We need to form an equation for the volume of the pyramid based on the height
Cross sectional area changes linearly with cross section of (756 ft)^2 on the bottom and 0 at 481 ft
h1 = 0, A1 = (756 ft) ^2 571536 ft^2
h2 = 481, A2 = 0
A = m*h + b
m = (0 - 571536) / (481 - 0 ) = -1188.2 ft^2 / ft
0 = -1188.2 * 481 + b
b = 571536
A = -1188.2 h + 571536
V = A*h = -1188.2 h^2 + 571536*h
Work = integral from h to hf (W * dh)
Work = integral from h to hf (density * g * V * dh)
Work = int from h to hf ( density * g * (-1188.2 h^2 + 571536*h) dh )
Work = density * g * (-369 h^3 + 285768 h^2) from h to hf
Now here's the tricky part with english units, mass is a slug, not a pound... weight = mass* g = slug * g = pound, since the denisty is already given in lbs/ft^3 you can ignore the g term... The correct term for g in english units is 32.2 ft/s^2.
Work = 3.75 * 10^12 ft*lb
FT*LB is a correct unit for work in the english system. If you want to covert it to BTU's or Joules, you can do so yourself.