Answer
$ y=r(x,y)=xi+yj+(4-y^2)k ; \\-2 \le y \le 2 \\ 0 \le x \le 2$
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 $ z=4-y^2 ; \\ y=r(x,y)=xi+yj+zk \implies r=xi+yj+(4-y^2)k\\ z=0, 0=4-y^2 $ or, $ y =\pm 2$
Hence: $ y=r(x,y)$
or, $ y=xi+yj+(4-y^2)k ; \\-2 \le y \le 2 \\ 0 \le x \le 2$
