Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 6 - Applications of the Integral - Chapter Review Exercises - Page 320: 41

Answer

$2 \pi (c+\dfrac{c^3 }{3}) $

Work Step by Step

The disk method to compute the volume of revolution: When the function $f(x)$ is continuous and $f(x) \geq 0$ 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: $V= \pi \int_{m}^{n} (R^2) \ dx$ Now, $V=2\pi \int_{-c}^{c} (\sqrt {1+x^2})^2 \ dx \\ =2\pi [x+\dfrac{1}{3}x^3]_{-c}^{c} \\=\pi [c+\dfrac{c^3}{3}-(-c+\dfrac{(-c^3)}{3}] \\=\pi [c+\dfrac{c^3}{3}+c +\dfrac{c^3}{3}]\\=2 \pi (c+\dfrac{c^3 }{3}) $
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