Answer
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} {\rho ^{ - 3}}{\rm{d}}V = \pi \ln 4$
Work Step by Step
We have $f\left( {x,y,z} \right) = {\rho ^{ - 3}}$ and the region ${\cal W}:2 \le {x^2} + {y^2} + {z^2} \le 4$.
Using ${\rho ^2} = {x^2} + {y^2} + {z^2}$, the inequality in spherical coordinates is $\sqrt 2 \le \rho \le 2$.
Thus, the region description of ${\cal W}$ in spherical coordinates is given by
${\cal W} = \left\{ {\left( {\rho ,\varphi ,\theta } \right)|\sqrt 2 \le \rho \le 2,0 \le \varphi \le \pi ,0 \le \theta \le 2\pi } \right\}$
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}^{} {\rho ^{ - 3}}{\rm{d}}V = \mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{\varphi = 0}^\pi \mathop \smallint \limits_{\rho = \sqrt 2 }^2 {\rho ^{ - 1}}\sin \varphi {\rm{d}}\rho {\rm{d}}\varphi {\rm{d}}\theta $
$ = \mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{\varphi = 0}^\pi \left( {\ln \rho |_{\sqrt 2 }^2} \right)\sin \varphi {\rm{d}}\varphi {\rm{d}}\theta $
$ = \mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{\varphi = 0}^\pi \left( {\ln 2 - \ln \sqrt 2 } \right)\sin \varphi {\rm{d}}\varphi {\rm{d}}\theta $
$ = \frac{1}{2}\ln 2\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{\varphi = 0}^\pi \sin \varphi {\rm{d}}\varphi {\rm{d}}\theta $
$ = - \frac{1}{2}\ln 2\mathop \smallint \limits_{\theta = 0}^{2\pi } \left( {\cos \varphi |_0^\pi } \right){\rm{d}}\theta $
$ = \ln 2\mathop \smallint \limits_{\theta = 0}^{2\pi } {\rm{d}}\theta = 2\pi \ln 2 = \pi \ln 4$
Thus, $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} {\rho ^{ - 3}}{\rm{d}}V = \pi \ln 4$.
