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: 26


$\dfrac{ 38 \pi}{5}$

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: $V=2 \pi \int_{m}^{n} (Radius) \times (height \ of \ the \ shell) \ dy=2 \pi \int_{m}^{n} (y) \times f(y) \ dy$ Now, $V=2\pi \int_{0}^{1} y (19-y^3-6y^2-12y) \ dy\\= 2 \pi [ \dfrac{19y^2}{2}-\dfrac{y^{5}}{5}-\dfrac{3y^4}{2}-4y^3]_0^1 \\= 2\pi [\dfrac{19}{2}-\dfrac{1}{5}-\dfrac{3}{2}-4-0] \\=\dfrac{ 38 \pi}{5}$
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