College Calculus: Level II Center of Mass

We are here to do a couple more example on centroids.0000

Our example 4 is to find the centroid of the region under the graph y = sin(x) from x=0 to x=π.0004

It is always good to draw a graph of these things.0016

There is the graph of y = sin(x).0021

There is x=0, and there is x=π.0025

We are trying to find the centroid of that region.0030

First of all, remember that the centroid formulas have an area in them.0034

We first need to find the area of that region.0040

The area is just the integral from 0 to π, of sin(x) dx.0045

The integral of sin is cos(x).0055

Evaluate that from x=0 to x=π.0060

Sorry, the integral of sin is -cos(x).0066

That gives us -, -1 -1, which gives us 2 for the area.0069

Now we want to find the two coordinates, the x and y coordinates of the centroid.0077

The x coordinate, if you will look at this region, it is completely symmetric around x = π/2.0082

That has to be the balance point in the x direction.0093

We know that the x coordinate, of the centroid must be π/2, just by symmetry.0096

Now we could work out the integral formula, but that would be a lot more work.0107

I am going to skip that.0111

Instead, I am going to look at the y coordinate of the centroid.0113

The formula that we have is 1/2 × the area, integral from a to b, of f(x)² dx.0117

So that is 1/2 × the area.0127

Well the area we figured out was 2, so this is 1/4.0133

× integral from 0 to π of sin²(x) dx.0138

Remember what we learned in the section on trigonometric integrals.0147

When you have an even power of sin or cosine, you want to use the half angle formula.0153

When you convert this sin²(x) into 1/2 × 1 - cosine(2x),0155

Then that is something much easier to integrate.0167

If I pulled that 1/2 out and combine it with that 1/4, I get 1/8.0171

Then, if I integrate 1, I get x.0177

Integral of cosine(2x) is 1/2sin(2x).0181

I am evaluating this from x=0 to x=π.0189

This is 1/8.0195

Now if you plug in x=π, you get π.0198

sin(2π) = 0, so that is just - 0.0204

Plug in x=0 and you get - 0, and then the sin(0) is 0 again.0208

So we just get π/8.0216

That was the y coordinate of the centroid.0219

If we package those two answers together,0225

x-bar, y-bar, we get (π/2, π/8) is the balance point of this region.0230

Finally, I would like to do an example with a triangular region.0000

The coordinates of the corners are (-1,0), (1,0), and (0,6).0004

Like the other examples, it is very useful to graph this region before we start.0012

So, (-1,0) is there.0020

(1,0) is there.0024

(0,6) is up there.0030

So, we are looking at this triangle.0035

We are trying to find the centroid of that region.0043

Again, we can exploit symmetry on this.0046

This region is symmetric in the x direction.0053

The x coordinate of the centroid must be exactly halfway in between the left and right extremes.0058

The x coordinate of the centroid must be 0, by symmetry.0065

The y coordinate is not so obvious, we actually have to do some calculus for that.0075

I think what I am going to do is find the centroid of the right hand triangle there.0081

That is the one I am outlining in red.0090

The point there is that if I can find the point where that balances.0092

Then that will be the same as where the entire triangle balances in the y direction.0099

If I just find the centroid of this triangle on the right here,0106

It will make the calculations a little bit easier and avoid some ugly negative numbers.0110

Let us just assume we are looking at this right hand triangle,0116

And find the y coordinate of the centroid for this.0122

So, as usual, we need to find the area of that region.0127

Well the area of a triangle is 1/2 base × height.0133

The base is 1, and the height is 6, so the area is just 3.0139

Then, our formula for the coordinates of the centroid tells us that y-bar is 1/2 × the area0144

× integral from a to b of f(x)² dx.0156

That means I have to figure out what f(x) is.0164

f(x) is that line right there.0167

We need to find the equation of that line.0170

That is not so hard, because that line has slope -6, because it is dropping down 6 units for 1 unit it goes over.0174

Its intercept is given by y = 6 right there.0183

So, the equation for that line is just y=-6x + 6, or 6-6x.0194

That is the f(x) we will be using in our formula.0202

Then the f values are x=0 to x=1.0206

Because we are only looking at the right hand triangle now.0212

Let us plug everything in here.0217

1/2a, our a was 3 so this is 1/6.0218

Our little a and b are 0 and 1.0224

Out f(x)² was 6-6x, so this is (6-6x)² dx.0229

The numbers are going to get a little bit if we leave those 6's in there.0240

So I think what I will do is pull a 6 out of that square.0244

When you pull that out, that will give us a 36.0249

I will pull that all the way out of the integral, that gives us 36/6 × integral from 0 to 1,0250

Of now there is just (1-x)² dx.0258

So, that in turn becomes 6 × integral from 0 to 1 of (1-x)²,0264

is x² - 2x + 1 dx.0270

That is a very tractable integral.0279

That is 6 × x³/3 - integral of 2x is x² + integral of 1 is just x.0282

We want to evaluate that from x=0 to x=1.0293

So that is 6 × 1/3 - 1 + 1.0299

Then if you plug in x=0, you just get a bunch of 0's.0304

These ones cancel and so you get 6 × 1/3, so that is just 2, as the y coordinate of the centroid.0310

So the centroid here is the x coordinate and the y coordinate packaged together into a point.0328

The x coordinate was 0 and the y coordinate was 2.0335

The balance point of that whole triangle is (0,2).0343

Again, what we did there was we looked at the question.0351

We immediately drew a graph of it because you do not really understand it until you draw a graph of it.0356

Just by looking at the graph we were able to realize the x coordinate of the centroid is 0 by symmetry.0362

Because the right and left sides obviously have the same mass so the x coordinate is going to be 0.0371

To find the y coordinate we used our formula for y-bar.0377

Which in turn required us to know the area.0381

Since the thing was symmetric, we just found the y coordinate for the triangle on the right,0385

But that would give us the same y coordinate for the entire triangle.0394

Once we plugged that in, we got a very easy integral to find the y coordinate.0398

So, we found the center of mass there.0403

That concludes the lecture on centroids and centers of mass.0406

Today we are going to learn how to use integration to calculate the center of mass of a region.0000

The idea here is that we will have a function y = f(x).0010

We will look at the region underneath it from x = a to x = b.0017

We are going to imagine that we cut out a thin plate that fills that region.0025

We want to figure out exactly where the center of mass is.0030

In other words, if we were going to balance this region on a particular point, where would it balance.0035

We want an x coordinate and a y coordinate of the center of mass.0043

We are going to call those coordinates x-bar and y-bar.0052

We want to figure out how to find those.0057

The center of mass is also known as the centroid.0060

On some of the examples you will see the word centroid.0066

We have 2 equations that both involve integrals that tell us how to find those 2 coordinates.0070

The x-bar is given by this integral equation0077

The y-bar is given by this integral equation.0081

The one thing that may not be clear here is that the a is equal to the area0085

Of course you can find the area by integrating from small a to small b of f(x) dx.0092

Let us try that out on some examples and see how that works out.0100

The first one we are given is to find the centroid of the region inside a semi-circle of radius 1.0105

We are definitely going to need to start with a graph there.0113

There is a semi-circle of radius 1 and we want to figure out where the centroid is.0119

We are trying to find the region inside it.0126

We want to find the centroid of that region.0131

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.0132

The x coordinate of the centroid is certainly going to be 0.0143

That is just by noticing the symmetry of the region.0150

The y coordinate is not going to be so easy.0157

We actually have to do some calculus for that.0159

First of all, remember that the area of the region, remember we need that for the formula.0165

Since it is a circle of radius 1, well the area of a circle of radius 1 is pi, but we just have half of that.0171

Then the y coordinate of the centroid is given by our integral formula.0178

I will copy that down here, 1/2a × the integral from a to b of f(x)² dx.0183

Now, we need to figure out what the function is for the circle.0195

That function is, well remember the function, or the defining equation for a circle is x² + y² = 1.0199

If we solve that for y in terms of x.0207

We get y = sqrt(1-x²)0210

In this case, if we plug that into our formula, we get 1/2 × area0215

Which is pi/2.0222

× integral our bounds on x are x=-1 and x=1 here, so the integral from -1 to 1.0226

f(x)² well since f(x) was a square root, this is just 1 - x² dx0241

This in turn becomes 1/pi ×0250

Now, if we integrate 1/x², that is easy, that is just 1-x³/3.0255

Evaluate that from x = -1 to x = 1.0265

This gives us 1/pi × 1 - 1/3 - (-1), so + 1, - (-1/3).0270

Sorry, minus, then there is another minus, then there is a third minus, so this whole thing is negative 1/3.0288

That gives us 1/pi × 2 - 2/3, which is 4/3.0300

That gives us 4/3 pi.0311

Remember that was just the y coordinate of the centroid0315

So, the centroid is the point located at x = 0 and y = 4 over 3pi.0319

So, that is the point at which this object would balance if you tried to balance it on a point.0333

Let us review there.0341

We were given a two dimensional shape and we want to use our integral formulas to find the centroid.0344

The x=0, that just came from the fact that both the left and the right hand sides of the circle look the same.0353

The area pi/2 just came from the formula for the area of a circle,0360

And then we use the integral formula to find the y coordinate of the centroid.0364

That turned out to be not too bad of an integral.0369

Our second example is to find the centroid of the region under the graph of y = 1/x from x = 1 to x = e,0375

Let me draw that, that should not cross the x axis, it is asymptotic to the x axis.0387

There is x=1 and x=e.0392

We are looking at that region and we want to find the center of mass of that region.0401

Let us start out by finding the area, that is just the integral from 1 to 3 of f(x).0408

That is 1/x dx0413

The integral of 1/x is ln(x).0417

Integrate that from x=1 to x=e0420

We get the ln(e) - ln(1), but the ln(e) is 1 and the ln(1) is just 0.0426

That is the first ingredient we needed for our formulas.0440

Now let us find the centroid.0441

x-bar, the x coordinate of the formula is 1/area × the integral from a to b of x(f(x)) dx.0445

The area is just 1 so that just turns out to be 10460

Now we want the integral from a to b, that is the integral from 1 to e.0465

f(x) is 1/x so f(f(x)) is just 1 dx.0470

You integrate that and you get x evaluated from 1 to e, so you get e-1.0480

So, that was the x coordinate.0486

The y coordinate is a little bit harder but not too much.0490

The y coordinate, again our main formula tells us it is 1/2 × the area, integral from a to b of f(x)² dx0493

The area was 1, so this is 1/2 integral from 1 to e0507

f(x)² is 1/x² dx.0514

So, this is 1/2, now the integral of x², you want to think about it as x^-20520

Its integral is x^-1/-1, or -1/x.0526

Evaluated from x=1 to x=e.0535

That is 1/2 -1/e + 10542

That could be written as 1/2 - 1/2e0550

That was the y coordinate of the centroid.0555

you put those together and you get the centroid, the balance point, is0558

Put the x and the y coordinate together to get an ordered pair0565

You get (e-1, 1/2 - 1/2e)0570

The point of that problem is that we had to figure out the area first0582

Then we plug that into the formula for the x coordinate of the centroid.0587

Work that out, and then plug that into the formula for the y coordinate of the centroid.0592

Then we sort of package them together to give us the coordinates of the centroid.0597

Let us do another example of that.0603

We want to find the centroid now of the region inside the unit circle, inside the first quadrant.0607

If we graph that, it is always good to start with a graph.0613

That is that part of the unit circle and we are trying to find the centroid of that region.0617

The first thing to notice here is that we can exploit some of our earlier work.0627

Remember in example 1 that we found the centroid of the semi-circle, or half disc.0632

The y coordinate of the centroid should be the same as the y coordinate of the centroid we had before.0646

Wherever this thing balance in the y direction is the same as where the half disc balances in the y direction.0659

Let us recall that that was y = 4/3pi.0663

Now the x coordinate of the centroid,0669

Again our formula is 1/area × the integral from a to b of x × f(x) dx.0673

We figured out before that the equation here is y = sqrt(1-x²)0684

We are now going from x=0 to x=1.0693

The area there, that is 1/4 of a circle.0697

A circle of radius 1 would have area pi.0701

The area is pi/4.0704

1/pi/4 integral from 0 to 1 of x × f(x) is 1 - x² dx.0708

Clean it up a little bit, we can flip over the pi/4 and get 4/pi × the integral from 0 to 1 of x × 1-x² dx.0721

At this point there are two ways you can proceed.0733

Actually you could have proceeded a different way earlier but I wanted to show how you can set up this method.0735

You can go ahead and work out this integral0740

It is not that bad.0745

What you can do is say u = 1 - x²,0748

Then du is -2x, dx0752

That is going to work pretty nicely because you already have an x there that provides your dx.0757

That would not be such a bad integral if you use a u substitution there.0767

On the other hand, you could also notice that this region is symmetric about the line y=x.0774

The x coordinate of the centroid should be the same as the y coordinate of the centroid0795

This region is symmetric in the x and y directions so it should balance at the same point in the x and y directions.0804

What you should be able to figure out just be exploiting the symmetry is that the x coordinate is also going to be 4/3pi.0812

If you had not noticed that, you could certainly work out this integral.0822

You should get 4/3pi by the integral formula as well.0827

Either way, you will get the same answer.0831

It is a little easier if you notice this, but if not you can still do the integral formula.0834

You end up with the two coordinate of the centroid being (4/3pi, 4/3pi).0840

Again here, you can use the formulas that we learned to calculate the coordinates of the centroid.0857

Or you can exploit the symmetry.0862

I guess you could not have exploited the symmetry if you did not already know the y coordinate.0868

We did use the formulas to calculate the y coordinate.0870

So you will have to do a bit of calculus at some point.0874

Having done it in the previous problem we can make our lives simpler in this problem and exploit that symmetry.0876

We will do some more problems later.0886