Answer
$0\leq \phi\leq \dfrac{\pi}{4}, 0\leq \rho\leq \cos \phi$
Work Step by Step
Given: $z=\sqrt {x^2+y^2}$
$\rho \cos \phi=\sqrt{(\rho \sin \phi \cos \theta)^2+(\rho \sin \phi \sin \theta)^2}$
$\rho \cos \phi =\rho \sin \phi \implies \cos \phi = \sin \phi$
Thus $\phi=\dfrac{\pi}{4}$
and $x^2+y^2+z^2=z$
or, $ \rho=\cos \phi$
Hence, $0\leq \phi\leq \dfrac{\pi}{4}, 0\leq \rho\leq \cos \phi$
