Answer
$ r(r, \theta)=(r \cos \theta) i+( r\sin \theta) j+2r k $ and $1 \le r \le 2 ; 0 \le \theta \le 2 \pi $
Work Step by Step
We know that $ r(r, \theta)=xi+yj+zk $ or, $ r^2=x^2+y^2+z^2$
$ z=2\sqrt{x^2+y^2} \implies z=2 \sqrt {r^2}=2r $
Apply cylindrical coordinates. $ x=r \cos \theta ;\\ y= r \sin \theta ;\\ z \gt 0$
and $ r=(r \cos \theta) i+( r\sin \theta) j+2r k $;
Also, $2 \le z \le 4 ;\\ 2 \le 2r \le 4$
or, $1 \le r \le 2$
Hence: $ r(r, \theta)=(r \cos \theta) i+( r\sin \theta) j+2r k $ and $1 \le r \le 2 ; 0 \le \theta \le 2 \pi $
