Acircular swimming pool has a diameter of 24ft. the sides are 5ft high and the depth of the water is 4ft. how much work is required to pump all of the water over the side. water weighs 62.5 lbs/ft3

Work done = mgh where h varies from 0 (at the bottom of the pool) to 5 ft (at the sides of the pool).

m= density*volume = 62.5*pi*24^2*d/4 = 9000πd lbs
h = 5-d (the particles at the top of the pool require less work as required to those at the bottom).
Then,
WD (E) = 9000πd*9.81*(5-d) = 88290πd(5-d) = 441450πd-88290πd^2
Integrating from 0-4 with respect to depth of pool, d
Total E = {441450π*d^2/2 - 88290π*d^3/} for d between 0 and 4 ft

E=220725π*d^2 -29430π*d^3
Substituting for d=0 ft and d=4 ft;

= {220725*π*4^2-29430π*4^3} - {220725π*0^2-29430*pi*2*0^3)
=5177596.021 J= 5.17 MJ

The answer is d) by requesting advice on science internet sites

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