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Welcome, to Calculus. I'm professor Grist.

We're about to begin Lecture 40 Bonus Material.

In an earlier lesson on Volumes of Revolution, we saw an example where the

volume seemed to be equal to the area of the cross sectional shape times what we

called the circumference of the center. That is the distance of the middle of

that disc traveled around the axis. Now, we have a little bit more expressive

language to say what we mean. This is, of course, is not the center of

the disc, but the centroid. I wonder if that holds in any more

generality. Well, indeed it does.

And the content of Pappus' theorem states that the volume of an object of rotation

is equal to the cross-sectional area times the distance that the centroid

travels. So, for example, if we took a more

interesting shape and rotated it about a non intersecting axis.

Then, computing its volume would not be so difficult.

Let's look at this specific example where we take an object that is formed by a

cutting out quarter circles from a square.

That is your cross-sectional area and then rotating that about an axis that is

a distance, capital R away from the middle.

Now, computing the centroid of this object is going to be pretty simple as is

computing the area. Let's assume that these quarter circles

all have radius, little a. Therefore the square that circumscribes

them has side length 2a. In this case, by symmetry, we know

exactly where the centroid is. It's right in the middle.

And so, the volume is going to be equal to 2piR.

That's the distance that the centroid travels about the axis times this cross

sectional area. Well, what's that, it's the area of the

square minus the area of these four quarter circles...

That's 4 minus pi times A squared. That is much simpler than setting up and

solving the associated integrals. What happens if instead of a solid volume

we take a curve and rotate that about an axis?

Is there anything we can say about the surface area of that surface of

revolution? Well, indeed there is.

Pappus' theorem works in this case as well.

The surface area is equal to the length of the curve that you're rotating times

the distance. That its centroid travels.

If we look at the specific example of rotating a semi-circle of radius r about

an axis to obtain a sphere. Then we'll see that in this case, as

indeed, in many cases involving curves in the plane.

The centroid is not located on the curve, but rather, at a point that is in the

plane, but not on the curve. In this case, because of symmetry, we

know that Y bar is equal to 0. But what is X bar?

Well, if we knew it. Then we could compute the surface area as

2 pi times x bar, that's the distance traveled by the centroid, times L, the

length of the curve. Now, it's interesting to note that in

this simple example, since we already know the surface area, and we know the

length of the curve. We could determine x bar in this way, but

let's do an example of computing this x bar from the definition.

This is the integral of x dl divided by the integral of 1 dl using the arc length

element. Of course, since the length is pi times

R. We know the denominator.

What about the numerator? Well, we have to integrate x dL.

That dL is somewhat complicated. It's the square root of 1 plus x squared

over a quantity R squared minus x squared dx.

That does not reflect a pleasant integral to do.

So, let's switch to a different coordinate system.

Let's think in terms of polar coordinates.

At any particular angle theta, the arc length element is R times d theta.

That's going to be a bit easier to work with.

In this case, theta is going from negative pi over 2 to pi over 2.

And we have to integrate x times dL. That is, x times Rd theta.

Now, the Rs cancel and we're left with the integral of xd theta in polar

coordinates. X equals R times cosine theta, and now

that's an integral that we can easily do. I'll leave it to you to check that after

dividing by pi out front and evaluating the integral we get 2R over pi.

Let's check that we didn't make any mistakes by plugging that in to Pappus'

formula for the surface area. When we do so, we get some cancellation

and obtain the familiar result of 4 pi R squared.

And you can imagine, how useful this would be, in the context of a more

interesting, looking curve rotated about an axis.

Centroids have a habit of cropping up in all sorts of computations.

For example, when we looked at the force of a fluid on the end cap of a tank, a

cylindrical tank of radius R. Then we computed that, that force was

equal to row the weight density times pi R cubed.

there's another way to interpret this. In general for a vertical submerged plate

that are fluid is pushing on from the side the force, the net force is equal to

row times the area A of that plate. Times x bar, that is the depth from the

top of the fluid to the centroid of the plate.

Let's check and see that, that indeed happened in our example.

The centroid of that disc is, of course, right in the middle and that is a

distance of capital R, the radius from the top.

So that, in this case, we would get row times R, the distance of that centroid

times the area of the disc. Pi R squared, indeed, that matches up

with what our more difficult integral computation gave.

In general, there are many examples of physical problems where knowing these

centroid or more generality, the center of mass is helpful.

Keep your eyes open and see if you can recognize some other examples, where

knowing a centroid will help.