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: 17


$$\bar{x}=0,\ \bar{y}=\frac{13}{31},\ \ I_y=\frac{7}{5} $$

Work Step by Step

Since \begin{align*} M&=\int_{-1}^{1} \int_{0}^{x^{2}}(7 y+1) d y d x\\ &=\int_{-1}^{1}\left(\frac{7 x^{4}}{2}+x^{2}\right) d x\\ &=\left(\frac{7 x^{5}}{10}+\frac{1}{3}x^{3}\right) \bigg|_{-1}^{1}\\ &=\frac{31}{15}\\ M_{x}&=\int_{-1}^{1} \int_{0}^{x^{2}} y(7 y+1) d y d x\\ &=\int_{-1}^{1}\left(\frac{7 x^{6}}{3}+\frac{x^{4}}{2}\right) d x\\ &=\frac{13}{15}\\ M_{y}&=\int_{-1}^{1} \int_{0}^{x^{2}} x(7 y+1) d y d x\\ &=\int_{-1}^{1}\left(\frac{7 x^{5}}{2}+x^{3}\right) d x=0 \end{align*} Then $$\bar{x}=0,\ \bar{y}=\frac{13}{31}$$ and \begin{align*} I_{y}&=\int_{-1}^{1} \int_{0}^{x^{2}} x^{2}(7 y+1) d y d x\\ &=\int_{-1}^{1}\left(\frac{7 x^{6}}{2}+x^{4}\right) d x\\ &=\left(\frac{7 x^{7}}{14}+\frac{1}{5}x^{5}\right)\bigg|_{-1}^{1} \\ &=\frac{7}{5} \end{align*}
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