Answer
$\dfrac{\pi kha^4}{2}$
Work Step by Step
Consider $I_z=\iiint_{E} (x^2+y^2) \rho(x,y,z) dV$
In the polar coordinates, we have $x=r \cos \theta ;y=r \sin \theta;z=z$ and $r^2=x^2+y^2 $
Thus, we have $k\int_{0}^{2 \pi} \int_{0}^{h}\int_{0}^{a}(r^2)r dr dz d\theta=k\int_{0}^{2 \pi} \int_{0}^{h}\int_{0}^{a}(r^3) dr dz d\theta $
This implies that
$k \int_{0}^{2 \pi} d\theta \int_{0}^{h} dz \int_{0}^{a}(r^3) dr=2\pi k [z]_0^h [\dfrac{r^4}{4}]_0^a $
Hence, $I_z=2\pi k (h-0) [\dfrac{a^4}{4}-0] =\dfrac{\pi kha^4}{2}$
