Answer
$2430 \pi$
Work Step by Step
$I=\iiint_Ediv \overrightarrow{F}dV =\int_0^{2 \pi}\int_0^3\int_{r^2}^{0} (10 z) r dz dr d\theta $
$=\int_0^{2 \pi}\int_0^3 [5z^2 r]_{r^2}^0 dr d\theta $
$=\int_0^{2 \pi}\int_0^3 405r-5r^5 dr d\theta $
$=\int_0^{2 \pi}[(\dfrac{405r^2}{2}-\dfrac{5r^6}{6}))]_0^3 d\theta $
$=\int_0^{2 \pi} \dfrac{405(3)^2}{2}-\dfrac{5 \cdot 3^{6}}{6} d\theta $
$=\int_0^{2 \pi} 1215 d \theta $
$=2430 \pi$
