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: 15


$$ \bar{x}=\frac{11}{3}, \ \ \bar{y}=\frac{14}{27}$$ $$I_{y} =432$$

Work Step by Step

Since \begin{align*} M&=\int_{0}^{1} \int_{0}^{6}(x+y+1) d x d y\\ &=\int_{0}^{1}(6 y+24) d y\\ &= (3y^2+24y)\bigg|_{0}^{1}=27\\ M_{x}&=\int_{0}^{1} \int_{0}^{6} y(x+y+1) d x d y\\ &=\int_{0}^{1} y(6 y+24) d y\\ &= 2y^3+ 12y^2\bigg|_{0}^{1}=14\\ M_{y}&=\int_{0}^{1} \int_{0}^{6} x(x+y+1) d x d y\\ &=\int_{0}^{1}(18 y+90) d y\\ &=(9 y^2+90y)\bigg|_{0}^{1} =99 \end{align*} Then $$ \bar{x}=\frac{11}{3}, \ \ \bar{y}=\frac{14}{27}$$ and \begin{align*} I_{y}&=\int_{0}^{1} \int_{0}^{6} x^{2}(x+y+1) d x d y\\ &=216 \int_{0}^{1}\left(\frac{y}{3}+\frac{11}{6}\right) d y\\ &=432 \end{align*}
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.