My suggestion is to integrate by having r go from the brown circle to the blue circle as θ goes from θ_1 to π, then double the answer.
Hi. I wanted to know if I did this right:
Using double integrals, set in polar coordinates, the integral that gives the area for the region
From the picture:
The pink region represents the area I want.
The intersection is given for x=3/2, where:
Tita 1 represents the limit on the angle for the circle that isn't centered at the origin. Is the limit for the brown region.
This is the limit on the angle for the orange region, for the circle centered at the origin.
And the area:
The first integral represents the entire circle, the second the brown region and the third the orange region, both multiplied on two by simmetry
In polar coordinates, the equation of the larger circle is obvious, hopefully.
Find the equation of the smaller circle in polar coords.
Solve for r using the quadratic formula or by completing the square.
We usually like to have r be positive, so use the positive sign with the square root. (Either sign will give the same polar graph.)
Great, didn't realized about that. That simplifies things pretty much. Anyway, the answer I give is right or did I made a mistake? Just wanna know, I know its horrendous, but anyway I'd like to know if instead of that its ok.
This is the solution I get using your suggestion: