Thread: Finding the polar region that is outside the graph r=2 and inside the graph r=4costheta

1. I'm really stumped here as usual. Here is what i've managed to figure out.

I'm given two equations.
r=2
r=4cos(theta)

I converted them both to rectangular coordinates to get an idea of what the graph would look like.

I need to find either the area in red or the area in green. (In this case, I tried to find the area in green) .If I have the area in red or green, I know that I can just subtract one of them from pi*r^2 which in this case would be 4pi to get the area of the other section.

The problem I am facing right now is setting up the integral the proper way. I was thinking of going about it like this.

So a few questions. First, did I approach this in a correct way? Second, how would I know how to set this up to solve the integral using polar coordinates instead of converting it all to rectangular form?

Thanks for taking the time to answer, I tried to make this as clear as possible to see where i'm making my mistake or what I should be doing differently.

2. I would prefer to do this problem using geometry rather than calculus.

The red area here is one half of the area that you are looking for. The two circles intersect at the points $(1, \pm\sqrt3)$ (by Pythagoras). So the red and green regions together form a $120^\circ$ sector equal to one-third of the left circle. Thus the red and green regions have a combined area $\dfrac{4\pi}3.$ The green triangle has area (half base times height) $\sqrt3.$ So the red region has area $\dfrac{4\pi}3 - \sqrt3.$ The area that you want is twice the area of the red region, namely $\dfrac{8\pi}3 - 2\sqrt3.$

So your answer is correct (and of course the calculus method is just as valid as the geometric method).

3. Hi,
You wanted to know how to do this using polar coordinates and polar areas. The attachment shows how to do this:

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