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: 9


$4 \pi \sqrt 5$

Work Step by Step

We need to compute the surface area. $$S= \int_{m}^{n} 2 \pi y \sqrt {1+(y')^2}$$ $$\\=(2 \pi) \times \int_{1}^{4}(\dfrac{x}{2}) \times \sqrt {1+\dfrac{1}{4}} \space dx \\=[\dfrac{\sqrt 5 \pi x^2}{4} ]_1^4\\=4 \pi \sqrt 5$$
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