Answer
$ r(u,v)=xi+3 \cos u \space j+3 \sin v \space k; \\ 0 \le u \le 3$ and $0 \le v \le 2\pi $
Work Step by Step
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^2+z^2=9$
Now, $ r(u,v)=xi+(3 \cos u)j+(3 \sin v)k $;
and $ y^2+z^2=9 ;\\ 0 \le u \le 3 \space $ and $\space 0 \le v \le 2\pi $
Hence:
$ r(u,v)=xi+3 \cos u \space j+3 \sin v \space k; \\ 0 \le u \le 3$ and $0 \le v \le 2\pi $
