Moment Of Inertia Of Cuboid!

My approach is laborious, but very straightforward. The general approach is the following:

1) Scan through the cuboid (in triple-integral fashion)
2) Determine the perpendicular distance from each infinitesimal mass to the body diagonal
3) Sum the infinitesimal moment contributions

Now we need a convenient way of determining a general formula for the distance from an infinitesimal mass element to the body diagonal. This reference has a nice formula for the perpendicular distance from a point to a line defined by two points. Call the start point of the body diagonal $A$ and the end point of the body diagonal $B$ . Call an arbitrary point inside the cuboid $P$ . Suppose the cuboid has dimensions $(L,W,H)$ .

$A = (0,0,0) \\ B = (L,W,H) \\ P = (x,y,z)$

The square of the distance from the infinitesimal mass to the body diagonal is:

$d^2 = \frac{ \Big | (x,y,z) \times (x-L, y-W, z - H) \Big |}{L^2 + W^2 + H^2}$

Computing the numerator cross product and laboriously simplifying gives:

$d^2 = \frac{ H^2 y^2 - 2 H W y z + W^2 z^2 + H^2 x^2 - 2 H L x z + L^2 z^2 + W^2 x^2 - 2 W L x y + L^2 y^2 }{L^2 + W^2 + H^2}$

The infinitesimal mass element is:

$dM = M \frac{dx \, dy \, dz}{L W H}$

The infinitesimal moment contribution is:

$dI = dM \, d^2 \\ = M \frac{dx \, dy \, dz}{L W H} \, \, \frac{ H^2 y^2 - 2 H W y z + W^2 z^2 + H^2 x^2 - 2 H L x z + L^2 z^2 + W^2 x^2 - 2 W L x y + L^2 y^2 }{L^2 + W^2 + H^2}$

The total moment is:

$I = \frac{M}{L W H \, (L^2 + W^2 + H^2)} \int_0^H \int_0^W \int_0^L \Big ( H^2 y^2 - 2 H W y z + W^2 z^2 + H^2 x^2 - 2 H L x z + L^2 z^2 + W^2 x^2 - 2 W L x y + L^2 y^2 \Big ) \, dx \, dy \, dz$

Evaluating the triple integral is trivial, but tedious. The result is:

$I = \frac{M}{L W H \, (L^2 + W^2 + H^2)} \Big (\frac{1}{6} H^3 L W^3 + \frac{1}{6} W H^3 L^3 + \frac{1}{6} H W^3 L^3 \Big ) \\ = \frac{M}{6 \, (L^2 + W^2 + H^2)} \Big ( H^2W^2 + H^2 L^2 + W^2 L^2 \Big )$

Plugging in numbers for this particular problem results in $49$ as the answer.