Answer
$\dfrac{1}{2}\pi kha^4$
Work Step by Step
Here, $I_z=\iiint_{E} (x^2+y^2) \rho(x,y,z) dV=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 $
$=k \int_{0}^{2 \pi} d\theta \times k\int_{0}^{h} dz \times k \int_{0}^{a}(r^3) dr $
$=(2\pi k) [z]_0^h [\dfrac{r^4}{4}]_0^a $
$=2\pi k (h-0) [\dfrac{a^4}{4}-0] $
$=2\pi k (h) (\dfrac{a^4}{4}) $
Hence, $I_z=\dfrac{\pi kha^4}{2}$