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: 52

Answer

The description of the given set in spherical coordinates: $0 \le \rho \le 1$, ${\ \ }$ $\theta = \frac{\pi }{4}$.

Work Step by Step

In spherical coordinates: 1. the radial coordinate $\rho $ Since ${\rho ^2} = {x^2} + {y^2} + {z^2}$, so we have ${\rho ^2} \le 1$. Since $\rho $ is always positive, we have $0 \le \rho \le 1$. 2. the angular coordinate $\theta$ satisfies $\tan \theta = \frac{y}{x} = \frac{y}{y} = 1$, ${\ \ }$ $\theta = \frac{\pi }{4},\frac{{5\pi }}{4}$ Since $x \ge 0$, $y \ge 0$, it is constrained in the first quadrant. So,the right choice is $\theta = \frac{\pi }{4}$. 3. the angular coordinate $\phi$ There is no constraint on $\phi$. Therefore, the description of the given set in spherical coordinates is $0 \le \rho \le 1$, ${\ \ }$ $\theta = \frac{\pi }{4}$.
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