Answer
$0\leq \phi\leq \dfrac{\pi}{4}, 0\leq \rho\leq \cos \phi$
Work Step by Step
The conversion of rectangular coordinates to spherical coordinates is given as:
$x=\rho \sin \phi \cos \theta; y=\rho \sin \phi \sin \theta;z=\rho \cos \phi$
Here, $\rho=\sqrt {x^2+y^2+z^2}$; $\phi =\cos^{-1} [\dfrac{z}{\rho}]; \theta=\cos^{-1}[\dfrac{x}{\rho \sin \phi}]$
Here, we have $z=\sqrt {x^2+y^2}$
This gives: $\rho \cos \phi=\sqrt{(\rho \sin \phi \cos \theta)^2+(\rho \sin \phi \sin \theta)^2}$
and $\rho \cos \phi =\rho \sin \phi \implies \cos \phi = \sin \phi$
Now, $\phi =\cos^{-1} [\dfrac{z}{\rho}]; \implies \phi=\dfrac{\pi}{4}$
and $x^2+y^2+z^2=z$
This gives: $\rho=\cos \phi$
Hence, we have $0\leq \phi\leq \dfrac{\pi}{4}, 0\leq \rho\leq \cos \phi$
