For reference and simplicity, first let us calculate the area of this dish as though it were a flat circle. 

The formula for the area of a circle is pi (3.1416) x r (radius) squared.
Or, r = 362 (1/2 of 724), so the area of a circle 724 inches wide, would be 2858.94 square feet or 265.60 sq. meters (0.09290304 sq. m./sq. ft.).
3.1416 x 131044 sq. in. = 411687.8304 sq. in.
411687.8304 sq. in. ÷  144 sq. in./sq.ft.

The area of the paraboloid is going to be larger than this number, but not by much. This number (265.60)  gives us a reference that was easy to calculate, so when we get into calculating the actual area of the paraboloid and we end up with a number significantly larger or smaller, we will know we made a mistake.

Let us break the formula down into parts.

D = 724 in.

D² = 524176 sq. in.

H = 143.38 in.

8H = 1147.04 in.

p = 456.98 (524176 ÷ 1147.04)

p² = 208832.044

p³ =  95432369.71

2pi ÷ 3p = 0.004583 (6.2832 / 1370.94)

D² ÷ 4 (524176 / 4) = 131044 + p² (208832.044) = 339876.043725

339876.043725 cubed = 39261027634527100.00 (ouch, big #, but we're ok)

The sq. root of 39261027634527100.00 is 198143956.85

198143956.85 minus 95432369.71 ( p³ ) = 102711587.14

102711587.14 x 0.004583 = 470739.35 square inches or 3269.02 sq. ft. (303.70 sq. meters).

303.70 sq. meters looks about right comparing it to the number we got for the area of the 724" circle (remember 265.60 sq. meters?).

Back Ground
The formula for finding dry film coverage per gallon per mil is based on volume and the non volatile contents (solids) of that volume. In paint chemistry terminology, solids is defined as all components that are non volatile, even those components that are in a liquid form and will eventually be part of the dry paint film.

If the contents of a gallon are 100% solids, then that gallon could be spread, 0.001 of an inch thick, over 1604.1664 square feet.

Formula
One gallon = 231 cubic inches.
231 cubic inches ÷ 0.001 inch = 231000 sq. inches.
231000 sq. inches ÷ 144 sq. ft. / inch = 1604.1666

If the calculated VOLUME solids in a gallon of coating is 46% (as it is  for Goldstone #6, see PDF ), then the area that the dried film would cover to a thickness of 0.001 inch (one mil) is 738 sq. ft. (1604.1664 x .46 = 737.916544 sq. ft.).

At the required DFT (dried film thickness) application of 1.5 mil, the theoretical coverage is 450 square feet, or:
737.916544 ÷ 1.5 = 490.944 square feet. 
The 450 sq. ft number contains a 9% fudge factor for loss.

The 18 meter dish has a surface area of 3269.02 sq. ft.
Therefore; you will need 8 gallons to cover the reflective surface of this 18 meter dish.

3269.02

 sq. ft ÷ 450sq. ft./gallon = 7.26 or 8 gallons.