to compute the mass moment of inertia tensor for a trapezoidal prism, can I break up the prism into 3 regions and then sum their inertias?

I want to find the mass moment of inertia tensor for a trapezoidal prism, like this:

My first step was to find the COM. First, I found the geometric centroid of a trapezoid.

$$begin{align}
bar{x} &= frac{b}{2} + frac{(2a+b)(c^2-d^2)}{6(b^2-a^2)}
bar{y} &= frac{b+2a}{3(a+b)}h
end{align}$$

Then, I reasoned the COM Should be

$$left(bar{x},bar{y},frac{ell}{2}right)$$

So, for a height $z$, so the coordinates for my initial trapezoid would become now

• $(0,0,z) rightarrow (-bar{x},-bar{y},z)$

• $(f,h,z) rightarrow (f-bar{x},h-bar{y},)$

• $(f+a,h,z) rightarrow (f+a-bar{x},h-bar{y},z)$

• $(b,0,z)rightarrow (b-bar{x}, -bar{y},z)$

I can draw this now.

So, I now have the trapezoidal prism centered on it’s center of mass.

I am assuming a uniform density $rho$. To find the moment of inertia tensor $mathbf{I}$, I need to do

$$mathbf{I} = rho iiint_{V} begin{bmatrix}
y^2 + z^2 & -xy & -xz
-xy & z^2 + x^2 & -yz
-xz & -yz & x^2 + y^2
end{bmatrix} dV$$

So I reasoned, I could break up the integration into 3 parts.

$$begin{align}
g_1(x) &= frac{(h-bar{y})-bar{y}}{(f-bar{x})-bar{x}}x
&= frac{h}{f}x
g_2(x) &= h
g_3(x) &= frac{(-bar{y})-(h-bar{y})}{(b-bar{x})-((f+a)-bar{x})}x
&= frac{-h}{b-(f+a)}x
end{align}$$

So, I thought I could break up the integration as follows:

$$mathbf{I}_{text{region 1}} = rho int_{-frac{-ell}{2}}^{frac{ell}{2}}int_{-bar{x}}^{f-bar{x}} int_{-bar{y}}^{frac{h}{f}x}begin{bmatrix}
y^2 + z^2 & -xy & -xz
-xy & z^2 + x^2 & -yz
-xz & -yz & x^2 + y^2
end{bmatrix} dydxdz$$

$$mathbf{I}_{text{region 2}} = rho int_{-frac{-ell}{2}}^{frac{ell}{2}}int_{f-bar{x}}^{(f+a)-bar{x}} int_{-bar{y}}^{h-bar{y}}begin{bmatrix}
y^2 + z^2 & -xy & -xz
-xy & z^2 + x^2 & -yz
-xz & -yz & x^2 + y^2
end{bmatrix} dydxdz$$

$$mathbf{I}_{text{region 3}} = rho int_{-frac{-ell}{2}}^{frac{ell}{2}}int_{(f+a)-bar{x}}^{b-bar{x}} int_{-bar{y}}^{frac{-h}{b-(f+a)}x}
begin{bmatrix}
y^2 + z^2 & -xy & -xz
-xy & z^2 + x^2 & -yz
-xz & -yz & x^2 + y^2
end{bmatrix} dydxdz$$

My question is, can I sum these up to get the overall inertia tensor for the whole?

$$mathbf{I}_{text{overall}} = mathbf{I}_{text{region 1}} +mathbf{I}_{text{region 2}} +mathbf{I}_{text{region 3}} $$

The problem is, when I tried to break a shape up and then sum the parts, it didn’t work in this question.

Is moment of inertia additive? If so, why doesn't adding two halves of solid box work?

Here, I tried to divide a cube into halves, find the mass moment of inertia for each half, and then add the inertia tensors. But it didn’t work because I neglected to take into account that the inertia for each individual part is for its own COM. So to find it overall, I have to move the tensors to the overall COM before solving.

So, would my approach actually give me the mass moment of inertia? Or am I making a similar mistake? If I am making a mistake in this summation approach, is there a simpler alternative strategy I could use for the integration besides just computing the tensors for the individual regions and then moving them to the overall COM?