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 304: 15


$256 \pi$

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: $V=\pi \int_{m}^{n} (R^2_{outside}-R^2_{inside}) \ dy$ Now, $V=\pi \int_{-2}^{2} \pi (10-x^2)^2-\pi (x^2+2)^2 \ dx \\ =\pi \int_{-2}^{2} \pi (960-24x^2) \ dx \\=2 \pi \int_0^2 (96-24x^2) \ dx \\=2 \pi [96-8x^3]_0^2 \\=2 \pi [96-8(2^3)]\\=256 \pi$
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