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