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 15.6 - Moments and Centers of Mass - Exercises 15.6 - Page 908: 8

Answer

$$\frac{\pi}{2}$$

Work Step by Step

\begin{align*} I_{y}&=\int_{\pi}^{2 \pi} \int_{0}^{\left(\sin ^{2} x\right) / x^{2}} x^{2} d y d x\\ &=\int_{\pi}^{2 \pi}\left(\sin ^{2} x-0\right) d x\\ &=\frac{1}{2} \int_{\pi}^{2 \pi}(1-\cos 2 x) d x\\ &=\frac{1}{2} \left(x- \frac{1}{2}\sin 2x\right)\bigg|_{\pi}^{2 \pi} \\ &=\frac{\pi}{2} \end{align*}
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