Answer
$384 \pi$
Work Step by Step
Consider $I=\iiint_E\sqrt{x^2+y^2} dV$
or, $=\int_0^{2\pi} \int_{-5}^{4}\int_0^{4} (r^2) dr dz d\theta$
or, $=\int_0^{2\pi} d\theta \times \int_{-5}^{4} dz \times \int_0^{4} r^2 dr$
or, $=[\theta]_0^{2\pi} \times [z]_{-5}^{4} \times [\dfrac{r^3}{3}]_0^{4}$
or, $=2\pi(4+5) \times [\dfrac{1}{3}(4)^3-0]$
or, $=18\pi (\dfrac{64}{3})$
or, $=384 \pi$
