Answer
$I_{mx}=\frac{2}{5}mb^2$
Work Step by Step
We can find the required moment of inertia as follows:
$m=\int_{-a}^a dm$
$\implies m=\int_{-a}^a \rho \pi y^2 dx$
$\implies m=\int_{-a}^a \rho \pi(1-\frac{x^2}{a^2})dx $
$\implies m=2\rho \pi b^2 (x-\frac{x^3}{3a^2})$
$\implies m=\frac{4}{3\rho \pi b^2 a}$
Now, $dI_{mx}=\frac{dm}{2}y^2$
$\implies dI_{mx}=\frac{dm}{2}b^2(1-\frac{x^2}{a^2})$
$\implies dI_{mx}=\frac{\rho \pi}{2}b^4(1-\frac{x^2}{a^2})^2dx$
We know that
$I_{mx}=\int dI_{mx}$
$\implies I_{mx}=\frac{\rho \pi}{2}b^4 \int_{-a}^a (1-\frac{x^2}{a^2})^2dx$
$\implies I_{mx}=\rho \pi b^4 (x+\frac{x^5}{5a^4}-\frac{2x^3}{3a^2})$
This simplifies to:
$I_{mx}=\frac{2}{5}mb^2$
