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: 14

Answer

$\frac{\pi}{24}(65\sqrt{65} - 17\sqrt{17})$

Work Step by Step

$x = 1+2y^{2}$ then $1+(dx/dy)^{2} = 1+(4y)^{2} = 1 + 16y^{2}$ So $S = 2\pi \int^{2}_{1} y \sqrt{1+16y^{2}} dy = \frac{\pi}{16} \int^{2}_{1} (16y^{2}+1)^{1/2} 32y dy = \frac{\pi}{24}(65\sqrt{65} - 17\sqrt{17})$
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