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: 13



Work Step by Step

$x = \frac{1}{3} (y^{2}+2)^{3/2}$ then $dx/dy = \frac{1}{2} (y^{2} + 2)^{1/2} (2y) = y \sqrt{y^{2} +2}$ and $1+(dx/dy)^{2} = 1+y^{2} (y^{2}+2) = (y^{2}+1)^{2}$ So $S = 2\pi \int^{2}_{1} y(y^{2}+1) dy = 2\pi [\frac{1}{4} y^{4} +\frac{1}{2} y^{2}]^{2}_{1} = 2\pi (4+2-\frac{1}{4} -\frac{1}{2}) = \frac{21\pi}{2}$
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