## Thomas' Calculus 13th Edition

$r(r, \theta)=(r \cos \theta) i+( r\sin \theta) j+2r k$ and $1 \le r \le 2 ; 0 \le \theta \le 2 \pi$
We know that $r(r, \theta)=xi+yj+zk$ or, $r^2=x^2+y^2+z^2$ $z=2\sqrt{x^2+y^2} \implies z=2 \sqrt {r^2}=2r$ Apply cylindrical coordinates. $x=r \cos \theta ;\\ y= r \sin \theta ;\\ z \gt 0$ and $r=(r \cos \theta) i+( r\sin \theta) j+2r k$; Also, $2 \le z \le 4 ;\\ 2 \le 2r \le 4$ or, $1 \le r \le 2$ Hence: $r(r, \theta)=(r \cos \theta) i+( r\sin \theta) j+2r k$ and $1 \le r \le 2 ; 0 \le \theta \le 2 \pi$