Answer
The volume of the solid is $\pi $.
Work Step by Step
Let ${\cal W}$ denote the cylinder ${x^2} + {y^2} = 1$ below the surface $z = {\left( {x + y} \right)^2}$ and above the surface $z = - {\left( {x - y} \right)^2}$. We can describe the solid region using cylindrical coordinates.
Using $x = r\cos \theta $, $y = r\sin \theta $, we get
${\cal W} = \left\{ {\left( {r,\theta ,z} \right)|0 \le r \le 1,0 \le \theta \le 2\pi , - {r^2}{{\left( {\cos \theta - \sin \theta } \right)}^2} \le z \le {r^2}{{\left( {\cos \theta + \sin \theta } \right)}^2}} \right\}$
Compute the volume of ${\cal W}$:
$V = \mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 0}^1 \mathop \smallint \limits_{z = - {r^2}{{\left( {\cos \theta - \sin \theta } \right)}^2}}^{{r^2}{{\left( {\cos \theta + \sin \theta } \right)}^2}} r{\rm{d}}z{\rm{d}}r{\rm{d}}\theta $
(1) ${\ \ \ \ \ }$ $V = \mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 0}^1 r\left( {z|_{ - {r^2}{{\left( {\cos \theta - \sin \theta } \right)}^2}}^{{r^2}{{\left( {\cos \theta + \sin \theta } \right)}^2}}} \right){\rm{d}}r{\rm{d}}\theta $
Evaluate $\left( {z|_{ - {r^2}{{\left( {\cos \theta - \sin \theta } \right)}^2}}^{{r^2}{{\left( {\cos \theta + \sin \theta } \right)}^2}}} \right)$:
$\left( {z|_{ - {r^2}{{\left( {\cos \theta - \sin \theta } \right)}^2}}^{{r^2}{{\left( {\cos \theta + \sin \theta } \right)}^2}}} \right) = {r^2}{\left( {\cos \theta + \sin \theta } \right)^2} + {r^2}{\left( {\cos \theta - \sin \theta } \right)^2}$
$ = {r^2}\left( {{{\cos }^2}\theta + 2\cos \theta \sin \theta + {{\sin }^2}\theta + {{\cos }^2}\theta - 2\cos \theta \sin \theta + {{\sin }^2}\theta } \right)$
$ = 2{r^2}$
So, $\left( {z|_{ - {r^2}{{\left( {\cos \theta - \sin \theta } \right)}^2}}^{{r^2}{{\left( {\cos \theta + \sin \theta } \right)}^2}}} \right) = 2{r^2}$.
Substituting the result $\left( {z|_{ - {r^2}{{\left( {\cos \theta - \sin \theta } \right)}^2}}^{{r^2}{{\left( {\cos \theta + \sin \theta } \right)}^2}}} \right) = 2{r^2}$ back in equation (1) gives
$V = 2\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 0}^1 {r^3}{\rm{d}}r{\rm{d}}\theta $
$ = \frac{1}{2}\mathop \smallint \limits_{\theta = 0}^{2\pi } \left( {{r^4}|_0^1} \right){\rm{d}}\theta $
$ = \frac{1}{2}\mathop \smallint \limits_{\theta = 0}^{2\pi } {\rm{d}}\theta = \pi $
So, the volume of the solid is $\pi $.
