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: 24


$2 \pi \int_{-\pi/2}^{\pi/2} \cos (x) [\sqrt {\sin^2 x +1}] dx $

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} \\ =(2 \pi)\int_{-\pi/2}^{\pi/2} (\cos x) \sqrt {\sin^2 x dx^2+dx^2} \\ =2 \pi \int_{-\pi/2}^{\pi/2} \cos (x) [\sqrt {\sin^2 x +1}] dx $
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