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: 5

Answer

$\displaystyle{V=162\pi}$

Work Step by Step

$\displaystyle{A\left(y\right)=\pi \left(2\sqrt{y}\right)^2}\\ \displaystyle{A\left(y\right)=\pi \left(4y\right)}\\$ $\displaystyle{V=\int_{0}^9A\left(y\right)\ dy}\\ \displaystyle{V=\int_{0}^9\pi \left(4y\right)\ dy}\\ \displaystyle{V=4\pi \int_{0}^9y\ dy}\\ \displaystyle{V=4\pi\left[\frac{1}{2}y^2\right]_{0}^9}\\ \displaystyle{V=4\pi\left(\left(\frac{1}{2}9^{2}\right)-\left(\frac{1}{2}0^{2}\right)\right)}\\ \displaystyle{V=162\pi}\\ $
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