Answer
$r(x,z)=xi+x^2j+zk$
$-\sqrt 2 \le x \le \sqrt 2$ and $0 \le z \le 3$
Work Step by Step
The spherical coordinates are: $x= l \sin \phi \cos \theta, y= l \sin \phi \sin \theta, z= l \cos \phi $ ; $0 \le \phi \le \pi$ and $0 \le \theta \le 2 \pi$
Here, we have $y=x^2$
Thus,
Thus, $r(x,z)=xi+yj+zk \implies r(x,z)=xi+x^2j+zk$;
and $y=2 \implies x =\pm \sqrt 2$
Hence, our answers are:
$r(x,z)=xi+x^2j+zk$;
$-\sqrt 2 \le x \le \sqrt 2$ and $0 \le z \le 3$
