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: 25

Answer

$4 \pi a^2$

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, $S =(2 \pi)\int_{-a}^{a} \sqrt {a^2-x^2} \times \sqrt {\dfrac{a^2}{a^2-x^2} } dx \\=\int_{-a}^{a} 2 \pi a dx \\=4 \pi a^2$
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