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 376: 29


$2\pi \space R \space h$

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_{a}^{a+h} [\sqrt {R^2-x^2}] [\dfrac{R}{\sqrt {r^2-x^2}}] dx \\=(2 \pi ) \int_{a}^{a+h} \space R \space dx\\= 2 \pi R (a+h-a)\\=2\pi \space R \space h$
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