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: 35


$\dfrac{824 \pi}{15}$

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, $R_{outside}=6-(x^2+2)=4-x^2$ and $ R_{inside}= 6$ Now, $V=\pi \int_0^2 [(6)^2 - (4-x^2)^2] \ dx \\ = \pi \int_0^2 [36-x^4 -16+8x^2] \ dx \\=\pi \int_0^2 [-x^4+8x^2+20] \ dx \\=\pi (-\dfrac{x^5}{5}+\dfrac{8x^3}{3}+20 x )_0^2 \ dx \\=\dfrac{824 \pi}{15}$
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