For more information, please see full course syllabus of College Calculus: Level II

For more information, please see full course syllabus of College Calculus: Level II

### Center of Mass

**Main formula:**

**Hints and tips:**

These formulas are

__not__symmetric. You don’t get from one to the other just by switching*x*’s and*y*’s!Often you will have to do another integral to find the area, . However, sometimes you can find it just using geometry, such as the standard formulas for areas of rectangles, triangles, and circles.

If the region has horizontal or vertical symmetry, you can use that to find the corresponding coordinate of the center of mass without integrating.

When it’s feasible, check that your answers make sense. Graph the region and locate the coordinates of the centroid that you found. Are they inside the region? Are they approximately where it looks like a plate of that shape would balance?

The centroid of a triangle with a horizontal base is always 1/3 of the way up from the base.

### Center of Mass

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

- Intro 0:00
- Important Equation 0:07
- Main Idea
- Centroid
- Area
- Lecture Example 1 1:44
- Lecture Example 2 6:13
- Lecture Example 3 10:04
- Additional Example 4
- Additional Example 5

### College Calculus 2 Online Course

I. Advanced Integration Techniques | ||
---|---|---|

Integration by Parts | 24:52 | |

Integration of Trigonometric Functions | 25:30 | |

Trigonometric Substitutions | 30:09 | |

Partial Fractions | 41:22 | |

Integration Tables | 20:00 | |

Trapezoidal Rule, Midpoint Rule, Left/Right Endpoint Rule | 22:36 | |

Simpson's Rule | 21:08 | |

Improper Integration | 44:18 | |

II. Applications of Integrals, part 2 | ||

Arclength | 23:20 | |

Surface Area of Revolution | 28:53 | |

Hydrostatic Pressure | 24:37 | |

Center of Mass | 25:39 | |

III. Parametric Functions | ||

Parametric Curves | 22:26 | |

Polar Coordinates | 30:59 | |

IV. Sequences and Series | ||

Sequences | 31:13 | |

Series | 31:46 | |

Integral Test | 23:26 | |

Comparison Test | 22:44 | |

Alternating Series | 25:26 | |

Ratio Test and Root Test | 33:27 | |

Power Series | 38:36 | |

V. Taylor and Maclaurin Series | ||

Taylor Series and Maclaurin Series | 30:18 | |

Taylor Polynomial Applications | 50:50 |

### Transcription: 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 sin ^{2}(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 sin ^{2}(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 area *0144

*× 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 x ^{2} - 2x + 1 dx.*0270

*That is a very tractable integral.*0279

*That is 6 × x ^{3}/3 - integral of 2x is x^{2} + 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 equation*0077

*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 area*0085

*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 x ^{2} + y^{2} = 1.*0199

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

*We get y = sqrt(1-x ^{2})*0210

*In this case, if we plug that into our formula, we get 1/2 × area*0215

*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 - x^{2} dx*0241

*This in turn becomes 1/pi × *0250

*Now, if we integrate 1/x ^{2}, that is easy, that is just 1-x^{3}/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 centroid*0315

*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 dx*0413

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

*Integrate that from x=1 to x=e*0420

*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 1*0460

*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} dx*0493

*The area was 1, so this is 1/2 integral from 1 to e *0507

*f(x) ^{2} is 1/x^{2} dx.*0514

*So, this is 1/2, now the integral of x ^{2}, you want to think about it as x^{-2}*0520

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

*Evaluated from x=1 to x=e.*0535

*That is 1/2 -1/e + 1 *0542

*That could be written as 1/2 - 1/2e*0550

*That was the y coordinate of the centroid.*0555

*you put those together and you get the centroid, the balance point, is*0558

*Put the x and the y coordinate together to get an ordered pair*0565

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

*The point of that problem is that we had to figure out the area first*0582

*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-x ^{2})*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 - x ^{2} 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-x ^{2} 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 integral*0740

*It is not that bad.*0745

*What you can do is say u = 1 - x ^{2},*0748

*Then du is -2x, dx*0752

*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 centroid*0795

*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

1 answer

Last reply by: Vasilios Sahinidis

Sat Feb 4, 2012 10:40 PM

Post by Jeffery Maynard on January 21, 2011

Hi,

I was just wondering how you solve these problems if you have two equations such as X^(1/2) and x. Would you just subtract g(x) (call it x) from f(x) (x^(1/2)) to make a new f(x).

Thanks