I'm stuck on this question. I'm not sure what the lower and upper limits area. Find the area of the region that lies inside both curves. r = 1+cos(theta), r = 1-cos(theta) Thanks.
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I suggest you to do a graph. The area is given by $\displaystyle 4\frac{1}{2}\int_0^\frac{\pi}{2}(1-\cos\theta)^2d\theta$.
Thanks for your help. I did draw out a graph but was still confused. How did you find the limits of integration?
Drawing out a graph or notice that $\displaystyle \cos x = \cos (2\pi - x)=-\cos (\pi+x)$. So $\displaystyle 1-\cos \theta$ it's symmetrical with x-axis and $\displaystyle 1+\cos \theta$ is $\displaystyle 1-\cos \theta$ reflection with y-axis.
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