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 - 11.7 Exercises - Page 810: 76

Answer

$$\left( {0, - \frac{{5\pi }}{6}, - 5} \right)$$

Work Step by Step

$$\eqalign{ & \left( {5, - \frac{{5\pi }}{6},\pi } \right) \cr & {\text{spherical}}\left( {\rho ,\theta ,\phi } \right):\left( {5, - \frac{{5\pi }}{6},\pi } \right) \to \rho = 5,{\text{ }}\theta = - \frac{{5\pi }}{6},{\text{ }}\phi = \pi \cr & {\text{Spherical to cylindrical }}\left( {r,\theta ,z} \right),{\text{ }}\left( {r \geqslant 0} \right){\text{ See page 807}} \cr & {r^2} = {\rho ^2}{\sin ^2}\phi ,{\text{ }}\theta = \theta ,{\text{ }}z = \rho \cos \phi \cr & {r^2} = {\left( 5 \right)^2}{\sin ^2}\left( \pi \right) = 0 \to r = 0 \cr & \theta = - \frac{{5\pi }}{6} \cr & z = 5cos\left( \pi \right) = - 5 \cr & {\text{The cylindrical }}\left( {r,\theta ,z} \right){\text{ coordinates are:}} \cr & \left( {0, - \frac{{5\pi }}{6}, - 5} \right) \cr} $$
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