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: 1

Answer

$$\displaystyle{V=\frac{26\pi }{3}}\\$$

Work Step by Step

$\displaystyle{A\left(x\right)=\pi \left(x+1\right)^2}\\ \displaystyle{A\left(x\right)=\pi \left(x^2+2x+1\right)}\\\\$ $\displaystyle{V=\int_0^2A\left(x\right)\ dx}\\ \displaystyle{V=\int_0^2\pi \left(x^2+2x+1\right)\ dx}\\ \displaystyle{V=\pi \int_0^2\left(x^2+2x+1\right)\ dx}\\ \displaystyle{V=\pi\left[\frac{1}{3}x^3+ x^2+ x\right]_0^2}\\ \displaystyle{V=\pi\left(\left(\frac{1}{3}2^3+ 2^2+ 2\right)-\left(0\right)\right)}\\ \displaystyle{V=\frac{26\pi }{3}}\\ $
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