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Welcome to calculus.

Â I'm professor Grist.

Â We're about to begin Lecture 40 on Centroids and Centers.

Â Averages appear in many guises and shapes and sizes.

Â In this lesson, we're going to look at how to compute average positions.

Â This will lead to the notion of centroids, and centers of mass.

Â 0:24

How do you compute an average location?

Â Well this starts off simple enough.

Â If I say, what's the average of two points?

Â Well, one draws the line segment and picks the point in the middle.

Â You could probably use some basic geometry to determine the average

Â of three locations, or the average of four locations.

Â But what would you do if asked to compute the average of an entire region?

Â This requires thinking in terms of calculus.

Â Breaking that region up into infinitesimal elements and

Â then computing an average, and that's exactly what we'll do in defining

Â the centroid is the average location in a domain, let's call that D.

Â In this case, if we set up x and y coordinates,

Â then we would characterize the centroid in terms of it's coordinates,

Â x r and y bar.

Â That notation is chosen to help you remember the definition.

Â X bar is the average x coordinates over the domain.

Â That is the integral of x over d divided by the integral of one over d.

Â Y bar correspondingly is the integral of y over D

Â divided by the integral of 1 over D.

Â In higher dimensional settings, I bet you can guess what the formula is.

Â We're going to use the perspectives from the bonus materials

Â in lecture 31 to compute these integrals.

Â Dividing the region not into strips, but

Â rather into infinitesimal rectangles of dimensions dx and

Â dy and then averaging over those.

Â The notation that is most useful to us is that of

Â a double integral, or an iterated integral.

Â The denominator in both of these cases, being the integral of one over the domain,

Â is really the integral with respect to the area element.

Â If you integrate dA, what do you get?

Â Well you get of course A, so one way to write these formula for

Â the centroid is as 1 over A times the integral over d

Â of x dA, or of y dA respectively.

Â 3:17

And in this case, the area element is going to be an infinitesimal rectangle

Â of dimensions, Dx and Dy.

Â Now in this case, if we look at the formula for x bar,

Â that is the double integral of x with respect to area over the area.

Â Then what do we get?

Â Well, dA is really the area

Â of this infinitesimal rectangle that is, dx times dy.

Â Now, we're gonna write that in the opposite order, dy times dx.

Â The reason for that is we're gonna perform these integrals one at a time.

Â Double integrals looks scary, they're really not.

Â You just do it one at a time.

Â The trick is determining the limits of integration.

Â When we integrate with respect to y, the lower limit Is g(x),

Â the upper limit is f(x).

Â When we integrate with respect to x, the limits are x goes from a to b.

Â 4:20

Now, when we do so, that's really integrating out to give a vertical

Â strip when you integrate with respect to y and then sweeping from left to right.

Â Let's perform those integrals.

Â It's not so bad.

Â What is the integral of x dy?

Â Well from the perspective of y, x is just a constant so we can move that outside.

Â And now all we have to do on the top and the bottom is integrate dy.

Â That of course gives us y.

Â We have to evaluate that from g(x) to f(x) when we

Â substitute that in, we get, simply, f(x)- g(x).

Â Now that's just the first integral that we've done.

Â We still need to do that outer integral with respect to x.

Â And so we obtain a formula for xr as shown.

Â It looks a little complicated, but notice that on the bottom, what we've really got

Â is the area between those two curves, our familiar formula.

Â 5:29

So we can rewrite this as x bar equals 1 over A times the integral as x goes from

Â a to b, of x times quantity f(x) minus g(x) dx.

Â This is a formula that is worth knowing.

Â You might think that the same formula holds for y bar.

Â But it's a little bit different there is one

Â point at which these computations differ dramatically.

Â That is in the numerator when we first integrate with respect to y.

Â We're integrating y dy instead of x dy.

Â 6:08

Of course, that integral is easy enough, it's y squared over two.

Â But it makes the final formula qualitatively different.

Â Of course, the denominator is the same, it integrates to give you the area.

Â But, when we consider what is happening in the numerator, our final formula

Â is y bar equals 1 over A times the integral as x goes from

Â a to b of 1/2 times quantity f(x)

Â squared minus g(x) squared dx.

Â So many students try to memorize this formula and mix it up or fail.

Â Remember how it's derived, remember that it's the one half

Â y squared evaluated from g to f and you'll be fine.

Â 7:00

Let's look at this in a specific example.

Â That is, we'll take a quarter disc of radius r in the plane.

Â We have to compute x bar and y bar.

Â Well, this is the region between two curves,

Â where on the top we have y equals root R squared minus x squared.

Â On the bottom, of course we have y equals zero.

Â To compute x bar, we take 1 over the area

Â that is 4 over pi R squared, since it's a quarter circle.

Â Times the integral as x goes from zero to R of x, times f of x, minus g of x.

Â That is x times root R squared minus x squared dx.

Â 7:49

Well, that is a doable integral, but let's wait a moment.

Â Because if we use the formula for y bar we obtain a different integral.

Â We get 4 over pi r squared times the integral from zero to r of 1/2

Â quantity root r squared minus x squared, squared dx.

Â In the case of y bar the integral is simpler, that square root goes away and

Â we can simply integrate our squared minus x squared dx.

Â That gives our squared x minus x cubed over 3 evaluating that from 0 to R and

Â accounting for the constant that we pulled out gives an answer of 4R over 3 pi.

Â 8:37

Well so much for y bar.

Â How are we going to compute the integral to get x bar?

Â Well if you consider what the shape looks like, and the symmetry that is present,

Â you can argue that the centroid would have to lie on the line where y=x and

Â so we have computed x bar as well as y bar.

Â 9:01

This points to a more general principle,

Â that centroids respect the symmetries implicit in a domain.

Â If you have an access of symmetry, the centroid lies along it.

Â If you have two axis of symmetry, the centroid lies in the intersection.

Â 9:23

Unfortunately, not all domains have nice symmetry properties.

Â Consider a triangle, let's say a right triangle, of height h and

Â length l, in the plane, defined by the hypotenuse.

Â Given by y equals h over l times quantity, l-x.

Â What's the centroid of that region?

Â Well we can write down our formulae for x bar and for y bar.

Â It's simple enough in this case.

Â What we have to integrate to obtain x bar.

Â Is 2 over hl, that's one over the area, times the integral.

Â As x goes from zero to l the h over l times quantity lx minus x squared dx.

Â That's a simple polynomial, easy enough to integrate.

Â Substituting in l gives us, after a little bit of simplification,

Â one third l so

Â the x coordinate of the centroid is one third of the way from the left.

Â If we compute the integral for y bar as well we'll see that we,

Â again, get a quadratic polynomial that we integrate to get a cubic.

Â There's a lot of substitution that's going on, but in the end it

Â simplifies to give us y bar equals 1/3 h.

Â Now, of course, this is as it ought to be,

Â because there really is something of a symmetry going on here.

Â In that if I rotated and flipped the triangle,

Â I would obtain a new triangle with h and l reversed.

Â And so there must be this relationship between x bar and y bar,

Â they've got to have the same formulae flipping out l and h.

Â 11:23

But in fact we can say more.

Â Let's say we share that triangle to the left or to the right.

Â What happens to the y coordinate of the centroid?

Â Well, really, nothing.

Â We're not changing any y coordinates of area elements, so the average is the same.

Â And what we obtain is that for any triangle, its centroid is located

Â one-third of the way in from each of the three sides.

Â That is where the centroid is at.

Â And considering it from a slightly different point of view,

Â this is what one obtains by focusing.

Â All of the mass, if you will, at the three vertices and

Â averaging the coordinates of those three vertices.

Â You can see how our formula falls out immediately.

Â From that, and in fact, this perspective of focusing mass

Â is common in physics when we're trying to represent locations of large objects or in

Â chemistry where we're trying to represent locations of very, very small objects.

Â 12:39

This leads us to the notion of a center of mass where we're

Â weighting the average of the locations by the mass element.

Â The formulae are very familiar.

Â The coordinates for x in the center of mass, x bar,

Â is the integral of x over d, divided by the integral of one over d.

Â But instead of integrating with respect to area,

Â we're integrating with respect to mass.

Â This would give us a formula of 1 over M, the mass, times the integral of x D M.

Â 13:18

Let's look at a simple example where it's one-dimensional and the mass element, d M,

Â is some linear density, row of x times d x.

Â Then in this case let's say x is going from zero to l and

Â the density varies quadratically in x.

Â You could imagine a situation in which the cross sectional

Â areas are getting bigger according to a quadratic relationship.

Â What would the center of mass be?

Â Well we have to compute the mass element that has thickness

Â dx and then density, some constant kappa, times x squared.

Â Plugging this into our formula for x bar, what do we get?

Â Well, we get a couple of integrals that are easy enough.

Â Notice that the constants, kappa, cancel,

Â and we obtain one-fourth x to the fourth from zero to l,

Â divided by one-third x cubed, from 0 to l.

Â Substituting that in, we obtain a center of mass of three-quarters l.

Â Notice how that's different than the centroid, which would be at l over two.

Â Physically, you've felt centers of mass before.

Â It's that particular location where the object would balance.

Â In this case, it's at three-quarters l.

Â 14:51

Thinking in terms of centers of mass can often help with centriod computations.

Â If you consider a region and break it up into pieces,

Â you can focus all of the areas of those individual pieces at their

Â centroids and weight it according to the area of the piece.

Â Then by computing the center of mass of this collection of weighted points,

Â you can sometimes make the problem much easier.

Â This even works with negative areas.

Â Thinking in terms of a negative Mass as well, that can allow you

Â to simplify a lot of computations as we'll see when we talk about moments.

Â >> Centroids and centers of mass are incredibly useful.

Â If you go to the bonus material for

Â this lecture you'll see how to put centroids to work.

Â In computing volumes, and forces.

Â We'll continue our exploration of solid bodies in our next lesson,

Â with moments of inertia.

Â