Answer
$\frac{9 \pi}{2}$
Work Step by Step
$\int_{R} \int \sqrt{-x^{2}-y^{2}+9}$
$R: 9=x^{2}+y^{2} \Rightarrow R=\left\{(r, \theta): 0 \leqslant r \leqslant 3,0 \leqslant \theta \leqslant \frac{\pi}{2}\right\}$
$\sqrt{9-x^{2}-y^{2}} d A=\int_{0}^{\frac{\pi}{2}} \int_{0}^{3} \sqrt{9-r^{2}} r d r d \theta$
integrate with respect to $r$ $=-\frac{1}{2} \int_{0}^{\frac{\pi}{2}}\left[\frac{\left(9-r^{2}\right)^{3 / 2}}{3 / 2}\right]_{0}^{3} d \theta$
$=-\frac{1}{3} \int_{0}^{\frac{\pi}{2}}\left[-(9-0)^{3 / 2}+(9-9)^{3 / 2}\right] d \theta$
$=-\frac{1}{3} \int_{0}^{\frac{\pi}{2}}(-27) d \theta=9 \int_{0}^{\frac{\pi}{2}} d \theta$
integrate
$9\left(-0+\frac{\pi}{2}\right)=\frac{9 \pi}{2}$
