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Welcome, to Calculus. I'm professor Grist.

Â We're about to begin Lecture 40 Bonus Material.

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

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

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

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

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

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

Â generality. Well, indeed it does.

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

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

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

Â interesting shape and rotated it about a non intersecting axis.

Â Then, computing its volume would not be so difficult.

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

Â cutting out quarter circles from a square.

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

Â a distance, capital R away from the middle.

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

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

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

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

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

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

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

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

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

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

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

Â we take a curve and rotate that about an axis?

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

Â revolution? Well, indeed there is.

Â Pappus' theorem works in this case as well.

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

Â the distance. That its centroid travels.

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

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

Â indeed, in many cases involving curves in the plane.

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

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

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

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

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

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

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

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

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

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

Â element. Of course, since the length is pi times

Â R. We know the denominator.

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

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

Â over a quantity R squared minus x squared dx.

Â That does not reflect a pleasant integral to do.

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

Â Let's think in terms of polar coordinates.

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

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

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

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

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

Â coordinates. X equals R times cosine theta, and now

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

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

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

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

Â and obtain the familiar result of 4 pi R squared.

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

Â interesting, looking curve rotated about an axis.

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

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

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

Â equal to row the weight density times pi R cubed.

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

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

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

Â top of the fluid to the centroid of the plate.

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

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

Â distance of capital R, the radius from the top.

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

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

Â with what our more difficult integral computation gave.

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

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

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

Â knowing a centroid will help.

Â