Answer
We show that the moment of inertia is $\frac{2}{5}M{R^2}$.
Work Step by Step
A sphere of radius $R$ of total mass $M$ with uniform mass density is given by
$\delta = \frac{M}{{\frac{4}{3}\pi {R^3}}} = \frac{{3M}}{{4\pi {R^3}}}$
By definition, the moment of inertia with respect to an axis $L$ is the integral of distance squared from the axis, weighted by mass density. Since the mass density is uniform, and the sphere is symmetric with respect to any axis that passes through its center, we can consider the $z$-axis as the $L$ axis. Thus, ${I_z}$ is equivalent to the moment of inertia of any axis $L$ that passes through the center of the sphere.
Let ${\cal W}$ be the region of the sphere of radius $R$. Using spherical coordinates, the description of ${\cal W}$:
${\cal W} = \left\{ {\left( {\rho ,\phi ,\theta } \right)|0 \le \rho \le R,0 \le \phi \le \pi ,0 \le \theta \le 2\pi } \right\}$
We have
$x = \rho \sin \phi \cos \theta $, ${\ \ \ }$ $y = \rho \sin \phi \sin \theta $, ${\ \ \ }$ $z = \rho \cos \phi $
Evaluate the moment of inertia ${I_z}$ in spherical coordinates:
${I_z} = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} \left( {{x^2} + {y^2}} \right)\delta \left( {x,y,z} \right){\rm{d}}V$
$ = \frac{{3M}}{{4\pi {R^3}}}\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{\phi = 0}^\pi \mathop \smallint \limits_{\rho = 0}^R \left( {{\rho ^2}{{\sin }^2}\phi } \right){\rho ^2}\sin \phi {\rm{d}}\rho {\rm{d}}\phi {\rm{d}}\theta $
$ = \frac{{3M}}{{4\pi {R^3}}}\left( {\mathop \smallint \limits_{\theta = 0}^{2\pi } {\rm{d}}\theta } \right)\left( {\mathop \smallint \limits_{\phi = 0}^\pi {{\sin }^3}\phi {\rm{d}}\phi } \right)\left( {\mathop \smallint \limits_{\rho = 0}^R {\rho ^4}{\rm{d}}\rho } \right)$
Using Eq. (5) of the Table of Trigonometric Integrals in Section 8.2:
$\smallint {\sin ^3}x{\rm{d}}x = - \frac{{{{\sin }^2}x\cos x}}{3} + \frac{2}{3}\smallint \sin x{\rm{d}}x$
we obtain
$\mathop \smallint \limits_{\phi = 0}^\pi {\sin ^3}\phi {\rm{d}}\phi = - \left( {\frac{{{{\sin }^2}x\cos x}}{3}|_0^\pi } \right) + \frac{2}{3}\mathop \smallint \limits_{\phi = 0}^\pi \sin \phi {\rm{d}}\phi $
$\mathop \smallint \limits_{\phi = 0}^\pi {\sin ^3}\phi {\rm{d}}\phi = - \frac{2}{3}\left( {\cos \phi |_0^\pi } \right) = \frac{4}{3}$
Thus,
${I_z} = \frac{{3M}}{{4\pi {R^3}}}\left( {2\pi } \right)\left( {\frac{4}{3}} \right)\left( {\frac{1}{5}{R^5}} \right) = \frac{2}{5}M{R^2}$
Thus, the moment of inertia of any axis that passes through the center of the sphere with uniform mass density is $\frac{2}{5}M{R^2}$.
