Answer
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} {x^2} + {y^2}{\rm{d}}V = \frac{{8\pi }}{{15}}$
Work Step by Step
We have $f\left( {x,y,z} \right) = {x^2} + {y^2}$ and the region $\rho \le 1$. Notice that the region is a sphere of radius $1$.
Since $x = \rho \sin \varphi \cos \theta $ and $y = \rho \sin \varphi \sin \theta $, in spherical coordinates:
$f\left( {\rho \sin \varphi \cos \theta ,\rho \sin \varphi \sin \theta ,\rho \cos \varphi } \right) = {\rho ^2}{\sin ^2}\varphi $
Since $\rho \ge 0$, we write the range of $\rho$ as $0 \le \rho \le 1$.
Thus, the region description of ${\cal W}$ in spherical coordinates is given by
${\cal W} = \left\{ {\left( {\rho ,\varphi ,\theta } \right)|0 \le \rho \le 1,0 \le \varphi \le \pi ,0 \le \theta \le 2\pi } \right\}$
Using Eq. (8), the triple integral of $f\left( {x,y,z} \right)$ over ${\cal W}$ in spherical coordinates:
$\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_{\varphi = 0}^\pi \mathop \smallint \limits_{\rho = 0}^1 {\rho ^4}{\sin ^3}\varphi {\rm{d}}\rho {\rm{d}}\varphi {\rm{d}}\theta $
$ = \frac{1}{5}\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{\varphi = 0}^\pi \left( {{\rho ^5}|_0^1} \right){\sin ^3}\varphi {\rm{d}}\varphi {\rm{d}}\theta $
$ = \frac{1}{5}\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{\varphi = 0}^\pi {\sin ^3}\varphi {\rm{d}}\varphi {\rm{d}}\theta $
Using the result in Example 1 of Section 8.2:
$\smallint {\sin ^3}x{\rm{d}}x = \frac{{{{\cos }^3}x}}{3} - \cos x + C$
we get
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} {x^2} + {y^2}{\rm{d}}V = \frac{1}{5}\mathop \smallint \limits_{\theta = 0}^{2\pi } \left( {\left( {\frac{{{{\cos }^3}\varphi }}{3} - \cos \varphi } \right)|_0^\pi } \right){\rm{d}}\theta $
$ = \frac{1}{5}\mathop \smallint \limits_{\theta = 0}^{2\pi } \left( { - \frac{1}{3} + 1 - \frac{1}{3} + 1} \right){\rm{d}}\theta $
$ = \frac{4}{{15}}\mathop \smallint \limits_{\theta = 0}^{2\pi } {\rm{d}}\theta = \frac{{8\pi }}{{15}}$
Thus, $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} {x^2} + {y^2}{\rm{d}}V = \frac{{8\pi }}{{15}}$.
