Compute the centroid of a 3D planar polygon

Just use the equations that you have twice, but the second time swap in z for y.

That is, calculate the centroids of the two projections, one onto the x-y plane, and the other onto the x-z plane. The centroids of the projections will be projections of the actual centroid, so the answer will be the x, y, and z values you find from these two calculations.

Stated more explicitly: If your points are (x1, y1, z1), (x2, y2, z2),... , to get the x-y centroid, (Cx, Cy), do a calculation using (x1, y1), (x2, y2),... and to get the x-z centroid, (Cx, Cz) use the points (x1, z1), (x2, z2),.... -- just do the second calculation with your same 2D formula, treating the z values as the y in the equation. Then your 3D centroid will be (Cx, Cy, Cz). This will work as long as your surface is flat and isn't parallel to the x-y, x-z, or y-z planes (but if it is parallel it's just the 2D equation).

Let the points be v₀, v₁, ..., v_N in counterclockwise, where v_i = (x_i, y_i, z_i).

Then the triplets (v₀, v₁, v₂), (v₀, v₂, v₃), ..., (v₀, v_i, v_i+1), ..., (v₀, v_N-1, v_N) forms N-1 triangles that create the polygon.

The area of each triangle is | (v_i − v₀) × (v_i+1 − v₀) | ÷ 2, where × is the cross product and | · | is vector length.

You may need to make the area negative to compensate for concave parts. A simple check is to compute (v_i − v₀) × (v_i+1 − v₀) · (v₁ − v₀) × (v₂ − v₀). The area should have the same sign as the result.

Since ratio of area of 2D figures is constant under parallel projection, you may want to choose a unit vector (e.g. z) not parallel to the plane, the treat (v_i − v₀) × (v_i+1 − v₀) · z as the area. With this, you don't need to perform the expensive square root, and the sign check is automatically taken care of.

The centroid of each triangle is (v₀ + v_i + v_i+1) ÷ 3.

Hence, the centroid of the whole polygon is, assuming uniform density,

                1       N-1
centroid = ——————————    ∑  ( centroid-of-triangle-i × area-of-triangle-i )
           total-area   i=1

(For dimensions ≥ 4D, the area needs to be computed with A_i = ½ |v_i−v₀| |v_i+1−v₀| sin θ_i, where cos θ_i = (v_i−v₀) · (v_i+1−v₀). )

If it's a planar surface, you can transform to a coordinate system local to the plane, calculate the centroid using the formulas you presented, and then transform back to get its coordinates in 3D space.