Answer
$\dfrac{32\pi}{3}$
Work Step by Step
$\iiint_E (x^2+y^2) dV=\int_{0}^{2 \pi}\int_0^{2} \int_{r^2}^{4} (r^2 \cos^2 \theta+r^2 \sin^2 \theta) r dz dr d\theta$
$=\int_{0}^{2 \pi}\int_0^{2} d\theta \times (\int_{r^2}^{4} r^3 dz dr) $
$=[\theta]_{0}^{2 \pi}\int_0^2 [zr^3]_{r^2}^4 dr$
$=(2 \pi) \times \int_{0}^{2} 4r^3-r^5 dr$
$=2 \pi[r^4-\dfrac{r^6}{6}]_0^{2}$
$=\dfrac{32\pi}{3}$
