>>13812283(x - 2)^2 + (y - 2)^2 = 4
y = 1/2 * x
(x - 2)^2 + (1/2 * x - 2)^2 = 4
x^2 - 4x + 4 + 1/4 * x^2 - 2x + 4 = 4
5/4 * x^2 - 6x + 4 = 0
5x^2 - 24x + 16 = 0
x = (24 +/- sqrt(576 - 4*5*16))/10 = (24 +/- sqrt(576 - 320)) / 10 = (24 +/- sqrt(256))/2 = (24 +/- 16)/10 => x = 4,4/5
x=4 => y=2 (intersection with diameter)
x=4/5 => y=2/5 (first intersection)
Decompose red area into two parts:
1) Triangle formed by y = 0, y = 1/2 * x, x = 4/5
A1 = (1/2)(4/5)(4/5*1/2) = (1/2)(4/5)(2/5) = (2/5)(2/5) = 4/25
2) y = 2 +/- sqrt(4 - (x - 2)^2)
Bottom half of the circle is: y = 2 - sqrt(4 - (x - 2))^2
Want area of y = 2 - sqrt(4 - (x - 2))^2 from x=4/5 to x=2.
int(4/5,2) (2 - sqrt(4 - (x - 2)^2)) dx
Let 2u = x - 2 => dx = 2du => x = 4/5 => u = -3/5; x = 2 => u = 0
int(-3/5,0) (2 - sqrt(4 - 4u^2)) 2du = 2*int(-3/5,0) 2(1 - sqrt(1 - u^2)) du = 4*int(-3/5,0) (1 - sqrt(1 - u^2)) du
u = sin(t) => du = cos(t)dt
u = -3/5 => t = arcsin(-3/5); u = 0 => t = arcsin(0) = 0
4*int(arcsin(-3/5),0) (1 - sqrt(1 - sin^2(t))*cos(t) dt = 4*int(arcsin(-3/5),0) (1 - sqrt(cos^2(t))*cos(t) dt = 4*int(arcsin(-3/5),0) (1 - cos(t))cos(t) dt = X + Y
X = 4*int(arcsin(-3/5),0) cos(t) dt = 4*[sin(t)]^{arcsin(-3/5),0} = 4*(sin(arcsin(-3/5)) - sin(0)) = 4*(3/5 - 0) = 12/5
Y = 4*int(arcsin(-3/5),0) cos^2(t) dt
We can use the trig identity that cos^2(t) = (1 + cos(2t))/2
Y = 4*int(arcsin(-3/5),0) 1/2 * (1 + cos(2t)) dt = 2 * int(arcsin(-3/5),0) (1 + cos(2t)) dt = 2*[t + sin(2t)/2]^{arcsin(-3/5),0} = 2*(arcsin(-3/5) + sin(2*arcsin(-3/5))/2)
A2 = X + Y = 12/5 + 2*(arcsin(-3/5) + sin(2*arcsin(-3/5))/2)
Therefore, the total area is: A1 + A2 = 4/25 + 12/5 + 2*(arcsin(-3/5) + sin(2*arcsin(-3/5))/2) = 64/25 + 2*(arcsin(-3/5) + sin(2*arcsin(-3/5))/2)
= 0.31299778...
Which seems to match:
>>13812481>>13812997 
