We are going to use Stokes’ 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.

The region $D$ for that we’ll need in converting the triple integral into iterated integrals is just the intersection of the two surfaces from the problem statement. This is,

$$0 = 6 - 2{x^2} - 2{y^2}\hspace{0.25in} \to \hspace{0.25in} {x^2} + {y^2} = 3$$

So, $D$ is a disk and so we’ll eventually be doing cylindrical coordinates for this integral. Here are the cylindrical limits for the region $E$.

$$\begin{array}{c}0 \le \theta \le 2\pi \\ 0 \le r \le \sqrt 3 \\ 0 \le z \le 6 - 2{x^2} - 2{y^2} = 6 - 2{r^2}\end{array}$$

Don’t forget to convert the $z$ limits into cylindrical coordinates as well!

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

$${\mathop{\rm div}\nolimits} \vec F = \frac{\partial }{{\partial x}}\left( {2xz} \right) + \frac{\partial }{{\partial y}}\left( {1 - 4x{y^2}} \right) + \frac{\partial }{{\partial z}}\left( {2z - {z^2}} \right) = 2 - 8xy$$ Show Step 3

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

$$\begin{align*}\iint\limits_{S}{{\vec F\centerdot d\vec S}} & = \iiint\limits_{E}{{{\mathop{\rm div}\nolimits} \vec F\,dV}}\\ & = \iiint\limits_{E}{{2 - 8xy\,dV}}\\ & = \int_{0}^{{2\pi }}{{\int_{0}^{{\sqrt 3 }}{{\int_{0}^{{6 - 2{r^2}}}{{\left( {2 - 8{r^2}\cos \theta \sin \theta } \right)r\,dz}}\,dr}}\,d\theta }}\\ & = \int_{0}^{{2\pi }}{{\int_{0}^{{\sqrt 3 }}{{\int_{0}^{{6 - 2{r^2}}}{{2r - 8{r^3}\cos \theta \sin \theta \,dz}}\,dr}}\,d\theta }}\end{align*}$$

Don’t forget to pick up the $r$ when converting the $dV$ to cylindrical coordinates.

Show Step 4

All we need to do then in evaluate the integral.

$$\begin{align*}\iint\limits_{S}{{\vec F\centerdot d\vec S}} & = \int_{0}^{{2\pi }}{{\int_{0}^{{\sqrt 3 }}{{\int_{0}^{{6 - 2{r^2}}}{{2r - 8{r^3}\cos \theta \sin \theta \,dz}}\,dr}}\,d\theta }}\\ & = \int_{0}^{{2\pi }}{{\int_{0}^{{\sqrt 3 }}{{\left. {\left( {2r - 8{r^3}\cos \theta \sin \theta } \right)z} \right|_0^{6 - 2{r^2}}\,dr}}\,d\theta }}\\ & = \int_{0}^{{2\pi }}{{\int_{0}^{{\sqrt 3 }}{{\left( {2r - 8{r^3}\cos \theta \sin \theta } \right)\left( {6 - 2{r^2}} \right)\,dr}}\,d\theta }}\\ & = \int_{0}^{{2\pi }}{{\int_{0}^{{\sqrt 3 }}{{12r - 4{r^3} - \left( {48{r^3} - 16{r^5}} \right)\cos \theta \sin \theta \,dr}}\,d\theta }}\\ & = \int_{0}^{{2\pi }}{{\left. {\left( {6{r^2} - {r^4} - \left( {12{r^4} - \frac{8}{3}{r^6}} \right)\cos \theta \sin \theta } \right)} \right|_0^{\sqrt 3 }\,d\theta }}\\ & = \int_{0}^{{2\pi }}{{9 - 36\cos \theta \sin \theta \,d\theta }}\\ & = \int_{0}^{{2\pi }}{{9 - 18\sin \left( {2\theta } \right)\,d\theta }}\\ & = \left. {\left( {9\theta + 9\cos \left( {2\theta } \right)} \right)} \right|_0^{2\pi } = \require{bbox} \bbox[2pt,border:1px solid black]{{18\pi }}\end{align*}$$