Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 11 - Vectors and the Geometry of Space - Review Exercises - Page 812: 72

Answer

$\left( {6\sqrt 3 , - \frac{\pi }{2}, - 6} \right)$

Work Step by Step

$$\eqalign{ & \left( {12, - \frac{\pi }{2},\frac{{2\pi }}{3}} \right) \cr & {\text{spherical}}\left( {\rho ,\theta ,\phi } \right):\left( {12, - \frac{\pi }{2},\frac{{2\pi }}{3}} \right) \to \rho = 12,{\text{ }}\theta = - \frac{\pi }{2},{\text{ }}\phi = \frac{{2\pi }}{3} \cr & {\text{Spherical to cylindrical }}\left( {r,\theta ,z} \right),{\text{ }}\left( {r \geqslant 0} \right){\text{ See page 807}} \cr & {r^2} = {\rho ^2}{\sin ^2}\phi ,{\text{ }}\theta = \theta ,{\text{ }}z = \rho \cos \phi \cr & {r^2} = {\left( {12} \right)^2}{\sin ^2}\left( {\frac{{2\pi }}{3}} \right) = 108 \to r = 6\sqrt 3 \cr & \theta = - \frac{\pi }{2} \cr & \cr & {\text{ }}z = \rho \cos \phi \cr & z = 12\cos\left( {\frac{{2\pi }}{3}} \right) = - 6 \cr & {\text{The cylindrical }}\left( {r,\theta ,z} \right){\text{ coordinates are:}} \cr & \left( {6\sqrt 3 , - \frac{\pi }{2}, - 6} \right) \cr} $$
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