In my limited experience with math of this caliber I see no spot for the cord of the wing and the amount of paint added to it.
I would note, if it is not clear, that each "part" can be defined the way it makes sense. You *could* do it with large assemblies like the wing, the tail, the fuselage, but, figure you have 40 wing ribs, that's 40 entries, each with a weight and a position (in 3 dimensions), you have 4 engine screws, that's 4 more, you have 4 blind nuts, that's 4 more, etc. Large parts would be subdivided into a bunch of sections, etc. That's how your summation gets hundreds of elements.
You break it down into as many pieces as necessary to get the accuracy you want. For instance, for the finish, you more-or-less know how many ounces the finish will be, say, 10 ounces when you consider everything from the tissue on out, whatever you think it is. You can pretty easily calculate the surface area. Divide it out, and you have the areal density in ounces/square inch. Then break the surface up into as many pieces as you want, say, 4 square inches, use the areal density to calculate the weight, determine the distance from the reference position to the center of the area , and you have the mass and position of each piece. Do that for each piece, and you have all the elements of the summation.
Example with made up, but reasonable, numbers. Say the total paint area is about 1800 square inches (600 for the top of the wing, 600 for the bottom, 150 for the top of the stab, 150 for the bottom, etc). Figure that including the tissue or carbon, it's about 10 ounces total. That means it weighs about 0.0056 ounces/square inch. Divide the entire surface into 5 square inch chunks, that's .0278 ounces for each chunk, and there are 360 of them, each with a center position from some reference. Thats 360 separate entries, and 360 elements of X*m in your summation.
Same thing for any large component, the fuselage sides might be 48" long and 6 " high, I might take that in 8 6"x6" long pieces. Figure 5 lbs/cu ft, 6"x6"x1/8" is 0.0026 cubic feet, so each section weighs ~.21 ounces, there are 8 of them, so each fuse side weighs 1.6ish ounces. Each peice weighs about the same but has a different position, sum them all up, get the CG.
Depending on how you do it, there may be 100 total items, or 1000, but the basic idea is the same. The sum of all the masses it the total mass, that's the denominator, the mass*position products all added together is the numerator, divide one by the other and the result is a distance from the "reference" position to the CG.
For purposes of computing the inertia, you can probably reasonably ignore the inertia about the center of each "chunk" and assume that each part is a point mass.
Brett