Answer
$0\leq \phi\leq \dfrac{\pi}{4}, 0\leq \rho\leq \cos \phi$
Work Step by Step
Conversion of rectangular to spherical coordinates is as follows:
$x=\rho \sin \phi \cos \theta; y=\rho \sin \phi \sin \theta;z=\rho \cos \phi$
and
$\rho=\sqrt {x^2+y^2+z^2}$;
$\cos \phi =\dfrac{z}{\rho}$; $\cos \theta=\dfrac{x}{\rho \sin \phi}$
Here, $z=\sqrt {x^2+y^2}$
or, $\rho \cos \phi=\sqrt{(\rho \sin \phi \cos \theta)^2+(\rho \sin \phi \sin \theta)^2}$
or, $\rho \cos \phi =\rho \sin \phi$
or, $ \cos \phi = \sin \phi$
This implies that $\phi=\dfrac{\pi}{4}$
and $x^2+y^2+z^2=z \implies \rho=\cos \phi$
Hence, $0\leq \phi\leq \dfrac{\pi}{4}, 0\leq \rho\leq \cos \phi$
