Calculus (3rd Edition)

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

Chapter 6 - Applications of the Integral - 6.4 The Method of Cylindrical Shells - Exercises - Page 311: 18


$2 \pi (\sqrt 5-1)$

Work Step by Step

The shell method to compute the volume of a region: The volume of a solid obtained by rotating the region under $y=f(x)$ over an interval $[m,n]$ about the y-axis is given by: $V=2 \pi \int_{m}^{n} (Radius) \times (height \ of \ the \ shell) \ dy=2 \pi \int_{m}^{n} (x) \times f(x) \ dx$ Now, $V=2\pi \int_{0}^{2} (x) (\dfrac{1}{\sqrt {x^2+1}}) \ dx\\= 2\pi \int_{0}^{2} (\dfrac{x}{\sqrt {x^2+1}}) \ dx \\=2 \pi [\dfrac{-4}{3x^3} +\dfrac{1}{2x^2}]_{-3}^{-1} \\= 2\pi [(1+x^2)^{1/2}]+0^2 \\=2 \pi (\sqrt 5-1)$
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