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: 20


$$ \bar{x}=0,\ \ \bar{y}=\frac{32}{45} $$$$ I_{0}=I_{x}+I_{y}=\frac{6}{5}$$

Work Step by Step

Since \begin{align*} M&=\int_{0}^{1} \int_{-y}^{y}\left(3 x^{2}+1\right) d x d y\\ &=\int_{0}^{1}\left(2 y^{3}+2 y\right) d y\\ &=\left(\frac{1}{2} y^{4}+ y^2\right)\bigg|_{0}^{1}\\ &=\frac{3}{2} \\ M_{x}&=\int_{0}^{1} \int_{-y}^{y} y\left(3 x^{2}+1\right) d x d y\\ &=\int_{0}^{1}\left(2 y^{4}+2 y^{2}\right) d y\\ &= \left(\frac{2}{5} y^{5}+\frac{2}{3} y^{3}\right)\bigg|_{0}^{1}\\ &=\frac{16}{15}\\ M_{y}&=\int_{0}^{1} \int_{-y}^{y} x\left(3 x^{2}+1\right) d x d y=0 \end{align*} Then $$ \bar{x}=0,\ \ \bar{y}=\frac{32}{45} $$ \begin{align*} I_{x}&=\int_{0}^{1} \int_{-y}^{y} y^{2}\left(3 x^{2}+1\right) d x d y\\ &=\int_{0}^{1}\left(2 y^{5}+2 y^{3}\right) d y\\ &=\left(\frac{1}{3} y^{6}+\frac{1}{2} y^{3}\right)\bigg|_{0}^{1}\\ &=\frac{5}{6}\\ I_{y}&=\int_{0}^{1} \int_{-y}^{y} x^{2}\left(3 x^{2}+1\right) d x d y\\ &=2 \int_{0}^{1}\left(\frac{3}{5} y^{5}+\frac{1}{3} y^{3}\right) d y\\ &=2\left(\frac{3}{30} y^{6}+\frac{1}{12} y^{4}\right)\bigg|_{0}^{1}\\ &=\frac{11}{30} \end{align*} Then $$ I_{0}=I_{x}+I_{y}=\frac{6}{5}$$
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.