Answer
$r(r, \theta)=(r \cos \theta) i+( r\sin \theta) j+(\dfrac{r}{2}) k$; $0 \le r \le 6$ and $0 \le \theta \le \dfrac{ \pi}{2}$
Work Step by Step
The cylindrical coordinates are: $x=r \cos \theta, y= r \sin \theta, z \gt 0$
Here, we have $z=\dfrac{\sqrt{x^2+y^2}}{2}$
or, $z=\dfrac{\sqrt{r^2}}{2}=\dfrac{r}{2}$
Thus, $r(r, \theta)=xi+yj+zk \implies r=(r \cos \theta) i+( r\sin \theta) j+(\dfrac{r}{2}) k$;
$0 \le z \le 3$ and $0 \le \dfrac{r}{2} \le 3 \implies 0 \le r \le 6$
Hence, our answers are:
$r(r, \theta)=(r \cos \theta) i+( r\sin \theta) j+(\dfrac{r}{2}) k$; $0 \le r \le 6$ and $0 \le \theta \le \dfrac{ \pi}{2}$
