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: 30

Answer

$11\pi \sqrt{3}$

Work Step by Step

$\int^{\pi/2}_{\pi/6} \int^{\pi/2}_{-\pi/2} \int^2_{\csc\phi}5p^4 sin\phi dp d\theta d\phi $ =$\int^{\pi/2}_{\pi/6} \int^{\pi/2}_{-\pi/2} (32-csc^5\phi) sin^3\phi $ $ d\theta $ $ d\phi $ =$\int^{\pi/2}_{\pi/6} \int^{\pi/2}_{-\pi/2} (32sin^3\phi-csc^2\phi) d\theta d\phi $ =$\pi \int^{\pi/2}_{[\pi/6}(32sin^3\phi-csc^2\phi)d\phi $ =$\pi[-\frac{32 sin^2\phi cos\phi}{3}]^{\pi/2}_{\pi/6}$ =$\pi(\frac{32\sqrt{3}}{24})-\frac{64\pi}{3}[cos\phi]^{\pi/2}_{\pi/6}-\pi(\sqrt{3})$ =$\frac{\sqrt{3}}{3}\pi+(\frac{64\pi}{3}(\frac{\sqrt{3}}{2}))$ =$\frac{33\pi\sqrt{3}}{3}$ =$11\pi \sqrt{3}$
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