Answer
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {x^2}{\rm{d}}S = \frac{\pi }{6}$
Work Step by Step
The surface is a sphere of radius $1$, so we can parametrize it by $G\left( {\theta ,\phi } \right) = \left( {\cos \theta \sin \phi ,\sin \theta \sin \phi ,\cos \phi } \right)$.
${{\bf{T}}_\theta } = \frac{{\partial G}}{{\partial \theta }} = \left( { - \sin \theta \sin \phi ,\cos \theta \sin \phi ,0} \right)$
${{\bf{T}}_\phi } = \frac{{\partial G}}{{\partial \phi }} = \left( {\cos \theta \cos \phi ,\sin \theta \cos \phi , - \sin \phi } \right)$
By Eq. (2), the normal vector is ${\bf{N}}\left( {\theta ,\phi } \right) = \sin \phi {{\bf{e}}_r}$. So, $||{\bf{N}}\left( {\theta ,\phi } \right)|| = \sin \phi $.
Since $x,y,z \ge 0$, the domain description in spherical coordinates is
${\cal D} = \left\{ {\left( {\theta ,\phi } \right):0 \le \theta \le \frac{\pi }{2},0 \le \phi \le \frac{\pi }{2}} \right\}$
Since $f\left( {x,y,z} \right) = {x^2}$, we obtain $f\left( {G\left( {\theta ,\phi } \right)} \right) = {\cos ^2}\theta {\sin ^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 $
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} f\left( {x,y,z} \right){\rm{d}}S = \mathop \smallint \limits_{\theta = 0}^{\pi /2} \mathop \smallint \limits_{\phi = 0}^{\pi /2} {\cos ^2}\theta {\sin ^3}\phi {\rm{d}}\theta {\rm{d}}\phi $
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} f\left( {x,y,z} \right){\rm{d}}S = \left( {\mathop \smallint \limits_{\theta = 0}^{\pi /2} {{\cos }^2}\theta {\rm{d}}\theta } \right)\left( {\mathop \smallint \limits_{\phi = 0}^{\pi /2} {{\sin }^3}\phi {\rm{d}}\phi } \right)$
From the Table of Integrals at the end of the book we obtain:
$\smallint {\cos ^2}u{\rm{d}}u = \frac{1}{2}u + \frac{1}{4}\sin 2u + C$
and
$\smallint {\sin ^3}u{\rm{d}}u = - \frac{1}{3}\left( {2 + {{\sin }^2}u} \right)\cos u + C$
So, the integral becomes
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} f\left( {x,y,z} \right){\rm{d}}S = \left( {\frac{1}{2}\theta + \frac{1}{4}\sin 2\theta } \right)|_0^{\pi /2}\left( { - \frac{1}{3}\left( {2 + {{\sin }^2}u} \right)\cos u} \right)|_0^{\pi /2}$
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} f\left( {x,y,z} \right){\rm{d}}S = \left( {\frac{\pi }{4}} \right)\left( {\frac{2}{3}} \right) = \frac{\pi }{6}$
So, $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {x^2}{\rm{d}}S = \frac{\pi }{6}$.
