Answer
$\dfrac{\pi}{96}$
Work Step by Step
$\int\int_{E}y^2 z^2 dV=\int_{0}^{2 \pi}\int_{0} ^{1}\int_{0} ^{1-r^2} r^4 \cos^2 \theta \sin^2 \theta(r) dx dr d\theta $
$=\int_{0}^{2 \pi}\int_{0} ^{1}\int_{0} ^{1-r^2} r^5 \cos^2 \theta \sin^2 \theta dx dr d\theta $
$=\int_{0}^{2 \pi}\int_{0} ^{1}(1-r^2) r^5 \cos^2 \theta \sin^2 \theta dr d\theta $
$=[ \int_{0}^{2 \pi} \cos^2 \theta \sin^2 \theta d\theta] \times [\dfrac{r^6}{6} -\dfrac{r^8}{8}]_0^1 $
$=(1/4) \int_{0}^{2 \pi} \sin^2 \theta d\theta \times [\dfrac{r^6}{6} -\dfrac{r^8}{8}]_0^1 $
Since, $\sin^2 x= \dfrac{1}{2}-\dfrac{\cos 2x}{2} $
$=(1/4) [\dfrac{ \theta}{2}-\dfrac{\sin 4 \theta}{8} ]_0^{2 \pi} ] \dfrac{1}{24}$
$=\dfrac{\pi}{96}$
