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