## Thomas' Calculus 13th Edition

$r(x,z)=xi+x^2j+zk ;\\-\sqrt 2 \le x \le \sqrt 2; \\ 0 \le z \le 3$
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: $r(x,z)=xi+x^2j+zk ;\\-\sqrt 2 \le x \le \sqrt 2; \\ 0 \le z \le 3$