Re: Area polar coordinates.

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.

Re: Area polar coordinates.

How would you do that? I tried but I couldn't get an expression for r.

Re: Area polar coordinates.

In polar coordinates, the equation of the larger circle is obvious, hopefully.

Find the equation of the smaller circle in polar coords.

$\displaystyle (x-2)^2+y^2=7$

$\displaystyle x^2-4x+4+y^2=7$

$\displaystyle r^2-4r\cos(\theta)-3=0$

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.)

Re: Area polar coordinates.

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:

$\displaystyle \displaystyle\int_0^{2\pi}\int_{\frac{-2\cos\theta+\sqrt[]{4\cos^2\theta+12}}{2}}^{3} rdrd\theta$

Nice :D

Re: Area polar coordinates.

You have some algebra error(s) in your solution to the quadratic equation.

Re: Area polar coordinates.

You're right, I've made some mistakes. The lower bound would be $\displaystyle \frac{4\cos\theta+\sqrt[]{16\cos^2\theta+12}}{2}$

Thank you very much SammyS.