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: 43



Work Step by Step

Since $ x=\frac{1}{2}, \, y=\frac{\sqrt3}{2}, \, z=\sqrt 3$ and $$\rho=\sqrt{x^2+y^2+z^2},\quad \theta=\tan^{-1}( y/x), \quad \phi=\cos^{-1}(z/\rho),$$ then we have $$\rho=\sqrt{\frac{1}{4}+\frac{3}{4}+3}= 2, $$ $$\theta=\tan^{-1}( \sqrt 3)=\frac{\pi}{3}, $$ $$\phi=\cos^{-1}(\sqrt 3/2)=\frac{\pi}{6}.$$ Hence, the spherical coordinates are $(2,\frac{\pi}{3},\frac{\pi}{6})$.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.