What is the area outside of $r=1$ and inside $r=2 \cos(3 \theta)$?

Question

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 $\pi$.

Answer 1

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 $3\cdot (\sqrt{3}/6+2\pi/9) = \sqrt{3}/2+\pi/3$.

Answer 2

You want$$ I=\frac12\int_0^{2\pi} |1^2-4\cos^2(3\theta)|\text d\theta $$ Note $$\begin{align}1-4\cos^2(3\theta)\ge 0\Longrightarrow |\cos(3\theta)|\le \frac12\end{align}$$ Also $$\int_0^{\pi/3}\cos^2(3\theta)\text d\theta=\int_{\pi/3}^{2\pi/3}\cos^2(3\theta)\text d\theta=\int_{2\pi/3}^{\pi}\cos^2(3\theta)\text d\theta=\int_\pi^{4\pi/3}\cos^2(3\theta)\text d\theta=\int_{4\pi/3}^{5\pi/3}\cos^2(3\theta)\text d\theta=\int_{5\pi/3}^{2\pi}\cos^2(3\theta)\text d\theta$$ So $$ \frac I6=\frac12\int_0^{\pi/3} |1^2-4\cos^2(3\theta)|\text d\theta $$ Substituting $x=3\theta$, $$\begin{align} I&=\int_0^{\pi}|1-4\cos^2x|\text dx\\ &=\int_0^{\pi/3}(4\cos^2x-1)\text dx+\int_{\pi/3}^{2\pi/3}(1-4\cos^2x)\text dx+\int_{2\pi/3}^{\pi}(4\cos^2x-1)\text dx\\ &=\int_0^{\pi/3}(2\cos2x+1)\text dx-\int_{\pi/3}^{2\pi/3}(2\cos 2x+1)\text dx+\int_{2\pi/3}^{\pi}(2\cos2x+1)\text dx\\ &=4\left(\int_0^{\pi/3}\cos 2x\text dx-\int_{\pi/3}^{\pi/2}\cos 2x\text dx\right)+\frac{\pi}3\\ &=2\left(\int_0^{2\pi/3}\cos x\text dx-\int_{2\pi/3}^{\pi}\cos x\text dx\right)+\frac{\pi}3\\ &=2\sqrt 3+\frac13\pi \end{align}$$