 We are at lecture 18 today, Math 241. Today we should wrap up Section 5 of Chapter 6, which is loaded with good applications. Today's applications are moments and center of mass. I think we could probably all make a real good guess if we've never dealt with center of mass, what that is, kind of the balancing point. But moment doesn't carry with it quite so much relevant knowledge that you bring to the page. So we'll talk about what moments are, how they enter into the picture as far as determining center of mass, and look at some center of mass problems. I think the best way to start with center of mass or in incorporating center of mass, we need the term moment. But the best way to start is one dimensionally. So we'll start kind of relative to a straight line. So this diagram is in your book, but I think it's very good in getting us started to what a center of mass is. And then one of the components really ends up being the moment of that particular mass where it's located. So we've got things located on a straight line, so one dimensionally. So the relevant knowledge that we all bring to the page is that this center of mass, x bar, is the balancing point. So we've got things on the left side, things on the right side, they're going to balance each other. So if M1, let's say, is me, pretty heavy, big and kind of fat, fatter than I'd like to be right now, working on that. But this is me, and this is my big old fat mass relative to the fulcrum here of this straight line. And this is somebody else, anybody else in this room who's a lot lighter than I am out here. So if we're going to balance each other, so we want the balance, we want center of mass, then obviously my distance from the fulcrum is going to have to be different from their distance to the fulcrum. So it's true from this picture that mass 1, M1, times the distance from its location, x1 to the balancing point, x bar minus x1, x bar being the center of mass of this one dimensional situation, and the product of M2, the lighter person's mass, and then their distance to the center of mass. So these two products are the same. That's what creates the balance. So if we'll take this equation, distribute M1 to both these pieces, distribute M2 to both these pieces, and solve for x bar, which is really what we're trying to find. So we do the distribution. We get the terms that have x bar on the same side, factor out x bar, and we end up with this equation. So this is a good start on what center of mass actually is. The numerator is called the moment with respect to, in this case, the distances to the balancing point. So this is the moment, the sum of the masses times their distance, really times their location. x1 is the location of M1, x2 is the location of M2. So that would be the moment or the sum of the moments in the numerator. This would be the sum of the masses. And basically, our final result today will be very, very close to that. Now, this is one dimensionally, and we want our results to be, most of the time, we deal with two dimensions. And I don't think in this section of the book, they deal with the center of mass of a 3D situation. So we're going to just try to take this, extend it to two dimensions, and see how it applies to some different lamina on a plane. So we want there not just to be some area under a curve. We want it to have the potential to have some uniform density laminate on top of it. It so happens that it doesn't really matter in the final answer as long as we've got a laminate of uniform density. But that's pretty key to us getting started. So what if, well, let me write that down just a little bit differently. Let's suppose that we want x bar, which we want the product of the masses times their location. We want the sum of all those things. That was the numerator. Let me write that down. I want to write that in a little different form. m sub i times x sub i from i equals 1 to however many we have in the picture. We've just got two here, so it'd be from 1 to 2. But in general, we want to be able to extend that to situations where we have more than two masses at more than two locations, one dimensionally. And whatever we have in the denominator, just be the sum of all the masses, or the mass of the system as it's referred to. So a lot of times, you'll just see big m in the denominator. So that's the mass of the system, the sum of all the masses. So that's x bar. We have a variety of different masses located at a variety of different places on this straight line one dimensional situation. Let's take that notation and that situation now and adapt it to two dimensions. And you'll see some similarities as we go from one to the other. So let's say we have m1 located here. That's x1. Now we're going to two dimensions. So it's located at x1 comma y1. Let's say we have m2 located here. So there's x2 and there's y2. Don't worry about things being positive or negative and how that's going to affect values. We'll take care of those when we. So there's our x3 and there's our y3. So we want to know of that now we have a two dimensional situation. We don't have all these things located on a single line. We've got things located in a two dimensional plane. So where is the center of mass of this thing two dimensionally? So we want an x bar comma y bar. So here's some terminology that is probably something you haven't used a lot to this point in time. M sub y is the moment of the system now that we have this two dimensional situation with lots of possibilities of different masses located at different points in the plane. The moment of the system about the y-axis. And it might look a little awkward at first to say that this is the moment of the system about the y-axis. And it looks like something that we've already had. Now why is it called the moment of the system about the y-axis? Because the distance of these masses from the y-axis really is their x values. And what the moment of the system about the y-axis tells us is its kind of torque or ability to twist with regard to the y-axis. So it makes sense that you ought to have the x values here because that's the moment with regard to the y-axis. So that probably doesn't look right the first time you see it, but x values are distances from the y-axis. So in the words of the author, it's the tendency of the system to rotate about the y-axis. So torque with regard to the y-axis. So there's a moment of the system about the y-axis. So what do you think the moment of the system about the x-axis is going to look like? It'll be m, x, and what's it going to look like? The sum of m. So that's the moment of the system about the x-axis. So again, in the words of the author, it is the tendency of the system to rotate about the x-axis. So we want those distances from the x-axis. What our distance is from the x-axis, they are y-values. All right, so if we want x-bar in this two-dimensional situation, we want to take m, y, and divide it by the sum of all the masses, the mass of the system. So there's the moment of the system divided by the sum of the masses for y-bar. And in this equation, if the thing in the middle bothers you, just toss it aside mentally and just skip from x-bar to this and from y-bar to. Now, how's that going to play out as far as what we need to put in the integrand? Because this is applications of the integral. So what we normally do is we look at one of the pieces of the region that we're dealing with. So let's do that here. So we've got this region from A to B. Let's look at one of the skinny little pieces, skinny little rectangle. Now, the awkward thing about looking at one of the skinny little rectangles, one of the pieces, is that it's so thin that really, and my diagram's not going to be illustrating this, but the left edge and the right edge are so close together that they're practically the same. Because we want the delta x, in this case, to be practically 0. And that way we can add them all together with integral calculus. So it's difficult to discern between the left edge and the right edge. So where's the center of mass of this particular piece in terms of its x value? Well, if the left edge and the right edge are practically the same, then whichever one you choose, if they're the same, that ends up being the x value of the center of mass. Is that correct? So this ends up being, if it's really, really thin, the x value x bar is the x value of this region. So this distance from here to here is really x bar or x sub i bar, if you want to think of it. Each region has its own x bar. So the second region is x sub 2 bar. The 11th region is x sub 11 bar. So this ends up being x sub i bar, sorry, all the way over, all the way over from the y axis over to that location. So it looks like the left edge and the right edge would have different x values. They're the same because of this. So if in the numerator, we want it to be that the sum of the masses times the distance of how far they are from the y axis, and we're going to eventually convert this, not just to this example, but so that we can add together all of the regions that we have identified here, I guess I better backtrack and do what the mass of, well, let's see if we can get by without it. So what are we going to need in the integrand? Well, the x sub i is just going to be an x value. For the mass associated with this, we're going to need, I guess I need to backtrack. I'm going to bring this diagram back up in just a second because what goes in here, we need to identify what that is. So the mass of the first region, the mass of the second region, and so on. So let's call r this region or any region that we identify in this picture. So we want the moment of this particular rectangle, the one that we've chosen. This is what I need to introduce that hasn't been brought up yet, but I have mentioned it. So we're talking about not just an area of this skinny little rectangle. We're talking about the mass of it because there's a distinct of a uniform density laminate on top of it. So it's something that has mass. So if it's something that has mass, this is the density of it. We want to take that uniform density and we want to multiply it by the area of that skinny little rectangle. That gives us the mass of that region. So this is mass. Don't worry about this little row out here because when it's all said and done, those are going to be gone. But right now, that's the density of the laminate of the region. So that density times the area gives us the mass. We want to take that by x sub i. That x sub i is how far this mass is away from the y axis. And we're doing m sub y. Is that OK? So we hadn't done this yet and we were needing it in the integrand. So that's where this is coming from. This is the mass of that particular m1. And then its distance from the y axis is x sub i. Actually, I guess you could call it x sub i bar. That's how far away it is. So let me go back to this. So what we need here is to put the delta x outside. Now we'll clean up this m in the denominator as well and you'll see what the mass of the system is, which is basically just the area of the whole region times the density of the laminate that's laying upon it. This ought to take us all the way from A to B. So we can factor that out in front. That's a number. We're going to have an x in here and an f of x and a dx. It's a lot easier to use than it is to develop, let me tell you. So once we get done with this, the problems are pretty quick and easy. Sometimes they involve some procedures that we haven't used for a while, but most of the time they're pretty quick and easy. Now this thing right here, the mass of the system, kind of the sum of the masses. So the mass of the system would be the area under the curve from A to B times the density of this laminate that's laying on the area. So that's the mass of the whole system. Now you can see what's going to happen. These are going to reduce. So we don't have to contend with that. If it is a laminate of uniform density, it's the same in the numerator as it is in the denominator. We don't have to mess with that for the center of mass. Now some of the problems that we're going to do will actually have to calculate the area under the curve. Others of them, it'll be a circle or a semicircle or a triangle. And we can just calculate it without using calculus. But the backup is we can find the area under the curve from A to B. Now that takes care of the x value of the center of mass. Now how about the y value? Well, if this is my skinny little rectangle, I'm estimating that the y value of the center of mass is right there. Do my picture again. So that's going to be y bar. So what do we know that y bar is? It is. Now the sum of the masses of the whole system is just going to be the area under the curve times this rho value. So that's the same. The denominators of each one are the same. So we want to be able to convert this to an integral. So for the mass of that, now what are we going to call that distance? If I have guessed right, that distance ought to be half of that entire distance. So for that skinny little rectangle, its center of mass, as far as y is concerned, is halfway up to where the y value is on the curve. So if this is f of that x value, this ought to be half of that. So the x value is easy. The left edge and the right edge are practically the same. So it's just that distance from the y axis. The center of mass, as far as y is concerned, is halfway up to the top of that skinny little rectangle. So it's half of f of x sub i. Let me try that again. And again, we've got the mass. Some of the masses, well, it's kind of the mass of the whole system is the area under the curve. Let's move things out that we can so that rho can come out. We've got f of x times f of x. So we've got f of x squared. You could take the half out. Denominator is the mass of the whole region. Take the area under the curve, multiply it by the rho value. You have the same rho value in the numerators you do in the denominator. So this is center of mass, as far as the y value is concerned. So these two, now that we're going to use, put them up here where we can. Now, if it happens to be a convenient figure, this is just the area under the curve. In fact, it's written in the book, this from A to B divided by the area. So you'll see that written that way. But sometimes we actually have to figure out the area under the curve. So it's probably better just to kind of hang onto it like that. The y value for the center of mass, or you'll see it written as that divided by the area. Not quite as easy to come up with, like outer radius squared minus inner radius squared. So you probably will just want to go directly to these formulas. They're not that different from one another. The x times the y here, the y value times itself, half, because we're going halfway up that skinny little rectangle to get to the center of mass as far as y is concerned. All right, I've got some examples picked out. There's a good example in the book. Let me at least mention it. Because of the symmetry involved in this, and we'll try to go through two or three examples, probably time-wise we'll get two today. On page 479 in your book, they have a semi-circle. It's kind of open because it's, we don't know the radius, it's just r. We don't want to work too hard at some of these if they're basically handed to us. What's the center of mass as far as x is concerned? What's going to balance this thing? We are a choice for x. Balance is it with regard to the y-axis. C-rep, right? So our center of mass of this region is somewhere on the y-axis, right? Now we don't know how far up it is. Using the lingo of the author here, this is kind of bottom-heavy, right? So we're not going to go whatever the radius is. We're not going to go halfway up and call that the center of mass because that's not the balancing point because it's bottom-heavy. So we would expect the center of mass to be somewhere right in here, right? Of a bottom-heavy figure down toward where it has most of its mass. And again, it's not just the region. Think of this as being covered with some laminate of uniform density. That doesn't affect the final answer, but that's the center of mass of not just this area, but this area that has some thickness to it. Therefore, it has some mass. So when we can use symmetry, we certainly want to use that. You can look at that for another example. Let's pick out some. I picked out one that doesn't have symmetry. This is problem 36. These might, in fact, be web-assigned problems because they do pick web-assigned problems from problems in the book. Sketch the region bounded by the curves, visually estimate the location of the centroid, then find the exact coordinates of the centroid. That's a new word, centroid. What is centroid? Gosh, maybe we did the wrong stuff today. Just another term for the center of mass is centroid. Gosh, I thought I did the wrong lesson today. So what's the line? 3x plus 2y equals 6, and we want it bounded by, we want basically the first quadrant. y equals 0, and x equals 0. It's probably pretty good to pay attention to directions that if we want the region bounded in the first quadrant, might simplify the work like a test question where it said take the region in the first quadrant. And some of you made more work for yourselves by ignoring that little assistance on that problem. You can choose to ignore it, but a lot of times it'll make more work for yourself. All right, so where are we here? It's all for y. What would y be equal to? 3 minus 3 halves x, is that right? So we'll come up here to 3. Give it a slope of minus 3 halves. That'll be easy. Come down 3 and to the right, 2. So there's the region, and I thought this would be a good first example, pretty simple. We can do some calculus for the denominator. We don't have to because this region is a right triangle, so we can find the area of it without using calculus. Center of mass, as far as x is concerned, what's the numerator going to look like? I don't expect you to be experts yet, but what do we have written down so far? x times 3 minus 3 halves x. x times f of x, is that it? dx. From this is integrated with respect to x, where are we going? 0 to 2. 0 to 2. And if we want to do this separately using calculus, we could find the area under this curve from 0 to 2. What's that going to be? 3. 1 half base times height, right? It's going to be 3, because the region itself is a right triangle. The height is 3. The base is 2. 1 half base times height is going to be 3. So we're going to get 3 in the denominator, but an x right there, that would help. Now this time, because we've got x times this f of x, we can distribute the x through here. So we've got 3x minus 3 halves x squared. So if we integrate, what do we get? Integral of 3x is? 3 half x squared. 3 halves x squared. Integrate 3 halves x squared, and we get? 1 half x cubed. Right, 1 half x cubed. So it would be 1 third x cubed. And these three values, numerator, denominator, knock each other out. So 1 half x cubed. From 0 to 2. So at 2, what do we get? 3 halves of 4, which is? 6. That sound right? And 1 half x cubed would be 1 half of 8. So that'd be 4. And then at 0, we're going to get 0 and 0, right? So that's not going to. So it looks as if the x bar value, the x coordinate of the centroid, is 2 thirds. Does that make sense with the diagram? So as far as the x value of the center of mass, this is a diagram that's what? Kind of left heavy, right? So if we went dead in the middle of our left to right positioning, it'd be at 1. So it should be a little bit left of there, because this diagram is left heavy. It's also what? Bottom heavy. So we should be a little bit under 1 and 1 half. When we find the y value for the center of mass, let's see. How's that numerator go? 1 half f of x squared. So what's that? 3 minus over the mass of the whole region, which we already figured out is 3. So if we square that, what do we get? 9. What's the middle term? 18 over 2 over 9. Negative 18 over 2. So negative 9 x, which would be twice their product. And the last term, 9 fourths x squared. So I don't think you're going to see the setup is really ridiculously difficult. In this case, we've got a square binomial that has constant and x in it. Even though we've got three terms, the integration is not awful. So what do we have? 9 x minus 9 halves x squared plus 9 12ths, which would be 3 fourths x cubed. 18 minus 18. Is that right? I could do that. Pretty limited, but I can handle that. And what do we have here? 6. So it looks like x bar, y bar, center of mass, where the centroid is 2 thirds 1. So for our region, if we go over 2 thirds and up one unit, that should be our balancing point for that region. Chandler. How would you estimate that just by looking at it? They want you to estimate by its bottom heavy. So it's underneath 1 and 1 half. So that's all the estimate needs to be. Underneath 1 and 1 half could be 1.3, 1.2. And then it's left heavy, so it'd be less than 2. So you could say it's 3 quarters, something like that. It's a very rough estimate. All right, next example. This one is a little bit tougher. Doesn't look that much different as far as the diagram is concerned. So y equals e to the x, something like that. y equals 0, x equals 0, and x equals 1. So there's our region. I don't know. Let's see if we can make a semi-intelligent guess here. Where are we right here? What is this y value? When x is 1, what's the y value on that curve? E. E. It's about 2.7, roughly. OK, guesses for center of mass, and why would you guess that? To the right, and up a bit. Closer to 2 thirds, up for the x value. OK, probably a little closer to the line y equals 1, because most of the region is over here. It's a little right heavy. And what? Up at the bottom. A little bottom heavy? Down a bit. Right? Down a bit. So what do you think? That's a little harder to estimate. Right. 2 thirds and 1.3. Let's see how this differs. Now, we're probably not going to be able to shortcut the area of this region and say, well, that's a such and such. That's a trapezoid. I can figure that off to the side. We're probably going to have to actually do that. So the area under e to the x from x equals 0 to x equals 1. That's the easy part. Tell me what the numerator, the integrand of the numerator is. What is it always for x bar? x times f of x. x times f of x. So it's x times f of x. Ah, I don't know about that. I don't know if I can handle that. Thank you, x. OK, so if you said let u equal x, what are you thinking? So you're going to get 1 dx. What's the method that you're thinking? I hope that's what you're thinking when you said let u equal x. Now, if you're thinking substitution, that's not going to work. If you let u equal x, how in the world do you accommodate for e to the x? You go e to u. We do table x. Could use a table of integrals. We don't have to do that. We're not going to do this. It's just a matter of, you've probably done this problem. There we go. This is an integration by parts problem. Yeah, I knew you'd like that. I could tell by how excited you were when that came across. Integration by parts. We get to use it. You're looking at me like you are an idiot. I know that. I was already told that. My wife told me that this morning. So let's not. So if we want to do off to the side this problem, we're just excited to get this reviewed integration by parts. So when somebody said let u equal x, I like the sound of that. But then we've got to have dv equals e to the x dx. So we've accommodated every bit of the integrand. So if we're going to choose u to be x, what's derivative of u? 1 dx or just dx. And if the derivative of v is e to the x, that's the derivative of e to the x. So how does that go? Something in the form of u, dv, uv, integral of v du. So we're hoping when we see that again, the v du part, that it's integrable because the first time we saw it in this form, it was not integrable. So for us in this problem, u, x, v, e to the x minus the integral of v du, what's the integral of our v du? e to the x. So let's go back to the previous page. We now have integrated this. It's x e to the x minus e to the x. So be prepared for stuff like that because sometimes we have an integrand. When you have x times f of x, sometimes you can't put them together like we did on the first example. You're kind of stuck with that. So be prepared for some of this resurrecting some of our earlier techniques. Let's go ahead and evaluate what's in the bottom. The integral of e to the x dx is itself. So numerator, at 1, what do we get? We get 1 times e to the 1 minus e to the 1. And at 0, we get 0 times whatever is 0, minus e to the 0, which is 1. And down here, we get e to the 1 minus e to the 0, which is e minus 1. Is that right? So we get 1 over e minus 1. Somebody give me an approximation of 1 over e minus 1. Right, 5, 8, 2. Thank you. Is that what we expected from our picture? All right, a little bit past the halfway point because it's right heavy. All right, y-bar. What's the numerator? Integral of 1 half x e to the x all squared. F of x squared, so f of x is e to the x. Denominator, we've already evaluated 1 half. e to the x squared is e to the 2x. What's the integrand need? Got e to the u. We better have du. What's the derivative of 2x? Needs a 2. Got another 1 half out in front. We've got a 1 fourth. This is now e to the u du. Integral of e to the u du is e to the u, so it's e to the 2x. So e to the 2 minus e to the 0. Could we reduce? Is that a factor of the numerator? It is, right? That's e plus 1, e minus 1. So the e minus 1's would knock out. We'd have an e plus 1. Or just we're going to have to approximate anyway. Just go ahead and approximate. What is that one? Did you run that through, Katie? 929. 9296. Sorry. I was trying to figure out how I could truncate that without rewriting it. All right, so we have our x bar, y bar, our centroid, our center of mass. So I did specifically choose that example so we could dig back in our memory of how to handle x, f of x, where we had to use integration by parts. Nicole? The problems that we have on this on the website just give you a set of data points. OK. How do you do that? You would go, that's really kind of easier. I don't know. They're about the same, I guess. So integration, remember what we had up here from the m sub i, x sub i, i equals 1 to n. That was our numerator. And then this was just the sum of the masses. So you would take, does it give you like a mass, the mass that's located there at that point? I don't know. It just says p and then m and it has two numbers, number, comma, number. OK. So you're going to actually add together each individual mass times the x values. So let's say you have a mass of 8 located at a certain point of, let's say, 2, 11. So you'd take 8 times 2. And maybe you have another mass of 13 located at a point 7, negative 5. You'd take 13 times 7 and so on. So you'd take the product of the mass times the x values all the way across for all of them. And then you would divide that through by the sum of the masses. So you'd have 8 plus 13 plus the others. So that's not integration. That's actually just number times distance, number times distance, add them together, divided by the sum of the masses. That's my guess if they have a problem like that. And how would you find the y's the same way? Yes, the same way. So for the y, you'd take 8 times 11, 13 times negative 5, and so on. So if we don't have the function we can't do it, can't use integral calculus on that. All right, we are done right on time. Have a nice weekend, and I will see you on Monday.