Answer
The volume of the region is $\frac{\pi }{2}$.
Work Step by Step
Let ${\cal W}$ denote the region between the graph of the function $f\left( {x,y} \right) = 1 - \left( {{x^2} + {y^2}} \right)$ and the $xy$-plane. In other words, ${\cal W}$ is lower bounded by $z=0$ and upper bounded by $z = f\left( {x,y} \right) = 1 - \left( {{x^2} + {y^2}} \right)$.
The projection of ${\cal W}$ onto the $xy$-plane ($z=0$) is the unit disk: ${\cal D}:{x^2} + {y^2} = 1$. In polar coordinates:
${\cal D} = \left\{ {\left( {r,\theta } \right)|0 \le r \le 1,0 \le \theta \le 2\pi } \right\}$
So, the description of ${\cal W}$ in cylindrical coordinates:
${\cal W} = \left\{ {\left( {r,\theta ,z} \right)|0 \le r \le 1,0 \le \theta \le 2\pi ,0 \le z \le 1 - {r^2}} \right\}$
Compute the volume of ${\cal W}$:
$V = \mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 0}^1 \mathop \smallint \limits_{z = 0}^{1 - {r^2}} r{\rm{d}}z{\rm{d}}r{\rm{d}}\theta $
$ = \mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 0}^1 r\left( {z|_0^{1 - {r^2}}} \right){\rm{d}}r{\rm{d}}\theta $
$ = \mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 0}^1 \left( {r - {r^3}} \right){\rm{d}}r{\rm{d}}\theta $
$ = \left( {\mathop \smallint \limits_{\theta = 0}^{2\pi } {\rm{d}}\theta } \right)\left( {\mathop \smallint \limits_{r = 0}^1 \left( {r - {r^3}} \right){\rm{d}}r} \right)$
$ = 2\pi \left( {\frac{1}{2}{r^2} - \frac{1}{4}{r^4}} \right)|_0^1 = 2\pi \left( {\frac{1}{4}} \right) = \frac{\pi }{2}$
So, the volume of the region is $\frac{\pi }{2}$.
