Answer
$ r(u,v)=2 \cos v \space i+u \space j+2 \sin v \space k;\\ -2 \le u \le 2;\space \\ 0 \le v \le \pi $
Work Step by Step
Apply polar coordinates in $ xz $ plane.
We know that $ r(r, \theta)=xi+yj+zk $ or, $ r^2=x^2+y^2+z^2$
We have $ x^2+z^2=4$
Now, $ r(x,z)=2 \cos \theta \space i+uj+2 \sin \theta \space k $
Consider $ y=u; \theta=v $
This yields: $ r(u,v)=2 \cos v \space i+u \space j+2 \sin v \space k $; $-2 \le u \le 2$ and $0 \le v \le \pi $
Hence:
$ r(u,v)=2 \cos v \space i+u \space j+2 \sin v \space k;\\ -2 \le u \le 2;\space \\ 0 \le v \le \pi $
