Answer
$\dfrac{64}{15}$
Work Step by Step
Set up the integral as follows:
Volume, $V=\iiint_{E} y z dV=\int_{0}^{ \pi }\int_{0} ^{2}\int_{0} ^{r \sin \theta } zr \sin \theta \times r dz dr d\theta $
$=(1/2) \int_{0}^{ \pi}\int_{0} ^{2} r^4 \sin^3 \theta dr d\theta $
$=(16/5) \int_{0}^{ \pi}\sin ^3 \theta d\theta $
Suppose that $ a= \cos \theta \implies du =-\sin \theta d\theta$
So, $Volume =\dfrac{16}{5} \int_{-1}^1 (1-u^2) du=\dfrac{16}{5} \times \dfrac{4}{3}=\dfrac{64}{15}$
