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: 69

Answer

$$\rho=2 \csc\phi\sqrt{ \sec\phi} .$$

Work Step by Step

We have $$4=x^2-y^2=\rho^2\sin^2\phi\cos^2\theta-\rho^2\sin^2\phi\sin^2\theta\\ =\rho^2\sin^2\phi \cos 2\theta .$$ We solve for $\rho$: $$\rho=\frac{2}{\sin \phi\sqrt{\cos 2\theta}}.$$
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