Answer
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} z{\rm{d}}V = 243\pi $
Work Step by Step
We have $f\left( {x,y,z} \right) = z$ and the region ${\cal W}:{x^2} + {y^2} \le z \le 9$.
In cylindrical coordinates, we get $f\left( {r\cos \theta ,r\sin \theta ,z} \right) = z$.
The inequality ${x^2} + {y^2} \le z \le 9$ implies that the region is bounded below by ${x^2} + {y^2} = z$ and bounded above by the plane $z=9$. 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 ${x^2} + {y^2} = z$ and the plane $z=9$.
Substituting $z=9$ in ${x^2} + {y^2} = z$ 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 ${x^2} + {y^2} \le z \le 9$, in cylindrical coordinates, we get ${r^2} \le z \le 9$.
Thus, the description of ${\cal W}$ in cylindrical coordinates:
${\cal W} = \left\{ {\left( {r,\theta ,z} \right)|0 \le r \le 3,0 \le \theta \le 2\pi ,{r^2} \le z \le 9} \right\}$
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 = {r^2}}^9 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}|_{{r^2}}^9} \right)r{\rm{d}}r{\rm{d}}\theta $
$ = \frac{1}{2}\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 0}^3 \left( {81r - {r^5}} \right){\rm{d}}r{\rm{d}}\theta $
$ = \frac{1}{2}\mathop \smallint \limits_{\theta = 0}^{2\pi } \left( {\left( {\frac{{81}}{2}{r^2} - \frac{1}{6}{r^6}} \right)|_0^3} \right){\rm{d}}\theta $
$ = \frac{1}{2}\mathop \smallint \limits_{\theta = 0}^{2\pi } \left( {\frac{{729}}{2} - \frac{{729}}{6}} \right){\rm{d}}\theta $
$ = \frac{{243}}{2}\mathop \smallint \limits_{\theta = 0}^{2\pi } {\rm{d}}\theta = 243\pi $
Thus, $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} z{\rm{d}}V = 243\pi $.
