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: 17


$\dfrac{ \pi (2\sqrt 2-1)}{9} $

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=\int_{0}^{1} 2 \pi \times \dfrac{y^3}{3} \times \sqrt {y^4+1} dx \\=(\dfrac{ \pi}{9})[ (y^4+1)^{3/2} ]_{0}^{1} \\=\dfrac{ \pi (2\sqrt 2-1)}{9} $
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