Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 8 - Further Applications of Integration - 8.2 Area of a Surface of Revolution - 8.2 Exercises: 2

Answer

(a) i) $ \int ^{2}_{1} 2\pi x^{-2} \sqrt{1+4x^{-6}} dx$ ii) $\int^{2}_{1} 2\pi x \sqrt{1+4x^{-6}} dx$ (b) i) 4.4566 ii) 11.7299

Work Step by Step

(a) i) $y = x^{-2}$ then $dy/dx = -2 x^{-3}$ and $ds = \sqrt{1+(dy/dx)^{2}}dx = \sqrt{1+4x^{-6}}dx$ So $S = \int 2\pi y ds = \int ^{2}_{1} 2\pi x^{-2} \sqrt{1+4x^{-6}} dx$ ii) $S = \int 2\pi x ds = \int^{2}_{1} 2\pi x \sqrt{1+4x^{-6}} dx$ (b) i) 4.4566 ii) 11.7299
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