Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 13 - Vector Geometry - 13.7 Cylindrical and Spherical Coordinates - Exercises - Page 700: 54


$$\phi=\pi/3, \quad \phi=2\pi/3.$$

Work Step by Step

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,$$\phi=\pi/3, \quad \phi=2\pi/3.$$
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