Prove that the area centroid of a planar region in $\mathbb{R}^3$ projects onto that of its projection onto the $xy$-plane Consider an arbitrary convex bounded region $\Omega'\in\mathbb{R}^2$ which is the projection onto the $xy$-plane of a region $\Omega$ that lies on a plane in $\mathbb{R}^3$ (that is not parallel to the $xy$-plane). Can someone help me prove that the area centroid* of $\Omega'$ is equal to the projection of the area centroid of $\Omega$ onto the $xy$-plane? Does the same hold for non-convex regions too?
*The employed definition of the area centroid $(x_c, y_c)$ of some region $\mathcal{R}\in\mathbb{R}^2$ is
$$ x_c = \frac{1}{A_{\mathcal{R}}}\!\!\iint_{\mathcal{R}}x\ dxdy,\quad y_c = \frac{1}{A_{\mathcal{R}}}\!\!\iint_{\mathcal{R}}y\ dxdy, $$
where $A_{\mathcal{R}}$ is the area of $\mathcal{R}$.
$\bf{Edit: }$ What I'm currently appealing to is that the 'shape' of $\Omega$ can be obtained by linearly scaling $\Omega'$ by some factor $c$ in the direction orthogonal to the intersection of the $xy$-plane and the plane containing $\Omega$. Therefore, since the area scales linearly as well, this gives
$$ x_{c,\Omega} = \frac{1}{cA_{\Omega'}}\!\!\iint_{\Omega'}x\ (c\ dxdy) = \frac{1}{A_{\Omega'}}\!\!\iint_{\Omega'}x\ dxdy,$$
and likewise for $y_{c,\Omega}$.
 A: I see nobody has answered so I'll give it a shot. I'll probably assume something I'm not allowed to in the process so if anyone catches anything let me know so I can adjust it. Anyway, I will only refer to the 'x' coordinate but the same applies for 'y'.
If the region consisted of discrete points, I think we can agree that the x coordinate of the centroid would be given by $x_c=\frac{\sum_{i=0}^{i=N}x_i}{N}$ where N is the number of points. Extending that to infinite points, we can define the centroid of the region in $R^3$ as $$x_c=\frac{\iint_SxdS}{S}$$ where $S$ is the surface area of the region in $R^3$. So, we integrate $x$ over the surface of the region. If you  agree with this definition of the centroid, we can prove what you asked as shown below. Given that the region is in the form of $z=f(x,y)$ we can write (you can look it up if you don't know this):
$$dS=\sqrt{f_x^2+f_y^2+1}\times dA$$
Note that the $A$ that showed up is the area of the projection of the region.Knowing this, the rest is simple. Let's set$\sqrt{f_x^2+f_y^2+1}=K$
1.$$S=\iint_S{dS}=\iint_A Kdxdy$$
2.$$x_c=\frac{\iint_SxdS}{S}=\frac{\iint_AxKdxdy}{\iint_A Kdxdy}=\frac{\iint_Axdxdy}{\iint_A dxdy}$$
Now, I repeat that this $A$ we ended up with is the area of the projection and since ${\iint_A dxdy}=A$ then :
$$x_c=\frac{1}{A}\iint_Axdxdy$$
which is also the centroid of the 2d projection.
Regarding the non-convex part of the question, it seems that a non-convex shape doesn't change the proof. Unless there is something in the proof we shouldn't do if the region is non-convex (or my proof is wrong - which is not impossible), then the same should hold for the centroid of a non-convex region.
