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February 24th, 2017, 02:44 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, 07:30 PM  #2 
Senior Member Joined: Sep 2015 From: USA Posts: 1,980 Thanks: 1027 
consider a disk of water at height $h$ of infinitesimal thickness $dh$ The radius of the disk is $r=\sqrt{16(4h)^2}$ as shown in the picture. Thus the volume of the disk is $dV = \pi (16(4h)^2~dh = \pi (8h  h^2)~dh$ That disk of water has to be lifted from height $h$ to $4~m$ The work done in doing this is $dW = g (4h)~dm = \rho_{water} g (4h)~ dV = \rho_{water} \pi g (4h)(8hh^2)~dh$ the total work is $\displaystyle{\int_0^2}~ = \rho_{water} \pi g (4h)(8hh^2)~dh = 36 \pi g \rho_{water} = 352800 \pi$ 
February 24th, 2017, 08:24 PM  #3 
Global Moderator Joined: Dec 2006 Posts: 19,059 Thanks: 1619 
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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calc, calc2, calculus, calculus 2, pump, required, slicing, tank, water, work, work done 
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