Calculus: Early Transcendentals 8th Edition

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

Chapter 6 - Section 6.3 - Volumes by Cylindrical Shells - 6.3 Exercises - Page 454: 14

Answer

$\displaystyle{V=\frac{27\pi}{2}}$

Work Step by Step

$\displaystyle{4-y=y^2-4y+4}\\ \displaystyle{0=y^2-4y+y}\\ \displaystyle{(y)(y-3)=0}\\ y=0\qquad y=3$ $\displaystyle{V=\int_{0}^{3}(2\pi y)\left(\left(4-y\right)-\left(y^2-4y+4\right)\right)\ dy}\\ \displaystyle{V=2\pi\int_{0}^{3}3y^2-y^3\ dy}\\ \displaystyle{V=2\pi\left[y^3-\frac{1}{4}y^4\right]_{0}^{3}}\\ \displaystyle{V=2\pi\left(\left((3)^3-\frac{1}{4}(3)^4\right)-(0)\right)}\\ \displaystyle{V=\frac{27\pi}{2}}$
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