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: 30

Answer

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

Work Step by Step

$$\eqalign{ & \left( { - 4,0,0} \right) \cr & \left( {x,y,z} \right):{\text{ }}\left( { - 4,0,0} \right) \to x = - 4,{\text{ }}y = 0,{\text{ }}z = 0 \cr & {\text{Rectangular to spherical }}\left( {\rho ,\theta ,\phi } \right) \cr & {\rho ^2} = {x^2} + {y^2} + {z^2} \to \rho = \sqrt {{{\left( { - 4} \right)}^2} + {{\left( 0 \right)}^2} + {{\left( 0 \right)}^2}} = 4 \cr & \tan \theta = \frac{y}{x} \to \theta = {\tan ^{ - 1}}\left( {\frac{0}{{ - 4}}} \right) = \pi \cr & \phi = \arccos \left( {\frac{z}{{\sqrt {{x^2} + {y^2} + {z^2}} }}} \right) \to \phi = \arccos \left( {\frac{0}{{ - 4}}} \right) = \frac{\pi }{2} \cr & {\text{The spherical }}\left( {\rho ,\theta ,\phi } \right){\text{ coordinates are:}} \cr & \left( {4,\pi ,\frac{\pi }{2}} \right) \cr} $$
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