Answer
$I_z=\dfrac{\pi kh^5}{10}$
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}^{z}(r^2)r dr dz d\theta $
$=k\int_{0}^{2 \pi} \int_{0}^{h}\int_{0}^{z}r^3 dr dz d\theta $
$=k \int_{0}^{2 \pi} d\theta \times k[\int_{0}^{h} \int_{0}^{z}(r^3) dr dz] $
$=(2\pi k) \times \int_0^h [\dfrac{z^4}{4}] dz $
$=2\pi k \times [\dfrac{z^5}{20}]_0^h $
$=(2\pi k) \times \dfrac{(h-0)^5}{20} $
Hence, $I_z=\dfrac{\pi kh^5}{10}$
