I think it is just area, but as you said before, you need to discount the edges of the 135 roll that don't consume developer because they aren't exposed.
If we suppose
* the roll film is 36 frames, each 36x24 mm and with an unexposed 2mm gap between each frame and the next;
* the perforations are 5mm along each edge, of which about half is the actual perforation hole: but they are so little exposed we may neglect the edge except at the leader.
* the length of film developed includes say 10 cm of leader, of which 5 cm is exposed. This does include exposed edges.
* the length of film developed includes 5 cm of tail, but this is completely unexposed.
Then the total area of the film, including that of the perforations, leader and tail, is
( no. of frames * (36mm of frame + 2mm of gap) + leader + tail ) * 35mm of width
A = ( 36 * ( .036 + .002) + .1 + .05 ) * .035 = .05313 square metre.
Of this, the area actually demanding anything of the developer is
( no. of frames * (frame length x height) + ( 50mm of leader * (frame height + 2 * 2.5mm for exposed edges)))
B = ( 36 + ( .036 * 024 ) + ( .05 * ( .024 + 2 * .0025) ) ) = .032554 square metre
For the sheet, suppose it is exactly 4 x 5 inches, and that 2 mm along each edge is unexposed because of the film holder.
Then its total area is C = 4 * 5 * 0.0254^2 = 0.0129032 square metre
And its exposed area is D = ( 4 * .0254 -.002 ) * ( 5 * .0254 - .002 ) = 0.01245 square metre
So by total area, the roll area is A/C = 4.11 times that of the sheet (or the sheet is 0.24 rolls)
But by exposed area it is B/D = 2.61 times that of the sheet (or a sheet is 0.38 rolls)
I know I could have extracted official values for some of my assumptions from the area table, but this does well enough to show that the unexposed edges make a big difference. There's no quarrel between area and capacity.