0:18

What other differential elements can we integrate to solve problems?

In this lesson,

we'll focus on several examples that range from physics to finance.

By now, I'm sure you've discerned our procedure for

computing the definite integral.

First determine the appropriate differential element

dU and then integrate to compute U.

And we've done this in several contexts,

area, volume, surface area, length, and work.

The question remains what else does this method apply to?

There are many, many things.

In this lesson, we'll take a look at a few.

1:17

Or, using calculus, we could set up a coordinate x

along the pen and consider slicing it into thin pieces.

And compute the mass of h.

This would give us what we might call the linear density row of this object.

The linear density is the rate of change of the mass as we change the coordinate.

Now for an actual pen, this might be a discontinuous function.

However, given that linear density rho is a function of x,

we could compute the mass element dM as rho of x,

that linear density, times dx, that length element.

And, of course, the mass is the integral of the mass element.

2:18

Let's consider a bigger problem.

What is the mass of the earth?

We're going to consider the density as a function of the radial coordinate r,

the distance to the center of the earth.

In this case, we know that the density changes as you move from the inner

core to the outer core to the mantle, then the crust and then the atmosphere.

This, however, is not a linear density, but a volumetric density.

It would be measured in, say,

grams per cubic centimeter or something of that form.

3:01

In this case, what is the mass element?

Well, if we choose a very small segment of the radial coordinate, that is dR.

Then we would have the volumetric density at that R value.

But it would not be times dR.

Rather, it would be times dV, the volume element,

since this is volumetric density.

And so the question is, what is the volume element?

Well, that is the surface area of the sphere of radius r,

4 pi r squared times the infinitesimal thickness dr.

This gives us the mass as the integral of the mass element.

That is the integral of 4 pi r squared times rho of r dr.

This would be measured, of course, as r

goes from 0 to 6,400 if we were doing this in terms of kilometers.

4:10

Let's switch to a different topic,

that of torque, which you've certainly experienced.

Even if you've never learned formally, if you extend your arm and

apply a force or a weight to it, you feel torque about your shoulder.

The magnitude of that torque depends on the magnitude of the force and

the distance from that force to your shoulder.

The torque is equal to the force,

really the perpendicular force perpendicular to your arm.

That force times the distance between where the force is applied and

your shoulder.

Now, that is true for a singular force of force applied at one point.

What happens if we have a weight that is variable

that is distributed across your entire arm?

Maybe we would specify this by some linear mass density

function rho, like we did in the case of the pen.

Then let's compute the torque in terms of the torque element dT.

5:21

dT is going to be the distance x times the force element, dF.

The amount of force applied at x.

What is that force element?

Well, force is mass times acceleration.

The acceleration is g, the gravitation constant.

The mass, however, is a mass element, dM.

5:48

And dM is computed as we have done before in the case

of the pen as the linear density, rho of x, times dx.

Putting all of these together, we can obtain an integral for

the torque as x times g times rho of x, dx integrated over the arm.

Now sometimes we'll collapse g and

rho of x together to give something called a weight density.

Now you don't have to memorize this formula, but

you have to know how to reason in terms of elements in this manner.

6:29

Let's think a little bit more about force.

And in particular, the force of a fluid on a tank.

Now, if we're given a weight density for

that fluid, let us call this rho as well.

Then if we set up a coordinate system, x,

that represents the depth from the top of the fluid in the tank,

then capital P Is going to represent the pressure.

What do we mean by pressure?

Well, there's a little bit of physics that goes on here.

For the moment,

you should think that pressure depends on the density of the fluid and the depth.

Now, I can argue that that pressure is really the product of these two.

You can see that that's reasonable.

If you think about how much pressure is at the bottom of the tank versus at the top,

you can see that by, say, poking some holes in the side of a tank and

seeing what happens to the water that is pushed out by the pressure.

All right, well, given that as a backdrop, how can we compute the force

that that water or fluid exerts on the side of the tank.

Well, that's going to depend on all these terms.

Pressure can be thought of as force per unit area, and

that leads us to the differential formulation

that the force element, dF is P, the pressure,

times the area element, dA, on the side of the tank.

We can expand that out further as rho times x times dA and

this allows us to integrate to obtain the force.

Force is the integral rho times x times dA.

8:30

Let's put this to use in an explicit problem.

Compute the net force on an end cap.

The full radius r cylindrical tank.

So consider a cylinder on its side, fill it with fluid.

What happens at the end?

Well the pressure is increasing as you go down,

but the area element is changing as well.

9:01

x is going to be our coordinate measured from the top of the tank.

We'll be integrating with respect to

dx obtaining a horizontal strip is an area element.

It's going to be easier if we change coordinates to u,

where u is equal to x minus R.

In this case, u is going to be centered at the middle of the disk,

but du will be equal to dx.

In this case, the area element is given by what?

Well, with a little bit of help from a right triangle,

we see that the area element is twice square root

of R squared- u squared times the thickness, du.

The force element is the pressure times the area element.

Recall that pressure is rho,

the weight density, times x, the distance from the top.

10:02

Now x, being u + R, and dA being twice root

R squared- u squared, du, gives us our entire force element.

To obtain the force,

we integrate the force element as u goes from negative R to R.

The resulting integral allows us to pull out the constant weight density

rho as we've seen in similar integrals in the past.

What is going to work best is to split this up into two

integrals distributing the multiplication over the addition and noticing that

one of these integrals has an odd integrand integrated from negative R to R.

Therefore, one of these integrals goes away.

And we're left with the second rho times R times the integral

from negative R to R of 2 root R squared- u squared du.

11:06

That is, in fact, the area element for the disc,

and so we obtain rho times R times the area of pi R squared,

yielding in that force of rho, pi R cubed.

Let's turn to a financial example.

The concept of present value, namely, how much is tomorrow's money worth today?

We can answer that if we reverse the question,

because if we assume a fixed interest rate of r and

a continuously compounded investment, we know from our work on simple

differential equations that money grows exponentially,

with an exponent, depending on r, the interest rate.

And so, if, instead of considering how money today grows,

we take money in the future and

say how much money would we need today to invest to get that?

Then, we can argue that a certain amount of money

at time t is worth that amount times e to the -rt right now.

And that allows us to discuss the present value of an income stream,

I(t), that depends on time given.

That rate of income then, what is the present value?

How much money would that be worth if you had it all today?

Well, we can express this in terms of the present value element dPV.

This is going to be e to the negative rt times I(t)dt.

And so integrating this, we get the present value.

13:02

This is simply a small collection of the many possible examples

of computing elements in order to integrate and solve problems.

We've gone through quite a number of different examples of applications.

But we've not covered everything.

There's no need to learn every possible application that's out there.

What one really needs to know is the procedure

by which we solve these problems using integrals.

In our next lesson, we're going to turn to a different class of problems that can be

solved through integrals, that of computing averages.