Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 15 - Review - Exercises - Page 1062: 31

Answer

$\dfrac{\pi}{96}$

Work Step by Step

Consider $I=\int\int_{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(r) dx dr d\theta \\=\int_{0}^{2 \pi}\int_{0} ^{1}(1-r^2) r^5 \cos^2 \theta \sin^2 \theta dr d\theta \\=[\dfrac{r^6}{6} -\dfrac{r^8}{8}]_0^1 [ \int_{0}^{2 \pi} \cos^2 \theta \sin^2 \theta d\theta] \\=\dfrac{1}{4}\times \int_{0}^{2 \pi} \sin^2 \theta d\theta \times [\dfrac{r^6}{6} -\dfrac{r^8}{8}]_0^1 $ Use the formula: $\sin^2 x= \dfrac{1-\cos 2x}{2} $ Now, $I=\dfrac{1}{4}\times [\dfrac{ \theta}{2}-\dfrac{\sin 4 \theta}{8} ]_0^{2 \pi} ] \dfrac{1}{24} =\dfrac{\pi}{96}$
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