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### Section 15.6 : Triple Integrals in Cylindrical Coordinates

1. Evaluate $$\displaystyle \iiint\limits_{E}{{4xy\,dV}}$$ where $$E$$ is the region bounded by $$z = 2{x^2} + 2{y^2} - 7$$ and $$z = 1$$.

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Okay, let’s start off with a quick sketch of the region $$E$$ so we can get a feel for what we’re dealing with.

We’ve given the sketches with a set of “traditional” axes as well as a set of “box” axes to help visualize the surface and region.

Show Step 2

So, from the sketch above it should be pretty clear that we’ll need to integrate $$z$$ first and so we’ll have the following limits for $$z$$.

$2{x^2} + 2{y^2} - 7 \le z \le 1$ Show Step 3

For this problem $$D$$ is the disk that “caps” the region sketched in Step 1. We can determine the equation of the disk by setting the two equations from the problem statement equal and doing a little rewriting.

$2{x^2} + 2{y^2} - 7 = 1\hspace{0.25in} \to \hspace{0.25in}2{x^2} + 2{y^2} = 8\hspace{0.25in} \to \hspace{0.25in}{x^2} + {y^2} = 4$

So, $$D$$ is the disk $${x^2} + {y^2} \le 4$$ and it should be pretty clear that we’ll need to use cylindrical coordinates for this integral.

Here are the cylindrical coordinates for this problem.

$\begin{array}{c}0 \le \theta \le 2\pi \\ 0 \le r \le 2\\ 2{r^2} - 7 \le z \le 1\end{array}$

Don’t forget to convert the $$z$$ limits from Step 2 into cylindrical coordinates as well.

Show Step 4

Plugging these limits into the integral and converting to cylindrical coordinates gives,

\begin{align*}\iiint\limits_{E}{{4xy\,dV}} & = \int_{0}^{{2\pi }}{{\int_{0}^{2}{{\int_{{2{r^2} - 7}}^{1}{{4\left( {r\cos \theta } \right)\left( {r\sin \theta } \right)r\,dz}}\,dr}}\,d\theta }}\\ & = \int_{0}^{{2\pi }}{{\int_{0}^{2}{{\int_{{2{r^2} - 7}}^{1}{{4{r^3}\cos \theta \sin \theta \,dz}}\,dr}}\,d\theta }}\end{align*}

Don’t forget to convert the $$x$$ and $$y$$’s into cylindrical coordinates and also don’t forget that $$dV = r\,dz\,dr\,d\theta$$ and so we pick up another $$r$$ when converting the $$dV$$ to cylindrical coordinates.

Show Step 5

Okay, now all we need to do is evaluate the integral. Here is the $$z$$ integration.

\begin{align*}\iiint\limits_{E}{{4xy\,dV}} & = \int_{0}^{{2\pi }}{{\int_{0}^{2}{{\left. {\left( {4{r^3}\cos \theta \sin \theta z} \right)} \right|_{2{r^2} - 7}^1\,dr}}\,d\theta }}\\ & = \int_{0}^{{2\pi }}{{\int_{0}^{2}{{4{r^3}\left( {8 - 2{r^2}} \right)\cos \theta \sin \theta \,dr}}\,d\theta }}\\ & = \int_{0}^{{2\pi }}{{\int_{0}^{2}{{\left( {32{r^3} - 8{r^5}} \right)\cos \theta \sin \theta \,dr}}\,d\theta }}\end{align*} Show Step 6

Next let’s do the $$r$$ integration.

\begin{align*}\iiint\limits_{E}{{4xy\,dV}} & = \int_{0}^{{2\pi }}{{\left. {\left( {8{r^4} - \frac{4}{3}{r^6}} \right)\cos \theta \sin \theta } \right|_0^2\,d\theta }}\\ & = \int_{0}^{{2\pi }}{{\frac{{128}}{3}\cos \theta \sin \theta \,d\theta }}\end{align*} Show Step 7

Finally, we’ll do the $$\theta$$ integration.

$\iiint\limits_{E}{{4xy\,dV}} = \int_{0}^{{2\pi }}{{\frac{{64}}{3}\sin \left( {2\theta } \right)\,d\theta }} = \left. { - \frac{{32}}{3}\cos \left( {2\theta } \right)} \right|_0^{2\pi } = \require{bbox} \bbox[2pt,border:1px solid black]{0}$

Note that we used the double angle formula for sine to simplify the integrand a little prior to the integration. We could also have done one of two substitutions for this step if we’d wanted to (and we’d get the same answer of course).