University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 6 - Section 6.1 - Volumes Using Cross-Sections - Exercises - Page 355: 11

Answer

$$10$$

Work Step by Step

We need to integrate the integral to compute the volume. $$ Volume = \int_{0}^{5} (\dfrac{6}{25}) (5-x)^2 dx \\= -(\dfrac{2}{25}) \times [(5-x)^3]_{0}^{5} \\ = (-\dfrac{2}{25}) \times (0-125) \\=10$$
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