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: 6

Answer

$$\left( {\frac{1}{4},\frac{{\sqrt 3 }}{4},8} \right)$$

Work Step by Step

$$\eqalign{ & {\text{We have the cylindrical coordinates }}\left( { - 0.5,\frac{{4\pi }}{3},8} \right) \cr & \left( {r,\theta ,z} \right):{\text{ }}\left( { - 0.5,\frac{{4\pi }}{3},8} \right) \to r = - 0.5,{\text{ }}\theta = \frac{{4\pi }}{3},{\text{ }}z = 8 \cr & {\text{Cylindrical to rectangular}} \cr & x = r\cos \theta \to x = - 0.5\cos \left( {\frac{{4\pi }}{3}} \right) = \frac{1}{4} \cr & y = r\sin \theta \to y = - 0.5\sin \left( {\frac{{4\pi }}{3}} \right) = \frac{{\sqrt 3 }}{4} \cr & z = z \to z = 8 \cr & {\text{The rectangular coordinates are:}} \cr & \left( {\frac{1}{4},\frac{{\sqrt 3 }}{4},8} \right) \cr} $$
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