calculuspolar coordinates
What is the area inside the polar curve $r = 2 \cos(3 \theta)$ but outside of the circle given by the polar equation $r = 1$?
A picture of the polar curve is at
http://www.wolframalpha.com/input/?i=polar+r%28t%29+%3D+2+Cos[3+t]
The area under the entire curve is $\int_0^\pi (2 Cos(3 t))^2/2 \, dt = \pi$, so the answer should be between $0$ and $\pi$.
HINT...You need to establish where the curves intersect in order to determine the limits, so solve $3\cos \theta=1+\cos \theta$.
So your limits are $\pm\frac {\pi}{3}$
If $$r_1(\theta) = 3a \cos \theta$$ is the equation of the circle, and $$r_2\theta) = a(1+\cos \theta)$$ is the equation for the cardioid, then these curves intersect when $r_1(\theta) = r_2(\theta)$, or $3a \cos \theta = a(1+\cos \theta)$, or $$2 \cos \theta = 1.$$ This gives us the intersection points $$\theta = \pm \frac{\pi}{3}.$$ There is another point of intersection at the origin, because the circle is traversed twice for a single traversal of the cardioid. But for our purposes, we can integrate on $\theta \in [-\pi/3, \pi/3]$. The area enclosed between $r_1$ and $r_2$ is then given by $$A = \int_{\theta = -\pi/3}^{\pi/3} \frac{1}{2}\left(r_1^2(\theta) - r_2^2(\theta)\right) \, d\theta.$$ Note this is not the same as $$\int_{\theta = -\pi/3}^{\pi/2} \frac{1}{2}\left(r_1(\theta) - r_2(\theta)\right)^2 \, d\theta.$$ The second integral is incorrect for the area enclosed.
Best Answer
If you're looking to integrate over the regions where $r<0$ as well then I refer you to Tim Ratigan's answer.
However, if you looking for the area in between the two curves on the graph, then we are trying to solve the following:
Note that $\cos(\pi/3)=1/2$ and $\cos(-\pi/3)=1/2$ and so two of our roots are $-\pi/9$ and $\pi/9$.
$$\int_{-\pi/9}^{\pi/9} \int_{1}^{2\cos(3\theta)} r\; dr\; d\theta \\ = 1/2 \cdot \int_{-\pi/9}^{\pi/9} (4\cos^2(3\theta) - 1)\; d\theta \\ = 1/2 \cdot \int_{-\pi/9}^{\pi/9} (2\cos(6\theta) + 1)\; d\theta \\ = 1/2 \cdot (1/3\cdot 2\cdot \sin(2\pi/3) + 2\pi/9) \\ = \sqrt(3)/6+\pi/9.$$
Hence, if we account for all three regions we have an area of $3\cdot (\sqrt{3}/6+2\pi/9) = \sqrt{3}/2+\pi/3$.