Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

enter image description here

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.

share|improve this question
    
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. –  user86418 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

1 Answer 1

up vote 0 down vote accepted

Can't be resolved in terms of calculus and that.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.