Calculus (3rd Edition)

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

Chapter 6 - Applications of the Integral - 6.4 The Method of Cylindrical Shells - Exercises - Page 311: 11



Work Step by Step

The shell method to compute the volume of a region: The volume of a solid obtained by rotating the region under $y=f(x)$ over an interval $[m,n]$ about the y-axis is given by: $V=2 \pi \int_{m}^{n} (Radius) \times (height \ of \ the \ shell) \ dy=2 \pi \int_{m}^{n} (x) \times f(x) \ dx$ Now, $V=2\pi \int_{0}^{2} (x)(2-2(x^2+1)^{-2}) \ dx\\= 2\pi \int_{0}^{2} [2x-\dfrac{2x}{(x^2+1)^2}] dx \\=2 \pi [x^2+\dfrac{1}{(x^2+1)}]_{0}^{2} \\=2 \pi [2^2+\dfrac{1}{(2^2+1)} ] \\=2 \pi (4+\dfrac{1}{5} ) \\=\dfrac{32\pi}{5}$
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.