Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 6 - Section 6.2 - Volume - 6.2 Exercises - Page 456: 12

Answer

$\displaystyle{V=\frac{3\pi }{4}}\\ $

Work Step by Step

$\displaystyle{A\left(x\right)=\pi \left(\frac{1}{x}\right)^2}\\ \displaystyle{A\left(x\right)=\pi \left(\frac{1}{x^2}\right)}\\ \displaystyle{A\left(x\right)=\pi x^{-2}}\\$ $\displaystyle{V=\int_1^4A\left(x\right)\ dx}\\ \displaystyle{V=\int_1^4\pi x^{-2}\ dx}\\ \displaystyle{V=\pi \int_1^4 x^{-2}\ dx}\\ \displaystyle{V=\pi\left[-x^{-1}\right]_1^4}\\ \displaystyle{V=\pi\left(\left(-4^{-1}\right)-\left(-1^{-1}\right)\right)}\\ \displaystyle{V=\frac{3\pi }{4}}\\ $
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