Answer
Please see the figure attached.
The triple integral that yields the volume of ${\cal W}$:
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} 1{\rm{d}}V = \mathop \smallint \limits_{y = - 1}^1 \mathop \smallint \limits_{x = 0}^{{y^2} + 2} \mathop \smallint \limits_{z = 1 - {y^2}}^{3 - x} {\rm{d}}z{\rm{d}}x{\rm{d}}y$
Work Step by Step
Please see the figure attached for the region ${\cal W}$ bounded by the surfaces given by $z = 1 - {y^2}$, $x=0$ and $z=0$, $z+x=3$.
The projection of ${\cal W}$ onto the $xy$-plane is the domain ${\cal D}$ obtained when $z=0$. We need to find the equation of the boundary curve of ${\cal D}$ in the $xy$-plane. A point $P$ on this curve is the projection of a point $Q$ on the intersection between the two surfaces: $z = 1 - {y^2}$ and $z+x=3$. The intersection occurs when they have the common $z$, thus
$z = 1 - {y^2} = 3 - x$
Thus, the boundary of ${\cal D}$ is the curve $x = {y^2} + 2$.
We choose to describe ${\cal D}$ as a horizontally simple region. Thus, the description of ${\cal D}$:
${\cal D} = \left\{ {\left( {x,y} \right)| - 1 \le y \le 1,0 \le x \le {y^2} + 2} \right\}$
Notice that ${\cal W}$ is a $z$-simple region bounded below by $z = 1 - {y^2}$ and bounded above by $z+x=3$. Thus, the region description of ${\cal W}$:
${\cal W} = \left\{ {\left( {x,y,z} \right)| - 1 \le y \le 1,0 \le x \le {y^2} + 2,1 - {y^2} \le z \le 3 - x} \right\}$
From the description we obtain the triple integral that yields the volume of ${\cal W}$:
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} 1{\rm{d}}V = \mathop \smallint \limits_{y = - 1}^1 \mathop \smallint \limits_{x = 0}^{{y^2} + 2} \mathop \smallint \limits_{z = 1 - {y^2}}^{3 - x} {\rm{d}}z{\rm{d}}x{\rm{d}}y$
