Answer
(a) $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} f\left( {x,y,z} \right){\rm{d}}V = \mathop \smallint \limits_{z = 0}^1 \mathop \smallint \limits_{x = 0}^4 \mathop \smallint \limits_{y = z}^{\left( {3 - z} \right)/2} f\left( {x,y,z} \right){\rm{d}}y{\rm{d}}x{\rm{d}}z$
(b) $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} {\rm{d}}V = 3$
(c) Using geometry, we show that the volume of ${\cal W}$ is 3.
This coincides with the result in part (b).
Work Step by Step
(a) We have the region ${\cal W}:y = z,2y + z = 3,z = 0,0 \le x \le 4$.
From the figure attached, we see that the projection of ${\cal W}$ onto the $xz$-plane, where $y=0$ is a rectangle. Using the equations $y=z$, $2y+z=3$ we obtain $z=1$. Since it is bounded by $z=0$ and $0 \le x \le 4$, the rectangle is $0 \le x \le 4$, $0 \le z \le 1$.
Denote ${\cal S}$, the projections of ${\cal W}$ onto the $xz$-plane. So,
${\cal S} = \left\{ {\left( {x,z} \right)|0 \le x \le 4,0 \le z \le 1} \right\}$
We also see that ${\cal W}$ is a $y$-simple region bounded below by $y=z$ and bounded above by $y = \frac{1}{2}\left( {3 - z} \right)$.
Thus, the description of ${\cal W}$:
${\cal W} = \left\{ {\left( {x,y,z} \right)|0 \le x \le 4,0 \le z \le 1,z \le y \le \frac{1}{2}\left( {3 - z} \right)} \right\}$
Expressing as an iterated integral in the order $dydxdz$, we get
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} f\left( {x,y,z} \right){\rm{d}}V = \mathop \smallint \limits_{z = 0}^1 \mathop \smallint \limits_{x = 0}^4 \mathop \smallint \limits_{y = z}^{\left( {3 - z} \right)/2} f\left( {x,y,z} \right){\rm{d}}y{\rm{d}}x{\rm{d}}z$
(b) We have $f\left( {x,y,z} \right) = 1$. So, the triple integral is the volume of ${\cal W}$:
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} {\rm{d}}V = \mathop \smallint \limits_{z = 0}^1 \mathop \smallint \limits_{x = 0}^4 \mathop \smallint \limits_{y = z}^{\left( {3 - z} \right)/2} {\rm{d}}y{\rm{d}}x{\rm{d}}z$
$ = \mathop \smallint \limits_{z = 0}^1 \mathop \smallint \limits_{x = 0}^4 \left( {y|_z^{\left( {3 - z} \right)/2}} \right){\rm{d}}x{\rm{d}}z$
$ = \mathop \smallint \limits_{z = 0}^1 \mathop \smallint \limits_{x = 0}^4 \left( {\frac{3}{2} - \frac{3}{2}z} \right){\rm{d}}x{\rm{d}}z$
$ = \frac{3}{2}\left( {\mathop \smallint \limits_{z = 0}^1 \left( {1 - z} \right){\rm{d}}z} \right)\left( {\mathop \smallint \limits_{x = 0}^4 {\rm{d}}x} \right)$
$ = \frac{3}{2}\left( {z - \frac{1}{2}{z^2}|_0^1} \right)\left( 4 \right)$
$ = 6\left( {1 - \frac{1}{2}} \right) = 3$
So, $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} {\rm{d}}V = 3$.
The volume of ${\cal W}$ is 3.
(c) From the figure attached, we see that the front face of ${\cal W}$ is a triangle with base $y = 1.5$ and height $z=1$. So, the area is $\frac{1}{2} \times 1.5 \times 1 = 0.75$. Since the length of ${\cal W}$ is $x=4$, the volume is $0.75 \times 4 = 3$. The result agrees with part (b).
