# Area of the shaded region

The numbers are to identify the circles

I've came out with this list of 4 inequalities(1 each circle), but I don't know if this is the best method to calculate it, neither how to solve it:

$(x+\frac{d}{2})^2+y^2\geq r_1^2 \\ (x-\frac{d}{2})^2+y^2\geq r_2^2 \\ (x+\frac{d}{2})^2+y^2\leq r_{1'}^2 \\ (x-\frac{d}{2})^2+y^2\leq r_{2'}^2$

The radius of the big circles can't be smaller than the small's.

The center of $1'$ is equal to the center of $1$, and the same with 2s.

-
I do not think that this problem has to be handled using inequalities. – Claude Leibovici Nov 27 '13 at 11:03
@ClaudeLeibovici The thing is that i want to calculate the area, yes. But i also want to take a formula for it, to graph it. That's why I choose inequalities. So, what is the good way to do it, then? – Arcotick Nov 27 '13 at 11:07
If all else fails, this looks doable (if laborious) with Green's theorem (in a form such as: the area of a region is the integral of $x\,dy$ over the boundary, traced counterclockwise), since each circle is easily parametrized, and the relevant angles for each boundary arc can be found in terms of $d$ and the radii. – Andrew D. Hwang Nov 27 '13 at 11:41
Let $B_1(\rho)$ and $B_2(\rho)$ be the balls with radius $\rho$ centered at $(-\frac{d}{2},0)$ and $(\frac{d}{2},0)$ respectively. Let $A(\rho_1,\rho_2)$ be the area of $B_1(\rho_1) \cap B_2(\rho_2)$. The area of the your shaped area is simply $$\frac12 \left( A(r_{1'}, r_{2'} ) - A(r_{1},r_{2'}) - A(r_{1'},r_{2}) + A(r_1,r_2) \right)$$ – achille hui Nov 27 '13 at 12:09