# Having problems with math problem

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Hi all,

Have been having some problems with the following math problem I was given last week:

A solid cube of mass M and side a has one corner at the origin and three sides along the positive x,y and z axes. The square of the distance of the mass element dM = pdV rom the z-axis is x^2 + y^2 so that the moment of inertia of the cube around the z-axis is given by

I = integral (x^2 + y^2) pdV

taken throughout the volume V of the cube.

If the cube has constant density, prove that I= (2/3) Ma^2

Any help/tips would be most appreciated!

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The square of the distance of the mass element dM = pdV rom the z-axis is x^2 + y^2
the distance of the mass? that makes no sense. And what do you mean by pdV?

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pdV is p (ro - density) times the volume element dV.

The square of the distance of the mass element from the z-axis - i assume means the distance from the mass element to the z-axis, all squared...

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Is x = y = a since this is a cube centered at the origin?

a^3 = V so 3a^2 da = dV. So the integral might be integral p(2a^2)(3a^2)da

Which becomes 6/5 pa^5 +C

M = pV = pa^3

Doing the calculations, I get 6/5 Ma^2 + C.

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Integral of $x^2 + y^2$ is $\frac{1}{3}x^3 + \frac{1}{3}y^2 + C$. Thought this might help.

Omit.

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