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: 1

Answer

$ \bar{x}=\frac{5}{14} $ and $ \bar{y}= \frac{38}{35}$

Work Step by Step

Set up the moments and center of mass: M=$\int^1_0 \int^{2-x^2}3 dy $ dx =$3 \int^1_0 (2-x^2-x) dx =\frac{7}{2}$ $ M_y =\int^1_0 \int^{2-x^2}_x 3x $ dy dx =$3 \int^1_0 [xy]^{2-x^2}_x $ dx =$3\int^1_0 (2x-x^3-x^2)dx=\frac{5}{4}$ $ M_x=\int^1_0 \int^{2-x^2}3y $ dy dx =$\frac{3}{2}\int^1_0 [y^2]^{2-x^2}_xdx $ =$\frac{3}{2}\int^1_0 (4-5x^2+x^4)dx=\frac{19}{5}$ Thus we have: $ \bar{x}=\frac{5}{14} $ and $ \bar{y}= \frac{38}{35}$
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