Answer
$r(r, \theta)=(r \cos \theta) i+( r\sin \theta) j+2r k$; $1 \le r \le 2$ and $0 \le \theta \le 2 \pi$
Work Step by Step
The cylindrical coordinates are: $x=r \cos \theta, y= r \sin \theta, z \gt 0$
Here, we have $z=2\sqrt{x^2+y^2}$
or, $z=2 \sqrt {r^2}=2r$
Thus, $r(r, \theta)=xi+yj+zk \implies r=(r \cos \theta) i+( r\sin \theta) j+2r k$;
Also, $2 \le z \le 4$ and $2 \le 2r \le 4 \implies 1 \le r \le 2$
Hence, our answers are:
$r(r, \theta)=(r \cos \theta) i+( r\sin \theta) j+2r k$; $1 \le r \le 2$ and $0 \le \theta \le 2 \pi$
