Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 7 - Section 7.2 - Trigonometric Integrals - 7.2 Exercises - Page 485: 63



Work Step by Step

$\displaystyle{A(x)=\pi(1-\sin x)^2-\pi(1-\cos x)^2}\\ \displaystyle{A(x)=\pi\left(2\cos x-2\sin x-\cos2x\right)}$ $\displaystyle{V=\int_0^{\frac{\pi}{4}}A(x)\ dx}\\ \displaystyle{V=\int_0^{\frac{\pi}{4}}\pi\left(2\cos x-2\sin x-\cos2x\right)\ dx}\\ \displaystyle{V=\pi\int_0^{\frac{\pi}{4}}2\cos x-2\sin x-\cos2x\ dx}\\ \displaystyle{V=\pi\left[2\sin x+2\cos x-\frac{1}{2}\sin2x\right]_0^{\frac{\pi}{4}}}\\ \displaystyle{V=\pi\left[2\sqrt2-\frac{5}{2}\right]}\\ \displaystyle{V=2\sqrt2\pi-\frac{5\pi}{2}}$
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