Answer
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} z\sqrt {{x^2} + {y^2}} {\rm{d}}V = \frac{{1024\pi }}{{15}}$
Work Step by Step
We have $f\left( {x,y,z} \right) = z\sqrt {{x^2} + {y^2}} $ and the region:
${\cal W}:{x^2} + {y^2} \le z \le 8 - \left( {{x^2} + {y^2}} \right)$
In cylindrical coordinates, we get $f\left( {r\cos \theta ,r\sin \theta ,z} \right) = rz$.
The inequality ${x^2} + {y^2} \le z \le 8 - \left( {{x^2} + {y^2}} \right)$ implies that the region is bounded below by ${x^2} + {y^2} = z$ and bounded above by $z = 8 - \left( {{x^2} + {y^2}} \right)$. 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 $z = 8 - \left( {{x^2} + {y^2}} \right)$.
Substituting ${x^2} + {y^2} = z$ in $z = 8 - \left( {{x^2} + {y^2}} \right)$ gives
$z=8-z$
So, $z=4$. Thus, the two surfaces intersect at the plane $z=4$.
Substituting $z=4$ in ${x^2} + {y^2} = z$ we obtain ${x^2} + {y^2} = 4$. Thus, ${\cal D}$ is a disk of radius $2$, namely ${x^2} + {y^2} \le 4$. So, the domain description of ${\cal D}$ in polar coordinates:
${\cal D} = \left\{ {\left( {r,\theta } \right)|0 \le r \le 2,0 \le \theta \le 2\pi } \right\}$
Since ${x^2} + {y^2} \le z \le 8 - \left( {{x^2} + {y^2}} \right)$, in cylindrical coordinates, we get ${r^2} \le z \le 8 - {r^2}$.
Thus, the description of ${\cal W}$ in cylindrical coordinates:
${\cal W} = \left\{ {\left( {r,\theta ,z} \right)|0 \le r \le 2,0 \le \theta \le 2\pi ,{r^2} \le z \le 8 - {r^2}} \right\}$
Using Eq. (5) of Theorem 2, we evaluate the integral of the function $f\left( {x,y,z} \right) = z\sqrt {{x^2} + {y^2}} $ over ${\cal W}$:
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} z\sqrt {{x^2} + {y^2}} {\rm{d}}V = \mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 0}^2 \mathop \smallint \limits_{z = {r^2}}^{8 - {r^2}} {r^2}z{\rm{d}}z{\rm{d}}r{\rm{d}}\theta $
$ = \frac{1}{2}\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 0}^2 {r^2}\left( {{z^2}|_{{r^2}}^{8 - {r^2}}} \right){\rm{d}}r{\rm{d}}\theta $
$ = \frac{1}{2}\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 0}^2 {r^2}\left( {64 - 16{r^2}} \right){\rm{d}}r{\rm{d}}\theta $
$ = \frac{1}{2}\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 0}^2 \left( {64{r^2} - 16{r^4}} \right){\rm{d}}r{\rm{d}}\theta $
$ = \frac{1}{2}\mathop \smallint \limits_{\theta = 0}^{2\pi } \left( {\left( {\frac{{64}}{3}{r^3} - \frac{{16}}{5}{r^5}} \right)|_0^2} \right){\rm{d}}\theta $
$ = \frac{1}{2}\mathop \smallint \limits_{\theta = 0}^{2\pi } \left( {\frac{{512}}{3} - \frac{{512}}{5}} \right){\rm{d}}\theta $
$ = \frac{{1024}}{{30}}\mathop \smallint \limits_{\theta = 0}^{2\pi } {\rm{d}}\theta = \frac{{1024\pi }}{{15}}$
Thus, $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} z\sqrt {{x^2} + {y^2}} {\rm{d}}V = \frac{{1024\pi }}{{15}}$.
