Answer
\[
3 \cos \theta=x, 3 \sin \theta=y, z=\sqrt{9-r^{2}} \text { where }\left\{\begin{array}{l}
0 \leq \theta \leq 2 \pi \\
0 \leq r \leq \sqrt{5}
\end{array}\right.
\]
Work Step by Step
If we use cylindrical coordinates
\[
z=z , \quad r \sin \theta=y, \quad r\cos \theta=x
\]
and then
\[
\begin{aligned}
x^{2}+y^{2}+z^{2}=9 & \Rightarrow r^{2} \cos ^{2} \theta+r^{2} \sin ^{2} \theta+z^{2} \\
& \Rightarrow r^{2}\left(\cos ^{2} \theta+\sin ^{2} \theta\right)+z^{2}=9 \\
& \Rightarrow r^{2}+z^{2}=9 \Rightarrow z=\pm \sqrt{9-r^{2}}
\end{aligned}
\]
So, the surface can be represented parametrically as:
\[
3 \cos \theta=x, 3 \sin \theta=y, z=\sqrt{9-r^{2}} \text { where }\left\{\begin{array}{l}
0 \leq \theta \leq 2 \pi \\
0 \leq r \leq \sqrt{5}
\end{array}\right.
\]
because
\[
2 \leq z \leq 3 \Rightarrow 4 \leq 9-r^{2} \leq 9 \Rightarrow 0 \leq r^{2} \leq 5
\]
