## Calculus: Early Transcendentals 8th Edition

$0\leq \phi\leq \dfrac{\pi}{4}, 0\leq \rho\leq \cos \phi$
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, $0\leq \phi\leq \dfrac{\pi}{4}, 0\leq \rho\leq \cos \phi$