Calculus (3rd Edition)

$$\phi=\pi/3, \quad \phi=2\pi/3.$$
Since $y^2+x^2=3z^2$, and \begin{aligned} & x=\rho \sin \phi \cos \theta\\ & y=\rho \sin \phi \sin \theta\\ & z=\rho \cos \phi, \end{aligned} Then we have $$\rho^2\sin^2\phi(\sin^2\theta+\cos^2\theta)=3 \rho^2\cos^2\phi$$ and hence $$\sin^2\phi=3 \cos^2\phi\Longrightarrow \tan\phi=\pm\sqrt{3}.$$ That is,$$\phi=\pi/3, \quad \phi=2\pi/3.$$