Answer
The mass is $\frac{{256}}{{15}}\pi $.
Work Step by Step
We have a mass density of $f\left( {x,y,z} \right) = {\left( {{x^2} + {y^2}} \right)^{1/2}}$.
Let ${\cal W}$ denote the solid region bounded by $z = 8 - {x^2} - {y^2}$ and $z = {x^2} + {y^2}$.
We find the boundary curve of ${\cal W}$ in cylindrical coordinates:
$z = 8 - {r^2} = {r^2}$
$2{r^2} = 8$, ${\ \ \ \ \ }$ $r=2$
So the boundary curve is a circle of radius $2$. Thus, the projection of ${\cal W}$ onto the $xy$-plane is the disk
${\cal D} = \left\{ {\left( {r,\theta } \right)|0 \le r \le 2,0 \le \theta \le 2\pi } \right\}$
Thus, the description of ${\cal W}$:
${\cal W} = \left\{ {\left( {r,\theta ,z} \right)|0 \le r \le 2,0 \le \theta \le 2\pi ,{r^2} \le z \le 8 - {r^2}} \right\}$
Using $x = r\cos \theta $, $y = r\sin \theta $, we get:
$f\left( {r\cos \theta ,r\sin \theta ,z} \right) = r$
Evaluate the mass in cylindrical coordinates:
mass $ = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} f\left( {x,y,z} \right){\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}{\rm{d}}z{\rm{d}}r{\rm{d}}\theta $
$ = \mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 0}^2 {r^2}\left( {z|_{{r^2}}^{8 - {r^2}}} \right){\rm{d}}r{\rm{d}}\theta $
$ = \mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 0}^2 {r^2}\left( {8 - 2{r^2}} \right){\rm{d}}r{\rm{d}}\theta $
$ = \left( {\mathop \smallint \limits_{\theta = 0}^{2\pi } {\rm{d}}\theta } \right)\left( {\mathop \smallint \limits_{r = 0}^2 \left( {8{r^2} - 2{r^4}} \right){\rm{d}}r} \right)$
$ = 2\pi \left( {\left( {\frac{8}{3}{r^3} - \frac{2}{5}{r^5}} \right)|_0^2} \right)$
$ = 2\pi \left( {\frac{{64}}{3} - \frac{{64}}{5}} \right) = \frac{{256}}{{15}}\pi $
So, the mass is $\frac{{256}}{{15}}\pi $.
