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I am trying to find the area between the following two curves given by the following polar equations: $r=\sqrt{3}\cos\theta$ and $r=1+\sin\theta$.

I did the following: First, I found the points of intersection: The curves intersect each other at the origin and when $\theta=\pi/6$. Then the area between the two curves is: $\int_{3\pi/2}^{2\pi+\pi/6}\frac{1}{2}(1+\sin\theta)^{2}d\ d\theta+\int_{\pi/6}^{\pi/2}\frac{1}{2}(\sqrt{3}\cos\theta)^{2}\ d\theta$. I am not sure if this is right or not. I appreciate if anyone can tell me whether there is a mistake somewhere in my solution. Also, if you know others methods of doing the problem, please let me know

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    $\begingroup$ That should work all right: you'll want just the area inside the cardioid up to $ \ \frac{\pi}{6} \ $ and then just the area inside the "circle" (actually a one-petal rosette) from $ \ \frac{\pi}{6} \ $ to $ \ \frac{\pi}{2} \ $ . You could also take the first integral from $ \ - \frac{\pi}{2} \ $ to $ \ \frac{\pi}{6} \ $ , but you should get the same result from yours. $\endgroup$ – colormegone Jun 19 '13 at 20:44
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A satisfactory answer can be found in the comments, hence this CW answer.

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