Answer
$I_y=\frac{2}{5}mr^2$
Work Step by Step
The required moment of inertia can be determined as follows:
We know that
$dV=\pi x^2 dy$
As $dm=\rho dV$
$\implies dm=\rho \pi x^2 dy$
Similarly $dI_y=\frac{1}{2}dmx^2$
$\implies dI_y=\frac{1}{2}(\rho \pi x^2 dy)x^2$
$\implies dI_y=\frac{1}{2}\pi \rho x^4 dy$
As $x^2=r^2-y^2$
$\implies dI_y=\frac{1}{2}\pi \rho (r^2-y^2)^2dy$
We integrate both sides to obtain:
$I_y=\int d I_y$
$\implies I_y=\int_0 ^r \frac{1}{2} \pi \rho (r^2-y^2)^2dy$
$\implies I_y=\frac{\pi \rho}{2}[r^4y-\frac{2}{3}r^2y^3+\frac{y^5}{5}]_{0}^r$
$\implies I_y=\frac{4\pi \rho}{15}r^5$ [eq(1)]
Similarly $m=\int dm=\int_0 ^r \pi \rho x^2 dy$
$\implies m=\int _0 ^r \pi \rho (r^2-y^2)dy$
$\implies m=\frac{2}{3} \pi \rho r^3$
We plug in this value in eq(1) to obtain:
$I_y=\frac{2}{5}mr^2$
