Welcome to calculus. I'm Professor Ghrist. We're about to begin
lecture 38 on the elements. In the past several lessons,
we've seen examples of area elements, volume elements,
length elements, and more. 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. Let's begin with mass. And consider, what is the mass of a pen? Well a pen is composed of many parts, we could compute the mass of each and
add them together. 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. 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. 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. 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. 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. 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. 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. 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. Let's take a look at the end
of that cylinder head on. It's a radius R disc. 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. 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. 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. 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.