Tensor of inertia - hollow cube.

[peripatein]

Hi,

Homework Statement

I have found the tensor of inertia of a rectangle of sides a and b and mass m, around its center, to be I₁₁=ma²/12, I₂₂=mb²/12, I₃₃=(ma² + mb²)/12. All other elements of that tensor are equal to zero. I would now like to use this result to determine the tensor of inertia of a hollow cube of side a around its center of mass.

Homework Equations

The Attempt at a Solution

I realize I have to use the parallel axis theorem. I have hence tried the following:
I₁₁=ma²/12 + m(a/2)², which yielded the wrong answer.
I know that the correct equation is I₁₁=I₂₂=I₃₃=ma²/12+ma²/12+ma²/6+4(ma²/12 + m(a/2)²)=5/3*ma²
I simply do not understand why this is correct. Could anyone please explain why this is the correct way to calculate the desired tensor of inertia? Also, why would I be summing all the diagonal elements in my tensor for the rectangle?

 

[TSny]

I realize I have to use the parallel axis theorem. I have hence tried the following:
I₁₁=ma²/12 + m(a/2)², which yielded the wrong answer.

Doesn't this give the contribution of only one of the sides of the cube? What about the other sides?