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: 29

Answer

$\frac{(4\sqrt{2}-5)}{\sqrt{2}}\pi $

Work Step by Step

$\int^1_0 \int^{\pi}_0 \int^{\pi/4}_0 12 p sin^3 \phi $ $ d\phi $ $ d\theta $ $ dp $ =$\int^1_0 \int^{\pi}_0 (12 p[\frac{-sin^2 \phi \cos \phi}{3}]^{\pi/4}_0+8p \int^{\pi/4}_0 \sin \phi d\phi)d\theta $ $ dp $ =$\int^1_0 \int^{\pi}_0 (-\frac{2p}{\sqrt{2}}-8p[cos\phi]^{\pi/4}_0)d\theta $ $ dp $ =$\int^1_0 \int^{\pi}_0 (8p-\frac{10p}{\sqrt{2}})d\theta dp $ =$\pi \int^1_0 (8p-\frac{10p}{\sqrt{2}})dp $ =$\pi [4p^2-\frac{5p^2}{\sqrt{2}}]^1_0$ =$\frac{(4\sqrt{2}-5)}{\sqrt{2}}\pi $
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