Example 4 A
tank in the shape of an inverted cone has a height of 15 meters and a base
radius of 4 meters and is filled with water to a depth of 12 meters. Determine the amount of work needed to pump
all of the water to the top of the tank.
Assume that the density of the water is 1000 kg/m3.
Okay, in this case we cannot just determine a force
function, that will work for us. So, we are going to need to approach this
from a different standpoint.
Let’s first set to be the lower end of the tank/cone and to be the top of the tank/cone. With this definition of our x’s
we can now see that the water in the tank will correspond to the interval .
So, let’s start off by dividing into n
subintervals each of width and let’s also let be any point from the ith subinterval where . Now, for each subinterval we will
approximate the water in the tank corresponding to that interval as a
cylinder of radius and height .
Here is a quick sketch of the tank. Note that the sketch really isn’t to scale
and we are looking at the tank from directly in front so we can see all the
various quantities that we need to work with.
The red strip in the sketch represents the “cylinder” of
water in the ith
subinterval. A quick application of
similar triangles will allow us to relate to (which we’ll need in a bit) as follows.
Okay, the mass,
of the volume of water, ,
for the ith subinterval is simply,
We know the
density of the water (it was given in the problem statement) and because we
are approximating the water in the ith
subinterval as a cylinder we can easily approximate the volume, , and hence the mass of the water in the ith subinterval.
The mass for
the ith subinterval is
To raise this
volume of water we need to overcome the force of gravity that is acting on
the volume and that is, ,
where is the gravitational acceleration. The force to raise the volume of water in
the ith subinterval is
Next, in order
to reach to the top of the tank the water in the ith subinterval will need to travel approximately to reach the top of the tank. Because the volume of the water in the ith subinterval is constant
the force needed to raise the water through any distance is also a constant
work to move the volume of water in the ith
subinterval to the top of the tank, i.e.
raise it a distance of ,
is then approximately,
The total amount of work required to raise all the water
to the top of the tank is then approximately the sum of each of the for . Or,
To get the
actual amount of work we simply need to take .
I.e. compute the following limit,
This limit of a
summation should look somewhat familiar to you. It’s probably been some time, but recalling
the definition of the definite integral
we can see that this is nothing more than the following definite integral,