Answer
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} z{\left( {{x^2} + {y^2} + {z^2}} \right)^{ - 3/2}}{\rm{d}}V = \pi $
Work Step by Step
We have $f\left( {x,y,z} \right) = z{\left( {{x^2} + {y^2} + {z^2}} \right)^{ - 3/2}}$ and the region ${\cal W}$ consists of the part of the ball ${x^2} + {y^2} + {z^2} \le 16$ defined by $z \ge 2$.
So, the ball is bounded below by the plane $z = 2$ and bounded above by the sphere of radius $4$, ${x^2} + {y^2} + {z^2} = 16$.
In spherical coordinates we have
$f\left( {\rho \sin \varphi \cos \theta ,\rho \sin \varphi \sin \theta ,\rho \cos \varphi } \right) = {\rho ^{ - 2}}\cos \varphi $
We convert the plane $z=2$ to spherical coordinates:
$z = \rho \cos \varphi = 2$
$\rho = \frac{2}{{\cos \varphi }}$
Since the ball of radius $4$ is $\rho = 4$, so $\frac{2}{{\cos \varphi }} \le \rho \le 4$.
Next, we find the range of $\varphi $. The ray starts from $\varphi = 0$ to the angle where it touches the intersection of the sphere $\rho = 4$ and the plane $\rho = \frac{2}{{\cos \varphi }}$. So, we find this angle by solving the following equation:
$\rho = \frac{2}{{\cos \varphi }} = 4$
$\cos \varphi = \frac{1}{2}$, ${\ \ \ \ }$ $\varphi = \frac{\pi }{3}$
Thus, the description of ${\cal W}$ in spherical coordinates:
${\cal W} = \left\{ {\left( {\rho ,\varphi ,\theta } \right)|\frac{2}{{\cos \varphi }} \le \rho \le 4,0 \le \varphi \le \frac{\pi }{3},0 \le \theta \le 2\pi } \right\}$
Using Eq. (8), the volume of ${\cal W}$ in spherical coordinates:
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} z{\left( {{x^2} + {y^2} + {z^2}} \right)^{ - 3/2}}{\rm{d}}V = \mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{\varphi = 0}^{\pi /3} \mathop \smallint \limits_{\rho = 2/\cos \varphi }^4 \cos \varphi \sin \varphi {\rm{d}}\rho {\rm{d}}\varphi {\rm{d}}\theta $
$ = \mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{\varphi = 0}^{\pi /3} \cos \varphi \sin \varphi \left( {\rho |_{2/\cos \varphi }^4} \right){\rm{d}}\varphi {\rm{d}}\theta $
$ = \mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{\varphi = 0}^{\pi /3} \cos \varphi \sin \varphi \left( {4 - \frac{2}{{\cos \varphi }}} \right){\rm{d}}\varphi {\rm{d}}\theta $
$ = \mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{\varphi = 0}^{\pi /3} \left( {2\sin 2\varphi - 2\sin \varphi } \right){\rm{d}}\varphi {\rm{d}}\theta $
$ = \mathop \smallint \limits_{\theta = 0}^{2\pi } \left( {\left( { - \cos 2\varphi + 2\cos \varphi } \right)|_0^{\pi /3}} \right){\rm{d}}\theta $
$ = \mathop \smallint \limits_{\theta = 0}^{2\pi } \left( {\frac{1}{2} + 1 + 1 - 2} \right){\rm{d}}\theta $
$ = \frac{1}{2}\mathop \smallint \limits_{\theta = 0}^{2\pi } {\rm{d}}\theta = \pi $
Thus, $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} z{\left( {{x^2} + {y^2} + {z^2}} \right)^{ - 3/2}}{\rm{d}}V = \pi $.
