# Moment of inertia of a (homogenous) triangle in R3

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 In a forgoing posthttps://community.wolfram.com/groups/-/m/t/2332921the moment of inertia of a triangle in R2 rotating around a point was calculated.Here a method is given to do this in R3, the triangle rotating around an axis thru x4 and with direction b (* square of distance of a point p to an axis thru x and direction b *) Clear[dist] dist2[p_, x_, b_] := Module[{aa, fL, vB}, aa = p - x; fL = b.aa/(b.b); vB = p - x - b fL; vB.vB] (* Moment of Inertia of a triangle given by x1, x2, x3 with respect to axis thru x4 and direction b *) Clear[fMI] fMI[x1_, x2_, x3_, x4_, b_] := Module[{d2, d3, fe}, d2 = x2 - x1; d3 = x3 - x1; fe = Cross[d2, d3]; fe = Sqrt[fe.fe]; (*surface-element of Triangle*) Integrate[ fe dist2[x1 + u d2 + v d3, x4, b], {v, 0, 1}, {u, 0, 1 - v}] ] Check with the result of forgoing post fMI[{4, 1, 0}, {1, 3, 0}, {2, 7, 0}, {.5, 5, 9}, {0, 0, 3}] Answer
 Hans showed very nice how to calculate the moment of inertia. If one is only interested in the result one can use the function provided by Mathematica which has a similiar interface compared to the function of Hans. Main difference is that not a list of coordinates has to be provided but rather a object in form of a region (here: triangle). With[{pointOfRotation = {0.5, 5, 9}, direction = {0, 0, 3}}, MomentOfInertia[Triangle[{{4, 1, 0}, {1, 3, 0}, {2, 7, 0}}], pointOfRotation, direction]] Answer