If I understand the question, Michael is asking how to calculate the cross sectional area of a _cable_, including several conductors, their insulation, and the over-all sheath, for purpose of calculating conduit fill.
For conduit fill purposes, the requirement for cables is that you treat them as a single large conductor. So if you had a cable with 5 conductors and a diameter of 3/4 inch, for purpose of conduit fill you would treat it as a single conductor 0.75 inch in diameter. As a _single_ conductor, it could fill 53% of the conduit, rather than using the normal 40% fill limit for 'more than 2' conductors.
There is an additional requirement for 'cables that have an elliptical cross section'. This requirement is that you find the major (largest) diameter of the ellipse, calculate the area of the _circle_ with that diameter, and then use the area of the _circle_ as the cross section of the cable for purposes of conduit fill.
The formula for the area of a circle is simply A = pi*r^2 . However if you are working in 'circular mils', then the formula is even simpler. The unit 'circular mil' already takes into account the pi factor.
Given a diameter in mils (0.001 inch), the area of a circle in circular mils is simply the diameter squared. Thus if you have a cable with a major diameter of 0.307 inches, you must treat is as having a circular cross section of 307^2 = 94249 circular mils.
As an aside, the requirement in note 9 applies to cables with an _elliptical_ cross section. An ellipse is a very specific shape, and most cables are _not_ ellipses...IMHO the requirement should probably be applied to any non-circular cable
-Jon