Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 6: Applications of Definite Integrals - Section 6.1 - Volumes Using Cross-Sections - Exercises 6.1 - Page 321: 9


$8 \pi $

Work Step by Step

We integrate the integral as follows: $V= \int_{0}^{2} (\dfrac{5\pi y^4}{4}) \cdot y^4 dy$ Now, $V= (\dfrac{5\pi}{4}) \times [(\dfrac{y^5}{5})]_{0}^{2}$ or, $= (\dfrac{5\pi}{4}) \times (\dfrac{32}{5})$ or, $=8 \pi $
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