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: 15

Answer

$\displaystyle{V=\frac{3\pi }{5}}\\ $

Work Step by Step

$\displaystyle{A\left(y\right)=\pi \left(2-\sqrt[3] y\right)^2-\pi \left(2-1\right)^2}\\ \displaystyle{A\left(y\right)=\pi \left(3+y^{\frac{2}{3}}-4y^{\frac{1}{3}}\right)}\\$ $\displaystyle{V=\int_0^1A\left(y\right)\ dy}\\ \displaystyle{V=\int_0^1\pi \left(3+y^{\frac{2}{3}}-4y^{\frac{1}{3}}\right)\ dy}\\ \displaystyle{V=\pi \int_0^13+y^{\frac{2}{3}}-4y^{\frac{1}{3}}\ dy}\\ \displaystyle{V=\pi\left[3y+\frac{3}{5}y^{\frac{5}{3}}-3y^{\frac{4}{3}}\right]_0^1}\\ \displaystyle{V=\pi\left(\left(3(1)+\frac{3}{5}(1)^{\frac{5}{3}}-3(1)^{\frac{4}{3}}\right)-\left(0\right)\right)}\\ \displaystyle{V=\frac{3\pi }{5}}\\ $
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