Answer
$r(r,\theta)=(r \cos \theta) i+( r\sin \theta) j+(9-r^2) k$
$-3 \le r \le 3$
$0 \le \theta \le 2 \pi$
Work Step by Step
The cylindrical coordinates are: $x=r \cos \theta, y= r \sin \theta, z \gt 0$
Here, we have $z=9-x^2-y^2$
or, $z=9-r^2$
Thus, $r=xi+yj+zk \implies r(r,\theta)=(r \cos \theta) i+( r\sin \theta) j+(9-r^2) k$
Now, $9-r^2 \gt 0$
This means that $r^2 \leq 9$
or, $-3 \le r \le 3$ and $0 \le \theta \le 2 \pi$
