Multivariable Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 0-53849-787-4
ISBN 13: 978-0-53849-787-9

Chapter 15 - Multiple Integrals - Review - Exercises - Page 1074: 25

Answer

$\dfrac{\pi}{96}$

Work Step by Step

Plug $y = r \cos \theta; z= r \sin \theta$ Use formula: $\sin^2 x= \dfrac{1}{2}-\dfrac{\cos 2x}{2} $ $\iiint_{E}y^2 z^2 dV=\int_{0}^{2 \pi}\int_{0} ^{1}\int_{0} ^{1-r^2} r^4 \cos^2 \theta \sin^2 \theta \times r \space dx dr d\theta \\ =\int_{0}^{2 \pi}\int_{0} ^{1}\int_{0} ^{1-r^2} r^5 \cos^2 \theta \sin^2 \theta \space dx dr d\theta \\ =[ \int_{0}^{2 \pi} \cos^2 \theta \sin^2 \theta \space d\theta] [\dfrac{r^6}{6} -\dfrac{r^8}{8}]_0^1 \\ =[(\dfrac{1}{4}) \int_{0}^{2 \pi} \sin^2 \theta d\theta] [\dfrac{r^6}{6} -\dfrac{r^8}{8}]_0^1 \\=(\dfrac{1}{4}) [\dfrac{ \theta}{2}-\dfrac{\sin 4 \theta}{8} ]_0^{2 \pi} \dfrac{1}{24}\\=\dfrac{\pi}{96}$
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