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: 11


$3 \pi \sqrt 5$

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_{1}^{3}(\dfrac{x}{2}+\dfrac{1}{2}) \cdot \sqrt {1+(\dfrac{1}{4})} dy\\= \sqrt 5 \pi \space [\dfrac{x^2}{4}+\dfrac{x^2}{2}] _{1}^{3}\\=3 \pi \sqrt 5$
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