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: 6


$$ \bar{x}=\frac{\pi}{2}\ \ \text{ and }\ \ \bar{y}=\frac{\pi}{8}$$

Work Step by Step

Since \begin{align*} M&=\int_{0}^{\pi} \int_{0}^{\sin x} d y d x\\ &=\int_{0}^{\pi} \sin x d x\\ &=2 ; \\ M_{x}&=\int_{0}^{\pi} \int_{0}^{\sin x} y d y d x\\ &=\frac{1}{2} \int_{0}^{\pi}\left[y^{2}\right]_{0}^{\sin x} d x\\ &=\frac{1}{2} \int_{0}^{\pi} \sin ^{2} x d x\\ &=\frac{1}{4} \int_{0}^{\pi}(1-\cos 2 x) d x\\ &=\frac{\pi}{4} \end{align*} Then $$ \bar{x}=\frac{\pi}{2}\ \ \text{ and }\ \ \bar{y}=\frac{\pi}{8}$$
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