Answer
\[
2 \int_{0}^{\frac{\pi}{2}} \int_{0}^{\cos \theta}\left(-r^{2}+1\right) r \operatorname{drd} \theta
\]
Work Step by Step
\[
\begin{array}{c}
1-x^{2}-y^{2}=z \\
1-r^{2}=z \\
0=x^{2}+y^{2}-x \\
r^{2}-r \cos \theta=0 \\
\cos \theta=r \text { or, } 0=r
\end{array}
\]
The integral is:
\[
2 \int_{0}^{\frac{\pi}{2}} \int_{0}^{\cos \theta}\left(-r^{2}+1\right) r \operatorname{drd} \theta
\]