University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 5 - Section 5.6 - Definite Integral Substitutions and the Area Between Curves - Exercises - Page 338: 49


The area of the shaded region is $4\sqrt2$.

Work Step by Step

The shaded region is bounded below by the curve $y=3\sin x\sqrt{1+\cos x}$ and above by the line $y=0$ from the point $x=-\pi$ to $x=0$. So the area of the region would be $$A=\int^0_{-\pi}\Big(0-3\sin x\sqrt{1+\cos x}\Big)dx=-3\int^0_{-\pi}\sin x\sqrt{1+\cos x}dx$$ We set $u=1+\cos x$, which means $$du=-\sin xdx$$ $$\sin xdx=-du$$ - For $x=-\pi$, we have $u=1+\cos(-\pi)=1-1=0$ - For $x=0$, we have $u=1+\cos0=1+1=2$ Therefore, $$A=-3\int^2_0\sqrt u(-du)=3\int^2_0\sqrt udu$$ $$A=3\times\frac{2u^{3/2}}{3}\Big]^2_0=2u^{3/2}\Big]^2_0$$ $$A=2(2^{3/2}-0)=2(2\sqrt2)$$ $$A=4\sqrt2$$
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