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 699: 8


$(\frac{3}{2}, \frac{3\sqrt 3}{2},9) \Longrightarrow (3, \frac{\pi}{3},9) $.

Work Step by Step

We have $$ x=\frac{3}{2}, \quad y =\frac{3\sqrt 3}{2}, \quad z=9.$$ Now, we have $$ r=\sqrt{x^2+y^2}=\sqrt{36/4}=3,$$ $$\theta =\tan^{-1}y/x=\tan^{-1}\frac{1}{\sqrt{3}}=\frac{\pi}{3},$$ $$ z=9.$$ So, $(\frac{3}{2}, \frac{3\sqrt 3}{2},9) \Longrightarrow (3, \frac{\pi}{3},9) $.
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