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: 40


$\dfrac{1953 \pi}{5}$

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, $V=\pi \int_0^3 [(15-x^2)^2 -(3+x)^2] \ dx \\ = \pi \int_0^3 [x^4+225-30x^2-9-x^2-6x] \ dx \\=\pi \int_0^3 [x^4-31x^2-6 x +216 ] \ dx \\=\pi (\dfrac{x^5}{5}-\dfrac{31 x^3}{3}-3x^2+216 x)_0^3 \ dx \\=\pi [\dfrac{(3^5)}{5}-\dfrac{31(3^3)}{3}-3(3^2)+216 (3) ]\\= \dfrac{1953 \pi}{5}$
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