WEBVTT mathematics/calculus-ii/murray
00:00:00.000 --> 00:00:10.000
Today we are going to learn how to use integration to calculate the center of mass of a region.
00:00:10.000 --> 00:00:17.000
The idea here is that we will have a function y = f(x).
00:00:17.000 --> 00:00:25.000
We will look at the region underneath it from x = a to x = b.
00:00:25.000 --> 00:00:30.000
We are going to imagine that we cut out a thin plate that fills that region.
00:00:30.000 --> 00:00:35.000
We want to figure out exactly where the center of mass is.
00:00:35.000 --> 00:00:43.000
In other words, if we were going to balance this region on a particular point, where would it balance.
00:00:43.000 --> 00:00:52.000
We want an x coordinate and a y coordinate of the center of mass.
00:00:52.000 --> 00:00:57.000
We are going to call those coordinates x-bar and y-bar.
00:00:57.000 --> 00:01:00.000
We want to figure out how to find those.
00:01:00.000 --> 00:01:06.000
The center of mass is also known as the centroid.
00:01:06.000 --> 00:01:10.000
On some of the examples you will see the word centroid.
00:01:10.000 --> 00:01:17.000
We have 2 equations that both involve integrals that tell us how to find those 2 coordinates.
00:01:17.000 --> 00:01:21.000
The x-bar is given by this integral equation
00:01:21.000 --> 00:01:25.000
The y-bar is given by this integral equation.
00:01:25.000 --> 00:01:32.000
The one thing that may not be clear here is that the a is equal to the area
00:01:32.000 --> 00:01:40.000
Of course you can find the area by integrating from small a to small b of f(x) dx.
00:01:40.000 --> 00:01:45.000
Let us try that out on some examples and see how that works out.
00:01:45.000 --> 00:01:53.000
The first one we are given is to find the centroid of the region inside a semi-circle of radius 1.
00:01:53.000 --> 00:01:59.000
We are definitely going to need to start with a graph there.
00:01:59.000 --> 00:02:06.000
There is a semi-circle of radius 1 and we want to figure out where the centroid is.
00:02:06.000 --> 00:02:11.000
We are trying to find the region inside it.
00:02:11.000 --> 00:02:12.000
We want to find the centroid of that region.
00:02:12.000 --> 00:02:23.000
Now one thing is obvious by symmetry, both sides of this semi-circle, or sorry this half disc, are going to weigh the same amount.
00:02:23.000 --> 00:02:30.000
The x coordinate of the centroid is certainly going to be 0.
00:02:30.000 --> 00:02:37.000
That is just by noticing the symmetry of the region.
00:02:37.000 --> 00:02:39.000
The y coordinate is not going to be so easy.
00:02:39.000 --> 00:02:45.000
We actually have to do some calculus for that.
00:02:45.000 --> 00:02:51.000
First of all, remember that the area of the region, remember we need that for the formula.
00:02:51.000 --> 00:02:58.000
Since it is a circle of radius 1, well the area of a circle of radius 1 is π, but we just have half of that.
00:02:58.000 --> 00:03:03.000
Then the y coordinate of the centroid is given by our integral formula.
00:03:03.000 --> 00:03:15.000
I will copy that down here, 1/2a × the integral from a to b of f(x)² dx.
00:03:15.000 --> 00:03:19.000
Now, we need to figure out what the function is for the circle.
00:03:19.000 --> 00:03:27.000
That function is, well remember the function, or the defining equation for a circle is x² + y² = 1.
00:03:27.000 --> 00:03:30.000
If we solve that for y in terms of x.
00:03:30.000 --> 00:03:35.000
We get y = sqrt(1-x²)
00:03:35.000 --> 00:03:42.000
In this case, if we plug that into our formula, we get 1/2 × area
00:03:42.000 --> 00:03:46.000
Which is π/2.
00:03:46.000 --> 00:04:01.000
× integral our bounds on x are x=-1 and x=1 here, so the integral from -1 to 1.
00:04:01.000 --> 00:04:10.000
f(x)² well since f(x) was a square root, this is just 1 - x² dx
00:04:10.000 --> 00:04:15.000
This in turn becomes 1/π ×
00:04:15.000 --> 00:04:25.000
Now, if we integrate 1/x², that is easy, that is just 1-x³/3.
00:04:25.000 --> 00:04:30.000
Evaluate that from x = -1 to x = 1.
00:04:30.000 --> 00:04:48.000
This gives us 1/π × 1 - 1/3 - (-1), so + 1, - (-1/3).
00:04:48.000 --> 00:05:00.000
Sorry, minus, then there is another minus, then there is a third minus, so this whole thing is negative 1/3.
00:05:00.000 --> 00:05:11.000
That gives us 1/π × 2 - 2/3, which is 4/3.
00:05:11.000 --> 00:05:15.000
That gives us 4/3 π.
00:05:15.000 --> 00:05:19.000
Remember that was just the y coordinate of the centroid
00:05:19.000 --> 00:05:33.000
So, the centroid is the point located at x = 0 and y = 4 over 3π.
00:05:33.000 --> 00:05:41.000
So, that is the point at which this object would balance if you tried to balance it on a point.
00:05:41.000 --> 00:05:44.000
Let us review there.
00:05:44.000 --> 00:05:53.000
We were given a two dimensional shape and we want to use our integral formulas to find the centroid.
00:05:53.000 --> 00:06:00.000
The x=0, that just came from the fact that both the left and the right hand sides of the circle look the same.
00:06:00.000 --> 00:06:04.000
The area π/2 just came from the formula for the area of a circle,
00:06:04.000 --> 00:06:09.000
And then we use the integral formula to find the y coordinate of the centroid.
00:06:09.000 --> 00:06:15.000
That turned out to be not too bad of an integral.
00:06:15.000 --> 00:06:27.000
Our second example is to find the centroid of the region under the graph of y = 1/x from x = 1 to x = e,
00:06:27.000 --> 00:06:32.000
Let me draw that, that should not cross the x axis, it is asymptotic to the x axis.
00:06:32.000 --> 00:06:41.000
There is x=1 and x=e.
00:06:41.000 --> 00:06:48.000
We are looking at that region and we want to find the center of mass of that region.
00:06:48.000 --> 00:06:53.000
Let us start out by finding the area, that is just the integral from 1 to 3 of f(x).
00:06:53.000 --> 00:06:57.000
That is 1/x dx
00:06:57.000 --> 00:07:00.000
The integral of 1/x is ln(x).
00:07:00.000 --> 00:07:06.000
Integrate that from x=1 to x=e
00:07:06.000 --> 00:07:20.000
We get the ln(e) - ln(1), but the ln(e) is 1 and the ln(1) is just 0.
00:07:20.000 --> 00:07:21.000
That is the first ingredient we needed for our formulas.
00:07:21.000 --> 00:07:25.000
Now let us find the centroid.
00:07:25.000 --> 00:07:40.000
x-bar, the x coordinate of the formula is 1/area × the integral from a to b of x(f(x)) dx.
00:07:40.000 --> 00:07:45.000
The area is just 1 so that just turns out to be 1
00:07:45.000 --> 00:07:50.000
Now we want the integral from a to b, that is the integral from 1 to e.
00:07:50.000 --> 00:08:00.000
f(x) is 1/x so f(f(x)) is just 1 dx.
00:08:00.000 --> 00:08:06.000
You integrate that and you get x evaluated from 1 to e, so you get e-1.
00:08:06.000 --> 00:08:10.000
So, that was the x coordinate.
00:08:10.000 --> 00:08:13.000
The y coordinate is a little bit harder but not too much.
00:08:13.000 --> 00:08:27.000
The y coordinate, again our main formula tells us it is 1/2 × the area, integral from a to b of f(x)² dx
00:08:27.000 --> 00:08:34.000
The area was 1, so this is 1/2 integral from 1 to e
00:08:34.000 --> 00:08:40.000
f(x)² is 1/x² dx.
00:08:40.000 --> 00:08:46.000
So, this is 1/2, now the integral of x², you want to think about it as x^-2
00:08:46.000 --> 00:08:55.000
Its integral is x^-1/-1, or -1/x.
00:08:55.000 --> 00:09:02.000
Evaluated from x=1 to x=e.
00:09:02.000 --> 00:09:10.000
That is 1/2 -1/e + 1
00:09:10.000 --> 00:09:15.000
That could be written as 1/2 - 1/2e
00:09:15.000 --> 00:09:18.000
That was the y coordinate of the centroid.
00:09:18.000 --> 00:09:25.000
you put those together and you get the centroid, the balance point, is
00:09:25.000 --> 00:09:30.000
Put the x and the y coordinate together to get an ordered pair
00:09:30.000 --> 00:09:42.000
You get (e-1, 1/2 - 1/2e)
00:09:42.000 --> 00:09:47.000
The point of that problem is that we had to figure out the area first
00:09:47.000 --> 00:09:52.000
Then we plug that into the formula for the x coordinate of the centroid.
00:09:52.000 --> 00:09:57.000
Work that out, and then plug that into the formula for the y coordinate of the centroid.
00:09:57.000 --> 00:10:03.000
Then we sort of package them together to give us the coordinates of the centroid.
00:10:03.000 --> 00:10:07.000
Let us do another example of that.
00:10:07.000 --> 00:10:13.000
We want to find the centroid now of the region inside the unit circle, inside the first quadrant.
00:10:13.000 --> 00:10:17.000
If we graph that, it is always good to start with a graph.
00:10:17.000 --> 00:10:27.000
That is that part of the unit circle and we are trying to find the centroid of that region.
00:10:27.000 --> 00:10:32.000
The first thing to notice here is that we can exploit some of our earlier work.
00:10:32.000 --> 00:10:46.000
Remember in example 1 that we found the centroid of the semi-circle, or half disc.
00:10:46.000 --> 00:10:59.000
The y coordinate of the centroid should be the same as the y coordinate of the centroid we had before.
00:10:59.000 --> 00:11:03.000
Wherever this thing balance in the y direction is the same as where the half disc balances in the y direction.
00:11:03.000 --> 00:11:09.000
Let us recall that that was y = 4/3π.
00:11:09.000 --> 00:11:13.000
Now the x coordinate of the centroid,
00:11:13.000 --> 00:11:24.000
Again our formula is 1/area × the integral from a to b of x × f(x) dx.
00:11:24.000 --> 00:11:33.000
We figured out before that the equation here is y = sqrt(1-x²)
00:11:33.000 --> 00:11:37.000
We are now going from x=0 to x=1.
00:11:37.000 --> 00:11:41.000
The area there, that is 1/4 of a circle.
00:11:41.000 --> 00:11:44.000
A circle of radius 1 would have area π.
00:11:44.000 --> 00:11:48.000
The area is π/4.
00:11:48.000 --> 00:12:01.000
1/π/4 integral from 0 to 1 of x × f(x) is 1 - x² dx.
00:12:01.000 --> 00:12:13.000
Clean it up a little bit, we can flip over the π/4 and get 4/π × the integral from 0 to 1 of x × 1-x² dx.
00:12:13.000 --> 00:12:15.000
At this point there are two ways you can proceed.
00:12:15.000 --> 00:12:20.000
Actually you could have proceeded a different way earlier but I wanted to show how you can set up this method.
00:12:20.000 --> 00:12:25.000
You can go ahead and work out this integral
00:12:25.000 --> 00:12:28.000
It is not that bad.
00:12:28.000 --> 00:12:32.000
What you can do is say u = 1 - x²,
00:12:32.000 --> 00:12:37.000
Then du is -2x, dx
00:12:37.000 --> 00:12:47.000
That is going to work pretty nicely because you already have an x there that provides your dx.
00:12:47.000 --> 00:12:54.000
That would not be such a bad integral if you use a u substitution there.
00:12:54.000 --> 00:13:15.000
On the other hand, you could also notice that this region is symmetric about the line y=x.
00:13:15.000 --> 00:13:24.000
The x coordinate of the centroid should be the same as the y coordinate of the centroid
00:13:24.000 --> 00:13:32.000
This region is symmetric in the x and y directions so it should balance at the same point in the x and y directions.
00:13:32.000 --> 00:13:42.000
What you should be able to figure out just be exploiting the symmetry is that the x coordinate is also going to be 4/3π.
00:13:42.000 --> 00:13:47.000
If you had not noticed that, you could certainly work out this integral.
00:13:47.000 --> 00:13:51.000
You should get 4/3π by the integral formula as well.
00:13:51.000 --> 00:13:54.000
Either way, you will get the same answer.
00:13:54.000 --> 00:14:00.000
It is a little easier if you notice this, but if not you can still do the integral formula.
00:14:00.000 --> 00:14:17.000
You end up with the two coordinate of the centroid being (4/3π, 4/3π).
00:14:17.000 --> 00:14:22.000
Again here, you can use the formulas that we learned to calculate the coordinates of the centroid.
00:14:22.000 --> 00:14:28.000
Or you can exploit the symmetry.
00:14:28.000 --> 00:14:30.000
I guess you could not have exploited the symmetry if you did not already know the y coordinate.
00:14:30.000 --> 00:14:34.000
We did use the formulas to calculate the y coordinate.
00:14:34.000 --> 00:14:36.000
So you will have to do a bit of calculus at some point.
00:14:36.000 --> 00:14:46.000
Having done it in the previous problem we can make our lives simpler in this problem and exploit that symmetry.
00:14:46.000 --> 00:25:39.000
We will do some more problems later.