## Thomas' Calculus 13th Edition

Published by Pearson

# Chapter 15: Multiple Integrals - Section 15.6 - Moments and Centers of Mass - Exercises 15.6 - Page 909: 18

#### Answer

$$\bar{x}=\frac{100}{9},\ \ \bar{y}=0,\ \ I_x= 20$$

#### Work Step by Step

Since \begin{align*} M&=\int_{0}^{20} \int_{-1}^{1}\left(1+\frac{x}{20}\right) d y d x\\ &=\int_{0}^{20}\left(2+\frac{x}{10}\right) d x\\ &=60 \\ M_{x}&=\int_{0}^{20} \int_{-1}^{1} y\left(1+\frac{x}{20}\right) d y d x\\ &=\int_{0}^{20}\left[\left(1+\frac{x}{20}\right)\left(\frac{y^{2}}{2}\right)\right]_{-1}^{1} d x\\ &=0\\ M_{y}&=\int_{0}^{20} \int_{-1}^{1} x\left(1+\frac{x}{20}\right) d y d x\\ &=\int_{0}^{20}\left(2 x+\frac{x^{2}}{10}\right) d x\\ &=\frac{2000}{3} \end{align*} Then $$\bar{x}=\frac{100}{9},\ \ \bar{y}=0$$ and \begin{align*} I_{x}&=\int_{0}^{20} \int_{-1}^{1} y^{2}\left(1+\frac{x}{20}\right) d y d x\\ &=\frac{2}{3} \int_{0}^{20}\left(1+\frac{x}{20}\right) d x=20 \end{align*}

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.