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