Answer
The mass of ${\cal W}$ is $\frac{{192}}{5}$.
Work Step by Step
We have the mass density of $\rho \left( {x,y,z} \right) = {z^2}$ and the region ${\cal W}$ defined by a portion of the half-cylinder ${x^2} + {y^2} \le 4$, $x \ge 0$ such that $0 \le z \le 3y$. The inequality $0 \le z \le 3y$ implies that $y \ge 0$. Thus, ${\cal W}$ is in the first quadrant.
Thus, the description of ${\cal W}$ in cylindrical coordinates:
${\cal W} = \left\{ {\left( {r,\theta ,z} \right)|0 \le r \le 2,0 \le \theta \le \frac{\pi }{2},0 \le z \le 3r\sin \theta } \right\}$
In cylindrical coordinates: $\rho \left( {r\cos \theta ,r\sin \theta ,z} \right) = {z^2}$.
Compute the mass:
mass $ = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} \rho \left( {x,y,z} \right){\rm{d}}V$
$ = \mathop \smallint \limits_{\theta = 0}^{\pi /2} \mathop \smallint \limits_{r = 0}^2 \mathop \smallint \limits_{z = 0}^{3r\sin \theta } {z^2}r{\rm{d}}z{\rm{d}}r{\rm{d}}\theta $
$ = \frac{1}{3}\mathop \smallint \limits_{\theta = 0}^{\pi /2} \mathop \smallint \limits_{r = 0}^2 \left( {{z^3}|_0^{3r\sin \theta }} \right)r{\rm{d}}r{\rm{d}}\theta $
$ = \frac{1}{3}\mathop \smallint \limits_{\theta = 0}^{\pi /2} \mathop \smallint \limits_{r = 0}^2 \left( {27{r^3}{{\sin }^3}\theta } \right)r{\rm{d}}r{\rm{d}}\theta $
$ = 9\left( {\mathop \smallint \limits_{\theta = 0}^{\pi /2} {{\sin }^3}\theta {\rm{d}}\theta } \right)\left( {\mathop \smallint \limits_{r = 0}^2 {r^4}{\rm{d}}r} \right)$
Using Eq. (5) of the Table of Trigonometric Integrals in Section 8.2:
$\smallint {\sin ^n}x{\rm{d}}x = - \frac{{{{\sin }^{n - 1}}x\cos x}}{n} + \frac{{n - 1}}{n}\smallint {\sin ^{n - 2}}x{\rm{d}}x$
we get
mass $ = 9\left( { - \frac{{{{\sin }^2}\theta \cos \theta }}{3}|_0^{\pi /2} + \frac{2}{3}\mathop \smallint \limits_{\theta = 0}^{\pi /2} \sin \theta {\rm{d}}\theta } \right)\left( {\frac{1}{5}{r^5}|_0^2} \right)$
$ = 9\left( { - \frac{2}{3}\cos \theta |_0^{\pi /2}} \right)\left( {\frac{{32}}{5}} \right)$
$ = 9\left( {\frac{2}{3}} \right)\left( {\frac{{32}}{5}} \right) = \frac{{192}}{5}$
So, the mass of ${\cal W}$ is $\frac{{192}}{5}$.
