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: 13


$$ \bar{x}=\frac{3}{8} \ \text{ and }\ \bar{y}=\frac{17}{16}$$

Work Step by Step

Since \begin{align*} M&=\int_{0}^{1} \int_{x}^{2-x}(6 x+3 y+3) d y d x\\ &=\int_{0}^{1}\left[6 x y+\frac{3}{2} y^{2}+3 y\right]_{x}^{2-x} d x\\ &=\int_{0}^{1}\left(12-12 x^{2}\right) d x\\ &=8 \end{align*} and \begin{align*} M_{y}&=\int_{0}^{1} \int_{x}^{2-x} x(6 x+3 y+3) d y d x\\ &=\int_{0}^{1}\left(12 x-12 x^{3}\right) d x\\ &=3 \\ M_{x}&=\int_{0}^{1} \int_{x}^{2-x} y(6 x+3 y+3) d y d x\\ &=\int_{0}^{1}\left(14-6 x-6 x^{2}-2 x^{3}\right) d x\\ &=\left(14x-3x^2-2 x^{3}-\frac{1}{3} x^{4}\right)\bigg|_{0}^{1} \\ &=\frac{17}{2} \end{align*} Then $$ \bar{x}=\frac{3}{8} \ \text{ and }\ \bar{y}=\frac{17}{16}$$
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