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## Homework Statement

I think I overdid this. Find area outside of the polar curve r = 2cosθ and inside r = 1

## The Attempt at a Solution

http://img651.imageshack.us/img651/2558/44170022.jpg [Broken]

It is the black region.

Here is what I did.

1) Find area of the semi-circle of r = 2cosθ

2) Find area of r = 2cosθ from 0 to π/3

3) Find area of r = 1 from π/3 to π/2

Do some geometry

Then the integral should be

Area = [tex]\int_{\frac{\pi}{3}}^\frac{\pi}{2} \frac{1}{2} \mathrm{d}\theta - \int_{0}^\frac{\pi}{2} \frac{1}{2} (2\cos\theta)^2 \mathrm{d}\theta + \int_{0}^\frac{\pi}{3} \frac{1}{2} (2\cos\theta)^2 \mathrm{d}\theta [/tex]

But this took quit a lot of work. Is there, in general, anyway to get around that curve?

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