a circular swimming pool has diameter of 24ft, 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 per cubic foot

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a circular swimming pool has diameter of 24ft, 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 per cubic foot

Mathematics
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dW = ρgAx dx
..... ρ∙g given as 62.5 lb/ft³ ..... A = πr²
so we are doing Force* distance = work. we are going to move cross sectional slabs of water with radius 12, pi*12^2, each slab has a different height to cover. for example the slab on the bottom has to cover 4 feet and then another foot. this is the x part. and all the slabs need to cover one foot so x + 1

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the 1 foot extra on top
so 62.5 * pi *12^2 integral ( x+1) from 0 to 4 is your work
whats important is that the force is given as pounds. dont worry about the pg part. im not so sure about that
well g is gravitational constant, and weight is mass * g , and you have density in there as well. but it all works out
you there?
yes just trying to make sense of it all. im still not understanding the x+1 since the bottom has to clear 5 ft not 1 ft
well it clears 4 feet, then +1 foot for the top . the next slab say has to clear 3 feet then +1 foot for top, so on and so forth. the toppest slab has to clear just +1 foot. but each slab has that extra +1 foot to clear
ah ok i see, thank you for clarifying

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