A circular pool has a radius of 12 ft, the sides are 5 ft high and the depth of the water is 4 ft. How much work is required to pump all of the water out over the side? (water weighs 62.5 lb/ft^3)

Question

anonymous · Answer

A = pi * r^2 = 144 * pi 

Say the thickness is dz

So

dV = A dz = 144 * pi dz
dF = (rho) dV = 144 * pi * (rho) dz
dW = d F z = 144 * pi * (rho) z dz

The upper layer of water must be lifted 1 ft over the top rim of the pool, whereas the layer at the bottom must be lifted 5 ft. Therefore, the work required to empty the
swimming pool of water is
$$W = \int\limits_{1}^{5} dW = \int\limits_{1}^{5} 144(\rho)(\pi)zdz$$
this is rho ---------^ the symbol that looks like a p

Solve integral and you'll get 1728(rho)(pi)
plug in density (62.5) for rho

and you'll get the answer
around 339,292 ft-lb.

anonymous · Answer

cheers ^_^

anonymous · Answer

Thanks. Wish I would have done that with the exam. I had all the components I just couldnt put it back together... fun and thanks!

anonymous · Answer

what is rho?

anonymous · Answer

rho is this :
$$\rho$$
it's a greek letter
usually meaning density