Answer
$ r=(r \cos \theta) i+( r\sin \theta) j+\sqrt {9-r^2} k $ and $0 \le r \le \dfrac{3}{\sqrt{2}}$ and $0 \leq \theta \leq 2 \pi $
Work Step by Step
Apply cylindrical coordinates. $ x=r \cos \theta ;\\ y= r \sin \theta ;\\ z \gt 0$
We know that $ r(r, \theta)=xi+yj+zk $ or, $ r^2=x^2+y^2+z^2$
We have $9=x^2+y^2+z^2 ;\\ z =\sqrt {9-r^2}$ and $ z \geq 0$
Now, $ r=(r \cos \theta) i+( r\sin \theta) j+\sqrt {9-r^2} k $;
or, $9=x^2+y^2+(\sqrt {x^2+y^2})^2 \implies 2(x^2+y^2)=9$
$2r^2 =9 $ and $ r=\dfrac{3}{\sqrt{2}}$
Thus, $0 \le r \le \dfrac{3}{\sqrt{2}}$
And $0 \leq \theta \leq 2 \pi $
