Answer
$ r(x,z)=xi+x^2j+zk ;\\-\sqrt 2 \le x \le \sqrt 2; \\ 0 \le z \le 3$
Work Step by Step
Use spherical coordinates as:
$ x= P \sin \phi \cos \theta; y=Pl \sin \phi \sin \theta; z=Pl \cos \phi $
$0 \le \phi \le \pi; \\0 \le \theta \le 2 \pi $
We know that $ r(r, \theta)=xi+yj+zk $ or, $ r^2=x^2+y^2+z^2$
We have $ y=x^2$
and $ r(x,z)=xi+x^2j+zk $;
Also, $ y=2$ So, $ x =\pm \sqrt 2$
Hence:
$ r(x,z)=xi+x^2j+zk ;\\-\sqrt 2 \le x \le \sqrt 2; \\ 0 \le z \le 3$
