Moment of inertia integral for a cube about its center

[fluidistic]

I've been trying to calculate the moment of inertia (with respect to the center of mass) of several rigid bodies (including 2 dimensional ones) but I never reached any good answer.
For example a cube whose edges are equal to $$a$$.
My work : From wikipedia and at least 2 physics books, $$I_{CM}=\int r^2 dm$$. From my notes it's equal to $$\int _{\Omega} \rho \zeta ^2 dV$$.
My alone work : $$dV=dxdydz$$ and as $$\rho$$ (the density) is constant I can write $$I_{CM}=\rho \int_{-\frac{a}{2}}^{\frac{a}{2}} x^2 dx \cdot \int_{-\frac{a}{2}}^{\frac{a}{2}} y^2 dy \cdot \int_{-\frac{a}{2}}^{\frac{a}{2}}z^2dz$$. Calculating this I get that $$I_{CM}=\rho \left( \frac{a^3}{12^3} \right)=\frac{M}{V}\cdot \frac{V}{12^3}=\frac{M}{12^3}$$.
I never took calculus III yet so I never dealt with triple integrals and even double ones. I'm guessing that I'm multiplying the 3 integrals and that instead I should be adding them or something like that, but I don't know why at all. I'd be glad if you could help me.
P.S. : I considered the origin of the system as being the center of mass of the cube.

 

[tiny-tim]

moment of inertia must be measured about an axis

Hi fluidistic!

(have an integral: ∫ and a rho: ρ and a squared: ² )

I've been trying to calculate the moment of inertia (with respect to the center of mass) of … a cube whose edges are equal to $$a$$.
…
$$I_{CM}=\rho \int_{-\frac{a}{2}}^{\frac{a}{2}} x^2 dx \cdot \int_{-\frac{a}{2}}^{\frac{a}{2}} y^2 dy \cdot \int_{-\frac{a}{2}}^{\frac{a}{2}}z^2dz$$. Calculating this I get that $$I_{CM}=\rho \left( \frac{a^3}{12^3} \right)=\frac{M}{V}\cdot \frac{V}{12^3}=\frac{M}{12^3}$$.

i] r² = x² + y² + z², not times, so your integral would be

∫∫∫x²dxdydz + ∫∫∫y²dxdydz + ∫∫∫z²dxdydz

(but that's still wrong, because:)

ii] moment of inertia must be measured about an axis …

I expect you've been misled by 2D questions which ask for the moment of inertia "about a point" … but they really mean about the axis coming straight out of the page through that point …

so the integral for the z-axis, say, would be ∫∫∫x²dxdydz + ∫∫∫y²dxdydz

 

[fluidistic]

Thank you very much tiny-tim.
I haven't been mislead by anything but myself.
I'll try to do it and do a lot of more complicated cases. I really like this part of Physics.
EDIT : I'm not sure I got it right. Can you confirm if the answer is $$\frac{Ma^2}{6}$$. I had not to forget about $$\rho$$.

 

[tiny-tim]

EDIT : I'm not sure I got it right. Can you confirm if the answer is $$\frac{Ma^2}{6}$$. I had not to forget about $$\rho$$.

Yes, that's right!

See wikipedia for some more examples: http://en.wikipedia.org/wiki/List_of_moments_of_inertia

 

[fluidistic]

Thank you very much once again.