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: 16


$\dfrac{1088 \pi }{15}$

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: $V=\pi \int_{m}^{n} (R^2_{outside}-R^2_{inside}) \ dy$ Now, $V=\pi \int_{-1}^{3} [(2x+3)^2-(x^2)^2] \ dx \\ =\pi \int_{-1}^{3} [4x^2+9+12x -x^4] \ dx \\= \pi [-\dfrac{x^5}{5}+\dfrac{4 x^3}{3}+6x^2+9x]_{-1}^3 \\=\dfrac{1088 \pi }{15}$
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