Answer
Please see the figure attached.
The triple integral that yields the volume of ${\cal W}$:
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} 1{\rm{d}}V = \mathop \smallint \limits_{z = 0}^1 \mathop \smallint \limits_{y = z - 1}^{1 - z} \mathop \smallint \limits_{x = - \sqrt z }^{\sqrt z } {\rm{d}}x{\rm{d}}y{\rm{d}}z$
Work Step by Step
Please see the figure attached for the region ${\cal W}$ bounded by the surfaces given by $z = {x^2}$, $z+y=1$, and $z-y=1$.
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 a triangle, bounded left by the line $z-y=1$ and bounded right by the line $z+y=1$.
So, the left and right boundaries are $y=z-1$ and $y=1-z$, respectively.
Thus, we describe ${\cal T}$ as a horizontally simple region defined by
${\cal T} = \left\{ {\left( {x,y} \right)|0 \le z \le 1,z - 1 \le y \le 1 - z} \right\}$
Referring to the figure attached, we see that ${\cal W}$ is located on both positive and negative $x$-axis. Using $z = {x^2}$, we obtain $x = \pm \sqrt z $. Thus, we can describe it as a $x$-simple region bounded below by $x = - \sqrt z $ and bounded above by $x = \sqrt z $.
Thus, the region description of ${\cal W}$:
${\cal W} = \left\{ {\left( {x,y,z} \right)|0 \le z \le 1,z - 1 \le y \le 1 - z, - \sqrt z \le x \le \sqrt z } \right\}$
From the description we obtain the triple integral that yields the volume of ${\cal W}$:
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} 1{\rm{d}}V = \mathop \smallint \limits_{z = 0}^1 \mathop \smallint \limits_{y = z - 1}^{1 - z} \mathop \smallint \limits_{x = - \sqrt z }^{\sqrt z } {\rm{d}}x{\rm{d}}y{\rm{d}}z$
