Calculus (3rd Edition)

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

Chapter 6 - Applications of the Integral - 6.3 Volumes of Revolution - Exercises - Page 305: 41


$\dfrac{1400 \pi}{3}$

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, $R_{outside}=16-2x$ and $ R_{inside}=6$ Now, $V=\pi \int_0^5 [(16-2x)^2 -(6)^2] \ dx \\ = \pi \int_0^5 [256+4x^2-64x-36] \ dx \\=\pi \int_0^5 [220+4x^2-64 x] \ dx \\=\pi (220+\dfrac{4x^3}{3}-32x^2]_0^5 \ dx \\= \dfrac{1400 \pi}{3}$
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