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: 15


$2 \pi$

Work Step by Step

We need to compute the surface area. $S= \int_{m}^{n} 2 \pi y \sqrt {1+(y')^2} \\=(2 \pi) \times \int_{1/2}^{3/2} \sqrt {2x-x^2} \times \sqrt {\dfrac{1}{2x-x^2}} \space dx\\=(2 \pi) \int_{1/2}^{3/2} x \space dx \\=2 \pi$
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