University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 6 - Section 6.4 - Areas of Surfaces of Revolution - Exercises - Page 375: 14

Answer

$\dfrac{28 \pi}{3}$

Work Step by Step

The formula to determine the surface area is as follows: $S= \int_{m}^{n} 2 \pi y \sqrt {1+(\dfrac{dy}{dx})^2}$ Now, $S =(2 \pi)\int_{(3/4)}^{(15/4)} \dfrac{1}{2} [\sqrt {4x+1}] dx \\= \dfrac{\pi}{6} [(4x+1)^{3/2}]_{3/4}^{15/4} \\=\dfrac{28 \pi}{3}$
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