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 312: 49


$0.3 \pi$

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, $V=2\pi \int_{0}^{1} x (\sqrt x-x^2) \ dx\\= 2 \pi \int_0^1 (x^{3/2}-x^3) \ dy \\=2 \pi [ \dfrac{x^{5/2}}{5/2}-\dfrac{x^{4}}{4}]_0^1 \\=2 \pi [ \dfrac{2}{5}-\dfrac{1}{4}] \\=0.3 \pi$
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