Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 6 - Applications of the Integral - 6.3 Volumes of Revolution - Exercises - Page 305: 27


$\dfrac{704 \pi}{15}$

Work Step by Step

The Washer method to compute the volume of revolution: When the function $f(x)$ is continuous and $f(x) \geq g(x) \geq 0$ on the interval $[m,n]$, then the volume of a solid obtained by rotating the region under $y=f(x)$ over an interval $[m,n]$ about the x-axis is given by: $V=\pi \int_{m}^{n} (R^2_{outside}-R^2_{inside}) \ dy$ $V=\pi \int_0^2 [(6)^2 -(x^2+2)^2] \ dx \\ = \pi \int_0^2 [36 -x^4 -4-4x^2 ] \ dx \\=\pi \int_0^2[-x^4 -4x^2+32] \ dx \\=\pi [-\dfrac{x^5}{5} -\dfrac{4 x^3}{3}+32 x]_0^2 \\=\dfrac{704 \pi}{15}$
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.