Answer
$\dfrac{64\pi}{3}$
Work Step by Step
The given integral is $I=\iiint_EZdV$
or, $=\int_0^{2\pi} \int_{0}^{2}\int_{r^2}^{4} (r)(z) dr dz d\theta$
This implies that
$I=\int_0^{2\pi} \int_{0}^{2}[\dfrac{z^2}{2}]_{r^2}^{4} \times ( r dr) dz d\theta$
and $I=\int_0^{2\pi} \int_{0}^{2}(8r-\dfrac{r^5}{2}) dr dz d\theta=\int_0^{2\pi}[\dfrac{8r^2}{2}-\dfrac{r^6}{12}]_{0}^{2} d\theta$
Hence, $\iiint_EZdV=\dfrac{64\pi}{3}$
