Answer
$$6 \pi $$
Work Step by Step
\[
S=\iint_{R} \sqrt{1+\left(\frac{\partial z}{\partial x}\right)^{2}+\left(\frac{\partial z}{\partial y}\right)^{2} d A}
\]
Take the partial derivatives for $x: 0+\sqrt{9-y^{2}}=z$ and $\mathrm{y}: \sqrt{9-y^{2}}=z$
\[
\begin{array}{l}
0=\frac{\partial z}{\partial x} \\
\frac{-y}{\sqrt{9-y^{2}}}=\frac{\partial z}{\partial y}
\end{array}
\]
Enter partial fractions in the first formula to obtain:
\[
S=\int_{0}^{2}\left[\int_{-3}^{3} \sqrt{1+\frac{y^{2}}{9-y^{2}}} d y\right] d x
\]
Integrate with respect to $y:$
\[
\left.3 \int_{0}^{2} \sin ^{-1}\left(\frac{1}{3}\right)\right|_{-3} ^{3} d x
\]
Integrate with respect to x.
$$6\pi$$
