# Area of the region inside $r = 1 - \cos(\theta)$ and also inside $r = \cos(\theta)$

Pretty simple polar integration question that I've been having trouble with...

The question says it all. I identified the limits of integration by setting $1 - \cos(\theta) = \cos(\theta)$ so that $\cos(\theta) = \frac{1}{2}$ and $\theta = \pm \frac{\pi}{3}$.

I've tried the integral

$$\frac{1}{2}\int_{-\frac{\pi}{3}}^{\frac{\pi}{3}} \cos^2(\theta) - (1 - \cos(\theta))^2 = \sqrt{3} - \frac{\pi}{3}$$

Back of the book says the answer is $\frac{7\pi}{12} - \sqrt{3}$.

I think the polar curves are messing me up. Am I interpreting $r = \cos(\theta)$ wrong or something? Do I need to change the bounds of integration?

## 1 Answer

You are calculating the area in the following graph and your calculation is correct.

But the question asks for the area "inside" both graphs which can be seen below: This area is then $$A=\int_0^{\pi/3}(1-\cos x)^2dx+\int_{\pi/3}^{\pi/2}\cos^2xdx$$ which is what the back of your book says.

• Could you please tell me what program can draw these graph? – user143993 Dec 18 '14 at 8:25
• @ user143993: PolarPlot function of Mathematica. – Math-fun Dec 18 '14 at 8:31
• Mathematica. Thanks a lot. – user143993 Dec 18 '14 at 10:01