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 909: 19


$$\bar{x}=0,\ \ \bar{y}=\frac{7}{10} $$ $$ I_{o}=I_{x}+I_{y}=\frac{6}{5}$$

Work Step by Step

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