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: 10

Answer

$\dfrac{128 \pi}{15}$

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, $V=2\pi \int_{0}^{2} (x)(4x-x^3) \ dx\\= 2\pi \int_{0}^{2} (4x^2-x^4) dx \\=2 \pi [\dfrac{4 x^3}{3}-\dfrac{x^5}{5}]_{0}^{2} \\=2 \pi [\dfrac{4 (2^3)}{3}-\dfrac{(2^5)}{5}] \\=\dfrac{128 \pi}{15}$
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