Last time we looked at three different application of integrals:
- A second type of volume integral
- A mass problem
- A work problem
In any case, I encourage you to think about starting with an integral,
of the general form
$A=\int{dA}$
But in each case, we have a different type of quantity that we're after:
- A volume integral: $V=\int{dV}$
- A mass problem: $M=\int{dM}$
- A work problem: $W=\int{dW}$
We ended last time on the work required to fill a spherical water tank with
water:
- The sphere is of radius 2 meters.
- The center of the sphere is 20 meters above the surface of the source of water.
- Water has a density of $1000kg/m^3$
What is the work required to lift the water into the tower?
- Work is technically defined as force times displacement: W=F*d
- In this case, we have the force of gravity lifting
layers of water up to the tower: dW=F*dh
Matt made a great observation at the end: that we
could find the work required to move the equal mass of water to 20 meters, and
we get the same answer!
As an example of using integrals to do practical or interesting things (such as
find the time to travel to exit 16), let me give you an example from my own
life. Sometimes it just provides a context for a way to think about a problem....
I was thinking this morning about my desire to help heal the damage that humans
have done to the Earth. But I do damage myself, even as I try to undo damage.
So I started this way:
Impact=Healing-Damage
But this is really a function of time, so that, at any moment,
Impact(t)=Healing(t)-Damage(t)
Sometimes we'll do short term damage, in order to make a long-term gain. So
it's really this function over my lifetime that is of interest (and perhaps
only my "conscious" lifetime, starting from some $t_0$ and ending at $t_f$:
$Impact=\frac{1}{t_f-t_0}\int_{t_0}^{t_f}(healing(t)-damage(t))dt$