Moment of Inertia - 4 semi-Circles with Hollowed Core

I calculated three double integrals: one for each of the green regions. Then I multiplied by four. I assumed that there was one unit of mass per unit area.

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import math

Num = 5000

I = 0.0

##########################################################

# Upper Quarter Disk

dr = 5.0/Num
dtheta = (math.pi/2.0)/Num

r = 0.0

while r <= 5.0:

    theta = 0.0

    while theta <= math.pi/2.0:

        y = 5.0 + r*math.sin(theta)
        dA = r*dr*dtheta

        dI = dA*(y**2.0)
        I = I + dI

        theta = theta + dtheta

    r = r + dr

##########################################################

# Lower Quarter Disk

dr = 5.0/Num
dtheta = (math.pi/2.0)/Num

r = 0.0

while r <= 5.0:

    theta = 0.0

    while theta <= math.pi/2.0:

        y = r*math.sin(theta)
        dA = r*dr*dtheta

        dI = dA*(y**2.0)
        I = I + dI

        theta = theta + dtheta

    r = r + dr

##########################################################

# Triangle

dx = 5.0/Num
dy = dx

x = 0.0

while x <= 5.0:

    y = 5.0

    while y >= -x + 5.0:

        dA = dx*dy

        dI = dA*(y**2.0)
        I = I + dI

        y = y - dy

    x = x + dx

##########################################################

# Multiply by 4 and print

I = I*4.0

print Num
print I

##########################################################

#>>> 
#1000
#5249.68358302
#>>> ================================ RESTART ================================
#>>> 
#2000
#5236.0317262
#>>> ================================ RESTART ================================
#>>> 
#4000
#5236.67885474
#>>> ================================ RESTART ================================
#>>> 
#5000
#5235.24995651
#>>> 

The solid figure (without the square cut) is comprised of four semicircles of radius $5$ units each, and a square of side length $10$ units. Applying the parallel axes theorem to the two semicircles whose diameters are parallel to the $x$ -axis, we get their geometric moments of inertia equal to $\dfrac{625π}{8}-\dfrac{5000}{9π}+\dfrac{25π}{2}\times \left(5+\dfrac{20}{3π}\right ) ^2$ each. The geometric moments of inertia of the other two semicircles are $\dfrac{625π}{8}$ each. The geometric moment of inertia of the square of side length $10$ units about the $x$ -axis is $\dfrac{2500}{3}$ , and of the cut off square about that axis is $\dfrac{625}{6}$ . Therefore the geometric moment of inertia of the given figure is $2\left (\dfrac{625π}{8}-\dfrac{5000}{9π}+\dfrac{25π}{2}\left (5+\dfrac{20}{3π}\right ) ^2+\dfrac{625π}{8}\right ) +\dfrac{2500}{3}-\dfrac{625}{6}\approx \boxed {5236.91}$ .