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Question:

Find the area of the region lying inside the polar curve $r=1+\cos\theta$ and outside the polar curve $r=2\cos\theta$.

Let $A_1 = \frac{1}{2}\int_0^{2\pi}(1+\cos\theta)^2d\theta = \frac{3\pi}{2}$ and $A_2=\pi(1)^2 = \pi$ because the radius of the curve $r=2\cos\theta$ is $1$. $A_1 - A_2 = \frac{\pi}{2}$, which is the final answer that I got.

However, when I tried to solve the problem on Wolfram Alpha with the input $\frac{1}{2}\int_0^{2\pi}((1+\cos\theta)^2-(2\cos\theta)^2)d\theta$, it gave me $-\frac{\pi}{2}$ even though the curve $r=1+\cos\theta$ looks bigger than $r=2\cos\theta$.

How do I reconcile these two answers? Any help would be appreciated!

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I see where your problem lies, in the integral around $2\cos t$ you integrate from $0$ to $2\pi$ but this goes around the circle twice, double counting the area, the right bounds should be from $0$ to $\pi$.

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  • $\begingroup$ Actually, since he was finding the difference in area between the two, he would square the individual parts. $\endgroup$ – Hrhm Mar 8 '17 at 16:29
  • $\begingroup$ If that's the case, does the setup $\frac{1}{2}\int_0^{2\pi}(1+\cos\theta - 2\cos\theta)^2d\theta$ seem correct? $\endgroup$ – Nikhil Srd Mar 8 '17 at 16:30
  • $\begingroup$ @Hrhm that still gives $-\pi/2$, you need $\frac 12 \int_0^{\pi}(2\cos t)^2 dt$ $\endgroup$ – Guacho Perez Mar 8 '17 at 17:24
  • $\begingroup$ @GuachoPerez Thank you for correcting my mistake. So it should be $$\frac{1}{2}\left[\int_0^{2\pi}\left(1+\cos\theta\right)^2-\int_0^\pi \left(2\cos\theta\right)^2\right]$$ $\endgroup$ – Hrhm Mar 8 '17 at 18:53
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Note that to find the area of $r=2\cos\theta$, you should compute $$2\int_{0}^\pi \cos^2\theta \,d\theta$$ instead of $$2\int_{0}^{2\pi} \cos^2\theta \,d\theta$$ The issue is that when you plugged in your expression into Wolfram Alpha, it subtracted the area of $r=2\cos\theta$ twice, yielding $\frac{3\pi}{2}-2\pi$.

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  • $\begingroup$ So, would it just be $\frac{3\pi}{2} - \pi$ then? $\endgroup$ – Nikhil Srd Mar 8 '17 at 16:31

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