Calculus (3rd Edition)

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

Chapter 6 - Applications of the Integral - Chapter Review Exercises - Page 320: 35


$\dfrac{56 \pi}{15}$

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: $V=\pi \int_{m}^{n} (R^2_{outside}-R^2_{inside}) \ dy$ where, $R_{outside}=-x^2+4x-3-(-1)=-x^2+4x-2 $ and $ R_{inside}=0-(-1)=1$ Now, $V=\pi \int_{1}^{3} [(-x^2+4x-2)^2 -1] \ dx \\ = \pi \int_1^3 [x^4-8x^3+20x^2-16x+3 ] \ dx \\=\pi [\dfrac{x^5}{5}-2x^4+\dfrac{20x^3}{3}-8x^2+3x]_{1}^{3} \ dx \\=\pi [\dfrac{18} {5}+\dfrac{2}{15}] \\=\dfrac{56 \pi}{15}$
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.