Answer
$r=(r \cos \theta) i+( r\sin \theta) j+\sqrt {9-r^2} k$; $0 \le r \le \dfrac{\sqrt 3}{2}$; $0\le\theta\le 2\pi$
Work Step by Step
The cylindrical coordinates are: $x=r \cos \theta, y= r \sin \theta, z \gt 0$
Here, we have $9=x^2+y^2+z^2$
or, $z^2=9-(x^2+y^2) \implies z =\sqrt {9-r^2}$; $z \geq 0$
Thus, $r(r, \theta)=xi+yj+zk \implies r=(r \cos \theta) i+( r\sin \theta) j+\sqrt {9-r^2} k$;
Now, $9=x^2+y^2+(\sqrt {x^2+y^2})^2 \implies 2(x^2+y^2)=9$
or, $2r^2 =9 \implies r=\dfrac{\sqrt 3}{2}$
Also, $0 \le r \le \dfrac{\sqrt 3}{2}$
Hence, our answers are:
$r=(r \cos \theta) i+( r\sin \theta) j+\sqrt {9-r^2} k$; $0 \le r \le \dfrac{\sqrt 3}{2}$; $0\le\theta\le 2\pi$
