Answer
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} z{\rm{d}}V = \frac{{243\pi }}{2}$
Work Step by Step
We have $f\left( {x,y,z} \right) = z$ and the region ${\cal W}:0 \le z \le {x^2} + {y^2} \le 9$.
In cylindrical coordinates, we get $f\left( {r\cos \theta ,r\sin \theta ,z} \right) = z$.
The inequality $0 \le z \le {x^2} + {y^2} \le 9$ implies that the region is bounded below by the plane $z=0$ and bounded above by the surface $z = {x^2} + {y^2}$ and also by the plane $z=9$. In addition, it is also bounded by the cylinder ${x^2} + {y^2} = 9$. This is illustrated in the figure attached.
We project ${\cal W}$ onto the $xy$-plane to obtain the domain ${\cal D}$. To find the boundary curve of ${\cal D}$, we find the intersection of the surfaces $z = {x^2} + {y^2}$ and the plane $z=9$.
Substituting $z=9$ in $z = {x^2} + {y^2}$ gives ${x^2} + {y^2} = 9$.
Thus, ${\cal D}$ is a disk of radius $3$, namely ${x^2} + {y^2} \le 9$. So, the domain description of ${\cal D}$ in polar coordinates:
${\cal D} = \left\{ {\left( {r,\theta } \right)|0 \le r \le 3,0 \le \theta \le 2\pi } \right\}$
Since $0 \le z \le {x^2} + {y^2} \le 9$ and the region ${\cal W}$ consists of all points lying between ${\cal D}$ and $z = {x^2} + {y^2}$; in cylindrical coordinates, the range of $z$ is $0 \le z \le {r^2}$.
Thus, the description of ${\cal W}$ in cylindrical coordinates:
${\cal W} = \left\{ {\left( {r,\theta ,z} \right)|0 \le r \le 3,0 \le \theta \le 2\pi ,0 \le z \le {r^2}} \right\}$
Using Eq. (5) of Theorem 2, we evaluate the integral of the function $f\left( {x,y,z} \right) = z$ over ${\cal W}$:
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} z{\rm{d}}V = \mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 0}^3 \mathop \smallint \limits_{z = 0}^{{r^2}} zr{\rm{d}}z{\rm{d}}r{\rm{d}}\theta $
$ = \frac{1}{2}\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 0}^3 \left( {{z^2}|_0^{{r^2}}} \right)r{\rm{d}}r{\rm{d}}\theta $
$ = \frac{1}{2}\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 0}^3 {r^5}{\rm{d}}r{\rm{d}}\theta $
$ = \frac{1}{{12}}\mathop \smallint \limits_{\theta = 0}^{2\pi } \left( {{r^6}|_0^3} \right){\rm{d}}\theta $
$ = \frac{{729}}{{12}}\mathop \smallint \limits_{\theta = 0}^{2\pi } {\rm{d}}\theta = \frac{{729\pi }}{6} = \frac{{243\pi }}{2}$
Thus, $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} z{\rm{d}}V = \frac{{243\pi }}{2}$.
