Answer
Please see the figure attached.
The triple integral that yields the volume of ${\cal W}$:
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} 1{\rm{d}}V = \mathop \smallint \limits_{y = 0}^1 \mathop \smallint \limits_{z = {y^2}}^{\sqrt y } \mathop \smallint \limits_{x = 0}^{4 - y - z} {\rm{d}}x{\rm{d}}z{\rm{d}}y$
Work Step by Step
Please see the figure attached for the region ${\cal W}$ bounded by the surfaces: $z = {y^2}$, $y = {z^2}$, and $x=0$, $x+y+z=4$.
We choose to project ${\cal W}$ onto the $yz$-plane and obtain the domain ${\cal T}$. From the figure attached, we see that ${\cal T}$ is bounded below by the curve $z = {y^2}$ and bounded above by the curve $y = {z^2}$. Thus, we can consider ${\cal T}$ as a vertically simple region.
To obtain the left and right boundaries of ${\cal T}$ in the $yz$-plane, we find the intersection of the two curves: $z = {y^2}$ and $y = {z^2}$ by solving the equation:
$z = {y^2} = \sqrt y $
Squaring both sides, we get ${y^4} = y$. So,
${y^4} - y = 0$
$y\left( {{y^3} - 1} \right) = 0$
So, $y=0$, $y=1$.
Thus, the description of ${\cal T}$:
${\cal T} = \left\{ {\left( {y,z} \right)|0 \le y \le 1,{y^2} \le z \le \sqrt y } \right\}$
Referring to the figure attached, we describe ${\cal W}$ as a $x$-simple region bounded below by $x=0$ and bounded above by $x=4-y-z$.
Thus, the region description of ${\cal W}$:
${\cal W} = \left\{ {\left( {x,y,z} \right)|0 \le y \le 1,{y^2} \le z \le \sqrt y ,0 \le x \le 4 - y - z} \right\}$
From the description we obtain the triple integral that yields the volume of ${\cal W}$:
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} 1{\rm{d}}V = \mathop \smallint \limits_{y = 0}^1 \mathop \smallint \limits_{z = {y^2}}^{\sqrt y } \mathop \smallint \limits_{x = 0}^{4 - y - z} {\rm{d}}x{\rm{d}}z{\rm{d}}y$
