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: 4


$ \bar{x}=1$ and $\bar{y}=1$

Work Step by Step

Since \begin{align*} M&=\int_{0}^{3} \int_{0}^{3-x} d y d x\\ &=\int_{0}^{3}(3-x) d x\\ &=\frac{9}{2} \\ M_{y}&=\int_{0}^{3} \int_{0}^{3-x} x d y d x\\ &=\int_{0}^{3}[x y]_{0}^{3-x} d x\\ &=\int_{0}^{3}\left(3 x-x^{2}\right) d x\\ &=\frac{9}{2} \end{align*} Then by symmetry $ \bar{x}=1$ and $\bar{y}=1$
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