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 375: 42

Answer

Area between $y= \sqrt{x}$, $y=0$ , $x=0$ and $x=4$ revolves about $y=3 $

Work Step by Step

Given $$V= \pi \int_1^4\left[3^2-(3-\sqrt{x})^2\right] d x$$ Compare with $$V= \pi \int_a^b [R^2(x )-r^2(x)]dx $$ where $R(x)$ is the outer radius and $r(x)$ is the inner, here \begin{aligned} R(x)&= 3\\ r(x)&= 3-\sqrt{x} \end{aligned} This means that the bounded area between $y= \sqrt{x}$, $y=0$ , $x=0$ and $x=4$ revolves about $y=3$.
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