Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 9 - Further Applications of the Integral and Taylor Polynomials - 9.3 Center of Mass - Exercises - Page 483: 7


$$\left(\frac{8}{5}, \frac{16}{7}\right)$$

Work Step by Step

Since \begin{align*} M_{x}&=\frac{1}{2} \rho \int_{0}^{2}\left(f(x)\right)^{2} d x\\ &=\frac{1}{2} \rho \int_{0}^{2}\left(x^{3}\right)^{2} d x\\ &= \frac{1}{2} \frac{x^7}{7}\bigg|_{0}^{2}\\ &=\frac{64 \rho}{7} \end{align*} and \begin{aligned} M_{y}&=\rho \int_{a}^{b} x f(x) d x \\ &=\rho \int_{0}^{2} x^{4} d x \\ &=\left.\rho\left(\frac{x^{5}}{5}\right)\right|_{0} ^{2} \\ &=\frac{32\rho}{5} \end{aligned} and \begin{align*} M&=\rho A\\ &=\rho \int_{0}^{2} x^{3} d x\\ &=\left.\rho\left(\frac{x^{4}}{4}\right)\right|_{0} ^{2}\\ &=4 \rho \end{align*} Then \begin{align*} (\bar{x}, \bar{y})&=\left(\frac{M_{y}}{M}, \frac{M_{x}}{M}\right)\\ &=\left(\frac{8}{5}, \frac{16}{7}\right) \end{align*}
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