Answer
$x=2 \sin \phi, \cos \theta; y=2 \sin \phi, \sin \theta;z=2 \cos \phi$
;$0 \leq \phi \leq \pi/4,0 \leq \theta \leq \pi/4$
or, $x=x,y=y,z=\sqrt{4-x^2-y^2}; x^2+y^2 \leq \sqrt 2$
Work Step by Step
Consider the parametric equations inside the cone are: $( r \cos \theta, r \sin \theta , z)$
The equation of sphere as $z=\sqrt{4-(x^2+y^2)}=\sqrt{4-(r^2 \sin^2 \theta+r^2 \cos \theta)}=\sqrt {4-r^2}$
Now, the points inside the sphere becomes: $[ r \cos \theta, r \sin \theta , \sqrt{4-(x^2+y^2)}] ; \theta \in [0, 2 \pi] $ and $r \in [\sqrt 2,2] $
Find the range of $r$.
$\sqrt{x^2+y^2}=\sqrt{4-(x^2+y^2)}$
$\implies x^2+y^2=4-(x^2+y^2)$
That is, $r=\sqrt 2$
So, the parametric equations or representation for the sphere:
$x=2 \sin \phi, \cos \theta; y=2 \sin \phi, \sin \theta;z=2 \cos \phi$
;$0 \leq \phi \leq \pi/4,0 \leq \theta \leq \pi/4$
or, $x=x,y=y,z=\sqrt{4-x^2-y^2}; x^2+y^2 \leq \sqrt 2$
