Answer
$r(\phi, \theta)=(2\sqrt 2\sin \phi \cos \theta) i+( 2\sqrt 2 \sin \phi \sin \theta) j+(2\sqrt 2 \cos \phi) k$
$0 \le \phi \le \dfrac{3 \pi}{4}$ and $0 \le \theta \le 2 \pi$
Work Step by Step
The spherical coordinates are: $x= l \sin \phi \cos \theta, y= l \sin \phi \sin \theta, z= l \cos \phi $ ; $0 \le \phi \le \pi$ and $0 \le \theta \le 2 \pi$
Here, we have $l=\sqrt{x^2+y^2+z^2} \implies l=2\sqrt 2$
$x= 2\sqrt 2\sin \phi \cos \theta, y= 2\sqrt 2 \sin \phi \sin \theta, z= 2\sqrt 2 \cos \phi $ ;
Thus,
Thus, $r(\phi, \theta)=xi+yj+zk \implies r=(2\sqrt 2\sin \phi \cos \theta) i+( 2\sqrt 2 \sin \phi \sin \theta) j+(2\sqrt 2 \cos \phi) k$;
and $-2 =2\sqrt 2 \cos \phi$
This gives: $\phi=\dfrac{3 \pi}{4}$
Also, $2\sqrt 2 =2\sqrt 2 \cos \phi$
This gives: $\phi=0$
Hence, our answers are:
$r(\phi, \theta)=(2\sqrt 2\sin \phi \cos \theta) i+( 2\sqrt 2 \sin \phi \sin \theta) j+(2\sqrt 2 \cos \phi) k$
$0 \le \phi \le \dfrac{3 \pi}{4}$ and $0 \le \theta \le 2 \pi$
