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: 13


$\dfrac{98 \pi}{81}$

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, $S =\int_{0}^{2} 2 \pi \dfrac{x^3}{9} \sqrt {1+(\dfrac{x}{3})^2} dx= \int_0^2 x^3 (\dfrac{2 \pi}{27}) \sqrt {x^4+9} dx$ Suppose $a =x^4+9$ and $da=4x^3 dx$ $ \dfrac{2 \pi}{27} \times \int_0^2 \dfrac{\sqrt a da}{4}= \dfrac{\pi}{54}[ (\dfrac{2}{3}) a^{(3/2)} ]_0^2=\dfrac{98 \pi}{81}$
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