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: 8


$$ \frac{4}{15}$$

Work Step by Step

By using Eq. (2), we get \begin{align*} M_{x}&= \rho \int_{a}^{b} y\left(g_{1}(y)-g_{2}(y)\right) d y\\ &=\int_{0}^{1} y \sqrt{1-y} d y \end{align*} Let $$ u^2=1-y\ \ \ \ \ \ \ \ \ 2udu=-dy$$ at $y=1$, $u=0$, $y=0\to u=1$, so \begin{align*} M_{x} &=\int_{1}^{0} y \sqrt{1-y} d y\\ &= 2\int_1^0 ( u^2-1)u^2du\\ &= 2\int_1^0 ( u^4-u^2)du\\ &=2\left(\frac{1}{5}u^5-\frac{1}{3}u^3\right)\bigg|_0^1\\ &=\frac{4}{15} \end{align*} By using Eq. (3), we get \begin{aligned} M_{x}&=\frac{1}{2} \rho \int_{a}^{b} f(x)^{2} d x \\ &=\frac{1}{2} \int_{0}^{1}\left(1-x^{2}\right)^{2} d x \\ &=\frac{1}{2} \int_{0}^{1}\left(1-2 x^{2}+x^{4}\right) d x \\ &=\left.\frac{1}{2}\left(x-\frac{2}{3} x^{3}+\frac{x^{5}}{5}\right)\right|_{0} ^{1} \\ &=\frac{1}{2}\left(\frac{8}{15}\right)= \frac{4}{15} \end{aligned}
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