Answer
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} {x^2} + {y^2}{\rm{d}}V = \frac{{405}}{2}\pi $
Work Step by Step
We have $f\left( {x,y,z} \right) = {x^2} + {y^2}$ and the region:
${\cal W} = \left\{ {\left( {x,y,z} \right)|{x^2} + {y^2} \le 9,0 \le z \le 5} \right\}$
In polar coordinates, we get $f\left( {r\cos \theta ,r\sin \theta ,z} \right) = {r^2}$.
Since the projection of ${\cal W}$ onto the $xy$-plane is a disk of radius $3$, the cylindrical description of ${\cal W}$:
${\cal W} = \left\{ {\left( {r,\theta ,z} \right)|0 \le r \le 3,0 \le \theta \le 2\pi ,0 \le z \le 5} \right\}$
Using Eq. (5) of Theorem 2, we evaluate the integral of the function $f\left( {x,y,z} \right) = {x^2} + {y^2}$ over ${\cal W}$:
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} {x^2} + {y^2}{\rm{d}}V = \mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 0}^3 \mathop \smallint \limits_{z = 0}^5 {r^3}{\rm{d}}z{\rm{d}}r{\rm{d}}\theta $
$ = \mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 0}^3 {r^3}\left( {z|_0^5} \right){\rm{d}}r{\rm{d}}\theta $
$ = 5\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 0}^3 {r^3}{\rm{d}}r{\rm{d}}\theta $
$ = \frac{5}{4}\mathop \smallint \limits_{\theta = 0}^{2\pi } \left( {{r^4}|_0^3} \right){\rm{d}}\theta $
$ = \frac{{405}}{4}\mathop \smallint \limits_{\theta = 0}^{2\pi } {\rm{d}}\theta = \frac{{405}}{2}\pi $
Thus, $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} {x^2} + {y^2}{\rm{d}}V = \frac{{405}}{2}\pi $.
