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: 39


$\dfrac{1872 \pi}{5}$

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, $R_{outside}=14-x$ and $ R_{inside}=x^2+2$ Now, $V=\pi \int_0^3 [(14-x)^2 -(x^2+2)^2] \ dx \\ = \pi \int_0^3 [196+x^2-28x -x^4-4-4x^2] \ dx \\=\pi \int_0^3 [-x^4-3x^2-28 x +192 ] \ dx \\=\pi (-\dfrac{x^5}{5}-x^3-14x^2+192 x)_0^3 \ dx \\=\dfrac{1872 \pi}{5}$
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