My Math Forum Calculus 2 : Work Required to Pump Tank of Water

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 February 24th, 2017, 03:44 PM #1 Newbie   Joined: Feb 2017 From: Kennesaw, Ga Posts: 1 Thanks: 0 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
February 24th, 2017, 08:30 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, 09:24 PM #3 Global Moderator   Joined: Dec 2006 Posts: 20,095 Thanks: 1905 Note that the person who replied on another website tried to use the same method, but set up the integral incorrectly.

 Tags calc, calc2, calculus, calculus 2, pump, required, slicing, tank, water, work, work done

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