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: 3


$$\Rightarrow \bar{x}=\frac{64}{35} \ \text{ and }\ \bar{y}=\frac{5}{7}$$

Work Step by Step

Since \begin{align*} M&=\int_{0}^{2} \int_{y^{2} / 2}^{4-y} d x d y\\ &=\int_{0}^{2}\left(4-y-\frac{y^{2}}{2}\right) d y\\ &=\frac{14}{3} \end{align*} and \begin{align*} M_{y}&=\int_{0}^{2} \int_{y^{2} / 2}^{4-y} x d x d y\\ &=\frac{1}{2} \int_{0}^{2}\left[x^{2}\right]_{y^{2} / 2}^{4-y} d y\\ & =\frac{1}{2} \int_{0}^{2}\left(16-8 y+y^{2}-\frac{y^{4}}{4}\right) dy\\ &=\frac{128}{3} \\ M_{x}&=\int_{0}^{2} \int_{y^{2} / 2}^{4-y} y d x d y\\ &=\int_{0}^{2}\left(4 y-y^{2}-\frac{y^{3}}{2}\right) d y\\ &=\frac{10}{3} \end{align*} Then $$\Rightarrow \bar{x}=\frac{64}{35} \ \text{ and }\ \bar{y}=\frac{5}{7}$$
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.