Answer
$$\frac{5}{32} \pi$$
Work Step by Step
$-x^{2}-y^{2}+1=z$
$-r^{2}+1=z$
$0=x^{2}+y^{2}-x$
$0=r^{2}-r \cos \theta$
$r=\cos \theta$ or, $r=0$
Thus, the integral is:
\[
\begin{array}{c}
2 \int_{0}^{\frac{\pi}{2}} \int_{0}^{\cos \theta}\left(-r^{2}+1\right) r \mathrm{d} \mathrm{rd} \theta \\
=2 \int_{0}^{\frac{\pi}{2}}\left[\frac{r^{2}}{2}-\frac{r^{4}}{4}\right]_{0}^{\cos \theta} \mathrm{d} \theta \\
=2 \int_{0}^{\frac{\pi}{2}}\left(\frac{\cos ^{2} \theta}{2}-\frac{\cos ^{4} \theta}{4}\right) \mathrm{d} \theta \\
=\frac{1}{16} \int_{0}^{\frac{\pi}{2}}(4 \cos 2 \theta-\cos 4 \theta+5) \mathrm{d} \theta \\
=\left[\frac{1}{8} \sin 2 \theta+\frac{1}{64} \sin 4 \theta+\frac{5}{16} \theta\right]_{0}^{\pi / 2} \\
=\frac{5}{32} \pi
\end{array}
\]
