Answer
$\dfrac{64}{15}$
Work Step by Step
Volume, $V=\iiint_{E} y z dV =\int_{0}^{ \pi }\int_{0} ^{2}\int_{0} ^{r \sin \theta } z \times r \sin \theta r dz dr \ d\theta $
or, $=\int_{0}^{ \pi }\int_{0} ^{2} r^2 \sin \theta[ (\dfrac{1}{2}) \times z^2 ]_0^{r \sin \theta}$
or, $=(\dfrac{1}{2})\times \int_{0}^{ \pi}\int_{0} ^{2} r^4 \sin^3 \theta dr d\theta $
or, $=(16/5) \int_{0}^{ \pi}\sin ^3 \theta d\theta
$ Set $ a= \cos \theta$ and $ du =-\sin \theta d\theta$
So, $Volume =\dfrac{16}{5} \times \int_{-1}^1 (1-u^2) du=(\dfrac{16}{5}) ( \dfrac{4}{3}) =\dfrac{64}{15}$
