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February 24th, 2017, 02:44 PM   #1
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Calculus 2 : Work Required to Pump Tank of Water

A water tank in the shape of a hemispherical bowl of radius 4 m is filled with water to a depth of 2 m. How much work is required to pump all the water over the top of the tank?
(The density of the water is 1000 kg/m3. Assume g = 9.8 m/s2.)

The answer is 352800π but I do not understand how the book came to this answer. Please I would like an explanation
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February 24th, 2017, 07:30 PM   #2
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consider a disk of water at height $h$ of infinitesimal thickness $dh$

The radius of the disk is $r=\sqrt{16-(4-h)^2}$ as shown in the picture.

Thus the volume of the disk is

$dV = \pi (16-(4-h)^2~dh = \pi (8h - h^2)~dh$

That disk of water has to be lifted from height $h$ to $4~m$

The work done in doing this is

$dW = g (4-h)~dm = \rho_{water} g (4-h)~ dV = \rho_{water} \pi g (4-h)(8h-h^2)~dh$

the total work is

$\displaystyle{\int_0^2}~ = \rho_{water} \pi g (4-h)(8h-h^2)~dh = 36 \pi g \rho_{water} = 352800 \pi$
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February 24th, 2017, 08:24 PM   #3
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Note that the person who replied on another website tried to use the same method, but set up the integral incorrectly.
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