Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 13 - Vector Geometry - 13.7 Cylindrical and Spherical Coordinates - Exercises - Page 700: 44

Answer

The spherical coordinates is $\left( {\rho ,\theta ,\phi } \right) = \left( {2,\frac{\pi }{4},\frac{\pi }{6}} \right)$.

Work Step by Step

We have $\left( {x,y,z} \right) = \left( {\frac{{\sqrt 2 }}{2},\frac{{\sqrt 2 }}{2},\sqrt 3 } \right)$. In spherical coordinates: 1. the radial coordinate is $\rho = \sqrt {{{\left( {\frac{{\sqrt 2 }}{2}} \right)}^2} + {{\left( {\frac{{\sqrt 2 }}{2}} \right)}^2} + {{\left( {\sqrt 3 } \right)}^2}} = \sqrt {\frac{1}{2} + \frac{1}{2} + 3} = 2$ 2. the angular coordinate $\theta$ satisfies $\tan \theta = \frac{y}{x} = \frac{{\sqrt 2 /2}}{{\sqrt 2 /2}} = 1$, ${\ \ }$ $\theta = \frac{\pi }{4}$ 3. the angular coordinate $\phi $ satisfies $\cos \phi = \frac{z}{\rho } = \frac{1}{2}\sqrt 3 $, ${\ \ }$ $\phi = \frac{\pi }{6}$ Therefore, the spherical coordinates is $\left( {\rho ,\theta ,\phi } \right) = \left( {2,\frac{\pi }{4},\frac{\pi }{6}} \right)$.
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