Answer
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {z^2}{\left( {{x^2} + {y^2} + {z^2}} \right)^{ - 1}}{\rm{d}}S = \frac{7}{3}\pi $
Work Step by Step
The surface is a sphere of radius $2$, so we can parametrize it by $G\left( {\theta ,\phi } \right) = \left( {2\cos \theta \sin \phi ,2\sin \theta \sin \phi ,2\cos \phi } \right)$.
${{\bf{T}}_\theta } = \frac{{\partial G}}{{\partial \theta }} = \left( { - 2\sin \theta \sin \phi ,2\cos \theta \sin \phi ,0} \right)$
${{\bf{T}}_\phi } = \frac{{\partial G}}{{\partial \phi }} = \left( {2\cos \theta \cos \phi ,2\sin \theta \cos \phi , - 2\sin \phi } \right)$
By Eq. (2), the normal vector is ${\bf{N}}\left( {\theta ,\phi } \right) = 4\sin \phi {{\bf{e}}_r}$. So, $||{\bf{N}}\left( {\theta ,\phi } \right)|| = 4\sin \phi $.
From $1 \le z \le 2$ we obtain
$1 \le 2\cos \phi \le 2$
$\frac{1}{2} \le \cos \phi \le 1$
Hence,
$0 \le \phi \le \frac{\pi }{3}$
From here we obtain the description of the domain in polar coordinates:
${\cal D} = \left\{ {\left( {\theta ,\phi } \right):0 \le \theta \le 2\pi ,0 \le \phi \le \frac{\pi }{3}} \right\}$
Since $f\left( {x,y,z} \right) = {z^2}{\left( {{x^2} + {y^2} + {z^2}} \right)^{ - 1}}$, we obtain $f\left( {G\left( {\theta ,\phi } \right)} \right) = \left( {4{{\cos }^2}\phi } \right)\cdot\frac{1}{4} = {\cos ^2}\phi $.
By Eq. (7) of Theorem 1, we have
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} f\left( {x,y,z} \right){\rm{d}}S = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} f\left( {G\left( {\theta ,\phi } \right)} \right)||{\bf{N}}\left( {\theta ,\phi } \right)||{\rm{d}}\theta {\rm{d}}\phi $
$ = 4\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{\phi = 0}^{\pi /3} {\cos ^2}\phi \sin \phi {\rm{d}}\theta {\rm{d}}\phi $
$ = 4\mathop \smallint \limits_{\theta = 0}^{2\pi } {\rm{d}}\theta \mathop \smallint \limits_{\phi = 0}^{\pi /3} {\cos ^2}\phi \sin \phi {\rm{d}}\phi $
$ = 8\pi \mathop \smallint \limits_{\phi = 0}^{\pi /3} {\cos ^2}\phi \sin \phi {\rm{d}}\phi $
Let $t = \cos \phi $, so ${\rm{d}}t = - \sin \phi {\rm{d}}\phi $. The integral becomes
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} f\left( {x,y,z} \right){\rm{d}}S = - 8\pi \mathop \smallint \limits_{t = 1}^{1/2} {t^2}{\rm{d}}t$
$ = - \frac{8}{3}\pi \left( {{t^3}|_1^{1/2}} \right) = - \frac{8}{3}\pi \left( {\frac{1}{8} - 1} \right) = \frac{7}{3}\pi $
So, $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {z^2}{\left( {{x^2} + {y^2} + {z^2}} \right)^{ - 1}}{\rm{d}}S = \frac{7}{3}\pi $.
