Calculus (3rd Edition)

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

Chapter 8 - Techniques of Integration - 8.1 Integration by Parts - Exercises - Page 396: 65

Answer

$\pi^3-4 \pi $

Work Step by Step

The volume of a revolution can be calculated by using integration by parts as follows: $V=\pi \int_0^{\pi} (x \sqrt {\sin x})^2\\=\pi \int_0^{\pi} x^2 \sin x \ dx \\=\pi (-\pi^2 \cos \pi+0) +2 \pi \int_0^{\pi} x \cos x \ dx \\=\pi^3 +2 \pi \int_0^{\pi} x \cos x \ dx \\=\pi^3 +2 \pi (x |\sin x|_0^{\pi} -\int_0^{\pi} \sin x \ dx) \\=\pi^3 -2 \pi \int_0^{\pi} \sin x \ dx \\=\pi^3 +2 \pi |\cos x|_0^{\pi} \\=\pi^3-4 \pi $
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