Calculus (3rd Edition)

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

Chapter 6 - Applications of the Integral - Chapter Review Exercises - Page 320: 34


$\dfrac{136 \pi}{15}$

Work Step by Step

The Washer method to compute the volume of revolution: When the function $f(x)$ is continuous and $f(x) \geq g(x) \geq 0$ on the interval $[m,n]$, then the volume of a solid obtained by rotating the region under $y=f(x)$ over an interval $[m,n]$ about the x-axis is given by: $V=\pi \int_{m}^{n} (R^2_{outside}-R^2_{inside}) \ dy$ where, $R_{outside}=4-(x^2-1)=5-x^2 $ and $ R_{inside}=4-(2x-1)=5-2x$ Now, $V=\pi \int_0^2 [(5-x)^2 -(5-2x)^2] \ dx \\ = \pi \int_0^2 [-14x^2+x^4+20x ] \ dx \\=\pi [ -\dfrac{14 x^3}{3}+\dfrac{x^5}{5}+10x^2 ]_0^2 \ dx \\=\pi [ -\dfrac{14 (2^3)}{3}+\dfrac{(2^5)}{5}+10(2^2)] \\=\dfrac{136 \pi}{15}$
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