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November 28, 2014

November 28, 2014

Posted by **COFFEE** on Monday, June 18, 2007 at 10:41pm.

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What is shown is just the tank (a hemisphere) with a radius of 5 ft.

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First I calculated the Volume of the hemisphere, V = (2/3)*pi*r^3

V = (2/3)*pi*125 = (250/3)*pi

Then I took the integral of: Volume*5y*dy from 0 to 5.

Which equals: ((250/3)*pi)*(5/2)y^2 evaluated at 5 and 0.

I came up with 16362.5 ft*lb.

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Am I using the wrong method?

The work required depends upon where the water is extracted. I assume you are pumping out the top.

I don't see why you claim that the energy is

Volume*5y*dy from 0 to 5

The work required is the weight of each differential slab of height dy, multiplied by the distance it must be lifted, 5 - y, integrated from 0 to 5. The area of each slab is different. It depends upon y.

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