Help with an unconventional center of mass question!

A standard US Mailbox has a cross-section shaped like a square of side S with a semicircle attached. Where is the center of mass of this cross-section? State your answer in coordinates with the origin at the center of the square.

This was all the information they gave me.

Thanks!

Since we are dealing with cross sections, we are basically dealing with area. With the origin centered on the center of the square, the center of mass will be on the X-axis, since the figure is symmetric around the X-axis.

So we want to find a point on the X-axis where the mass to the left is equal to the mass to the right. Since we are dealing with a cross section, this is the same as solving for a point on X-axis where the area to the left is equal to the area on the right.

The equation gets very messy and is hard to display clearly via text, so please bear with me.

Side of square is S. This means, the semicircle that is attached to the right of the square has diameter of S, or radius of S/2.

We are solving for x such that:
Area to the left of x = Area to the right of x
S(0.5S+x) = S(0.5S - x) + 0.5pi* r^2
S(0.5S+x) = S(0.5S - x) + 0.5 pi (S/2)^2

Dividing both sides by S gets you:
0.5S + x = 0.5S - X + 0.5 pi (S/4)

Let's bring all the X terms to the left, and remaining terms to the right

2x = 0.5 pi (S/4)
2x = S*pi/8
x = S*pi/16

Hope that helps =/

Edited for corrections