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: 34

Answer

$80.6095$

Work Step by Step

$y = x^{1/2}$ then $y' = \frac{1}{2} x^{-1/2}$ and $1+(y')^{2} = 1+1/4x$ So $S = \int^{4}_{0} 2\pi (4-\sqrt{x}) \sqrt{1_1/(4x)} dx$ by using CAS $S = 2\pi \ln{(\sqrt{17} +4)} + \frac{\pi}{6}(31\sqrt{17} + 1) \approx 80.6095$
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