#### Answer

$0 \le r \le 3$ and $0 \le \theta \le 2 \pi $

#### Work Step by Step

We have $ z=9-x^2-y^2 \implies z=9-r^2$
Apply cylindrical coordinates. $ x=r \cos \theta ;\\ y= r \sin \theta \\ z \gt 0$
and $ r=(r \cos \theta) i+( r\sin \theta) j+(9-r^2) k $
But $9-r^2 \ge 0 \implies r^2 \leq 9$
So, $-3 \le r \le 3$ and $0 \le \theta \le 2 \pi $
We can restrict $r$ to $0 \le r \le 3$ (since the negative values repeat the same points).