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: 10


$$8 \pi \sqrt 5$$

Work Step by Step

We need to compute the surface area. $$S= \int_{m}^{n} 2 \pi y \sqrt {1+(y')^2}$$ $$\\=(2 \pi)\int_{0}^{2}(2y) \times \sqrt {1+(2)^2} dy \\=[4 \pi \sqrt 5y]_{0}^{2}\\=8 \pi \sqrt 5$$
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