See Chapter 9 (70-711 in 2011 NEC) Notes (2), (4), (5), and (9).
Note (2) says if it is sleeve (not part of conduit system) then table 1 does not apply.
Note (4) says if it is a nipple (part of a complete conduit system (and 24" or less) then you can fill it to 60%.
Note (5) says if it is other than a conductor listed in Chapter 9, Table 5, you use the actual dimension.
Note (9) says if it is elliptical, use the major (largest) dimension as its diameter.
So for each cable, you were supplied its major diameter. Just as a guess let's say 14/2 has a major diameter of 1/2 inch, 12/2 5/8 inch, and 10/2 3/4 inch.
Since diameter = 2*radius.
The calculated area of 14/2 would be PI*R(squared) or 3.14*(1/4*1/4) or 3.14*0.0625 or 0.196
The calculated area of 12/2 would be PI*R(squared) or 3.14*(5/16*5/16) or 3.14*0.0977 or 0.307.
The calculated area of 10/2 would be PI*R(squared) or 3.14*(3/8*3/8) or 3.14*0.141 or 0.442
3 14/2 3*0.196 or 0.588
2 12/2 2*0.307 or 0.614
1 10/2 1*0.442 or 0.442
total is 1.644 sq inches
If it is a sleeve you can fill it to 100% I guess, so according to table 4 PVC 40 the smallest conduit with that or greater diameter under the 100% column is 1-1/2 inch
If it is a nipple you can fill it to 60% so according to table 4 PVC 40 the smallest conduit with that or greater diameter under the 60% coluumn is 2 inch




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