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: 7


$\displaystyle{V=\frac{4\pi }{21}}$

Work Step by Step

$\displaystyle{x^3=x}\\ \displaystyle{x^3-x=0}\\ \displaystyle{x\left(x^2-1\right)=0}\\ x=1\qquad x=0$ $\displaystyle{A\left(x\right)=\pi \left(x\right)^2-\pi \left(x^3\right)^2}\\ \displaystyle{A\left(x\right)=\pi \left(x^2-x^6\right)}\\$ $\displaystyle{V=\int_0^1A\left(x\right)\ dx}\\ \displaystyle{V=\int_0^1\pi\left(x^2-x^6\right)\ dx}\\ \displaystyle{V=\pi \int_0^1x^2-x^6\ dx}\\ \displaystyle{V=\pi\left[\frac{1}{3}x^3-\frac{1}{7}x^7\right]_0^1}\\ \displaystyle{V=\pi\left(\left(\frac{1}{3}\times1^3-\frac{1}{7}\times1^7\right)-\left(0\right)\right)}\\ \displaystyle{V=\frac{4\pi }{21}}$
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