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: 68

Answer

$$\left( {4\sqrt 2 ,\frac{\pi }{3},\pi } \right)$$

Work Step by Step

$$\eqalign{ & \left( { - 4,\frac{\pi }{3},4} \right) \cr & \left( {r,\theta ,z} \right):\left( { - 4,\frac{\pi }{3},4} \right) \to r = - 4,{\text{ }}\theta = \frac{\pi }{3},{\text{ }}z = 4 \cr & {\text{Cylindrical to spherical }}\left( {\rho ,\theta ,\phi } \right),{\text{ }}\left( {r \geqslant 0} \right){\text{ See page 807}} \cr & \rho = \sqrt {{r^2} + {z^2}} ,{\text{ }}\theta = \theta ,{\text{ }}\phi = \arccos \left( {\frac{z}{{\sqrt {{r^2} + {z^2}} }}} \right) \cr & \rho = \sqrt {{{\left( { - 4} \right)}^2} + {{\left( 4 \right)}^2}} = \sqrt {32} = \sqrt {16 \cdot 2} = 4\sqrt 2 \cr & \theta = \frac{\pi }{3} \cr & \phi = \arccos \left( {\frac{4}{{ - 4}}} \right) = \pi \cr & {\text{The spherical }}\left( {\rho ,\theta ,\phi } \right){\text{ coordinates are:}} \cr & \left( {4\sqrt 2 ,\frac{\pi }{3},\pi } \right) \cr} $$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.