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 809: 5

Answer

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

Work Step by Step

$$\eqalign{ & {\text{We have the cylindrical coordinates }}\left( {4,\frac{{7\pi }}{6},3} \right) \cr & \left( {r,\theta ,z} \right):{\text{ }}\left( {4,\frac{{7\pi }}{6},3} \right) \to r = 4,{\text{ }}\theta = \frac{{7\pi }}{6},{\text{ }}z = 3 \cr & {\text{Cylindrical to rectangular}} \cr & x = r\cos \theta \to x = 4\cos \left( {\frac{{7\pi }}{6}} \right) = - 2\sqrt 3 \cr & y = r\sin \theta \to y = 4\sin \left( {\frac{{7\pi }}{6}} \right) = - 2 \cr & z = z \to z = 3 \cr & {\text{The rectangular coordinates are:}} \cr & \left( { - 2\sqrt 3 , - 2,3} \right) \cr} $$
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