Answer
$r(u,v)=(2 \cos v) i+uj+(2 \sin v)k$; $-2 \le u \le 2$ and $0 \le v \le \pi$
Work Step by Step
Here, we have $x^2+z^2=4$
Thus,
Thus, $r(x,z)=xi+yj+zk \implies r(x,z)=(2 \cos \theta) i+uj+(2 \sin \theta)k$
Then, we have $r(u,v)=(2 \cos v) i+uj+(2 \sin v)k$; $-2 \le u \le 2$ and $0 \le v \le \pi$
