## Thomas' Calculus 13th Edition

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