0:18

What other differential elements can we integrate to solve problems?

Â In this lesson,

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

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

Â computing the definite integral.

Â First determine the appropriate differential element

Â dU and then integrate to compute U.

Â And we've done this in several contexts,

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

Â The question remains what else does this method apply to?

Â There are many, many things.

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

Â 1:17

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

Â along the pen and consider slicing it into thin pieces.

Â And compute the mass of h.

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

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

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

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

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

Â that linear density, times dx, that length element.

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

Â 2:18

Let's consider a bigger problem.

Â What is the mass of the earth?

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

Â the distance to the center of the earth.

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

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

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

Â It would be measured in, say,

Â grams per cubic centimeter or something of that form.

Â 3:01

In this case, what is the mass element?

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

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

Â But it would not be times dR.

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

Â since this is volumetric density.

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

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

Â 4 pi r squared times the infinitesimal thickness dr.

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

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

Â This would be measured, of course, as r

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

Â 4:10

Let's switch to a different topic,

Â that of torque, which you've certainly experienced.

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

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

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

Â the distance from that force to your shoulder.

Â The torque is equal to the force,

Â really the perpendicular force perpendicular to your arm.

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

Â your shoulder.

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

Â What happens if we have a weight that is variable

Â that is distributed across your entire arm?

Â Maybe we would specify this by some linear mass density

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

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

Â 5:21

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

Â The amount of force applied at x.

Â What is that force element?

Â Well, force is mass times acceleration.

Â The acceleration is g, the gravitation constant.

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

Â 5:48

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

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

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

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

Â Now sometimes we'll collapse g and

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

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

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

Â 6:29

Let's think a little bit more about force.

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

Â Now, if we're given a weight density for

Â that fluid, let us call this rho as well.

Â Then if we set up a coordinate system, x,

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

Â then capital P Is going to represent the pressure.

Â What do we mean by pressure?

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

Â For the moment,

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

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

Â You can see that that's reasonable.

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

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

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

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

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

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

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

Â that leads us to the differential formulation

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

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

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

Â this allows us to integrate to obtain the force.

Â Force is the integral rho times x times dA.

Â 8:30

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

Â Compute the net force on an end cap.

Â The full radius r cylindrical tank.

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

Â What happens at the end?

Â Well the pressure is increasing as you go down,

Â but the area element is changing as well.

Â 9:01

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

Â We'll be integrating with respect to

Â dx obtaining a horizontal strip is an area element.

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

Â where u is equal to x minus R.

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

Â but du will be equal to dx.

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

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

Â we see that the area element is twice square root

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

Â The force element is the pressure times the area element.

Â Recall that pressure is rho,

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

Â 10:02

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

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

Â To obtain the force,

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

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

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

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

Â integrals distributing the multiplication over the addition and noticing that

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

Â Therefore, one of these integrals goes away.

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

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

Â 11:06

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

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

Â yielding in that force of rho, pi R cubed.

Â Let's turn to a financial example.

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

Â We can answer that if we reverse the question,

Â because if we assume a fixed interest rate of r and

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

Â differential equations that money grows exponentially,

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

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

Â we take money in the future and

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

Â Then, we can argue that a certain amount of money

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

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

Â I(t), that depends on time given.

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

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

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

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

Â And so integrating this, we get the present value.

Â 13:02

This is simply a small collection of the many possible examples

Â of computing elements in order to integrate and solve problems.

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

Â But we've not covered everything.

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

Â What one really needs to know is the procedure

Â by which we solve these problems using integrals.

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

Â solved through integrals, that of computing averages.

Â