Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 15: Multiple Integrals - Section 15.7 - Triple Integrals in Cylindrical and Spherical Coordinates - Exercises 15.7 - Page 920: 28

Answer

$\frac{28\pi}{3\sqrt{3}}$

Work Step by Step

$\int^{\pi/3}_{\pi/6} \int^{2csc\phi}_{csc\phi} \int^{2\pi}_0 p^2\sin \phi d\theta dp $ $ d\phi $ =$2\pi \int^{\pi/3}_{\pi/6} \int^{2csc\phi}_{csc\phi} p^2sin\phi dp d\phi $ =$\frac{2\pi}{3}\int^{\pi/3}_{\pi/6} [p^3sin\phi]^{2csc\phi}_{csc\phi}$ =$\frac{14\pi}{3}\int^{\pi/3}_{\pi/6} csc^2\phi d\phi $ =$\frac{28\pi}{3\sqrt{3}}$
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