Answer
Please see the figure attached.
The triple integral that yields the volume of ${\cal W}$:
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} 1{\rm{d}}V = \mathop \smallint \limits_{x = - 1}^1 \mathop \smallint \limits_{z = 0}^{1 - {x^2}} \mathop \smallint \limits_{y = 0}^{3 - {x^2} - {z^2}} {\rm{d}}y{\rm{d}}z{\rm{d}}x$
Work Step by Step
Please see the figure attached for the region ${\cal W}$ bounded by the surfaces: $z = 1 - {x^2}$, $z=0$, and $y = 3 - {x^2} - {z^2}$.
We choose to project ${\cal W}$ onto the $xz$-plane and obtain the domain ${\cal S}$. From the figure attached, we see that ${\cal S}$ is bounded by the curve $z = 1 - {x^2}$ and by $z=0$.
We obtain the left and right boundaries of ${\cal S}$ by solving the equation:
$z = 0 = 1 - {x^2}$
$x = \pm 1$
Thus, we describe ${\cal S}$ as a vertically simple region defined by
${\cal S} = \left\{ {\left( {x,z} \right)| - 1 \le x \le 1,0 \le z \le 1 - {x^2}} \right\}$
Referring to the figure attached, we can describe ${\cal W}$ as a $y$-simple region bounded below by $y=0$ and bounded above by $y = 3 - {x^2} - {z^2}$.
Thus, the region description of ${\cal W}$:
${\cal W} = \left\{ {\left( {x,y,z} \right)| - 1 \le x \le 1,0 \le z \le 1 - {x^2},0 \le y \le 3 - {x^2} - {z^2}} \right\}$
From the description we obtain the triple integral that yields the volume of ${\cal W}$:
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} 1{\rm{d}}V = \mathop \smallint \limits_{x = - 1}^1 \mathop \smallint \limits_{z = 0}^{1 - {x^2}} \mathop \smallint \limits_{y = 0}^{3 - {x^2} - {z^2}} {\rm{d}}y{\rm{d}}z{\rm{d}}x$
