Sphere Moment - Non-Uniform Mass Distribution

Using symmetry, we can make life a little easier and consider a density of $x^2+y^2$ ; let's not forget to divide by 2 at the end. Thus $I_z=\frac{1}{2}\int_W (x^2+y^2)^2dV = \frac{1}{2} \int_0^{2\pi}\int_0^{\pi}\int_0^1 r^4(\sin^4\phi) r^2\sin\phi drd\phi d\theta=\pi\times \frac{1}{7}\times \frac{16}{15}\approx \boxed{0.47872}$

Take an elementary disc with centre on the x axis at a distance $x$ from origin.

Radius of disc = $\sqrt {1-x^2}$

Mass of disc (dm) = $x^2$ × Volume of disc = $Area × dx×x^2$

= $π(1-x^2)(x^2)dx$

Moment of inertia of disc about diameter = $\frac {MR^2}{4}$

Using parallel axis theorem, the moment of inertia of our elementary disc about z-axis = $\frac {dm(1-x^2)}{4} + dm(x^2)$

$\therefore, dI=dm(\frac {1+3x^2}{4}$

$I=\int ^{1} _{0} π(1-x^2)(x^2)(\frac {1+3x^2}{4})dx$

$2I=\frac {π}{2}(\int ^{1}_{0} (-3x^6+2x^4+x^2)dx)$

Integrate this and put the limits to get the answer as $≈0.47847$

(We calculate $2I$ to count in the second half also)