How do I interpret this integral answer?

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SUMMARY

The discussion focuses on calculating the work required to pump water out of a swimming pool with dimensions 30ft by 16ft by 10ft. The first part of the problem yields a work value of 1,497,600 ft-lbs when the pump is at the water level. The second part involves integrating the height of the pool, resulting in a work calculation of 2,995,200 ft-lbs when the pump is at the bottom. The integration of the constant height (10) represents the total distance all water must be lifted, clarifying why more work is needed when the pump is positioned at the bottom.

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Homework Statement



There is a swimming pool with length 30ft, width 16ft, and height 10ft. How much work is done to pump all of the water out of the pool assuming that the density of water is 62.4 and that the pump of the hose floats at the level of water at all time.

Then find the answer if the pump hose lies at the bottom of the pool.

Homework Equations



Work=Force*Distance

The Attempt at a Solution



The first part is straight forward enough: 16*30*62.4*(100/2)=1497600 ft-lbs

But I'm having trouble understanding the solution to the second part:

16*30*62.4 \int^{10}_{0} 10dh

Can someone explain to me why you would integrate 10? Thanks for the help.
 
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where did you get the 10 inside the integral?
 
that's the answer in my book and I'm trying to figure out why it is. So I am basically asking the same question as you.
 
Ohh i gotcha now.

well, if we multiply 30, 16, and 10, (the dimensions of the pool), we get a volume in cubic feet. this is achieved by pulling the 10 out of the integral, since its a constant.
this is a Volume. Now, what units are the density? this is probably the key.

also, think about the integral. if you pull the 10 out like i said above, what does it become? what does it represent?
 
density is 62.4 lbs/ft^3. so if I pull the ten out of the integral I get volume times density and then times the antiderivative of dh evaluated from zero to 10 which is just 10. that would then give me volume times density times ten: 16*30*10*62.4*10=2995200. Tis doesn't make sense though since it means there's more work required than for the first part.
 
That's right. If the pump sits at the top of the water as it pumps, some water has to go barely any distance, since its already closer to the top of the water as its pumped out. If the pump is at the bottom however, all the water will have to go the whole depth of the pool to get pumped out, hence more work.
 
But wouldn't the water at the bottom of the pool have to travel less distance to get to the pump if its at the bottom than the water at the top?
 
I think you're missing what the work in this case means. The work is the energy the pump puts into move the water out of the pool, not the energy required to get the water to the pump.
 
I got it. By the pump being at the bottom, I was thinking that it was essentially draining from the bottom. In the case it would be the work done by gravity to empty it. But I got it now thanks.
 

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