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 322: 42


$\dfrac{ 2\pi}{3}$

Work Step by Step

Area $=R^2\pi- r^2 \pi=\pi (1^2-(1-y)^2) =\pi (2y-y^2)$ We integrate the integral to calculate the volume as follows: $V= \pi \times \int_{0}^{1} (2y-y^2) dy$ Now, $V= \pi [y^2-\dfrac{y^3}{3}]_{0}^{1}$ or, $= \pi (1-\dfrac{1}{3})$ or, $=\dfrac{ 2\pi}{3}$
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