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.