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: 30

Answer

$$ \left( f(x)=4-x^{2}, \quad y =0, \quad x=-1, \quad x=1 \right) $$ The volume of the solid formed by revolving the function, $f(x)=4-x^{2} $ about the x-axis from $x=-1$ to $x=1$, is $$ \begin{aligned} V &=\int_{a}^{b} \pi [f(x)]^{2}dx\\ &=\pi \int_{-1}^{1}\left(4-x^{2}\right)^{2} d x\\ &=\frac{406 \pi}{15} \\ &\approx 85.03 \end{aligned} $$

Work Step by Step

$$ \left( f(x)=4-x^{2}, \quad y =0, \quad x=-1, \quad x=1 \right) $$ If $f(x) $ is nonnegative and $R$ is the region between $f(x) $ and the x-axis from $x=a$ to $x=b$, the volume of the solid formed by rotating $R$ about the x-axis is given by $$ \int_{a}^{b} \pi [f(x)]^{2}dx $$ So,the volume of the solid formed by revolving the function, $f(x)=4-x^{2} $ about the x-axis from $x=-1$ to $x=1$, is $$ \begin{aligned} V &=\int_{a}^{b} \pi [f(x)]^{2}dx\\ &=\pi \int_{-1}^{1}\left(4-x^{2}\right)^{2} d x \\ &=\pi \int_{-1}^{1}\left(16-8 x^{2}+x^{4}\right) d x \\ &=\left.\pi\left(16 x-\frac{8 x^{3}}{3}+\frac{x^{5}}{5}\right)\right|_{-1} ^{1} \\ &=\pi\left(16-\frac{8}{3}+\frac{1}{5}+16-\frac{8}{3}+\frac{1}{5}\right) \\ &=\pi\left(32-\frac{16}{3}+\frac{2}{5}\right) \\ &=\frac{406 \pi}{15} \\ &\approx 85.03 \end{aligned} $$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.