# Find the area of intersection determined by three circles (Green's Thm)

I'm looking to find the shared area between these three circles using Green's Theorem:

$$x^2+y^2=1$$ $$(x-1)^2 + y^2 = 1$$ $$\left(x-\frac{1}{2}\right)^2 + \left(y - \frac{\sqrt{3}}{2}\right)^2 = 1$$

So far I know:

The centres of the three circles form an equilateral triangle with vertices $(0,0)$, $(1,0)$, and $\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)$. The area I'm concerned with is the area of that triangle plus the area underneath three arcs surrounding the triangle.

Green's Theorem states that $A = \int_Cxdy$ so I'll get:

$$A = \int_{C_1}xdy + \int_{C_2}xdy + \int_{C_3}xdy$$

$C_1$: $(0,0) \implies x = cost$ and $y = sint$

$C_2$: $(1,0) \implies x = 1 + cost$ and $y = sint$

$C_3$: $\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right) \implies x = \frac{1}{2} + cost$ and $y = \frac{\sqrt{3}}{2} + sint$

So I have:

$$A = \int_{C_1}cos^2tdt + \int_{C_2}cost+cos^2tdt + \int_{C_3}\frac{cost}{2}+cos^2tdt$$

The part I'm stuck on is finding the angles of the arcs, or the bounds on the integrals.

Also, I don't know whether this equation will give me the total area I'm looking for or just the area under the three arcs. If it's just the area of the three arcs than I can calculate the area of the triangle easily, but I don't want to double count it.

In my opinion, solving this problem by line integrals and Green's theorem is overkill. The area of the region of intersection can be found by extremely simple geometry. Note that the triangle formed by the vertices $\{\big(0,0\big), \big(1,0\big), \big(\frac{1}{2}, \frac{\sqrt{3}}{2}\big)\}$ is an equilateral triangle with unit side length, so its area is $\frac{\sqrt{3}}{4}$. Next note that the area of this triangle plus one of the circular segments is a unit circle sector with angle $60^{\circ}$, or in other words, $\frac16$ the area of a unit circle, i.e., $\frac{\pi}{6}$. Thus, the area of one circular segment must be $\frac{\pi}{6}-\frac{\sqrt{3}}{4}$. Altogether, the area of the triangle plus the area of three circular segments is,
$$A=\frac{\sqrt{3}}{4}+3\left(\frac{\pi}{6}-\frac{\sqrt{3}}{4}\right)=\frac{\pi-\sqrt{3}}{2}.$$
If I may offer a contradicting opinion to that of @DavidH, using Green's theorem to evaluate the area in general provide a major simplification of the general problem. If you want to compute the areas directly, you'll find it requires $18$ different geometrical configurations - that is, $18$ different possible integration limits. Using Green's theorem, that number is halved.