Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 5 - Applications of Integration - 5.2 Volumes - 5.2 Exercises - Page 374: 8

Answer

$\displaystyle{V=\frac{384\pi }{5}}\\ $

Work Step by Step

$\displaystyle{6-x^2=2}\\ \displaystyle{x^2=4}\\ \displaystyle{x=2 \qquad x=-2}\\$ $\displaystyle{A\left(x\right)=\pi \left(6-x^2\right)^2-\pi\left(2\right)^2}\\ \displaystyle{A\left(x\right)=\pi \left(x^4-12x+32\right)}\\$ $\displaystyle{V=\int_{-2}^2A\left(x\right)\ dx}\\ \displaystyle{V=\int_{-2}^2\pi \left(x^4-12x+32\right)\ dx}\\ \displaystyle{V=\pi \int_{-2}^2\left(x^4-12x+32\right)\ dx}\\ \displaystyle{V=\pi\left[\frac{1}{5}x^5-6x^2+ 32x\right]_{-2}^2}\\ \displaystyle{V=\pi\left(\left(\frac{1}{5}\times2^5-6\times 2^2+ 32\times2\right)-\left(\frac{1}{5}\times{-2}^5-6\times {-2}^2+ 32\times-2\right)\right)}\\ \displaystyle{V=\frac{384\pi }{5}}\\ $
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