Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 6 - Section 6.2 - Volume - 6.2 Exercises - Page 446: 21



Work Step by Step

$\displaystyle{A(y)=\pi\left(1-y\right)^2}\\ \displaystyle{A(y)=\pi\left(y^2-2y+1\right)}\\$ $\begin{aligned} V &=\int_{0}^{1} A(y) \ dy \\ V &=\int_{0}^{1} \pi\left(y^2-2y+1\right) \ dy \\ V &=\pi \int_{0}^{1} y^2-2y+1 \ dy \\ V &=\pi\left[\frac{1}{3} y^{3}-y^2+y\right]_{0}^{1} \\ V &=\pi\left(\left(\frac{1}{3} (1)^{3}-(1)^2+1\right)-\left(\frac{1}{3} 0^{3}-0^2+0\right)\right) \\ V &=\frac{\pi}{3} \end{aligned}$
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