Area of intersection between 4 circles centered at the vertices of a square

Solution 1:

I think this question becomes much easier if we can know the four intersection points. And we can do this through observing the equilateral triangles if we join lines between centers of the circles.

If my calculation right, C $(\frac{\sqrt3}{2}r,\frac{1}{2}r)$, B $(\frac{1}{2}r,\frac{\sqrt3}{2}r)$, and you can work out the leftmost and downmost ones because of symmetry about the center $(\frac{1}{2}r,\frac{1}{2}r)$ but not even necessary.

One fourth of the area (suppose the upperright quarter) can be represented as:

$$\frac{S}{4} = \pi r^2\times\frac{30}{360}-\frac{(\frac{\sqrt3}{2}r-\frac{1}{2}r)\frac{1}{2}r}{2}\times 2$$

Simplified:

$$S=\frac{\pi}{3}r^2-(\sqrt3-1)r^2$$

Graph and the hilarious font :) Look! This is integration-free!

Solution 2:

The intersection point at the east is where $$x^2 + y^2 = 1$$ and $y = \frac{1}{2}$, so $x = \frac{\sqrt{3}}{2}$.

So by symmetry the quantity you desire is $$ 400 \int_{\frac{1}{2}}^{\frac{\sqrt{3}}{2}} \sqrt{1-x^2} - \frac{1}{2} dx $$ which Wolfram Alpha says is $$ \frac{100\pi}{3} - 100(\sqrt{3} - 1). $$

Solution 3:

lets define some points: (Assuming a radius of 1)

O = (0,0)
A = (0.5, 0.5)
B = (cos(30°), sin(30°))
C = (cos(60°), sin(60°))

The rounded "triangle" ABC is a quarter of the area you are looking for.

The rounded triangle OBC is a sector with radius 1 and angle 30°, so its area is 1/12 of a disk, $\pi/12$.

From this sector we substract the two triangles OBA and OAC which are congruent and have a base of cos(30°) - 0.5 and a height of 0.5, so their total area is 0.5*(cos(30°)-0.5)

the area of ABC is then $$\pi/12 - (\sqrt{3}/2 - 1/2)/2$$ and the total area four times as much: $$\pi/3 - \sqrt{3} + 1$$