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!