Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 8 - Section 8.2 - Area of a Surface of Revolution - 8.2 Exercises - Page 573: 11

Answer

$\frac{\pi}{6}$ $($$17$$\sqrt 17$ $-$ $5$$\sqrt 5$$)$

Work Step by Step

$x$ $=$ $y^{2}$ $-$$1$ $\frac{dx}{dy}$ $=$ $2y$ $2$$\pi$$\int_{0}^{3}$ $\sqrt {x+1}$ $=$$\int_{0}^{3}$ $2 \pi$ $\sqrt {x+\frac{5}{4}}$$dx$ $=$$\frac{4}{3}$ $\times$ $\frac{1}{8}$$\pi$$($$17$$\sqrt 17$ $-$ $5$$\sqrt 5$$)$ $=$ $\frac{\pi}{6}$ $($$17$$\sqrt 17$ $-$ $5$$\sqrt 5$$)$
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