Answer
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} {y^2}{\rm{d}}V = \frac{{128}}{{15}}$
Work Step by Step
We have $f\left( {x,y,z} \right) = {y^2}$ and the region ${\cal W}$ is within the cylinder ${x^2} + {y^2} = 4$ and $0 \le z \le y$.
Referring to the figure attached, we see that ${\cal W}$ is a region between the plane $z=0$ and the surface $z=y$. So, it is a $z$-simple region which lies over ${\cal D}$, the upper hemisphere of radius $2$ in the $xy$-plane. Since the limit involves square root, and the integrand is ${y^2}$; therefore for convenience, we consider ${\cal D}$ as a vertically simple region defined by
${\cal D} = \left\{ {\left( {x,y} \right)| - 2 \le x \le 2,0 \le y \le \sqrt {4 - {x^2}} } \right\}$
Thus, the triple integral is equal to the iterated integral:
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} f\left( {x,y,z} \right){\rm{d}}V = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} \left( {\mathop \smallint \limits_{z = 0}^y {y^2}{\rm{d}}z} \right){\rm{d}}A$
$ = \mathop \smallint \limits_{x = - 2}^2 \mathop \smallint \limits_{y = 0}^{\sqrt {4 - {x^2}} } \left( {\mathop \smallint \limits_{z = 0}^y {y^2}{\rm{d}}z} \right){\rm{d}}y{\rm{d}}x$
$ = \mathop \smallint \limits_{x = - 2}^2 \mathop \smallint \limits_{y = 0}^{\sqrt {4 - {x^2}} } {y^2}\left( {z|_0^y} \right){\rm{d}}y{\rm{d}}x$
$ = \mathop \smallint \limits_{x = - 2}^2 \mathop \smallint \limits_{y = 0}^{\sqrt {4 - {x^2}} } {y^3}{\rm{d}}y{\rm{d}}x$
$ = \frac{1}{4}\mathop \smallint \limits_{x = - 2}^2 \left( {{y^4}|_0^{\sqrt {4 - {x^2}} }} \right){\rm{d}}x$
$ = \frac{1}{4}\mathop \smallint \limits_{x = - 2}^2 {\left( {4 - {x^2}} \right)^2}{\rm{d}}x$
$ = \frac{1}{4}\mathop \smallint \limits_{x = - 2}^2 \left( {16 - 8{x^2} + {x^4}} \right){\rm{d}}x$
$ = \frac{1}{4}\left( {16x - \frac{8}{3}{x^3} + \frac{1}{5}{x^5}} \right)|_{ - 2}^2$
$ = \frac{1}{4}\left( {32 - \frac{{64}}{3} + \frac{{32}}{5} + 32 - \frac{{64}}{3} + \frac{{32}}{5}} \right)$
$ = \frac{{128}}{{15}}$
Thus, $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} {y^2}{\rm{d}}V = \frac{{128}}{{15}}$.
