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.
