Now we need to convert the integral into cylindrical coordinates. Let’s first deal with the limits.

We are integrating $z$ first in the integral set up to use Cartesian coordinates and so we’ll integrate that first in the integral set up to use cylindrical coordinates as well. It is easy to convert the $z$ limits to cylindrical coordinates as follows.

$${r^2} - 11 \le z \le 9 - 3{r^2}$$ Show Step 3

Now, the $x$ and $y$ limits. These are the two “outer” integrals in the original integral and so they also define $D$. So, let’s see if we can determine what $D$ is first. Once we have that we should be able to determine the $r$ and $\theta$ limits for our integral in cylindrical coordinates.

The lower $y$ limit is $y = - \sqrt {5 - {x^2}}$ and we can see that $D$ will be at most the lower portion of the disk of radius $\sqrt 5$ centered at the origin.

From the $x$ limits we see that $x$ must be positive and so $D$ is the portion of the disk of radius $\sqrt 5$ that is in the 4^th quadrant.

We now know what $D$ is here so the full set of limits for the integral is,

$$\begin{array}{c}\displaystyle \frac{{3\pi }}{2} \le \theta \le 2\pi \\ 0 \le r \le \sqrt 5 \\ {r^2} - 11 \le z \le 9 - 3{r^2}\end{array}$$ Show Step 4

Okay, let’s convert the integral into cylindrical coordinates.

$$\begin{align*}\int_{0}^{{\sqrt 5 }}{{\int_{{ - \sqrt {5 - {x^{\,2}}} }}^{0}{{\int_{{{x^{\,2}} + {y^{\,2}} - 11}}^{{9 - 3{x^{\,2}} - 3{y^{\,2}}}}{{\,\,\,2x - 3y\,\,\,dz}}\,dy}}\,dx}} & = \int_{{\frac{{3\pi }}{2}}}^{{2\pi }}{{\int_{0}^{{\sqrt 5 }}{{\int_{{{r^{\,2}} - 11}}^{{9 - 3{r^{\,2}}}}{{\,\,\,\left( {2r\cos \theta - 3r\sin \theta } \right)r\,dz}}\,dr}}\,d\theta }}\\ & = \int_{{\frac{{3\pi }}{2}}}^{{2\pi }}{{\int_{0}^{{\sqrt 5 }}{{\int_{{{r^{\,2}} - 11}}^{{9 - 3{r^{\,2}}}}{{\,\,\,{r^2}\left( {2\cos \theta - 3\sin \theta } \right)\,dz}}\,dr}}\,d\theta }}\end{align*}$$

Don’t forget that the $dz\,dy\,dz$ in the Cartesian form of the integral comes from the $dV$ in the original triple integral. We also know that, in terms of cylindrical coordinates we have $dV = r\,dz\,dr\,d\theta$ and so we know that $dz\,dy\,dx = r\,dz\,dr\,d\theta$ and we’ll pick up an extra $r$ in the integrand.

Show Step 5

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

$$\begin{align*}\int_{0}^{{\sqrt 5 }}{{\int_{{ - \sqrt {5 - {x^{\,2}}} }}^{0}{{\int_{{{x^{\,2}} + {y^{\,2}} - 11}}^{{9 - 3{x^{\,2}} - 3{y^{\,2}}}}{{\,\,\,2x - 3y\,\,\,dz}}\,dy}}\,dx}} & = \int_{{\frac{{3\pi }}{2}}}^{{2\pi }}{{\int_{0}^{{\sqrt 5 }}{{\left. {\left( {{r^2}\left( {2\cos \theta - 3\sin \theta } \right)z} \right)} \right|_{{r^2} - 11}^{9 - 3{r^2}}\,dr}}\,d\theta }}\\ & = \int_{{\frac{{3\pi }}{2}}}^{{2\pi }}{{\int_{0}^{{\sqrt 5 }}{{{r^2}\left( {2\cos \theta - 3\sin \theta } \right)\left( {20 - 4{r^2}} \right)\,dr}}\,d\theta }}\\ & = \int_{{\frac{{3\pi }}{2}}}^{{2\pi }}{{\int_{0}^{{\sqrt 5 }}{{\left( {2\cos \theta - 3\sin \theta } \right)\left( {20{r^2} - 4{r^4}} \right)\,dr}}\,d\theta }}\end{align*}$$ Show Step 6

Next let’s do the $r$ integration.

$$\begin{align*}\int_{0}^{{\sqrt 5 }}{{\int_{{ - \sqrt {5 - {x^{\,2}}} }}^{0}{{\int_{{{x^{\,2}} + {y^{\,2}} - 11}}^{{9 - 3{x^{\,2}} - 3{y^{\,2}}}}{{\,\,\,2x - 3y\,\,\,dz}}\,dy}}\,dx}} & = \int_{{\frac{{3\pi }}{2}}}^{{2\pi }}{{\left. {\left( {\left( {2\cos \theta - 3\sin \theta } \right)\left( {\frac{{20}}{3}{r^3} - \frac{4}{5}{r^5}} \right)} \right)} \right|_0^{\sqrt 5 }\,d\theta }}\\ & = \int_{{\frac{{3\pi }}{2}}}^{{2\pi }}{{\frac{{40}}{3}\sqrt 5 \left( {2\cos \theta - 3\sin \theta } \right)\,d\theta }}\end{align*}$$ Show Step 7

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

$$\int_{0}^{{\sqrt 5 }}{{\int_{{ - \sqrt {5 - {x^{\,2}}} }}^{0}{{\int_{{{x^{\,2}} + {y^{\,2}} - 11}}^{{9 - 3{x^{\,2}} - 3{y^{\,2}}}}{{\,\,\,2x - 3y\,\,\,dz}}\,dy}}\,dx}} = \left. {\left( {\frac{{40}}{3}\sqrt 5 \left( {2\sin \theta + 3\cos \theta } \right)} \right)} \right|_{\frac{{3\pi }}{2}}^{2\pi } = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{200}}{3}\sqrt 5 }}$$