WEBVTT 00:00:01.000 --> 00:00:03.000 The following content is provided under a Creative 00:00:03.000 --> 00:00:05.000 Commons license. Your support will help MIT 00:00:05.000 --> 00:00:08.000 OpenCourseWare continue to offer high quality educational 00:00:08.000 --> 00:00:13.000 resources for free. To make a donation or to view 00:00:13.000 --> 00:00:18.000 additional materials from hundreds of MIT courses, 00:00:18.000 --> 00:00:23.000 visit MIT OpenCourseWare at ocw.mit.edu. 00:00:23.000 --> 00:00:28.000 Yesterday we saw how to define double integrals and how to 00:00:28.000 --> 00:00:33.000 start computing them in terms of x and y coordinates. 00:00:33.000 --> 00:00:41.000 We have defined the double integral over a region R and 00:00:41.000 --> 00:00:45.000 plane of a function f of x, y dA. 00:00:45.000 --> 00:00:51.000 You cannot hear me? Is the sound working? 00:00:51.000 --> 00:00:57.000 Can you hear me in the back now? Can we make the sound louder? 00:00:57.000 --> 00:01:01.000 Does this work? People are not hearing me in 00:01:01.000 --> 00:01:05.000 the back. Is it better? 00:01:05.000 --> 00:01:09.000 People are still saying make it louder. 00:01:09.000 --> 00:01:11.000 Is it better? OK. 00:01:11.000 --> 00:01:18.000 Great. Thanks. 00:01:18.000 --> 00:01:22.000 That's not a reason to start chatting with your friends. 00:01:22.000 --> 00:01:27.000 Thanks. When we have a region in the x, 00:01:27.000 --> 00:01:31.000 y plane and we have a function of x and y, 00:01:31.000 --> 00:01:36.000 we are defining the double integral of f over this region 00:01:36.000 --> 00:01:40.000 by taking basically the sum of the values of a function 00:01:40.000 --> 00:01:44.000 everywhere in here times the area element. 00:01:44.000 --> 00:01:48.000 And the definition, actually, is we split the region into 00:01:48.000 --> 00:01:52.000 lots of tiny little pieces, we multiply the value of a 00:01:52.000 --> 00:01:55.000 function at the point times the area of a little piece and we 00:01:55.000 --> 00:01:59.000 sum that everywhere. And we have seen, 00:01:59.000 --> 00:02:07.000 actually, how to compute these things as iterated integrals. 00:02:07.000 --> 00:02:14.000 First, integrating over dy and then over dx, 00:02:14.000 --> 00:02:21.000 or the other way around. One example that we did, 00:02:21.000 --> 00:02:25.000 in particular, was to compute the double 00:02:25.000 --> 00:02:29.000 integral of a quarter of a unit disk. 00:02:29.000 --> 00:02:35.000 That was the region where x squared plus y squared is less 00:02:35.000 --> 00:02:40.000 than one and x and y are positive, of one minus x squared 00:02:40.000 --> 00:02:43.000 minus y squared dA. Well, hopefully, 00:02:43.000 --> 00:02:48.000 I kind of convinced you that we can do it using enough trig and 00:02:48.000 --> 00:02:52.000 substitutions and so on, but it is not very pleasant. 00:02:52.000 --> 00:02:56.000 And the reason for that is that using x and y coordinates here 00:02:56.000 --> 00:03:04.000 does not seem very appropriate. In fact, we can use polar 00:03:04.000 --> 00:03:16.000 coordinates instead to compute this double integral. 00:03:16.000 --> 00:03:22.000 Remember that polar coordinates are about replacing x and y as 00:03:22.000 --> 00:03:28.000 coordinates for a point on a plane by instead r, 00:03:28.000 --> 00:03:31.000 which is the distance from the origin to a point, 00:03:31.000 --> 00:03:35.000 and theta, which is the angle measured 00:03:35.000 --> 00:03:40.000 counterclockwise from the positive x-axis. 00:03:40.000 --> 00:03:48.000 In terms of r and theta, you have x equals r cosine 00:03:48.000 --> 00:03:54.000 theta, y equals r sine theta. The claim is we are able, 00:03:54.000 --> 00:03:59.000 actually, to do double integrals in polar coordinates. 00:03:59.000 --> 00:04:06.000 We just have to learn how to. Just to draw a quick picture -- 00:04:06.000 --> 00:04:12.000 When we were integrating in x, y coordinates, 00:04:12.000 --> 00:04:16.000 in rectangular coordinates, we were slicing our region by 00:04:16.000 --> 00:04:20.000 gridlines that were either horizontal or vertical. 00:04:20.000 --> 00:04:23.000 And we used that to set up the iterated integral. 00:04:23.000 --> 00:04:28.000 And we said dA became dx dy or dy dx. 00:04:28.000 --> 00:04:35.000 Now we are going to actually integrate, in terms of the polar 00:04:35.000 --> 00:04:41.000 coordinates, r and theta. Let's say we will integrate in 00:04:41.000 --> 00:04:45.000 the order with r first and then theta. 00:04:45.000 --> 00:04:50.000 That is the order that makes the most sense usually when you 00:04:50.000 --> 00:04:53.000 do polar coordinates. What does that mean? 00:04:53.000 --> 00:04:58.000 It means that we will first focus on a slice where we fix 00:04:58.000 --> 00:05:02.000 the value of theta and we will let r vary. 00:05:02.000 --> 00:05:06.000 That means we fix a direction, we fix a ray out from the 00:05:06.000 --> 00:05:11.000 origin in a certain direction. And we will travel along this 00:05:11.000 --> 00:05:15.000 ray and see which part of it, which values of r are in our 00:05:15.000 --> 00:05:18.000 region. Here it will be actually pretty 00:05:18.000 --> 00:05:24.000 easy because r will just start at zero, and you will have to 00:05:24.000 --> 00:05:28.000 stop when you exit this quarter disk. 00:05:28.000 --> 00:05:31.000 Well, what is the equation of this circle in polar 00:05:31.000 --> 00:05:34.000 coordinates? It is just r equals one. 00:05:34.000 --> 00:05:40.000 So, we will stop when r reaches one. 00:05:40.000 --> 00:05:43.000 But what about theta? Well, the first ray that we 00:05:43.000 --> 00:05:48.000 might want to consider is the one that goes along the x-axis. 00:05:48.000 --> 00:05:51.000 That is when theta equals zero. And we will stop when theta 00:05:51.000 --> 00:05:55.000 reaches pi over two because we don't care about the rest of the 00:05:55.000 --> 00:06:00.000 disk. We only care about the first 00:06:00.000 --> 00:06:05.000 quadrant. We will stop at pi over two. 00:06:05.000 --> 00:06:11.000 Now, there is a catch, though, which is that dA is not 00:06:11.000 --> 00:06:15.000 dr d theta. Let me explain to you why. 00:06:15.000 --> 00:06:19.000 Let's say that we are slicing. What it means is we are cutting 00:06:19.000 --> 00:06:24.000 our region into little pieces that are the elementary, 00:06:24.000 --> 00:06:26.000 you know, what corresponds to a small 00:06:26.000 --> 00:06:28.000 rectangle in the x, y coordinate system, 00:06:28.000 --> 00:06:36.000 here would be actually a little piece of circle between a given 00:06:36.000 --> 00:06:42.000 radius r and r plus delta r. And given between an angle 00:06:42.000 --> 00:06:44.000 theta and theta plus delta theta. 00:06:44.000 --> 00:06:48.000 I need to draw, actually, a bigger picture of 00:06:48.000 --> 00:06:53.000 that because it makes it really hard to read. 00:06:53.000 --> 00:06:58.000 Let's say that I fix an angle theta and a slightly different 00:06:58.000 --> 00:07:02.000 one where I have added delta theta to it. 00:07:02.000 --> 00:07:11.000 And let's say that I have a radius r and I add delta r to 00:07:11.000 --> 00:07:14.000 it. Then I will have a little piece 00:07:14.000 --> 00:07:20.000 of x, y plane that is in here. And I have to figure out what 00:07:20.000 --> 00:07:26.000 is its area? What is delta A for this guy? 00:07:26.000 --> 00:07:29.000 Well, let's see. This guy actually, 00:07:29.000 --> 00:07:33.000 you know, if my delta r and delta theta are small enough, 00:07:33.000 --> 00:07:35.000 it will almost look like a rectangle. 00:07:35.000 --> 00:07:37.000 It is rotated, but it is basically a 00:07:37.000 --> 00:07:39.000 rectangle. I mean these sides, 00:07:39.000 --> 00:07:44.000 of course, are curvy, but they are short enough and 00:07:44.000 --> 00:07:50.000 it is almost straight. The area here should be this 00:07:50.000 --> 00:07:55.000 length times that length. Well, what is this length? 00:07:55.000 --> 00:08:00.000 That one is easy. It is delta r. 00:08:00.000 --> 00:08:03.000 What about that length? Well, it is not delta theta. 00:08:03.000 --> 00:08:05.000 It is something slightly different. 00:08:05.000 --> 00:08:13.000 It is a piece of a circle of radius r corresponding to angle 00:08:13.000 --> 00:08:18.000 delta theta, so it is r delta theta. 00:08:18.000 --> 00:08:26.000 So, times r delta theta. That means now, 00:08:26.000 --> 00:08:32.000 even if we shrink things and take smaller and smaller 00:08:32.000 --> 00:08:37.000 regions, dA is going to be r dr d theta. 00:08:37.000 --> 00:08:38.000 That is an important thing to remember. 00:08:38.000 --> 00:08:44.000 When you integrate in polar coordinates, you just set up 00:08:44.000 --> 00:08:50.000 your bounds in terms of r and theta, but you replace dA by r 00:08:50.000 --> 00:08:55.000 dr d theta, not just dr d theta. And then, of course, 00:08:55.000 --> 00:08:57.000 we have some function that we are integrating. 00:08:57.000 --> 00:09:06.000 Let's say that I call that thing f then it is the same f 00:09:06.000 --> 00:09:12.000 that I put up here. Concretely, how do I do it here? 00:09:12.000 --> 00:09:18.000 Well, my function f was given as one minus x squared minus y 00:09:18.000 --> 00:09:20.000 squared. And I would like to switch that 00:09:20.000 --> 00:09:24.000 to polar coordinates. I want to put r and theta in 00:09:24.000 --> 00:09:26.000 there. Well, I have formulas for x and 00:09:26.000 --> 00:09:30.000 y in polar coordinates so I could just replace x squared by 00:09:30.000 --> 00:09:34.000 r squared cosine squared theta, y squared by r squared sine 00:09:34.000 --> 00:09:37.000 squared theta. And that works just fine. 00:09:37.000 --> 00:09:44.000 But maybe you can observe that this is x squared plus y 00:09:44.000 --> 00:09:46.000 squared. It is just the square of a 00:09:46.000 --> 00:09:49.000 distance from the origin, so that is just r squared. 00:09:49.000 --> 00:09:52.000 That is a useful thing. You don't strictly need it, 00:09:52.000 --> 00:09:56.000 but it is much faster if you see this right away. 00:09:56.000 --> 00:10:07.000 It saves you writing down a sine and a cosine. 00:10:07.000 --> 00:10:13.000 Now we just end up with the integral from zero to pi over 00:10:13.000 --> 00:10:20.000 two, integral from zero to one of one minus r squared r dr d 00:10:20.000 --> 00:10:24.000 theta. Now, if I want to compute this 00:10:24.000 --> 00:10:30.000 integral, so let's first do the inner integral. 00:10:30.000 --> 00:10:37.000 If I integrate r minus r cubed, I will get r squared over two 00:10:37.000 --> 00:10:43.000 minus r squared over four between zero and one. 00:10:43.000 --> 00:10:47.000 And then I will integrate d theta. 00:10:47.000 --> 00:10:51.000 What is this equal to? Well, for r equals one you get 00:10:51.000 --> 00:10:54.000 one-half minus one-quarter, which is going to be just 00:10:54.000 --> 00:10:56.000 one-quarter. And when you plug in zero you 00:10:56.000 --> 00:10:59.000 get zero. So, it is the integral from 00:10:59.000 --> 00:11:02.000 zero to pi over two of one-quarter d theta. 00:11:02.000 --> 00:11:11.000 And that just integrates to one-quarter times pi over two, 00:11:11.000 --> 00:11:18.000 which is pi over eight. That is a lot easier than the 00:11:18.000 --> 00:11:23.000 way we did it yesterday. Well, here we were lucky. 00:11:23.000 --> 00:11:26.000 I mean usually you will switch to polar coordinates either 00:11:26.000 --> 00:11:28.000 because the region is easier to set up. 00:11:28.000 --> 00:11:31.000 Here it is indeed easier to set up because the bounds became 00:11:31.000 --> 00:11:34.000 very simple. We don't have that square root 00:11:34.000 --> 00:11:38.000 of one minus x squared anymore. Or because the integrant 00:11:38.000 --> 00:11:40.000 becomes much simpler. Here our function, 00:11:40.000 --> 00:11:43.000 well, it is not very complicated in x, 00:11:43.000 --> 00:11:46.000 y coordinates, but it is even simpler in r 00:11:46.000 --> 00:11:50.000 theta coordinates. Here we were very lucky. 00:11:50.000 --> 00:11:52.000 In general, there is maybe a trade off. 00:11:52.000 --> 00:11:55.000 Maybe it will be easier to set up bounds but maybe the function 00:11:55.000 --> 00:11:58.000 will become harder because it will have all these sines and 00:11:58.000 --> 00:12:01.000 cosines in it. If our function had been just 00:12:01.000 --> 00:12:04.000 x, x is very easy in x, y coordinates. 00:12:04.000 --> 00:12:08.000 Here it becomes r cosine theta. That means you will have a 00:12:08.000 --> 00:12:10.000 little bit of trig to do in the integral. 00:12:10.000 --> 00:12:14.000 Not a very big one, not a very complicated 00:12:14.000 --> 00:12:21.000 integral, but imagine it could get potentially much harder. 00:12:21.000 --> 00:12:25.000 Anyway, that is double integrals in polar coordinates. 00:12:25.000 --> 00:12:30.000 And the way you set up the bounds in general, 00:12:30.000 --> 00:12:37.000 well, in 99% of the cases you will integrate over r first. 00:12:37.000 --> 00:12:40.000 What you will do is you will look for a given theta what are 00:12:40.000 --> 00:12:42.000 the bounds of r to be in the region. 00:12:42.000 --> 00:12:46.000 What is the portion of my ray that is in the given region? 00:12:46.000 --> 00:12:49.000 And then you will put bounds for theta. 00:12:49.000 --> 00:12:51.000 But conceptually it is the same as before. 00:12:51.000 --> 00:12:55.000 Instead of slicing horizontally or vertically, 00:12:55.000 --> 00:12:59.000 we slice radially. We will do more examples in a 00:12:59.000 --> 00:13:03.000 bit. Any questions about this or the 00:13:03.000 --> 00:13:21.000 general method? Yes? 00:13:21.000 --> 00:13:27.000 That is a very good question. Why do I measure the length 00:13:27.000 --> 00:13:30.000 inside instead of outside? Which one do I want? 00:13:30.000 --> 00:13:34.000 This one. Here I said this side is r 00:13:34.000 --> 00:13:36.000 delta theta. I could have said, 00:13:36.000 --> 00:13:38.000 actually, r delta theta is the length here. 00:13:38.000 --> 00:13:41.000 Here it is slightly more, r plus delta r times delta 00:13:41.000 --> 00:13:43.000 theta. But, if delta r is very small 00:13:43.000 --> 00:13:46.000 compared to r, then that is almost the same 00:13:46.000 --> 00:13:48.000 thing. And this is an approximation 00:13:48.000 --> 00:13:51.000 anyway. I took this one because it 00:13:51.000 --> 00:13:56.000 gives me the simpler formula. If you take the limit as delta 00:13:56.000 --> 00:14:01.000 r turns to zero then the two things become the same anyway. 00:14:01.000 --> 00:14:04.000 The length, whether you put r or r plus delta r in here, 00:14:04.000 --> 00:14:09.000 doesn't matter anymore. If you imagine that this guy is 00:14:09.000 --> 00:14:15.000 infinitely small then, really, the lengths become the 00:14:15.000 --> 00:14:17.000 same. We will also see another proof 00:14:17.000 --> 00:14:20.000 of this formula, using changes of variables, 00:14:20.000 --> 00:14:23.000 next week. But, I mean, 00:14:23.000 --> 00:14:28.000 hopefully this is at least slightly convincing. 00:14:28.000 --> 00:14:34.000 More questions? No. 00:14:34.000 --> 00:14:40.000 OK. Let's see. 00:14:40.000 --> 00:14:42.000 We have seen how to compute double integrals. 00:14:42.000 --> 00:14:49.000 I have to tell you what they are good for as well. 00:14:49.000 --> 00:14:53.000 The definition we saw yesterday and the motivation was in terms 00:14:53.000 --> 00:14:57.000 of finding volumes, but that is not going to be our 00:14:57.000 --> 00:15:00.000 main preoccupation. Because finding volumes is fun 00:15:00.000 --> 00:15:02.000 but that is not all there is to life. 00:15:02.000 --> 00:15:05.000 I mean, you are doing single integrals. 00:15:05.000 --> 00:15:08.000 When you do single integrals it is usually not to find the area 00:15:08.000 --> 00:15:13.000 of some region of a plane. It is for something else 00:15:13.000 --> 00:15:16.000 usually. The way we actually think of 00:15:16.000 --> 00:15:19.000 the double integral is really as summing the values of a function 00:15:19.000 --> 00:15:22.000 all around this region. We can use that to get 00:15:22.000 --> 00:15:26.000 information about maybe the region or about the average 00:15:26.000 --> 00:15:29.000 value of a function in that region and so on. 00:15:29.000 --> 00:15:39.000 Let's think about various uses of double integrals. 00:15:39.000 --> 00:15:43.000 The first one that I will mention is actually something 00:15:43.000 --> 00:15:47.000 you thought maybe you could do with a single integral, 00:15:47.000 --> 00:15:51.000 but it is useful very often to do it as a double integral. 00:15:51.000 --> 00:15:59.000 It is to find the area of a given region r. 00:15:59.000 --> 00:16:06.000 I give you some region in the plane and you want to know just 00:16:06.000 --> 00:16:08.000 its area. In various cases, 00:16:08.000 --> 00:16:12.000 you could set this up as a single integral, 00:16:12.000 --> 00:16:16.000 but often it could be useful to set it up as a double integral. 00:16:16.000 --> 00:16:20.000 How do you express the area as a double integral? 00:16:20.000 --> 00:16:22.000 Well, the area of this region is the sum of the areas of all 00:16:22.000 --> 00:16:28.000 the little pieces. It means you want to sum one dA 00:16:28.000 --> 00:16:37.000 of the entire region. The area R is the double 00:16:37.000 --> 00:16:46.000 integral over R of a function one. 00:16:46.000 --> 00:16:48.000 One way to think about it, if you are really still 00:16:48.000 --> 00:16:51.000 attached to the idea of double integral as a volume, 00:16:51.000 --> 00:16:54.000 what this measures is the volume below the graph of a 00:16:54.000 --> 00:16:56.000 function one. The graph of a function one is 00:16:56.000 --> 00:16:59.000 just a horizontal plane at height one. 00:16:59.000 --> 00:17:07.000 What you would be measuring is the volume of a prism with base 00:17:07.000 --> 00:17:11.000 r and height one. And the volume of that would 00:17:11.000 --> 00:17:12.000 be, of course, base times height. 00:17:12.000 --> 00:17:16.000 It would just be the area of r again. 00:17:16.000 --> 00:17:18.000 But we don't actually need to think about it that way. 00:17:18.000 --> 00:17:24.000 Really, what we are doing is summing dA over the entire 00:17:24.000 --> 00:17:28.000 region. A related thing we can do, 00:17:28.000 --> 00:17:33.000 imagine that, actually, this is some physical 00:17:33.000 --> 00:17:35.000 object. I mean, it has to be a flat 00:17:35.000 --> 00:17:38.000 object because we are just dealing with things in the plane 00:17:38.000 --> 00:17:41.000 so far. But you have a flat metal plate 00:17:41.000 --> 00:17:45.000 or something and you would like to know its mass. 00:17:45.000 --> 00:17:50.000 Well, its mass is the sum of the masses of every single 00:17:50.000 --> 00:17:52.000 little piece. You would get that by 00:17:52.000 --> 00:17:57.000 integrating the density. The density for a flat object 00:17:57.000 --> 00:18:09.000 would be the mass per unit area. So, you can get the mass of a 00:18:09.000 --> 00:18:23.000 flat object with density. Let's use delta for density, 00:18:23.000 --> 00:18:29.000 which is the mass per unit area. 00:18:29.000 --> 00:18:34.000 Each little piece of your object will have a mass, 00:18:34.000 --> 00:18:40.000 which will be just the density, times its area for each small 00:18:40.000 --> 00:18:45.000 piece. And you will get the total mass 00:18:45.000 --> 00:18:51.000 by summing these things. The mass will be the double 00:18:51.000 --> 00:18:56.000 integral of the density times the area element. 00:18:56.000 --> 00:18:58.000 Now, if it has constant density, 00:18:58.000 --> 00:19:00.000 if it is always the same material then, 00:19:00.000 --> 00:19:03.000 of course, you could just take the density 00:19:03.000 --> 00:19:07.000 out and you will get density times the total area if you know 00:19:07.000 --> 00:19:10.000 that it is always the same material. 00:19:10.000 --> 00:19:13.000 But if, actually, it has varying density maybe 00:19:13.000 --> 00:19:17.000 because it is some metallic thing with various metals or 00:19:17.000 --> 00:19:21.000 with varying thickness or something then you can still get 00:19:21.000 --> 00:19:24.000 the mass by integrating the density. 00:19:24.000 --> 00:19:26.000 Of course, looking at flat objects might be a little bit 00:19:26.000 --> 00:19:28.000 strange. That is because we are only 00:19:28.000 --> 00:19:30.000 doing double integrals so far. In a few weeks, 00:19:30.000 --> 00:19:33.000 we will be triple integrals. And then we will be able to do 00:19:33.000 --> 00:19:36.000 solids in space, but one thing at a time. 00:19:55.000 --> 00:20:08.000 Another useful application is to find the average value of 00:20:08.000 --> 00:20:16.000 some quantity in a region. What does it mean to take the 00:20:16.000 --> 00:20:19.000 average value of some function f in this region r? 00:20:19.000 --> 00:20:22.000 Well, you know what the average of a finite set of data is. 00:20:22.000 --> 00:20:24.000 For example, if I asked you to compute your 00:20:24.000 --> 00:20:26.000 average score on 18.02 problem sets, 00:20:26.000 --> 00:20:30.000 you would just take the scores, add them and divide by the 00:20:30.000 --> 00:20:33.000 number of problem sets. What if there are infinitely 00:20:33.000 --> 00:20:35.000 many things? Say I ask you to find the 00:20:35.000 --> 00:20:37.000 average temperature in this room. 00:20:37.000 --> 00:20:39.000 Well, you would have to measure the temperature everywhere. 00:20:39.000 --> 00:20:42.000 And then add all of these together and divide by the 00:20:42.000 --> 00:20:45.000 number of data points. But, depending on how careful 00:20:45.000 --> 00:20:47.000 you are, actually, there are potentially 00:20:47.000 --> 00:20:49.000 infinitely many points to look at. 00:20:49.000 --> 00:20:54.000 The mathematical way to define the average of a continuous set 00:20:54.000 --> 00:20:58.000 of data is that you actually integrate the function over the 00:20:58.000 --> 00:21:02.000 entire set of data, and then you divide by the size 00:21:02.000 --> 00:21:06.000 of the sample, which is just the area of the 00:21:06.000 --> 00:21:10.000 region. In fact, the average of f, 00:21:10.000 --> 00:21:17.000 the notation we will use usually for that is f with a bar 00:21:17.000 --> 00:21:23.000 on top to tell us it is the average f. 00:21:23.000 --> 00:21:31.000 We say we will take the integral of f and we will divide 00:21:31.000 --> 00:21:38.000 by the area of the region. You can really think of it as 00:21:38.000 --> 00:21:44.000 the sum of the values of f everywhere divided by the number 00:21:44.000 --> 00:21:48.000 of points everywhere. And so that is an average where 00:21:48.000 --> 00:21:51.000 everything is, actually, equally likely. 00:21:51.000 --> 00:21:55.000 That is a uniform average where all the points on the region, 00:21:55.000 --> 00:21:59.000 all the little points of the region are equally likely. 00:21:59.000 --> 00:22:02.000 But maybe if want to do, say, an average of some solid 00:22:02.000 --> 00:22:06.000 with variable density or if you want to somehow give more 00:22:06.000 --> 00:22:10.000 importance to certain parts than to others then you can actually 00:22:10.000 --> 00:22:14.000 do a weighted average. What is a weighted average? 00:22:14.000 --> 00:22:21.000 Well, in the case of taking the 00:22:21.000 --> 00:22:23.000 average your problem sets, if I tell you problem set one 00:22:23.000 --> 00:22:25.000 is worth twice as much as the others, 00:22:25.000 --> 00:22:29.000 then you would count twice that score in the sum and then you 00:22:29.000 --> 00:22:33.000 would count it as two, of course, when you divide. 00:22:33.000 --> 00:22:36.000 The weighted average is the sum of the values, 00:22:36.000 --> 00:22:39.000 but each weighted by a certain coefficient. 00:22:39.000 --> 00:22:43.000 And then you will divide by the sum of the weight. 00:22:43.000 --> 00:22:48.000 It is a bit the same idea as when we replace area by some 00:22:48.000 --> 00:22:53.000 mass that tells you how important a given piece. 00:22:53.000 --> 00:23:02.000 We will actually have a density. Let's call it delta again. 00:23:02.000 --> 00:23:07.000 We will see what we divide by, but what we will take is the 00:23:07.000 --> 00:23:13.000 integral of a function times the density times the area element. 00:23:13.000 --> 00:23:18.000 Because this would correspond to the mass element telling us 00:23:18.000 --> 00:23:22.000 how to weight the various points of our region. 00:23:22.000 --> 00:23:27.000 And then we would divide by the total weight, 00:23:27.000 --> 00:23:34.000 which is the mass of a region, as defined up there. 00:23:34.000 --> 00:23:39.000 If a density is uniform then, of course, the density gets out 00:23:39.000 --> 00:23:44.000 and you can simplify and reduce to that if all the points are 00:23:44.000 --> 00:23:47.000 equally likely. Why is that important? 00:23:47.000 --> 00:23:49.000 Well, that is important for various applications. 00:23:49.000 --> 00:23:53.000 But one that you might have seen in physics, 00:23:53.000 --> 00:23:58.000 we care about maybe where is the center of mass of a given 00:23:58.000 --> 00:24:01.000 object? The center of mass is basically 00:24:01.000 --> 00:24:05.000 a point that you would say is right in the middle of the 00:24:05.000 --> 00:24:06.000 object. But, of course, 00:24:06.000 --> 00:24:10.000 if the object has a very strange shape or if somehow part 00:24:10.000 --> 00:24:14.000 of it is heavier than the rest then that takes a very different 00:24:14.000 --> 00:24:17.000 meaning. Strictly speaking, 00:24:17.000 --> 00:24:20.000 the center of mass of a solid is the point where you would 00:24:20.000 --> 00:24:24.000 have to concentrate all the mass if you wanted it to behave 00:24:24.000 --> 00:24:28.000 equivalently from a point of view of mechanics, 00:24:28.000 --> 00:24:31.000 if you are trying to do translations of that object. 00:24:31.000 --> 00:24:37.000 If you are going to push that object that would be really 00:24:37.000 --> 00:24:42.000 where the equivalent point mass would lie. 00:24:42.000 --> 00:24:44.000 The other way to think about it, 00:24:44.000 --> 00:24:47.000 if I had a flat object then the center of mass would basically 00:24:47.000 --> 00:24:50.000 be the point where I would need to hold it so it is perfectly 00:24:50.000 --> 00:24:52.000 balanced. And, of course, 00:24:52.000 --> 00:24:56.000 I cannot do this. Well, you get the idea. 00:24:56.000 --> 00:24:59.000 And the center of mass of this eraser is somewhere in the 00:24:59.000 --> 00:25:00.000 middle. And so, in principle, 00:25:00.000 --> 00:25:03.000 that is where I would have to put my finger for it to stay. 00:25:03.000 --> 00:25:11.000 Well, it doesn't work. But that is where the center of 00:25:11.000 --> 00:25:19.000 mass should be. I think it should be in the 00:25:19.000 --> 00:25:22.000 middle. Maybe I shouldn't call this 00:25:22.000 --> 00:25:25.000 three. I should call this 2a, 00:25:25.000 --> 00:25:31.000 because it is really a special case of the average value. 00:25:31.000 --> 00:25:47.000 How do we find the center of mass of a flat object with 00:25:47.000 --> 00:25:57.000 density delta. If you have your object in the 00:25:57.000 --> 00:26:00.000 x, y plane then its center of mass 00:26:00.000 --> 00:26:03.000 will be at positions that are actually just the coordinates of 00:26:03.000 --> 00:26:09.000 a center of mass, will just be weighted averages 00:26:09.000 --> 00:26:14.000 of x and y on the solid. So, the center of mass will be 00:26:14.000 --> 00:26:17.000 a position that I will call x bar, y bar. 00:26:17.000 --> 00:26:21.000 And these are really just the averages, the average values of 00:26:21.000 --> 00:26:26.000 x and of y in the solid. Just to give you the formulas 00:26:26.000 --> 00:26:33.000 again, x bar would be one over the mass times the double 00:26:33.000 --> 00:26:42.000 integral of x times density dA. And the same thing with y. 00:26:42.000 --> 00:26:53.000 y bar is the weighted average of a y coordinate in your 00:26:53.000 --> 00:26:56.000 region. You see, if you take a region 00:26:56.000 --> 00:26:59.000 that is symmetric and has uniform density that will just 00:26:59.000 --> 00:27:01.000 give you the center of the region. 00:27:01.000 --> 00:27:05.000 But if the region has a strange shape or if a density is not 00:27:05.000 --> 00:27:08.000 homogeneous, if parts of it are heavier then 00:27:08.000 --> 00:27:12.000 you will get whatever the weighted average will be. 00:27:12.000 --> 00:27:15.000 And that will be the point where this thing would be 00:27:15.000 --> 00:27:19.000 balanced if you were trying to balance it on a pole or on your 00:27:19.000 --> 00:27:20.000 finger. 00:27:56.000 --> 00:28:12.000 Any questions so far? Yes. 00:28:12.000 --> 00:28:18.000 No. Here I didn't set this up as a 00:28:18.000 --> 00:28:23.000 iterated integral yet. The function that I am 00:28:23.000 --> 00:28:28.000 integrating is x times delta where density will be given to 00:28:28.000 --> 00:28:30.000 me maybe as a function of x and y. 00:28:30.000 --> 00:28:33.000 And then I will integrate this dA. 00:28:33.000 --> 00:28:36.000 And dA could mean dx over dy, it could mean dy over dx, 00:28:36.000 --> 00:28:40.000 it could be mean r dr d theta. I will choose how to set it up 00:28:40.000 --> 00:28:43.000 depending maybe on the shape of the region. 00:28:43.000 --> 00:28:46.000 If my solid is actually just going to be round then I might 00:28:46.000 --> 00:28:49.000 want to use polar coordinates. If it is a square, 00:28:49.000 --> 00:28:51.000 I might want to use x, y coordinates. 00:28:51.000 --> 00:28:56.000 If it is more complicated, well, I will choose depending 00:28:56.000 --> 00:29:01.000 on how I feel about it. Yes? 00:29:01.000 --> 00:29:05.000 Delta is the density. In general, it is a function of 00:29:05.000 --> 00:29:08.000 x and y. If you imagine that your solid 00:29:08.000 --> 00:29:11.000 is not homogenous then its density will depend on which 00:29:11.000 --> 00:29:15.000 piece of it you are looking at. Of course, to compute this, 00:29:15.000 --> 00:29:18.000 you need to know the density. If you have a problem asking 00:29:18.000 --> 00:29:20.000 you to find the center of mass of something and you have no 00:29:20.000 --> 00:29:23.000 information about the density, assume it is uniform. 00:29:23.000 --> 00:29:26.000 Take the density to be a constant. 00:29:26.000 --> 00:29:29.000 Even take it to be a one. That is even easier. 00:29:29.000 --> 00:29:30.000 I mean it is a general fact of math. 00:29:30.000 --> 00:29:34.000 We don't care about units. If density is constant, 00:29:34.000 --> 00:29:36.000 we might as well take it to be one. 00:29:36.000 --> 00:29:41.000 That just means our mass unit becomes the area unit. 00:29:41.000 --> 00:29:53.000 Yes? That is a good question. 00:29:53.000 --> 00:29:57.000 No, I don't think we could actually find the center of mass 00:29:57.000 --> 00:30:00.000 in polar coordinates by finding the average of R or the average 00:30:00.000 --> 00:30:02.000 of theta. For example, 00:30:02.000 --> 00:30:05.000 take a disk center at the origin, well, 00:30:05.000 --> 00:30:09.000 the center of mass should be at the origin. 00:30:09.000 --> 00:30:12.000 But the average of R is certainly not zero because R is 00:30:12.000 --> 00:30:14.000 positive everywhere. So, that doesn't work. 00:30:14.000 --> 00:30:18.000 You cannot get the polar coordinates of a center of mass 00:30:18.000 --> 00:30:21.000 just by taking the average of R and the average of theta. 00:30:21.000 --> 00:30:23.000 By the way, what is the average of theta? 00:30:23.000 --> 00:30:26.000 If you take theta to from zero to 2pi, the average theta will 00:30:26.000 --> 00:30:28.000 be pi. If you take it to go from minus 00:30:28.000 --> 00:30:30.000 pi to pi, the average theta will be zero. 00:30:30.000 --> 00:30:34.000 So, there is a problem there. That actually just doesn't 00:30:34.000 --> 00:30:38.000 work, so we really have to compute x bar and y bar. 00:30:38.000 --> 00:30:41.000 But still we could set this up and then switch to polar 00:30:41.000 --> 00:30:44.000 coordinates to evaluate this integral. 00:30:44.000 --> 00:30:58.000 But we still would be computing the average values of x and y. 00:30:58.000 --> 00:31:04.000 We are basically re-exploring mechanics and motion of solids 00:31:04.000 --> 00:31:10.000 here. The next thing is moment of 00:31:10.000 --> 00:31:15.000 inertia. Just to remind you or in case 00:31:15.000 --> 00:31:18.000 you somehow haven't seen it in physics yet, 00:31:18.000 --> 00:31:23.000 the moment of inertia is basically to rotation of a solid 00:31:23.000 --> 00:31:26.000 where the mass is to translation. 00:31:26.000 --> 00:31:30.000 In the following sense, the mass of a solid is what 00:31:30.000 --> 00:31:34.000 makes it hard to push it. How hard it is to throw 00:31:34.000 --> 00:31:36.000 something is related to its mass. 00:31:36.000 --> 00:31:41.000 How hard it is to spin something, on the other hand, 00:31:41.000 --> 00:31:44.000 is given by its moment of inertia. 00:31:44.000 --> 00:31:51.000 Maybe I should write this down. Mass is how hard it is to 00:31:51.000 --> 00:31:59.000 impart a translation motion to a solid. 00:31:59.000 --> 00:32:06.000 I am using fancy words today. And the moment of inertia -- 00:32:06.000 --> 00:32:13.000 The difference with a mass is that the moment of inertia is 00:32:13.000 --> 00:32:17.000 defined about some axis. You choose an axis. 00:32:17.000 --> 00:32:19.000 Then you would try to measure how hard it is to spin your 00:32:19.000 --> 00:32:21.000 object around that axis. For example, 00:32:21.000 --> 00:32:24.000 you can try to measure how hard it is to spin this sheet of 00:32:24.000 --> 00:32:27.000 paper about an axis that is in the center of it. 00:32:27.000 --> 00:32:30.000 We would try to spin it light that and see how much effort I 00:32:30.000 --> 00:32:35.000 would have to make. Well, for a sheet of paper not 00:32:35.000 --> 00:32:43.000 very much. That would measure the same 00:32:43.000 --> 00:32:58.000 thing but it would be rotation motion about that axis. 00:32:58.000 --> 00:33:02.000 Maybe some of you know the definition but I am going to try 00:33:02.000 --> 00:33:05.000 to derive it again. I am sorry but it won't be as 00:33:05.000 --> 00:33:07.000 quite as detailed as the way you have probably seen it in 00:33:07.000 --> 00:33:09.000 physics, but I am not trying to replace your physics teachers. 00:33:09.000 --> 00:33:16.000 I am sure they are doing a great job. 00:33:16.000 --> 00:33:19.000 What is the idea for the definition to find a formula for 00:33:19.000 --> 00:33:21.000 moment of inertia? The idea is to think about 00:33:21.000 --> 00:33:24.000 kinetic energy. Kinetic energy is really when 00:33:24.000 --> 00:33:28.000 you push something or when you try to make it move and you have 00:33:28.000 --> 00:33:32.000 to put some inertia to it. Then it has kinetic energy. 00:33:32.000 --> 00:33:38.000 And then, if you have the right device, you can convert back 00:33:38.000 --> 00:33:41.000 that kinetic energy into something else. 00:33:41.000 --> 00:33:46.000 If you try to look at the kinetic energy of a point mass, 00:33:46.000 --> 00:33:53.000 so you have something with mass m going at the velocity v, 00:33:53.000 --> 00:33:57.000 well, that will be one-half of a mass times the square of the 00:33:57.000 --> 00:34:00.000 speed. I hope you have all seen that 00:34:00.000 --> 00:34:04.000 formula some time before. Now, let's say instead of just 00:34:04.000 --> 00:34:07.000 trying to push this mass, I am going to make it spin 00:34:07.000 --> 00:34:12.000 around something. Instead of just somewhere, 00:34:12.000 --> 00:34:20.000 maybe I will have the origin, and I am trying to make it go 00:34:20.000 --> 00:34:29.000 around the origin in a circle at a certain angular velocity. 00:34:29.000 --> 00:34:40.000 For a mass m at distance r, let's call r this distance. 00:34:40.000 --> 00:34:47.000 And angular velocity, let's call the angular velocity 00:34:47.000 --> 00:34:50.000 omega. I think that is what physicists 00:34:50.000 --> 00:34:53.000 call it. Remember angular velocity is 00:34:53.000 --> 00:34:57.000 just the rate of the change of the angle over time. 00:34:57.000 --> 00:35:02.000 It is d theta dt, if you want. Well, what is the kinetic 00:35:02.000 --> 00:35:05.000 energy now? Well, first we have to find out 00:35:05.000 --> 00:35:07.000 what the speed is. What is the speed? 00:35:07.000 --> 00:35:10.000 Well, if we are going on a circle of 00:35:10.000 --> 00:35:16.000 radius r at angular velocity omega that means that in unit 00:35:16.000 --> 00:35:22.000 time we rotate by omega and we go by a distance of r times 00:35:22.000 --> 00:35:26.000 omega. The actual speed is the radius 00:35:26.000 --> 00:35:32.000 times angular velocity. And so the kinetic energy is 00:35:32.000 --> 00:35:38.000 one-half mv squared, which is one-half m r squared 00:35:38.000 --> 00:35:41.000 omega squared. And so, 00:35:41.000 --> 00:35:47.000 by similarity with that formula, 00:35:47.000 --> 00:35:51.000 the coefficient of v squared is the mass, 00:35:51.000 --> 00:35:53.000 and here we will say the coefficient of omega squared, 00:35:53.000 --> 00:35:57.000 so this thing is the moment of inertia. 00:35:57.000 --> 00:36:16.000 That is how we define moment of inertia. 00:36:16.000 --> 00:36:20.000 Now, that is only for a point mass. 00:36:20.000 --> 00:36:23.000 And it is kind of fun to spin just a small bowl, 00:36:23.000 --> 00:36:26.000 but maybe you would like to spin actually a larger solid and 00:36:26.000 --> 00:36:29.000 try to define this moment of inertia. 00:36:29.000 --> 00:36:33.000 Well, the moment inertia of a solid will be just the sum of 00:36:33.000 --> 00:36:36.000 the moments of inertia of all the little pieces. 00:36:36.000 --> 00:36:45.000 What we will do is just cut our solid into little chunks and 00:36:45.000 --> 00:36:51.000 will sum this thing for each little piece. 00:36:51.000 --> 00:37:00.000 For a solid with density delta, each little piece has mass 00:37:00.000 --> 00:37:07.000 which is the density times the amount of area. 00:37:07.000 --> 00:37:12.000 This is equal actually. And the moment of inertia of 00:37:12.000 --> 00:37:16.000 that small portion of a solid will be delta m, 00:37:16.000 --> 00:37:18.000 the small mass, times r squared, 00:37:18.000 --> 00:37:25.000 the square of a distance to the center of the axis along which I 00:37:25.000 --> 00:37:29.000 am spinning. That means if I sum these 00:37:29.000 --> 00:37:35.000 things together, well, it has moment of inertia 00:37:35.000 --> 00:37:42.000 delta m times r squared, which is r squared times the 00:37:42.000 --> 00:37:48.000 density times delta A. And so I will be summing these 00:37:48.000 --> 00:37:52.000 things together. And so, the moment of inertia 00:37:52.000 --> 00:37:56.000 about the origin will be the double integral of r squared 00:37:56.000 --> 00:37:59.000 times density times dA. 00:38:28.000 --> 00:38:36.000 The final formula for the moment of inertia about the 00:38:36.000 --> 00:38:46.000 origin is the double integral of a region of r squared density 00:38:46.000 --> 00:38:48.000 dA. If you are going to do it in x, 00:38:48.000 --> 00:38:51.000 y coordinates, of course, r squared becomes x 00:38:51.000 --> 00:38:56.000 squared plus y squared, it is the square of the 00:38:56.000 --> 00:39:02.000 distance from the origin. When you integrate this, 00:39:02.000 --> 00:39:05.000 that tells you how hard it is to spin that solid about the 00:39:05.000 --> 00:39:09.000 origin. The motion that we try to do -- 00:39:09.000 --> 00:39:15.000 We keep this fixed and then we just rotate around the origin. 00:39:15.000 --> 00:39:20.000 Sorry. That is a pretty bad picture, 00:39:20.000 --> 00:39:26.000 but hopefully you know what I mean. 00:39:26.000 --> 00:39:29.000 And the name we use for that is I0. 00:39:29.000 --> 00:39:37.000 And then the rotational kinetic energy is one-half times this 00:39:37.000 --> 00:39:46.000 moment of inertia times the square of the angular velocity. 00:39:46.000 --> 00:39:54.000 So that shows as that this replaces the mass for rotation 00:39:54.000 --> 00:39:57.000 motions. OK. 00:39:57.000 --> 00:40:03.000 What about other kinds of rotations? 00:40:03.000 --> 00:40:06.000 In particular, we have been rotating things 00:40:06.000 --> 00:40:13.000 about just a point in the plane. What you could imagine also is 00:40:13.000 --> 00:40:19.000 instead you have your solid. What I have done so far is I 00:40:19.000 --> 00:40:22.000 have skewered it this way, and I am rotating around the 00:40:22.000 --> 00:40:25.000 axis. Instead, I could skewer it 00:40:25.000 --> 00:40:27.000 through, say, the horizontal axis. 00:40:27.000 --> 00:40:36.000 And then I could try to spin about the horizontal axis so 00:40:36.000 --> 00:40:46.000 then it would rotate in space in that direction like that. 00:40:46.000 --> 00:40:51.000 Let's say we do rotation about the x-axis. 00:40:51.000 --> 00:40:53.000 Well, the idea would still be the same. 00:40:53.000 --> 00:40:58.000 The moment of inertia for any small piece of a solid would be 00:40:58.000 --> 00:41:02.000 its mass element times the square of a distance to the x 00:41:02.000 --> 00:41:06.000 axes because that will be the radius of a trajectory. 00:41:06.000 --> 00:41:12.000 If you take this point here, it is going to go in a circle 00:41:12.000 --> 00:41:16.000 like that centered on the x-axis. 00:41:16.000 --> 00:41:21.000 So the radius will just be this distance here. 00:41:21.000 --> 00:41:24.000 Well, what is this distance? It is just y, 00:41:24.000 --> 00:41:34.000 or maybe absolute value of y. Distance to x-axis is absolute 00:41:34.000 --> 00:41:39.000 value of y. What we actually care about is 00:41:39.000 --> 00:41:44.000 the square of a distance, so it will just be y squared. 00:41:44.000 --> 00:41:51.000 The moment of inertia about the x-axis is going to be obtained 00:41:51.000 --> 00:41:57.000 by integrating y squared times the mass element. 00:41:57.000 --> 00:42:00.000 It is slightly strange but I have y in inertia about the 00:42:00.000 --> 00:42:03.000 x-axis. But, if you think about it, 00:42:03.000 --> 00:42:07.000 y tells me how far I am from the x-axis, so how hard it will 00:42:07.000 --> 00:42:11.000 be to spin around the x-axis. And I could do the same about 00:42:11.000 --> 00:42:17.000 any axis that I want. Just I would have to sum the 00:42:17.000 --> 00:42:23.000 square of a distance to the axis of rotation. 00:42:23.000 --> 00:42:31.000 Maybe I should do an example. Yes? 00:42:31.000 --> 00:42:36.000 Same thing as above, distance to the x-axis, 00:42:36.000 --> 00:42:39.000 because that is what we care about. 00:42:39.000 --> 00:42:47.000 For the moment of inertia, we want the square of a 00:42:47.000 --> 00:42:52.000 distance to the axis of rotation. 00:42:52.000 --> 00:42:57.000 Let's do an example. Let's try to figure out if we 00:42:57.000 --> 00:43:03.000 have just a uniform disk how hard it is to spin it around its 00:43:03.000 --> 00:43:08.000 center. That shouldn't be very hard to 00:43:08.000 --> 00:43:16.000 figure out. Say that we have a disk of 00:43:16.000 --> 00:43:29.000 radius a and we want to rotate it about its center. 00:43:29.000 --> 00:43:32.000 And let's say that it is of uniform density. 00:43:32.000 --> 00:43:36.000 And let's take just the density to be a one so that we don't 00:43:36.000 --> 00:43:40.000 really care about the density. What is the moment of inertia 00:43:40.000 --> 00:43:45.000 of that? Well, we have to integrate of 00:43:45.000 --> 00:43:51.000 our disk r squared times the density, which is one, 00:43:51.000 --> 00:43:55.000 times dA. What is r squared? 00:43:55.000 --> 00:43:58.000 You have here to resist the urge to say the radius is just 00:43:58.000 --> 00:44:00.000 a. We know the radius is a. 00:44:00.000 --> 00:44:05.000 No, it is not a because we are looking at rotation of any point 00:44:05.000 --> 00:44:07.000 inside this disk. And, when you are inside the 00:44:07.000 --> 00:44:09.000 disk, the distance to the origin is not a. 00:44:09.000 --> 00:44:13.000 It is less than a. It is actually anything between 00:44:13.000 --> 00:44:16.000 zero and a. Just to point out a pitfall, 00:44:16.000 --> 00:44:18.000 r here is really a function on this disk. 00:44:18.000 --> 00:44:20.000 And we are going to integrate this function. 00:44:20.000 --> 00:44:28.000 Don't plug r equals a just yet. What coordinates do we use to 00:44:28.000 --> 00:44:31.000 compute this integral? They are probably polar 00:44:31.000 --> 00:44:35.000 coordinates, unless you want a repeat of what happened already 00:44:35.000 --> 00:44:39.000 with x and y. That will tell us we want to 00:44:39.000 --> 00:44:42.000 integrate r squared time r dr d theta. 00:44:42.000 --> 00:44:47.000 And the bounds for r, well, r will go from zero to a. 00:44:47.000 --> 00:44:51.000 No matter which direction I go from the origin, 00:44:51.000 --> 00:44:56.000 if I fixed it, r goes from zero to r equals a. 00:44:56.000 --> 00:45:02.000 The part of this ray that lives inside the disk is always from 00:45:02.000 --> 00:45:05.000 zero to a. And theta goes from, 00:45:05.000 --> 00:45:11.000 well, zero to 2 pi for example. And now you can compute this 00:45:11.000 --> 00:45:14.000 integral. Well, I will let you figure it 00:45:14.000 --> 00:45:18.000 out. But the inner integral becomes 00:45:18.000 --> 00:45:25.000 a to the four over four and the outer multiplies things by 2pi, 00:45:25.000 --> 00:45:30.000 so you get pi a to the four over two. 00:45:30.000 --> 00:45:33.000 OK. That is how hard it is to spin 00:45:33.000 --> 00:45:37.000 this disk. Now, what about instead of 00:45:37.000 --> 00:45:43.000 spinning it about the center we decided to spin it about a point 00:45:43.000 --> 00:45:46.000 on a second point. For example, think of a Frisbee. 00:45:46.000 --> 00:45:50.000 A Frisbee has this rim so you can actually try to make it 00:45:50.000 --> 00:45:55.000 rotate around the point on the circumference by holding it near 00:45:55.000 --> 00:45:59.000 the rim and spinning it there. How much harder is that than 00:45:59.000 --> 00:46:02.000 around the center? Well, we will try to compute 00:46:02.000 --> 00:46:05.000 now the moment of inertia about this point. 00:46:05.000 --> 00:46:08.000 We have two options. One is we keep the system of 00:46:08.000 --> 00:46:12.000 coordinates centers here. But then the formula for 00:46:12.000 --> 00:46:15.000 distance to this point becomes harder. 00:46:15.000 --> 00:46:18.000 The other option, which is the one I will choose, 00:46:18.000 --> 00:46:21.000 is to change the coordinate so that this point become the 00:46:21.000 --> 00:46:23.000 origin. Let's do that. 00:46:50.000 --> 00:46:58.000 About a point on the circumference, 00:46:58.000 --> 00:47:13.000 what I would have to do maybe is set up my region like that. 00:47:13.000 --> 00:47:17.000 I have moved the origin so that it is on the circumference of a 00:47:17.000 --> 00:47:21.000 disk, and I will again try to find 00:47:21.000 --> 00:47:27.000 the moment of inertia of this disk about the origin. 00:47:27.000 --> 00:47:31.000 It is still, for the the double integral of 00:47:31.000 --> 00:47:36.000 r squared dA. But now I want to find out how 00:47:36.000 --> 00:47:40.000 to set up the integral. I could try to use x, 00:47:40.000 --> 00:47:43.000 y coordinates and it would work. 00:47:43.000 --> 00:47:47.000 Or I can use polar coordinates, and it works a little bit 00:47:47.000 --> 00:47:52.000 better that way. But both are doable. 00:47:52.000 --> 00:47:56.000 Let's say I do it this way. I have to figure out how to set 00:47:56.000 --> 00:48:00.000 up my bounds. What are the bounds for r? 00:48:00.000 --> 00:48:06.000 Well, if I fix a value for theta, which means I chose an 00:48:06.000 --> 00:48:12.000 angle here, now I am shooting a ray from the origin in that 00:48:12.000 --> 00:48:16.000 direction. I enter my region at r equals 00:48:16.000 --> 00:48:19.000 zero. That hasn't changed. 00:48:19.000 --> 00:48:23.000 The question is where do I exit the region? 00:48:23.000 --> 00:48:33.000 What is that distance? Maybe you have seen it in 00:48:33.000 --> 00:48:37.000 recitation, maybe not. Let's see. 00:48:37.000 --> 00:48:40.000 Actually, I should have written down the radius of a circle is 00:48:40.000 --> 00:48:47.000 a. So this distance here is 2a. 00:48:47.000 --> 00:48:52.000 If you draw this segment in here, you know that here you 00:48:52.000 --> 00:48:56.000 have a right angle. You have a right triangle. 00:48:56.000 --> 00:48:59.000 The hypotenuse here has length 2a. 00:48:59.000 --> 00:49:08.000 This angle is theta. Well, this length is 2a cosine 00:49:08.000 --> 00:49:14.000 theta. The polar coordinates equation 00:49:14.000 --> 00:49:22.000 of this circle passing through the origin is r equals 2a cosine 00:49:22.000 --> 00:49:27.000 theta. So, r will go from zero to 2a 00:49:27.000 --> 00:49:33.000 cosine theta. That is the distance here. 00:49:33.000 --> 00:49:37.000 Now, what are the bounds for theta? 00:49:37.000 --> 00:49:39.000 It is not quite zero to 2pi because, actually, 00:49:39.000 --> 00:49:42.000 you see in this direction, if I shoot a ray in this 00:49:42.000 --> 00:49:44.000 direction I will never meet my region. 00:49:44.000 --> 00:49:47.000 We have to actually think a bit more. 00:49:47.000 --> 00:49:52.000 Well, the directions in which I will actually hit my circle are 00:49:52.000 --> 00:49:56.000 all the directions in the right half of a plane. 00:49:56.000 --> 00:49:58.000 I mean, of course, if I shoot very close to the 00:49:58.000 --> 00:50:00.000 axis, you might think, oh, I won't be in there. 00:50:00.000 --> 00:50:03.000 But, actually, that is not true because here 00:50:03.000 --> 00:50:05.000 the circle is tangent to the axis. 00:50:05.000 --> 00:50:09.000 No matter which direction I take, I will still have a little 00:50:09.000 --> 00:50:13.000 tiny piece. The angle actually goes from 00:50:13.000 --> 00:50:15.000 minus pi over two to pi over two. 00:50:15.000 --> 00:50:20.000 If you compute that you will get, 00:50:20.000 --> 00:50:25.000 well, the inner integral will be r to the four over four 00:50:25.000 --> 00:50:28.000 between zero and 2a cosine theta, 00:50:28.000 --> 00:50:34.000 which will turn out to be 4a to the four cosine to the four 00:50:34.000 --> 00:50:39.000 theta. And now you will integrate that 00:50:39.000 --> 00:50:43.000 for minus pi over two to pi over two. 00:50:43.000 --> 00:50:47.000 And that is, again, the evil integral that 00:50:47.000 --> 00:50:50.000 we had yesterday. Either we remember the method 00:50:50.000 --> 00:50:53.000 from yesterday or we remember from yesterday that actually 00:50:53.000 --> 00:50:56.000 there are formulas in the notes to help you. 00:50:56.000 --> 00:50:58.000 On homework, you can use these formulas. 00:50:58.000 --> 00:51:04.000 In the notes at the beginning of section 3b there are formulas 00:51:04.000 --> 00:51:08.000 for these particular kinds of integrals. 00:51:08.000 --> 00:51:13.000 And that will end up being three-halves of pi a to the 00:51:13.000 --> 00:51:15.000 four. In case you wanted to know, 00:51:15.000 --> 00:51:18.000 it is three times harder to spin a Frisbee about a point on 00:51:18.000 --> 00:51:20.000 a circumference than around the center. 00:51:20.000 --> 00:51:26.000 We got three times the moment of inertia about the center. 00:51:26.000 --> 00:51:27.000 OK. That is it. 00:51:27.000 --> 00:51:29.000 Have a nice weekend.