How Is the Moment of Inertia Tensor Calculated for a Cuboid?

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SUMMARY

The moment of inertia tensor for a cuboid with constant mass density and edges along the x, y, and z axes can be calculated using the formula: I = M * [[(1/3)(b^2+c^2), -(ab/4), -(ac/4)], [-(ab/4), (1/3)(a^2+c^2), -(bc/4)], [-(ac/4), -(bc/4), (1/3)(a^2+b^2)]]. This tensor is symmetric, which is a necessary property. The equality involving the integral of the density function is crucial for deriving the tensor, specifically the expression ∫_V ρ(𝑟)(r²δjk - xjxk)dV = ∫_V ρ(x,y,z) * [[y²+z², -xy, -xz], [-xy, z²+x², -yz], [-xz, -yz, x²+y²]]dxdydz.

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Logarythmic
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Homework Statement


Compute the moment of inertia tensor I with respect to the origin for a cuboid of constant mass density whose edges (of lengths a, b, c) are along the x,y,z-axes, with one corner at the origin.

The Attempt at a Solution


I get

[tex]I = M \left( <br /> \begin{array}{ccc}<br /> \frac{1}{3} (b^2+c^2) & -\frac{ab}{4} & -\frac{ac}{4}\\<br /> -\frac{ab}{4} & \frac{1}{3} (a^2 + c^2) & -\frac{bc}{4}\\<br /> -\frac{ac}{4} & -\frac{bc}{4} & \frac{1}{3} (a^2 + b^2)<br /> \end{array}<br /> \right)[/tex]

Can this be right?
 
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I think it may be right, since the matrix is supposed to be symmetric.
 
Yeah, but it can be symmetric in many ways. ;)

Can someone please explain the equality

[tex]\int_V \rho(\vec{r}) (r^2 \delta_{jk} - x_jx_k) dV = \int_V \rho(x,y,z)<br /> \left(<br /> \begin{array}{ccc}<br /> y^2+z^2 & -xy & -xz\\<br /> -xy & z^2+x^2 & -yz\\<br /> -xz & -yz & x^2+y^2<br /> \end{array}<br /> \right)dxdydz[/tex]

for me? I think this is the most important step in my understanding for this problem.
 

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