# Common area between two circles.

1. Mar 13, 2009

### alalall2

I am currently trying to figure out a problem in polar coordinates:
Find the area common to the two circles x2 + y2 = 4, x2 + y2 = 6x.

Using polar coordinates I know the two equations of the circles are r=2 and r=6 cos(theta) respectively. What I tried to do was find the area over the x-axis first then double the result to provide the entire area.

What I thought would be this top area would be the sum of the double integrals from theta limits 0->acos(1/3) and r limits 6cos(theta)->4/6 for r dr d(theta)+ theta limits 0->acos(1/3) and r limits 4/6->2sin(theta) for r dr d(theta).

However I don't think this is right as both integrals provide a result that is either negative or too large to be the value within the designated area.

2. Mar 13, 2009

### Staff: Mentor

Is this what you have?
$$\int_{\theta = 0}^{arccos(1/3)} \int_{r = 0}^2 r dr d\theta + \int_{\theta = arccos(1/3)}^{\pi/2} \int_{r = 6 cos(\theta)}^0 r dr d\theta$$

3. Mar 13, 2009

### HallsofIvy

Staff Emeritus
I think the best thing to do would be to find the area of the portion of the second circle that is outside the first circle and subtract that from the area of the second circle. I don't see why you would need to do a double integral. Find the $\theta$ values for the intersections of the two circles and integrate $(6cos(\theta)- 2)(rd\theta)$ over that interval.

4. Mar 13, 2009

### alalall2

That makes much more sense to me. Thank you!

5. Mar 13, 2009

### alalall2

Originally I thought I had done this question correctly using cartesians, but my previous answers don't equate to the same values as the ones through polar coordinates. i used double integrals for this solution as well as a sum from between the y-axis and x=2/3 and between x=2/3 and x=2. Is there a better way I am not seeing?