Moment of Inertia of a regular hexagonal plate?

[CrimsonFlash]

Homework Statement

How do you calculate the moment of inertia of a regular hexagonal plate of side a and mass M along an axis passing through its opposite vertices?

Homework Equations

Moment of inertia for a right triangle with an axis running along its base would be I = m h² /6 where h and m are the height and the mass of the triangle respectively.

The Attempt at a Solution

I tried breaking up the hexagon into 4 congruent right triangles and a rectangle. So there would be 2 triangles at the top, the rectangle in the middle and the two triangles below.
The moment of inertia of the triangle would be = [itex] \frac{m a^2} {8} [/itex] because the height will be = asin60
Mass of this triangular bit = M/12
Density of the hexagon = 2M/(a²3√3)
So the rectangular bit in the middle will have moment of inertia equal to this density multiplied by the double integral of x^2 with x running from a√3 to -a√3 and y running from a/2 to -a/2.
This gives moment of inertia for the rectangular bit = 4Ma²/3.
I thought then you finally add all these up to get moment of inertia for the hexagon because integration is a linear operator or whatever. But I get the wrong answer.

 

[haruspex]

This gives moment of inertia for the rectangular bit = 4Ma²/3.

Seems too much. Pls post your working.

 

[CrimsonFlash]

So the rectangular bit in the middle will have moment of inertia equal to this density multiplied by the double integral of x^2 with x running from a√3 to -a√3 and y running from a/2 to -a/2.
This gives moment of inertia for the rectangular bit = 4Ma²/3.
I thought then you finally add all these up to get moment of inertia for the hexagon because integration is a linear operator or whatever. But I get the wrong answer.

Silly silly me. The integral for x runs from (a√3)/2 to -(a√3)/2 and this gives the right answer.