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I've got no problems with calculating the moment of inertia/tensor of inertia of a cube using an integral over the lamina of a cube. However, I must be missing something obvious or making some sort of silly mistake but doing an integral over all three dimensions like

$$ \int_V \rho r^2 dV = \int_V \rho (x^2 + y^2 + z^2) dxdydz,$$

does not give the correct answer. Is there something I am missing in why this is not actually valid?

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    $\begingroup$ The "$r$" in $r^2dm$ is the distance from the axis, not the origin. See e.g. here. $\endgroup$
    – J.G.
    May 8, 2022 at 13:02

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The moment of inertia is defined for a specific rotation axis. The radius $r$ is the distance from this axis, not from the origin point. For example, for the moment of inertia around the $z$-axis the radius is $r=\sqrt{x^2+y^2}$, but not $r=\sqrt{x^2+y^2+z^2}$.

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