Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter 8 - Further Techniques and Applications of Integration - Chapter Review - Review Exercises - Page 455: 27

Answer

$$ \left( f(x)=\sqrt {x-4}, \quad y =0, \quad x=13 \right) $$ If $f(x)=\sqrt {x-4} $ is nonnegative and $R$ is the region between $f(x) $ , the x-axis and $x=13 $ , from the x-intercepts we can find lower bound $a$. If $y=0 $ then $x=4 $. So, the volume of the solid formed by rotating $R$ about the x-axis is given by $$ \begin{aligned} V &=\int_{a}^{b} \pi [f(x)]^{2}dx \\ &=\pi \int_{4}^{13}(\sqrt{x-4})^{2} d x \\ &=\frac{81}{2} \pi \\ &\approx 127.2 \end{aligned} $$

Work Step by Step

$$ \left( f(x)=\sqrt {x-4}, \quad y =0, \quad x=13 \right) $$ If $f(x) $ is nonnegative and $R$ is the region between $f(x) $ , the x-axis and from $x=a$ to $x=b$, the volume of the solid formed by rotating $R$ about the x-axis is given by $$ \int_{a}^{b} \pi [f(x)]^{2}dx $$ If $f(x)=\sqrt {x-4} $ is nonnegative and $R$ is the region between $f(x) $ , the x-axis and $x=13 $ , from the x-intercepts we can find lower bound $a$. If $y=0 $ then $x=4 $. So, the volume of the solid formed by rotating $R$ about the x-axis is given by $$ \begin{aligned} V &=\pi \int_{4}^{13}(\sqrt{x-4})^{2} d x \\ &=\pi \int_{4}^{13}(x-4) d x \\ &=\left.\pi\left(\frac{x^{2}}{2}-4 x\right)\right|_{4} ^{13} \\ &=\pi\left[\left(\frac{169}{2}-52\right)-(8-16) |\right.\\ &=\pi\left(\frac{65}{2}+8\right)=\\ &=\frac{81}{2} \pi \\ &\approx 127.2 \end{aligned} $$
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