Let the points where the four quarter-circles intersect (from top to bottom, left to right) be A, B, C and D.
Let the bottom left corner be the origin, O. These are the equations of the two bottom circles which intersect at A:
(1) x^2 + y^2 = 1
(2) (x - 1)^2 + y^2 = 1
Subtract (1) from (2) to get:
(x - 1)^2 - x^2 = 0
x^2 - 2x + 1 - x^2 = 0
-2x + 1 = 0
2x = 1
x = 1/2
We can then find y by substituting in x = 1/2.
(1/2)^2 + y^2 = 1
(1/4) + y^2 = 1
y^2 = 1 - 1/4 = 3/4
y = √3/2
So the coordinates of the point where these two circles intersect is (1/2,√3/2)
We can then easily find the other coordinates as the square is of side length 1.
A(1/2,√3/2)
B(1 - √3/2, 1/2)
C(√3/2,1/2)
D(1/2,1 - √3/2)
The largest square that can fit in the shaded area will be the square formed by joining the four points A, B, C and D with straight lines. All these lines will be equal due to symmetry. We can use Pythagoras's theorem to work out the length of this line.
√3/2 - 1/2 = (√3 - 1)/2
This is the length of MA and MC where M is the centre of the square. We can use Pythagoras's theorem to find the length of AC.
This turns out to be √(2-√3). To find the area of the square we just need to square it, which is 2-√3.
This isn't a good estimate though, the real value will be higher.
To find the exact area, we can draw two lines to the points A and C from the origin O, which will give a sector of the circle. To find the area of this sector we need the angle, which can be found by bisecting the sector, as this creates two right angled triangles.
Sin(θ/2) = (√(2-√3))/2 divided by 1. (opposite/hypotenuse)
Which gives θ/2 = π/12 so θ = π/6.
Area of a sector => 1/2(r^2)(π/6) = π/12
Now we need to take away the two triangles OAM and OCM. The base of these triangles is 1, and the height can be found by using trigonometry.
OAC is an equilateral triangle, so angle OAC = (180 - 30)/2 = 75. Triangle AMC is an isosceles right angled triangle so angle MAC = 45, so angle OAM = 75 - 45 = 30.
Sin(30) = h/(√3 - 1)/2
Sin(30) = 1/2 so h = (√3 - 1)/4
So the area of the triangle is (√3 - 1)/4 * 1 * 1/2 = (√3 - 1)/8.
So the area of one quarter of the shaded region is π/12 - (√3 - 1)/8 - (√3 - 1)/8 = π/12 - (√3 - 1)/4. Multiplying this by four gives the full shaded area which is π/3 - (√3 - 1) = π/3 - √3 + 1
Another method that can be used is integration of the circle between x = 1/2 and x = √3/2.
x^2 + y^2 = 1
y = √(1 - x^2)
To integrate this, substitute x = sin(u).
x = sin(u)
dx/du = cos(u)
dx = cos(u) du
int √(1 - sin(u)^2) cos(u) du
√(1 - sin(u)^2) = cos(u)
So we just need to integrate cos^2(u) du.
To integrate this, use the trig identity cos^2(u) = 1/2(cos2u + 1).
1/2 int cos2u + 1 = 1/2[1/2sin2u + u]
u = arcsin(x), substituting this gives:
1/2[1/2sin2(arcsin x) + arcsin x]
We can simplify sin 2(arcsin x) by using the following trig identity:
sin2θ = 2sinθcosθ
sin(arcsinx) = x
cos(arcsinx) = √(1-x^2) [by sin^2 + cosx^2 = 1]
So sin2(arcsin x) = 2x√(1-x^2)
This gives:
1/2[x√(1-x^2) + arcsin x] + c
Substituting in √3/2 and 1/2 and subtracting gives:
1/2[√3/4 + π/3] - 1/2[√3/4 + π/6] = 1/2[π/6] = π/12
This is the area under the curve between 1/2 and √3/2, to find the area of the quadrant we must be subtract the rectangle below.
The base of the rectangle is √3/2 - 1/2 = (√3 - 1)/2 and the height is 1/2, so the area is (√3 - 1)/2 x 1/2 = (√3 - 1)/4
So the area of the quadrant = π/12 - (√3 - 1)/4
Multiplying this by 4 gives the full area of the shaded region which is π/3 - (√3 - 1) = π/3 - √3 + 1 as before.