Answer
The equation is symmetric in terms of $x$, $y$, and $z$. Therefore, we conclude that
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {x^2}{\rm{d}}S = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {y^2}{\rm{d}}S = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {z^2}{\rm{d}}S$
We show that
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {x^2}{\rm{d}}S = \frac{4}{3}\pi {R^4}$
Work Step by Step
The sphere $S$ of radius $R$ centered at the origin has equation:
${x^2} + {y^2} + {z^2} = {R^2}$
The equation is symmetric in terms of $x$, $y$, and $z$. Therefore, we conclude that
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {x^2}{\rm{d}}S = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {y^2}{\rm{d}}S = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {z^2}{\rm{d}}S$
Next, we evaluate $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} \left( {{x^2} + {y^2} + {z^2}} \right){\rm{d}}S$.
In spherical coordinates the integral becomes
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} \left( {{x^2} + {y^2} + {z^2}} \right){\rm{d}}S = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {R^2}{\rm{d}}S = {R^2}\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {\rm{d}}S$
The surface area of the sphere is $4\pi {R^2}$. Therefore,
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} \left( {{x^2} + {y^2} + {z^2}} \right){\rm{d}}S = 4\pi {R^4}$
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {x^2}{\rm{d}}S + \mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {y^2}{\rm{d}}S + \mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {z^2}{\rm{d}}S = 4\pi {R^4}$
Since $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {x^2}{\rm{d}}S = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {y^2}{\rm{d}}S = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {z^2}{\rm{d}}S$, so
$3\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {x^2}{\rm{d}}S = 4\pi {R^4}$
Hence, $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {x^2}{\rm{d}}S = \frac{4}{3}\pi {R^4}$.