Answer
$ r(u,v)=\lt \sqrt 3 \sin u \cos v,\sqrt 3 \sin u \sin v, \sqrt 3 \cos u \gt $ and $\dfrac{\pi}{3} \le u \le \dfrac{2\pi}{3};\\0 \le v \le 2 \pi $
Work Step by Step
Use spherical coordinates as:
$ x= l \sin \phi \cos \theta; y= l \sin \phi \sin \theta; z= l \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: $ x^2+y^2+z^2 =3$
$ x= \sqrt 3 \sin u \cos v, y= \sqrt 3 \sin u \sin v, z= \sqrt 3 \cos u $
So,
$ r(u,v)=\lt \sqrt 3 \sin u \cos v,\sqrt 3 \sin u \sin v, \sqrt 3 \cos u \gt $ and $\dfrac{\pi}{3} \le u \le \dfrac{2\pi}{3};\\0 \le v \le 2 \pi $
