Math Homework help needed

Question

A rectangular swimming pool 50 ft long, 20 ft wide, and 16 ft deep is filled with water to a depth of 15 ft. Use an integral to find the work required to pump all the water out over the top. (Take as the density of water �=62.4lb/ft3.)

Answer 1

Let

S = 50 ft × 20 ft = 1000 ft² be the cross section of the pool,

D = 16 ft be the depth of the pool,

H = 1 ft be the original distance between the water level and the edge,

ρ = 62.4 lb/ft³ be density of water

g = 32 ft/s² be gravitational acceleration.

During the pumping, consider a time when the water level is some distance h from the edge of the pool.

And consider pumping out a thin layer of surface water of thickness dh.

That layer has

volume = Sdh

mass = ρSdh

weight = gρSdh

The work necessary to lift this layer to the edge is then hgρSdh.

The work necessary to lift all the layers to the edge is the sum of the work

for the individual layers, i.e. the definite integral from H to D

∫hgρSdh

The indefinite integral

W(h) = ∫hgρSdh = gρS∫hdh = gρSh²/2

The definite integral is

W(D) - W(H) = gρSD²/2 - gρSH²/2 = gρS(D² - H²)/2

Substituting actual numbers the total work is

32×62.4×1000×(16² - 1²)/2 = 254,592,000 lb-ft

You might think that I am making an unreasonable assumption that it is

always the top layer of water being pumped out.

After all, the opening of the inlet hose might be placed at the bottom of the pool.

In that case, the pressure of the water above the inlet pushes the water up the hose,

so the trip from the bottom to the surface requires no work.

Only once the water reaches the surface, is there work required.