## Thomas' Calculus 13th Edition

$r=(r \cos \theta) i+( r\sin \theta) j+\sqrt {9-r^2} k$ and $0 \le r \le \dfrac{3}{\sqrt{2}}$ and $0 \leq \theta \leq 2 \pi$
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 $9=x^2+y^2+z^2 ;\\ z =\sqrt {9-r^2}$ and $z \geq 0$ Now, $r=(r \cos \theta) i+( r\sin \theta) j+\sqrt {9-r^2} k$; or, $9=x^2+y^2+(\sqrt {x^2+y^2})^2 \implies 2(x^2+y^2)=9$ $2r^2 =9$ and $r=\dfrac{3}{\sqrt{2}}$ Thus, $0 \le r \le \dfrac{3}{\sqrt{2}}$ And $0 \leq \theta \leq 2 \pi$