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Section 17.6 : Divergence Theorem

3. Use the Divergence Theorem to evaluate \( \displaystyle \iint\limits_{S}{{\vec F\centerdot d\vec S}}\) where \(\vec F = 2xz\vec i + \left( {1 - 4x{y^2}} \right)\,\vec j + \left( {2z - {z^2}} \right)\,\vec k\) and \(S\) is the surface of the solid bounded by \(z = 6 - 2{x^2} - 2{y^2}\) and the plane \(z = 0\) . Note that both of the surfaces of this solid included in \(S\).

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Start Solution

Let’s start off with a quick sketch of the surface we are working with in this problem.

We included a sketch with traditional axes and a sketch with a set of “box” axes to help visualize the surface. The bottom “cap” of the elliptic paraboloid is also included in the surface but isn’t shown.

Note as well here that because we are including both of the surfaces shown above that the surface does enclose (or is the boundary curve if you want to use that terminology) the region.

Show Step 2

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.

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