Answer
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} z{\rm{d}}V = 2$
Work Step by Step
We have $f\left( {x,y,z} \right) = z$.
Referring to Figure 13, we can consider the region ${\cal W}$ as a $x$-simple region. The projection of ${\cal W}$ onto the $yz$-plane is the domain ${\cal T}$ bounded by $y=4$, and the line $z = \frac{1}{4}y$. Thus, the domain description of ${\cal T}$ is given by
${\cal T} = \left\{ {\left( {x,y} \right)|0 \le y \le 4,0 \le z \le \frac{1}{4}y} \right\}$
The region ${\cal W}$ consists of all points lying between ${\cal T}$ and the front face $x=3$. Thus, the $x$-coordinate satisfies $0 \le x \le 3$. Thus,
${\cal W} = \left\{ {\left( {x,y,z} \right)|0 \le y \le 4,0 \le z \le \frac{1}{4}y,0 \le x \le 3} \right\}$
We evaluate the triple integral as an iterated integral:
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} z{\rm{d}}V = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal T}^{} \left( {\mathop \smallint \limits_{x = 0}^3 z{\rm{d}}x} \right){\rm{d}}A$
$ = \mathop \smallint \limits_{y = 0}^4 \mathop \smallint \limits_{z = 0}^{y/4} z\left( {x|_0^3} \right){\rm{d}}z{\rm{d}}y$
$ = 3\mathop \smallint \limits_{y = 0}^4 \mathop \smallint \limits_{z = 0}^{y/4} z{\rm{d}}z{\rm{d}}y$
$ = \frac{3}{2}\mathop \smallint \limits_{y = 0}^4 \left( {{z^2}|_0^{y/4}} \right){\rm{d}}y$
$ = \frac{3}{{32}}\mathop \smallint \limits_{y = 0}^4 {y^2}{\rm{d}}y$
$ = \frac{1}{{32}}\left( {{y^3}|_0^4} \right) = 2$
Thus, $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} z{\rm{d}}V = 2$.
