute the divergence. $$\nabla\cdot\mathbf F=\frac{\partial(y^3)}{\partial x}+\frac{\partial(x^3)}{\partial y}+\frac{\partial(z^3)}{\partial z}=0+0+3z^2=3z^2.$$ The $x$- and $y$-components have no dependence on $x,y$ respectively, so both cross-partials vanish — only the $z^3$ term survives.
2. Find the cross-sectional area. The full ellipse $x^2+4y^2\leq1$ has semi-axes $a=1$ (along $x$) and $b=\tfrac12$ (along $y$), so its area is $\pi ab=\tfrac{\pi}{2}$. The region $x\geq0,y\geq0$ is exactly one quadrant of it, so $$\text{Area}=\frac14\cdot\frac{\pi}{2}=\frac{\pi}{8}.$$
3. Integrate over the solid. Since $3z^2$ depends only on $z$, the triple integral separates into (cross-sectional area) $\times$ ($z$-integral): $$\iiint_V 3z^2\,dV=\left(\frac{\pi}{8}\right)\int_0^4 3z^2\,dz=\frac{\pi}{8}\cdot\left[z^3\right]_0^4=\frac{\pi}{8}\cdot64=8\pi.$$

$$\oiint_S\mathbf F\cdot d\mathbf S=\boxed{8\pi}$$