We are going to use the Divergence Theorem in the following direction.

$$\iint\limits_{S}{{\vec F\centerdot d\vec S}} = \iiint\limits_{E}{{{\mathop{\rm div}\nolimits} \vec F\,dV}}$$

where $E$ is just the solid shown in the sketches from Step 1.

Because $E$ is a portion of a sphere we’ll be wanting to use spherical coordinates for the integration. Here are the spherical limits we’ll need to use for this region.

$$\begin{array}{c}\pi \le \theta \le 2\pi \\ \displaystyle \frac{\pi }{2} \le \varphi \le \pi \\ 0 \le \rho \le 4\end{array}$$

One of the restrictions on the region in the problem statement was $y \le 0$. This means that if we look at this from above we’d see the portion of the circle of radius 4 that is below the $x$ axis and so we need the given range of $\theta$ above to cover this region.

We were also told in the problem statement that $z \le 0$ and so we only want the portion of the sphere that is below the $xy$-plane. We therefore need the given range of $\varphi$ to make sure we are only below the $xy$-plane.

We’ll also need the divergence of the vector field so here is that.

$${\mathop{\rm div}\nolimits} \vec F = \frac{\partial }{{\partial x}}\left( {y{x^2}} \right) + \frac{\partial }{{\partial y}}\left( {x{y^2} - 3{z^4}} \right) + \frac{\partial }{{\partial z}}\left( {{x^3} + {y^2}} \right) = 4xy$$ Show Step 3

Now let’s apply the Divergence Theorem to the integral and get it converted to spherical coordinates while we’re at it.

$$\begin{align*}\iint\limits_{S}{{\vec F\centerdot d\vec S}} & = \iiint\limits_{E}{{{\mathop{\rm div}\nolimits} \vec F\,dV}}\\ & = \iiint\limits_{E}{{4xy\,dV}}\\ & = \int_{\pi }^{{2\pi }}{{\int_{{\frac{1}{2}\pi }}^{\pi }{{\int_{0}^{4}{{4\left( {\rho \sin \varphi \cos \theta } \right)\left( {\rho \sin \varphi \sin \theta } \right)\left( {{\rho ^2}\sin \varphi } \right)\,d\rho }}\,d\varphi }}\,d\theta }}\\ & = \int_{\pi }^{{2\pi }}{{\int_{{\frac{1}{2}\pi }}^{\pi }{{\int_{0}^{4}{{4{\rho ^4}{{\sin }^3}\varphi \cos \theta \sin \theta \,d\rho }}\,d\varphi }}\,d\theta }}\end{align*}$$

Don’t forget to pick up the ${\rho ^2}\sin \varphi$ when converting the $dV$ to spherical coordinates.

Show Step 4

All we need to do then in evaluate the integral.

$$\begin{align*}\iint\limits_{S}{{\vec F\centerdot d\vec S}} & = \int_{\pi }^{{2\pi }}{{\int_{{\frac{1}{2}\pi }}^{\pi }{{\int_{0}^{4}{{4{\rho ^4}{{\sin }^3}\varphi \cos \theta \sin \theta \,d\rho }}\,d\varphi }}\,d\theta }}\\ & = \int_{\pi }^{{2\pi }}{{\int_{{\frac{1}{2}\pi }}^{\pi }{{\left. {\left( {\frac{4}{5}{\rho ^5}{{\sin }^3}\varphi \cos \theta \sin \theta } \right)} \right|_0^4\,d\varphi }}\,d\theta }}\\ & = \int_{\pi }^{{2\pi }}{{\int_{{\frac{1}{2}\pi }}^{\pi }{{\frac{{4096}}{5}\sin \varphi \left( {1 - {{\cos }^2}\varphi } \right)\cos \theta \sin \theta \,d\varphi }}\,d\theta }}\\ & = \int_{\pi }^{{2\pi }}{{\left. {\left( { - \frac{{4096}}{5}\left( {\cos \varphi - \frac{1}{3}{{\cos }^3}\varphi } \right)\cos \theta \sin \theta } \right)} \right|_{\frac{1}{2}\pi }^\pi \,d\theta }}\\ & = \int_{\pi }^{{2\pi }}{{\frac{{8192}}{{15}}\cos \theta \sin \theta \,d\theta }}\\ & = \int_{\pi }^{{2\pi }}{{\frac{{4096}}{{15}}\sin \left( {2\theta } \right)\,d\theta }}\\ & = \left. {-\frac{{2048}}{{15}}\cos \left( {2\theta } \right)} \right|_\pi ^{2\pi } = \require{bbox} \bbox[2pt,border:1px solid black]{0}\end{align*}$$

Make sure you can do use your trig formulas as we did here to deal with these kinds of integrals!