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Welcome to calculus.

I'm professor Grist.

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

Averages appear in many guises and shapes and sizes.

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

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

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How do you compute an average location?

Well this starts off simple enough.

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

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

You could probably use some basic geometry to determine the average

of three locations, or the average of four locations.

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

This requires thinking in terms of calculus.

Breaking that region up into infinitesimal elements and

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

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

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

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

x r and y bar.

That notation is chosen to help you remember the definition.

X bar is the average x coordinates over the domain.

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

Y bar correspondingly is the integral of y over D

divided by the integral of 1 over D.

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

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

in lecture 31 to compute these integrals.

Dividing the region not into strips, but

rather into infinitesimal rectangles of dimensions dx and

dy and then averaging over those.

The notation that is most useful to us is that of

a double integral, or an iterated integral.

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

is really the integral with respect to the area element.

If you integrate dA, what do you get?

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

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

of x dA, or of y dA respectively.

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

of dimensions, Dx and Dy.

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

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

Then what do we get?

Well, dA is really the area

of this infinitesimal rectangle that is, dx times dy.

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

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

Double integrals looks scary, they're really not.

You just do it one at a time.

The trick is determining the limits of integration.

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

the upper limit is f(x).

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

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

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

Let's perform those integrals.

It's not so bad.

What is the integral of x dy?

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

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

That of course gives us y.

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

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

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

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

And so we obtain a formula for xr as shown.

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

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

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

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

This is a formula that is worth knowing.

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

But it's a little bit different there is one

point at which these computations differ dramatically.

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

We're integrating y dy instead of x dy.

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

But it makes the final formula qualitatively different.

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

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

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

a to b of 1/2 times quantity f(x)

squared minus g(x) squared dx.

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

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

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

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Let's look at this in a specific example.

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

We have to compute x bar and y bar.

Well, this is the region between two curves,

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

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

To compute x bar, we take 1 over the area

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

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

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

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

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

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

quantity root r squared minus x squared, squared dx.

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

we can simply integrate our squared minus x squared dx.

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

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

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Well so much for y bar.

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

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

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

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

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This points to a more general principle,

that centroids respect the symmetries implicit in a domain.

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

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

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Unfortunately, not all domains have nice symmetry properties.

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

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

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

What's the centroid of that region?

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

It's simple enough in this case.

What we have to integrate to obtain x bar.

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

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

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

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

one third l so

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

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

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

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

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

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

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

In that if I rotated and flipped the triangle,

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

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

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

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But in fact we can say more.

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

What happens to the y coordinate of the centroid?

Well, really, nothing.

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

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

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

That is where the centroid is at.

And considering it from a slightly different point of view,

this is what one obtains by focusing.

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

averaging the coordinates of those three vertices.

You can see how our formula falls out immediately.

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

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

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

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

weighting the average of the locations by the mass element.

The formulae are very familiar.

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

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

But instead of integrating with respect to area,

we're integrating with respect to mass.

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

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

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

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

the density varies quadratically in x.

You could imagine a situation in which the cross sectional

areas are getting bigger according to a quadratic relationship.

What would the center of mass be?

Well we have to compute the mass element that has thickness

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

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

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

Notice that the constants, kappa, cancel,

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

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

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

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

Physically, you've felt centers of mass before.

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

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

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

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

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

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

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

you can sometimes make the problem much easier.

This even works with negative areas.

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

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

>> Centroids and centers of mass are incredibly useful.

If you go to the bonus material for

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

In computing volumes, and forces.

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

with moments of inertia.