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 $ and
$0 \le \phi \le \dfrac{3 \pi}{4}$ and $0 \le \theta \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 $ P=\sqrt{x^2+y^2+z^2} \implies P=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 $ ;
and $ r=(2\sqrt 2\sin \phi \cos \theta) i+( 2\sqrt 2 \sin \phi \sin \theta) j+(2\sqrt 2 \cos \phi) k \\ \implies -2 =2\sqrt 2 \cos \phi \\ \implies \phi=\dfrac{3 \pi}{4}$
Also, $2\sqrt 2 =2\sqrt 2 \cos \phi \\ \implies \phi=0$
Hence: $ 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 $ and
$0 \le \phi \le \dfrac{3 \pi}{4}$ and $0 \le \theta \le 2 \pi $
