I am wondering if there is a "trick" I can use to ease the equation level for this problem.
Say I have a solid (constant density) prism which is bounded by the following 6 "planes" (I use subscript to mark the different functions, function "i" doesn't sound correct). $$y_0 = 0$$ $$y_1 = C$$ $$x_0 = f_0(y)$$ $$x_1 = f_1(y)$$ $$z_0 = f_2(x, y)$$ $$z_1 = f_3(x, y)$$
So basically the prism has parallel vertical sides in the x-z projection, 2 more sides are vertical, and the last pair are in a "near horizontal" plane.
Now the moment the centroid (shown for x only here) is calculated using $$m = \rho \iiint_E \mathrm{dV} = \rho \int_0^{C}\int_{f_0}^{f_1}\int_{f_2}^{f_3}\mathrm{dz \ dx \ dy}$$ $$\bar{x} = \frac{\rho \iiint_Ex \mathrm{dV}}{m} = \frac{ \int_0^{C}\int_{f_0}^{f_1}\int_{f_2}^{f_3}x \ \mathrm{dz \ dx \ dy}} { \int_0^{C}\int_{f_0}^{f_1}\int_{f_2}^{f_3}\mathrm{dz \ dx \ dy}}$$
Now to get the moment of inertia I shift the axis system, so I get new functions for the planes. and then solve: $$I_x = \rho \iiint_E(y^2 + z^2) \mathrm{dV} = \int_{0 - \bar{y}}^{C - \bar{y}}\int_{f_0(y + \bar{y}) - \bar{x}}^{f_1(y + \bar{y}) - \bar{x} }\int_{f_2(x + \bar{x}, y + \bar{y}) - \bar{z}}^{f_3(x + \bar{x},y + \bar{y}) - \bar{z}}(y^2 + z^2) \mathrm{dz \ dx \ dy}$$
Now I wonder, is this "long winded road" the easiest? Or can I take a shortcut (IE: not calculate centroid/ignore it). Also my colleguea should've brought me the numbers, but I haven't received them - is there a way to calculate this & make a formula I can fit into a script? Considering everything is a linear function