Answer
$ r(r, \theta)=(r \cos \theta) i+( r\sin \theta) j+(\dfrac{r}{2}) k $ and $0 \le r \le 6$; $0 \le \theta \le \dfrac{ \pi}{2}$
Work Step by Step
We have $ z=\dfrac{\sqrt{x^2+y^2}}{2} \implies z=\dfrac{\sqrt{r^2}}{2}=\dfrac{(r^{1/2})^2}{2}=\dfrac{r}{2}$
Now, $ r(r, \theta)=xi+yj+zk ;\\
r=(r \cos \theta) i+( r\sin \theta) j+(\dfrac{r}{2}) k $;
and $0 \le z \le 3;\\ 0 \le \dfrac{r}{2} \le 3 \implies 0 \le r \le 6$
Hence,
$ r(r, \theta)=(r \cos \theta) i+( r\sin \theta) j+(\dfrac{r}{2}) k $ and $0 \le r \le 6$; $0 \le \theta \le \dfrac{ \pi}{2}$
