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: 50

Answer

$\rho=1$ , $0\leq \phi\leq \pi/2$, $0\le \theta \le 2\pi$.

Work Step by Step

The set $x^2+y^2+z^2=1$, $z\geq 0$ is the upper hemisphere in $R^3$, centered at the origin. Since $$ \begin{aligned} &x=\rho \sin \phi \cos \theta\\ &y=\rho \sin \phi \sin \theta\\ &z=\rho \cos \phi, \end{aligned} $$we have $$x^2+y^2+z^2=\rho^2(\sin^2\phi\cos^2\theta+\sin^2\phi\sin^2\theta+\cos^2\phi)\\ =\rho^2$$ That is, $\rho^2= 1$ and since $\rho$ is non negative then $\rho=1$ and since $z\geq 0$, then $\rho\cos \phi \geq0$ i.e, $0\leq \phi\leq \pi/2$. There are no restrictions on $\theta$, so: $0\le \theta \le 2\pi$
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