Answer
$\implies I_x=\frac{2}{5}mr^2$
Work Step by Step
We can determine the required moment of inertia as follows:
We know that
$dV=\pi y^2dx=\pi (r^2-x^2)dx$
$\implies dm=\rho dV$
$\implies dm=\rho \pi(r^2-x^2)dx$
$\implies m=\rho \pi \int_{-r}^r (r_2-x^2)dx$
$\implies m=\frac{4}{3}\pi \rho r^3$ [eq(1)]
We also know that
$dI=\frac{1}{2}y^2dm$
$\implies dI=\frac{1}{2}\rho \pi(r^2-x^2)^2dx$
$\implies I_x=\frac{1}{2}\pi \rho \int_{-r}^r (r^2-x^2)dx$
$\implies I_x=\frac{8}{15}\pi \rho r^5$
This can be rearranged as:
$I_x=(\frac{4}{3}\pi \rho r^3)\frac{2}{5}r^2$
From eq(1) $m=\frac{4}{3}\pi \rho r^3$
$\implies I_x=\frac{2}{5}mr^2$
