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: 16


$$ \bar{x}=0,\ \ \ \bar{y}=\frac{9}{14}$$ $$I_{y} =\frac{16}{35}$$

Work Step by Step

Since \begin{align*} M&=\int_{-1}^{1} \int_{x^{2}}^{1}(y+1) d y d x\\ &=-\int_{-1}^{1}\left(\frac{x^{4}}{2}+x^{2}-\frac{3}{2}\right) d x\\ &=-\left(\frac{x^{5}}{10}+\frac{1}{3}x^{3}-\frac{3}{2}x\right) \bigg|_{-1}^{1}\\ &=\frac{32}{15} \\ M_{x}&=\int_{-1}^{1} \int_{x^{2}}^{1} y(y+1) d y d x\\ &=\int_{-1}^{1}\left(\frac{5}{6}-\frac{x^{6}}{3}-\frac{x^{4}}{2}\right) d x\\ &=\left(\frac{5}{6}x-\frac{x^{7}}{21}-\frac{x^{5}}{10}\right)\bigg|_{-1}^{1}\\ &=\frac{48}{35}\\ M_{y}&=\int_{-1}^{1} \int_{x^{2}}^{1} x(y+1) d y d x\\ &=\int_{-1}^{1}\left(\frac{3 x}{2}-\frac{x^{5}}{2}-x^{3}\right) d x=0 \end{align*} Then $$ \bar{x}=0,\ \ \ \bar{y}=\frac{9}{14}$$ and \begin{align*} I_{y}&=\int_{-1}^{1} \int_{x^{2}}^{1} x^{2}(y+1) d y d x\\ &=\int_{-1}^{1}\left(\frac{3 x^{2}}{2}-\frac{x^{6}}{2}-x^{4}\right) d x\\ &=\left(\frac{3 x^{3}}{6}-\frac{x^{7}}{14}-\frac{1}{5}x^{5}\right)\bigg|_{-1}^{1}\\ &=\frac{16}{35} \end{align*}
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