Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 6 - Applications of Integration - 6.3 Volume by Slicing - 6.3 Exercises - Page 431: 23

Answer

$\dfrac{\pi^2 }{6} \ unit \ cube$

Work Step by Step

Since, the vertical slices solid are circular discs , so we will use the disk method to compute the volume of revolution of the curve. The volume of revolution of the curve can be expressed as: $V=\pi \int_p^q [f(x)]^2 dx\\=\pi \int_{0}^{1/2} [\dfrac{1}{\sqrt [4] {1-x^2}}]^{2} \ dx \\= \int_{0}^{1/2} [\dfrac{1}{\sqrt {-x^2}}] \ dx \\= \pi [\sin^{-1} x]_0^{1/2} \\=\pi [arcsin (1/2) -arcsin (0)]\\=\dfrac{\pi^2 }{6} \ unit \ cube$
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