Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter 8 - Further Techniques and Applications of Integration - 8.2 Volume and Average Value - 8.2 Exercises - Page 439: 3

Answer

$\frac{364\pi}{3}$

Work Step by Step

Use the formula given above for the volume, with a= 0, b=4 and f(x) =2x+1 $V=\int^{4}_{0} \pi (2x+1)^{2}dx=\int^{4}_{0}\pi(4x^{2} +4x+1)dx=\pi [\frac{4x^{3}}{3}+2x^{2}+x]|^{4}_{0}=\frac{364\pi}{3}$
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