Answer
$I_{y}=\frac{2}{5}mr^{5}$
Work Step by Step
$m=\int \rho dV=\rho \int dV= \rho \int \pi x^2 dy=\rho \pi \int (r^2-y^2)dy=\rho \pi(r^3-\frac{1}{3}r^3)=\frac{2}{3} \rho \pi r^3 $
$I_{y}=\int dm(\frac{1}{2} x^2 )=\frac{\rho}{2} \int x^4 \pi dy=\frac{\rho}{2} \int (r^2-y^2)^2 \pi dy=\frac{\rho \pi}{2}(r^4y-\frac{2}{3}r^2y^3+\frac{y^5}{5})$
Now fill in $r$ for $y$ (integral).
$I_{y}=\frac{2}{5} \rho \pi r^5$
$I_{y}=\frac{2}{5}mr^2$
