Work, of filling a water tower:
Section 5.4
Andy Long

Suppose the water tower is spherical, located 20 meters up, and is of radius 2 meters. How much work against gravity is required to fill the tank, ignoring losses, the column of water in the pipe up to the tower, etc. Assume that the water starts from the ground. The density of water is ρ=1000 kg/m^3.

W=∫dW

Work is force times displacement. What’s a dW look like?

Newton says that F=ma

a=g=9.81m/s/s is the acceleration due to gravity. 

Consider a circular slice of water at a height of y, where y runs from 18 to 22 meters. 

dW	= dF * y
	= dM * g * y
	= ρ * dV * g * y
	= ρ * A(y) * dy * g * y

where A(y) is the cross-sectional area of the circle. The equation of the circle is x^2+(y-20)^2=2^2, so x=Sqrt[4-(y-20)^2]. So, finally, 

dW	= ρ * ℼ * ( 4-(y-20)^2)* dy * g * y

So the method of Disks gives:

ρ=1000
g=9.81
work=Integrate[ρ*Pi*(4-(y-20)^2)*g*y,{y,18,22}]

Matt noticed that this is the same of doing the work of moving all the mass to 20 meters: he made use of symmetry! Good work, Matt.

mass=4/3 Pi (2)^3 * ρ;
force=mass* g;
displacement= 20.0;
work=force * displacement

For every bit of the mass moved to 
20+h meters, 
there is an equal mass moved to 
20-h meters. 
Since we’re simply adding up all the work, the h’s cancel, and it’s the same as if we move everything to 20 meters!