## Thomas' Calculus 13th Edition

$y=r(x,y)=xi+yj+(4-y^2)k ; \\-2 \le y \le 2 \\ 0 \le x \le 2$
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, $y=xi+yj+(4-y^2)k ; \\-2 \le y \le 2 \\ 0 \le x \le 2$