1 00:00:01 --> 00:00:03 The following content is provided under a Creative 2 00:00:03 --> 00:00:05 Commons license. Your support will help MIT 3 00:00:05 --> 00:00:08 OpenCourseWare continue to offer high quality educational 4 00:00:08 --> 00:00:13 resources for free. To make a donation or to view 5 00:00:13 --> 00:00:18 additional materials from hundreds of MIT courses, 6 00:00:18 --> 00:00:23 visit MIT OpenCourseWare at ocw.mit.edu. 7 00:00:23 --> 00:00:28 I guess last time on Friday we went over the first half of the 8 00:00:28 --> 00:00:33 class very quickly. And so today we are going to go 9 00:00:33 --> 00:00:38 over the second half of the class very quickly. 10 00:00:38 --> 00:00:45 And so that was stuff about double and triple integrals and 11 00:00:45 --> 00:00:51 vector calculus in the plane and in space. 12 00:00:51 --> 00:00:55 As usual, what is on the final is basically what was on the 13 00:00:55 --> 00:00:57 other tests, exactly the same stuff. 14 00:00:57 --> 00:01:03 Well, not the same problem, unfortunately. 15 00:01:03 --> 00:01:13 The first thing we learned about was double integrals in 16 00:01:13 --> 00:01:25 the plane and how to set up the bounds and how to evaluate them. 17 00:01:25 --> 00:01:30 Just to remind you quickly, the important thing with 18 00:01:30 --> 00:01:34 iterated integrals is when you integrate a function f of x, 19 00:01:34 --> 00:01:38 y, say dy dx for example, 20 00:01:38 --> 00:01:41 is that you have to draw a picture of a region. 21 00:01:41 --> 00:01:45 Unless it is completely obvious you should really draw some 22 00:01:45 --> 00:01:48 picture of the domain of integration. 23 00:01:48 --> 00:01:53 And once you have that picture you can use it to find the 24 00:01:53 --> 00:01:57 bounds. Remember the general method is 25 00:01:57 --> 00:02:03 that we first look at the inner integral, here integral of f dy. 26 00:02:03 --> 00:02:07 And in this inner integral the outer variable here, 27 00:02:07 --> 00:02:10 x, is fixed. That means we are slicing our 28 00:02:10 --> 00:02:13 region by a vertical line corresponding to a fixed value 29 00:02:13 --> 00:02:16 of x. We fix a value of x. 30 00:02:16 --> 00:02:20 And what we have to find out is the bounds for y, 31 00:02:20 --> 00:02:24 so the value of y at this point, the value of y at that 32 00:02:24 --> 00:02:28 point. Let me call that y some bottom 33 00:02:28 --> 00:02:34 of x, in general depends on x. And this one will be y at that 34 00:02:34 --> 00:02:43 top, and it also depends on x. And then the bounds for y would 35 00:02:43 --> 00:02:46 be this. And then, when you look at the 36 00:02:46 --> 00:02:48 outer bound, things are different. 37 00:02:48 --> 00:02:51 Because there you expect to have just numbers, 38 00:02:51 --> 00:02:53 no longer functions of anything. 39 00:02:53 --> 00:02:56 And what you do is look at the shadow of your region. 40 00:02:56 --> 00:03:00 We are doing it by shadow so you just project to the x-axis. 41 00:03:00 --> 00:03:04 If you project to the x-axis your region will look like this. 42 00:03:04 --> 00:03:08 Its shadow is going to be this integral form, 43 00:03:08 --> 00:03:12 some minimum value of x to some maximum value of x. 44 00:03:12 --> 00:03:24 And that will give us the bounds for the outer integral. 45 00:03:24 --> 00:03:27 And then, to evaluate, we evaluate the usual way. 46 00:03:27 --> 00:03:29 Speaking of evaluation, what you need to know for the 47 00:03:29 --> 00:03:31 final, well, essentially the same kind 48 00:03:31 --> 00:03:35 of evaluation techniques that we were supposed to know for the 49 00:03:35 --> 00:03:40 other tests. That means the usual functions, 50 00:03:40 --> 00:03:47 substitutions, basic trig, stuff like that. 51 00:03:47 --> 00:03:51 Well, I don't expect that you would need integration by parts, 52 00:03:51 --> 00:03:54 although I still hope that some of you remember it from single 53 00:03:54 --> 00:03:59 variable calculus. If there is a need to integrate 54 00:03:59 --> 00:04:06 some big power of cosine or sine then the formula will be given 55 00:04:06 --> 00:04:11 to you the way it is in the notes. 56 00:04:11 --> 00:04:17 And, of course, we know also how to set up 57 00:04:17 --> 00:04:23 these integrals in polar coordinates. 58 00:04:23 --> 00:04:28 And then the area element becomes r dr d theta. 59 00:04:28 --> 00:04:32 And because you integrate first over r, 60 00:04:32 --> 00:04:36 well, first of all you should remember the polar coordinate 61 00:04:36 --> 00:04:42 formulas, namely x equals r cosine theta 62 00:04:42 --> 00:04:50 and y equals r sine theta. And second you should remember 63 00:04:50 --> 00:04:53 that what we do, when we have our region, 64 00:04:53 --> 00:04:56 is for a fixed value of theta we look for the bounds for r, 65 00:04:56 --> 00:04:59 just like before, so the way we are slicing the 66 00:04:59 --> 00:05:02 region is now we are actually shooting rays straight from the 67 00:05:02 --> 00:05:04 origin. And, in a given direction, 68 00:05:04 --> 00:05:08 we are asking ourselves how far does my region go? 69 00:05:08 --> 00:05:12 You have to find a bound and you have to find whatever the 70 00:05:12 --> 00:05:15 value of r will be out here as a function of theta. 71 00:05:15 --> 00:05:19 And ways to do that can be geometric or they can be by 72 00:05:19 --> 00:05:22 starting from the x, y equation of whatever curve 73 00:05:22 --> 00:05:26 you have and then expressing it in terms of r and theta and 74 00:05:26 --> 00:05:28 solving for r. For example, 75 00:05:28 --> 00:05:33 just to illustrate it, we have seen that one of our 76 00:05:33 --> 00:05:39 classics has been the circle of radius one centered at one, 77 00:05:39 --> 00:05:42 zero. This guy, you have two 78 00:05:42 --> 00:05:45 different ways of getting its polar coordinate equation. 79 00:05:45 --> 00:05:49 One is to argue geometrically that you have a right angle in 80 00:05:49 --> 00:05:51 here. And this length is two, 81 00:05:51 --> 00:05:54 this angle is theta, this length is r, 82 00:05:54 --> 00:05:59 so the polar equation is r equals two cosine theta. 83 00:05:59 --> 00:06:01 The other way to do it, if somehow you are missing the 84 00:06:01 --> 00:06:03 geometric trick, is to start from the x, 85 00:06:03 --> 00:06:05 y equation. What is the x, 86 00:06:05 --> 00:06:10 y equation of this guy? Well, it is x minus one squared 87 00:06:10 --> 00:06:16 plus y squared equals one. If you expand that you will get 88 00:06:16 --> 00:06:22 x squared minus two x plus one plus y squared equals one. 89 00:06:22 --> 00:06:27 The ones simplify. X squared plus y squared 90 00:06:27 --> 00:06:34 becomes r squared minus two x becomes r cosine theta equals 91 00:06:34 --> 00:06:36 zero. That gives you, 92 00:06:36 --> 00:06:40 when you simplify by r, r equals two cosine theta. 93 00:06:40 --> 00:06:49 Two ways to get the same polar equation. 94 00:06:49 --> 00:06:53 I should say this is an example, in case you were 95 00:06:53 --> 00:07:01 wondering what I was doing. We have also actually seen how 96 00:07:01 --> 00:07:14 to change variables to more complicated coordinate systems. 97 00:07:14 --> 00:07:16 Let's say u, v coordinates. But, of course, 98 00:07:16 --> 00:07:20 you can call them whatever you want. 99 00:07:20 --> 00:07:24 The main thing to remember is that you have to look for the 100 00:07:24 --> 00:07:30 Jacobian which will give you the conversion ratio between dx dy 101 00:07:30 --> 00:07:32 and du dv. For example, 102 00:07:32 --> 00:07:36 if you know u and v as functions of x and y then you 103 00:07:36 --> 00:07:42 will write du dv equals absolute value of the Jacobian partial u, 104 00:07:42 --> 00:07:50 v over partial x, y times dx dy. Or, if it is easier for you, 105 00:07:50 --> 00:07:52 you can do the Jacobian the other way around. 106 00:07:52 --> 00:07:56 And this Jacobian, remember, is the determinant by 107 00:07:56 --> 00:08:00 a two by two matrix that you obtain by putting the partial 108 00:08:00 --> 00:08:03 derivatives of u and v with respect to x and y. 109 00:08:03 --> 00:08:08 Then, when we have that, we can change the integrant, 110 00:08:08 --> 00:08:13 f of x, y, into something involving u and v possibly. 111 00:08:13 --> 00:08:15 And then we have to find the bounds. 112 00:08:15 --> 00:08:19 And to find the bounds perhaps the easiest is to draw a picture 113 00:08:19 --> 00:08:21 of a region in u, v coordinates. 114 00:08:21 --> 00:08:24 Maybe you have some picture in the x, 115 00:08:24 --> 00:08:28 y plane that might actually be really hard to draw and maybe in 116 00:08:28 --> 00:08:32 terms of u and v the picture will become much simpler. 117 00:08:32 --> 00:08:35 It might just become a rectangle. 118 00:08:35 --> 00:08:37 Of course, if you see immediately what the bounds are 119 00:08:37 --> 00:08:39 in terms of u and v, and they turn out to be very 120 00:08:39 --> 00:08:41 easy, then maybe you don't even have 121 00:08:41 --> 00:08:45 to draw this picture. But if it is not completely 122 00:08:45 --> 00:08:49 obvious then that might be a helpful way of figuring out what 123 00:08:49 --> 00:08:52 the bounds will be when you switch from x, 124 00:08:52 --> 00:08:55 y to u, v. We have seen some problems like 125 00:08:55 --> 00:09:00 that and there are more in the notes in case you need more. 126 00:09:00 --> 00:09:07 Questions? Yes? 127 00:09:07 --> 00:09:10 That is the second time you've asked for something real quick 128 00:09:10 --> 00:09:13 in these review sessions. You are in a hurry. 129 00:09:13 --> 00:09:23 Take your time. Partial u, v over partial x, 130 00:09:23 --> 00:09:26 y is just going to be the determinant of u sub x, 131 00:09:26 --> 00:09:29 u sub y, v sub x, v sub y. 132 00:09:29 --> 00:09:37 That is the definition. That is pretty direct. 133 00:09:37 --> 00:09:39 And, of course, a general common sense thing 134 00:09:39 --> 00:09:43 that applies to actually all the integrals that we are going to 135 00:09:43 --> 00:09:45 see, there are two things in an integral. 136 00:09:45 --> 00:09:49 One is whatever you integral is called the integrant. 137 00:09:49 --> 00:09:52 It could be a function here. It is a vector field in some of 138 00:09:52 --> 00:09:55 the flux things and so on. There is another thing which is 139 00:09:55 --> 00:09:57 the region over which you integrate. 140 00:09:57 --> 00:10:02 And the two have strictly nothing to do with each other. 141 00:10:02 --> 00:10:05 When you are given a piece of data in the statement of a 142 00:10:05 --> 00:10:07 problem, you have to figure out whether 143 00:10:07 --> 00:10:09 that is part of a function to be integrated or whether that is 144 00:10:09 --> 00:10:11 part of the region of integration. 145 00:10:11 --> 00:10:14 If it is the region of integration then it will go into 146 00:10:14 --> 00:10:17 the bounds of the integral and maybe in the choice of the 147 00:10:17 --> 00:10:19 coordinate system that you use for integrating. 148 00:10:19 --> 00:10:23 While the function that you are integrating goes before the dx 149 00:10:23 --> 00:10:26 dy and not into the bounds or anything like that. 150 00:10:26 --> 00:10:32 I know it sounds kind of silly but it is a good safety check. 151 00:10:32 --> 00:10:34 Ask yourselves, when you have a piece of data, 152 00:10:34 --> 00:10:36 where in my formula should this go. 153 00:10:36 --> 00:10:47 Yes? I case you want the bounds for 154 00:10:47 --> 00:10:50 this region in polar coordinates, indeed it would be 155 00:10:50 --> 00:10:53 double integral. For a fixed theta, 156 00:10:53 --> 00:10:57 r goes from zero to whatever it is on that curve. 157 00:10:57 --> 00:11:04 So it would be zero to two cosine theta of whatever the 158 00:11:04 --> 00:11:10 function is r dr d theta. And the bounds on theta would 159 00:11:10 --> 00:11:15 be from negative pi over two to pi over two. 160 00:11:15 --> 00:11:20 We have seen that one several times, so hopefully by now it is 161 00:11:20 --> 00:11:21 clearer. OK. 162 00:11:21 --> 00:11:28 Let me move on a bit because we have a lot of other kinds of 163 00:11:28 --> 00:11:32 integrals to see. Other kinds of integrals we 164 00:11:32 --> 00:11:35 have seen are triple integrals. And I am not doing things in 165 00:11:35 --> 00:11:39 the order that we did them in the class just so you can see 166 00:11:39 --> 00:11:43 parallels between stuff in the plane and in space. 167 00:11:43 --> 00:11:47 When we do triple integrals in space, well, it is the same kind 168 00:11:47 --> 00:11:50 of story, except now we have, of course, more coordinate 169 00:11:50 --> 00:11:53 systems. We have rectangular 170 00:11:53 --> 00:11:59 coordinates, we have cylindrical coordinates and we have 171 00:11:59 --> 00:12:05 spherical coordinates. And cylindrical coordinates 172 00:12:05 --> 00:12:09 only mean that we are, instead of x, y and z, 173 00:12:09 --> 00:12:15 we are replacing x and y by the polar coordinate in the x, 174 00:12:15 --> 00:12:17 y plane, so the angle theta and the 175 00:12:17 --> 00:12:21 distance r. So R is somehow the distance 176 00:12:21 --> 00:12:24 from the z-axis and z is the height. 177 00:12:24 --> 00:12:28 Usually you don't have to choose between rectangular and 178 00:12:28 --> 00:12:31 cylindrical until somewhat late in the process, 179 00:12:31 --> 00:12:33 especially if you integrate first of all z, 180 00:12:33 --> 00:12:36 because then the choice will come up mostly when you try to 181 00:12:36 --> 00:12:39 figure out what are the bounds for the shadow of your region. 182 00:12:39 --> 00:12:43 I mean the z part looks exactly the same in rectangular and in 183 00:12:43 --> 00:12:45 cylindrical. Spherical is, 184 00:12:45 --> 00:12:49 on the other hand, a little bit more annoying 185 00:12:49 --> 00:12:52 because it looks quite different. 186 00:12:52 --> 00:12:56 You should think of it as doing polar coordinates not only in 187 00:12:56 --> 00:12:59 the horizontal direction but also in the vertical direction 188 00:12:59 --> 00:13:02 at the same time. You have this angle phi. 189 00:13:02 --> 00:13:06 That measures the angle down from the positive z-axis. 190 00:13:06 --> 00:13:10 And you have rho which is the distance from the origin. 191 00:13:10 --> 00:13:19 And if I project to the z-axis, r becomes rho sine phi and z 192 00:13:19 --> 00:13:24 becomes rho cosine phi. I hope that you all know these 193 00:13:24 --> 00:13:27 two formulas, but if you ever have a small 194 00:13:27 --> 00:13:31 somehow memory lapse during the final then you should consider 195 00:13:31 --> 00:13:35 drawing this kind of picture because it will let you check 196 00:13:35 --> 00:13:42 very quickly which one is sine, which one is cosine. 197 00:13:42 --> 00:13:44 Now, of course, we have to have formulas for dv 198 00:13:44 --> 00:13:47 in all these coordinate systems. Here, for example, 199 00:13:47 --> 00:13:52 that might be dz r dr d theta or r dr d theta dz or anything 200 00:13:52 --> 00:13:58 like that. Here it might be rho squared 201 00:13:58 --> 00:14:04 times phi times d rho d phi d theta. 202 00:14:04 --> 00:14:07 And the general method for setting up bounds is pretty much 203 00:14:07 --> 00:14:10 the same as in the plane, just there is one more step. 204 00:14:10 --> 00:14:14 If you are doing rectangular or cylindrical coordinates with z 205 00:14:14 --> 00:14:16 first, for example, that is the most common. 206 00:14:16 --> 00:14:22 Well, if you do z first then you have to actually start by 207 00:14:22 --> 00:14:28 figuring out for a given value of x and y or r and theta what 208 00:14:28 --> 00:14:31 is the portion of a vertical line above x, 209 00:14:31 --> 00:14:37 y that lies within my region? That will go from z on the 210 00:14:37 --> 00:14:44 bottom of my solid which depends on x and y to z at the top of my 211 00:14:44 --> 00:14:49 solid which also usually will depend on x and y. 212 00:14:49 --> 00:14:52 And so that will give me the bounds for dz. 213 00:14:52 --> 00:14:56 And then I will be left with the shadow of my region in the 214 00:14:56 --> 00:15:02 x, y plane. And that one I will set up like 215 00:15:02 --> 00:15:09 a double integral over there. Strictly-speaking, 216 00:15:09 --> 00:15:12 if you are curious, we could also change to weird 217 00:15:12 --> 00:15:15 coordinate systems using Jacobian with three variables at 218 00:15:15 --> 00:15:18 the same time. But we haven't seen that so it 219 00:15:18 --> 00:15:24 won't be on the final. But it would work just the same 220 00:15:24 --> 00:15:30 way, just with more pictures to do. 221 00:15:30 --> 00:15:33 And, in fact, I just wanted to say this rho 222 00:15:33 --> 00:15:37 squared sine phi is actually the Jacobian for the change of 223 00:15:37 --> 00:15:42 variables for rectangular to spherical coordinates. 224 00:15:42 --> 00:15:48 OK. Let's not think too much about 225 00:15:48 --> 00:15:51 that. Applications. 226 00:15:51 --> 00:15:57 Well, we have seen how to use double integrals to find the 227 00:15:57 --> 00:16:03 area of a volume of a piece of a plane or a piece of space, 228 00:16:03 --> 00:16:08 and find also the mass. Remember, area is just double 229 00:16:08 --> 00:16:12 integral of one dA, volume is triple integral of 230 00:16:12 --> 00:16:14 one dV. Sometimes if it's the volume 231 00:16:14 --> 00:16:16 between the x, y plane and the graph of some 232 00:16:16 --> 00:16:19 function, you can just set it up directly as a double integral. 233 00:16:19 --> 00:16:23 But there is no harm in doing it as a triple integral if you 234 00:16:23 --> 00:16:27 feel better about that. And mass will be double or 235 00:16:27 --> 00:16:31 triple integral, depending on how many 236 00:16:31 --> 00:16:36 dimensions you have, of whatever density function 237 00:16:36 --> 00:16:45 you have, dA or dV. Then there is how to find the 238 00:16:45 --> 00:16:53 average value of some function. Well, let me do the 239 00:16:53 --> 00:16:58 three-dimensional case. You will just replace volume by 240 00:16:58 --> 00:17:02 area and dV by dA and so on, if need be. 241 00:17:02 --> 00:17:09 That would be one over volume of the solid times the triple 242 00:17:09 --> 00:17:18 integral of f dV, or if it's a weighted average, 243 00:17:18 --> 00:17:24 one over mass times the triple integral of a function times 244 00:17:24 --> 00:17:33 density times dV. If you don't have a density or 245 00:17:33 --> 00:17:44 if the density is constant then that reduces to that one. 246 00:17:44 --> 00:17:46 In particular, we have seen the notion of 247 00:17:46 --> 00:17:50 center of mass. The center of mass is just 248 00:17:50 --> 00:17:54 given by taking the average values of the coordinates, 249 00:17:54 --> 00:17:56 x bar, y bar, z bar. 250 00:17:56 --> 00:18:00 It is just this formula but taking x, y or z as the 251 00:18:00 --> 00:18:08 function. There are moments of inertia. 252 00:18:08 --> 00:18:12 For example, the moment of inertia about the 253 00:18:12 --> 00:18:17 z-axis is the triple integral of x squared plus y squared density 254 00:18:17 --> 00:18:19 dV. Or, if you have just a 255 00:18:19 --> 00:18:22 two-dimensional object, it is the same formula, 256 00:18:22 --> 00:18:23 but, of course, with dA. 257 00:18:23 --> 00:18:26 And then we call that the polar moment of inertia because we 258 00:18:26 --> 00:18:30 thought of it as rotating the plane about the origin, 259 00:18:30 --> 00:18:34 but the origin is just where the z-axis hits the x, 260 00:18:34 --> 00:18:38 y plane so it is really the same thing. 261 00:18:38 --> 00:18:42 And we have also seen gravitational attraction in 262 00:18:42 --> 00:18:47 space, and I will let you look at your notes for that. 263 00:18:47 --> 00:18:57 It is just one formula to remember. 264 00:18:57 --> 00:19:02 Questions about iterated integrals, things like that? 265 00:19:02 --> 00:19:18 Yes? The formula that you should 266 00:19:18 --> 00:19:22 know for gravitational attraction is that if yu have a 267 00:19:22 --> 00:19:28 point mass at the origin and you have some solid centered on the 268 00:19:28 --> 00:19:33 z-axis that is attracting it then the force will be given by 269 00:19:33 --> 00:19:38 G times the mass times the triple integral of density times 270 00:19:38 --> 00:19:42 cosine phi over rho squared times dV. 271 00:19:42 --> 00:19:45 And, of course, you will actually do that in 272 00:19:45 --> 00:19:50 spherical coordinates because it is easier that way. 273 00:19:50 --> 00:19:55 That is the formula I have in mind. 274 00:19:55 --> 00:19:59 But, see, all these formulas just give you examples of things 275 00:19:59 --> 00:20:01 to integrate. And how to set up the bounds 276 00:20:01 --> 00:20:03 and so on does not depend on what you are actually 277 00:20:03 --> 00:20:08 integrating. It is done always using the 278 00:20:08 --> 00:20:22 same methods. Let's move on to work and line 279 00:20:22 --> 00:20:28 integrals. We have seen how to do that in 280 00:20:28 --> 00:20:35 the plane and in space. And it looks very similar 281 00:20:35 --> 00:20:41 somehow. Remember, you have to know how 282 00:20:41 --> 00:20:48 to set up and evaluate a line integral of this form. 283 00:20:48 --> 00:20:49 Let me do it in the plane this time. 284 00:20:49 --> 00:20:53 If you are in the plane you have two components, 285 00:20:53 --> 00:20:58 and then this becomes the line integral of M dx plus N dy. 286 00:20:58 --> 00:21:01 If you have a space curve then you will have a third component 287 00:21:01 --> 00:21:05 here. You will add that guy times dz. 288 00:21:05 --> 00:21:08 Now, how do we evaluate that? Well, it is very different from 289 00:21:08 --> 00:21:11 there because here we are just on a curve so there should be 290 00:21:11 --> 00:21:15 only one degree of freedom. One variable should be enough 291 00:21:15 --> 00:21:21 to know where we are. We will have to express x and y 292 00:21:21 --> 00:21:28 in terms of -- Well, I should and z optionally if 293 00:21:28 --> 00:21:32 there is one, in terms of a single 294 00:21:32 --> 00:21:36 parameters. And that might be just one of 295 00:21:36 --> 00:21:38 the coordinates. If you are told y equals z 296 00:21:38 --> 00:21:41 squared, that is easy. You just substitute y equals x 297 00:21:41 --> 00:21:44 squared and dy equals two x dx into everything, 298 00:21:44 --> 00:21:48 and you are left with an integral over x. 299 00:21:48 --> 00:22:01 Maybe it will be something in terms of time or in terms of an 300 00:22:01 --> 00:22:05 angle. We express everything in terms 301 00:22:05 --> 00:22:09 of a single parameter, and that will give us a usual 302 00:22:09 --> 00:22:14 single integral. Any questions about that? 303 00:22:14 --> 00:22:23 Yes? If you cannot parameterize the 304 00:22:23 --> 00:22:27 curve then it is really, really hard to evaluate the 305 00:22:27 --> 00:22:31 line integral. Well, you might be able to 306 00:22:31 --> 00:22:35 evaluate it numerically into a computer, but that is the 307 00:22:35 --> 00:22:40 easiest way to describe a curve. Indeed it could be that in the 308 00:22:40 --> 00:22:43 plane you have an equation in terms of x and y given by some 309 00:22:43 --> 00:22:46 completed formulas defining some curve. 310 00:22:46 --> 00:22:49 Then actually there are ways you can use basically 311 00:22:49 --> 00:22:52 differentials and constrained partials to figure out what the 312 00:22:52 --> 00:22:55 tangent vector to the curve is and so on. 313 00:22:55 --> 00:22:57 But we haven't really seen how to do that. 314 00:22:57 --> 00:23:00 That would be a really nice topic for tying together the end 315 00:23:00 --> 00:23:02 of the second unit that we discussed last time, 316 00:23:02 --> 00:23:04 constrained partials, with this stuff. 317 00:23:04 --> 00:23:08 But that is not going to be part of our topics. 318 00:23:08 --> 00:23:11 Basically, all the curves we have seen in this class, 319 00:23:11 --> 00:23:14 there is a way to express the position of a point in terms of 320 00:23:14 --> 00:23:23 a parameter. We haven't seen any curves that 321 00:23:23 --> 00:23:34 are so complicated that you cannot do that. 322 00:23:34 --> 00:23:37 The other thing we have seen is that there are some special 323 00:23:37 --> 00:23:40 cases of vector fields where we don't actually have to compute 324 00:23:40 --> 00:23:43 this thing because maybe we know that it is the gradient of some 325 00:23:43 --> 00:23:48 potential function. And then we have a fundamental 326 00:23:48 --> 00:23:55 theorem that gives us a way to compute this without computing 327 00:23:55 --> 00:24:04 it. We've seen about gradient 328 00:24:04 --> 00:24:16 fields and path independence. The thing to check is whether 329 00:24:16 --> 00:24:18 the curl of our vector field is zero. 330 00:24:18 --> 00:24:27 And remember in the plane that is one condition, 331 00:24:27 --> 00:24:33 Nx equals My. In 3D in space that is actually 332 00:24:33 --> 00:24:39 three conditions because you have to check all the mixed 333 00:24:39 --> 00:24:44 partials of the various components. 334 00:24:44 --> 00:24:47 If the curl of f is zero that tells us we are likely to have a 335 00:24:47 --> 00:24:52 gradient field. Strictly-speaking, 336 00:24:52 --> 00:25:06 I should mention and F is defined in a simply-connected 337 00:25:06 --> 00:25:16 region. Then F is a gradient field. 338 00:25:16 --> 00:25:25 That means that we can find a potential function. 339 00:25:25 --> 00:25:30 You can write F as the gradient of little f for some potential 340 00:25:30 --> 00:25:39 function little f. And we have seen how to find 341 00:25:39 --> 00:25:46 the potential. In fact, we have seen two 342 00:25:46 --> 00:25:50 methods for that. And we have seen them twice. 343 00:25:50 --> 00:25:53 We have seen them once for functions of two variables, 344 00:25:53 --> 00:25:55 once for functions of three variables. 345 00:25:55 --> 00:25:58 They look very much the same. I encourage you to compare your 346 00:25:58 --> 00:26:03 notes for the two side by side to see where they differ. 347 00:26:03 --> 00:26:05 Where they differ, roughly-speaking, 348 00:26:05 --> 00:26:09 well, I never know if it is the first or the second, 349 00:26:09 --> 00:26:14 but one of the two methods was to compute a line integral. 350 00:26:14 --> 00:26:18 In the plane, what we did is we set up and 351 00:26:18 --> 00:26:23 evaluated a line integral along our favorite path from the 352 00:26:23 --> 00:26:27 origin to a point with coordinates say x1, 353 00:26:27 --> 00:26:32 y1. And then we had to evaluate the 354 00:26:32 --> 00:26:38 line integral for the work done along this path. 355 00:26:38 --> 00:26:43 And that will give us the value of potential at that point. 356 00:26:43 --> 00:26:46 If we are doing it with three variables, that method remains 357 00:26:46 --> 00:26:49 very similar. The only difference is now we 358 00:26:49 --> 00:26:52 have to go also up in space to some point x1, 359 00:26:52 --> 00:26:55 y1, z1. And so we actually sum three 360 00:26:55 --> 00:26:59 pieces together. But on each piece it is the 361 00:26:59 --> 00:27:01 same story, only one variable changes. 362 00:27:01 --> 00:27:06 Here it is only x that changes, it is only y that changes, 363 00:27:06 --> 00:27:11 and on the third one only z would be changing. 364 00:27:11 --> 00:27:18 That is one possibility. And the other possibility for 365 00:27:18 --> 00:27:23 finding the potential is that we start with the condition that 366 00:27:23 --> 00:27:28 the first component of our vector field should be equal to 367 00:27:28 --> 00:27:32 f sub x for the unknown potential function. 368 00:27:32 --> 00:27:36 What we do is integrate with respect to x, 369 00:27:36 --> 00:27:40 and we will get our potential function up to an integration 370 00:27:40 --> 00:27:44 constant. And that integration constant 371 00:27:44 --> 00:27:48 typically depends on the remaining variables that might 372 00:27:48 --> 00:27:53 be y or equal in space y and z. And then what we have to do is 373 00:27:53 --> 00:27:56 take the partial of this with respect to y and compare it to 374 00:27:56 --> 00:28:00 what we want it to be, namely the y component of a 375 00:28:00 --> 00:28:02 vector field, and match them to get some 376 00:28:02 --> 00:28:08 information about this guy. And if we have three variables 377 00:28:08 --> 00:28:13 then there is a third step because there you will still 378 00:28:13 --> 00:28:18 have an unknown function of z that you need to get by 379 00:28:18 --> 00:28:23 comparing the partials with respect to z. 380 00:28:23 --> 00:28:30 I see a lot of very quiet faces somehow. 381 00:28:30 --> 00:28:36 Well, hopefully that is because you know that stuff. 382 00:28:36 --> 00:28:42 If it is because you are hopelessly confused then please 383 00:28:42 --> 00:28:49 review a lot before the final, but I really hope that is not 384 00:28:49 --> 00:28:52 the case. And so, 385 00:28:52 --> 00:29:22 386 00:29:22 --> 00:29:26 in particular, what we have seen is once we 387 00:29:26 --> 00:29:32 have the potential then we can use the fundamental theorem of 388 00:29:32 --> 00:29:38 calculus to tell us that if we have a line integral to compute 389 00:29:38 --> 00:29:44 for work along a curve that goes from some point P zero to some 390 00:29:44 --> 00:29:49 point P one then the line integral for the work done by 391 00:29:49 --> 00:29:54 gradient F is actually going just to be the change in value 392 00:29:54 --> 00:29:58 of a potential. And, in particular, 393 00:29:58 --> 00:30:02 that does not depend on how we got from P zero to P one. 394 00:30:02 --> 00:30:10 That is why we say that we have path independence. 395 00:30:10 --> 00:30:22 Next topic is flux in plane and space. 396 00:30:22 --> 00:30:26 Flux looks quite different in the plane and in space because, 397 00:30:26 --> 00:30:29 in the plane, it is just another kind of line 398 00:30:29 --> 00:30:32 integral, while in space it is a surface 399 00:30:32 --> 00:30:34 integral. If you were in four-dimensional 400 00:30:34 --> 00:30:35 space it would be a triple integral. 401 00:30:35 --> 00:30:43 Generally, you do flux for something that is somehow a wall 402 00:30:43 --> 00:30:50 that separates regions of space from each other. 403 00:30:50 --> 00:30:56 In the plane, the way we do it is we have a 404 00:30:56 --> 00:31:02 curve C and we look at its tangent vector, 405 00:31:02 --> 00:31:09 let's call that T, and we rotate it by 90 degrees 406 00:31:09 --> 00:31:13 clockwise. That is our convention to get a 407 00:31:13 --> 00:31:18 unit normal vector that points to the right of the curve as we 408 00:31:18 --> 00:31:21 move along the curve. That is our convention for 409 00:31:21 --> 00:31:24 orienting curves. And we are always going to be 410 00:31:24 --> 00:31:33 using that one. N equals T rotated 90 degrees 411 00:31:33 --> 00:31:37 clockwise. In particular, 412 00:31:37 --> 00:31:44 that means that n ds, which will be what we integrate 413 00:31:44 --> 00:31:52 against when we try to compute flux, will just end up being dy, 414 00:31:52 --> 00:31:58 negative dx. Concretely, when we have to 415 00:31:58 --> 00:32:03 evaluate a line integral of F dot n ds, 416 00:32:03 --> 00:32:07 geometrically we could try to take the dot product of our 417 00:32:07 --> 00:32:11 field with the normal vector and then sum the length element 418 00:32:11 --> 00:32:14 along the curve. And, in some cases, 419 00:32:14 --> 00:32:16 for example, if you know that the vector 420 00:32:16 --> 00:32:19 field is tangent to the curve or if a dot product is constant or 421 00:32:19 --> 00:32:22 things like that then that might actually give you a very easy 422 00:32:22 --> 00:32:24 answer. But, in general, 423 00:32:24 --> 00:32:28 the most efficient way to do it will be to say that if your 424 00:32:28 --> 00:32:31 vector field has components, I don't know, 425 00:32:31 --> 00:32:36 let's call them P and Q, then that will be just the line 426 00:32:36 --> 00:32:39 integral of PQ dot dy, negative dx, 427 00:32:39 --> 00:32:44 which means negative Q dx plus P dy. 428 00:32:44 --> 00:32:49 And, from that point onward, you evaluate it exactly the 429 00:32:49 --> 00:32:54 same way as you would for a work integral. 430 00:32:54 --> 00:32:56 But, of course, the geometric meaning is very 431 00:32:56 --> 00:32:57 different. It is the same meaning that we 432 00:32:57 --> 00:33:08 have always seen for flux. It measures how much a vector 433 00:33:08 --> 00:33:22 field goes across the curve. Now, if we are in space then 434 00:33:22 --> 00:33:30 you take flux for a surface, not for a curve. 435 00:33:30 --> 00:33:33 And the way it will work is that you have to choose an 436 00:33:33 --> 00:33:37 orientation of a surface, which just means choosing one 437 00:33:37 --> 00:33:40 of the two possible unit normal vectors. 438 00:33:40 --> 00:33:51 And then you will do a surface integral for F dot n dS. 439 00:33:51 --> 00:34:00 That is the surface i element. The setup for this surface 440 00:34:00 --> 00:34:07 integral is that first we have to express n and dS in some way. 441 00:34:07 --> 00:34:14 One possibility is that we can express the normal vector n dS 442 00:34:14 --> 00:34:18 geometrically. That is, for example, 443 00:34:18 --> 00:34:23 what we do when we look at, say, a horizontal plane or a 444 00:34:23 --> 00:34:27 vertical plane or a sphere or a cylinder. 445 00:34:27 --> 00:34:31 Then we have some geometric idea of why the normal vector is 446 00:34:31 --> 00:34:34 what it is and we have some formula for dS. 447 00:34:34 --> 00:34:39 Or, we can use one of the standard formulas. 448 00:34:39 --> 00:34:43 Basically, we have seen two formulas that work in fairly 449 00:34:43 --> 00:34:50 generate situations. One of them says -- If S is 450 00:34:50 --> 00:35:00 given by an equation z equals some function of x, 451 00:35:00 --> 00:35:06 y then you can just say n dS equals minus f sub x, 452 00:35:06 --> 00:35:09 minus f sub y, one, dx dy. 453 00:35:09 --> 00:35:14 And I need to rewrite that because I am running out of 454 00:35:14 --> 00:35:16 space. But, while I erase, 455 00:35:16 --> 00:35:20 I would like to point out the most important there in here. 456 00:35:20 --> 00:35:24 When I say n dS equals blah, blah, blah times dx dy, 457 00:35:24 --> 00:35:28 dx dy is not the same thing as dS at all. 458 00:35:28 --> 00:35:31 If you make that mistake you are going to get into trouble 459 00:35:31 --> 00:35:34 the next time that you try to buy real estate in a region 460 00:35:34 --> 00:35:37 which hills or cliffs or things like that. 461 00:35:37 --> 00:35:40 dS is the area on the slanted surface. 462 00:35:40 --> 00:35:46 dx dy is the area on the map that shows the x, 463 00:35:46 --> 00:35:49 y plane. And these are not the same 464 00:35:49 --> 00:35:51 thing. In particular, 465 00:35:51 --> 00:36:00 you cannot just take one piece of it and not the other piece. 466 00:36:00 --> 00:36:06 Let me give you formulas for n and for dS separately just to 467 00:36:06 --> 00:36:14 convince you. That way, if you feel that you 468 00:36:14 --> 00:36:22 need them, then you will have them. 469 00:36:22 --> 00:36:26 N is minus f sub x, minus f sub y, 470 00:36:26 --> 00:36:31 one, but scaled down to unit length. 471 00:36:31 --> 00:36:35 This is not a unit vector. It is actually divided by the 472 00:36:35 --> 00:36:38 length of this guy which is fx squared plus fy squared plus 473 00:36:38 --> 00:36:46 one. And dS is that same vector 474 00:36:46 --> 00:36:52 times dx dy. And so the square roots cancel 475 00:36:52 --> 00:36:56 out when you multiply them together. 476 00:36:56 --> 00:37:00 But it would be completely wrong to just say I will replace 477 00:37:00 --> 00:37:03 n dS by minus f sub x, minus f sub y and one. 478 00:37:03 --> 00:37:07 Then I end up again with the dS and I do something else with dS. 479 00:37:07 --> 00:37:16 That is a pretty bad conceptual mistake because it gives you the 480 00:37:16 --> 00:37:20 wrong answer. Another option more general 481 00:37:20 --> 00:37:24 than that. If we have not seen how to 482 00:37:24 --> 00:37:32 solve for z, how to express z as a function of x and y, 483 00:37:32 --> 00:37:40 well, maybe we still know some normal vector to the surface. 484 00:37:40 --> 00:37:49 Then there is another formula for n dS which is up to sine, 485 00:37:49 --> 00:37:55 N divided by N dot k dx dy. And, see, that projection 486 00:37:55 --> 00:37:58 formula works also if you have to project to another coordinate 487 00:37:58 --> 00:37:59 plane. For example, 488 00:37:59 --> 00:38:02 if you want to project to the x, z coordinate plane, 489 00:38:02 --> 00:38:10 the relation between n dS and dx dz is given by N over N dot 490 00:38:10 --> 00:38:13 j, because j is the direction 491 00:38:13 --> 00:38:19 perpendicular to the xz plane. But this one is more useful. 492 00:38:19 --> 00:38:24 What is a good example of that? If you have a slanted plane 493 00:38:24 --> 00:38:28 given to you, you can easily find its normal 494 00:38:28 --> 00:38:31 vector. That is just given by the 495 00:38:31 --> 00:38:33 coefficients of x, y, z in the equation. 496 00:38:33 --> 00:38:38 Another situation where that might happen is if your surface 497 00:38:38 --> 00:38:41 is given by an equation of a form of g of x, 498 00:38:41 --> 00:38:44 y, z equals zero. If that is the case then you 499 00:38:44 --> 00:38:47 know this is a level set of g. And we know how to find a 500 00:38:47 --> 00:38:50 normal vector to the level set, namely the gradient vector is 501 00:38:50 --> 00:38:52 always perpendicular to the level set. 502 00:38:52 --> 00:38:59 You would take the gradient of g to be your big N. 503 00:38:59 --> 00:39:06 OK. Now, these are basically all 504 00:39:06 --> 00:39:10 the integrals we have seen how to set up. 505 00:39:10 --> 00:39:15 Now we have a bunch of theorems relating them. 506 00:39:15 --> 00:39:23 Let me think about how I am going to organize that. 507 00:39:23 --> 00:39:32 Let me try like this. This part of the board will be 508 00:39:32 --> 00:39:35 work, this part of the board will be 509 00:39:35 --> 00:39:40 about flux and the left part of the board will be about things 510 00:39:40 --> 00:39:46 in the plane and the right one will be about things in space. 511 00:39:46 --> 00:39:53 What have we seen? Well, we have seen Green's 512 00:39:53 --> 00:39:57 theorem for work. That doesn't work so well 513 00:39:57 --> 00:39:59 because that is too small, so I am going to actually use 514 00:39:59 --> 00:40:01 more blackboards to do that. 515 00:40:01 --> 00:40:26 516 00:40:26 --> 00:40:32 This side will be space, this side will be the plane and 517 00:40:32 --> 00:40:37 we are going to start with theorems about work. 518 00:40:37 --> 00:40:41 And we will see theorems about flux pretty soon. 519 00:40:41 --> 00:40:47 We have two theorems about work. In the plane that is called 520 00:40:47 --> 00:40:51 Green's theorem. In space that is called Stokes' 521 00:40:51 --> 00:40:56 theorem. Green's theorem says if I have 522 00:40:56 --> 00:41:03 a closed curve in the plane going counterclockwise enclosing 523 00:41:03 --> 00:41:11 entirely some region R then the line integral along C for the 524 00:41:11 --> 00:41:20 work of F is equal to the double integral of a region inside of 525 00:41:20 --> 00:41:27 the curl of F dA. Concretely, if my components of 526 00:41:27 --> 00:41:35 F are called M and N that is the line integral of M dx plus N dy 527 00:41:35 --> 00:41:43 is equal to the double integral of R of N sub x minus M sub y 528 00:41:43 --> 00:41:46 dA. This side here is a usual line 529 00:41:46 --> 00:41:49 integral. This side here is a usual 530 00:41:49 --> 00:41:53 double integral in the plane. And somehow their values end up 531 00:41:53 --> 00:41:58 being magically related. Well, not quite magically. 532 00:41:58 --> 00:42:03 We actually have seen how to prove it. 533 00:42:03 --> 00:42:10 And now the analog of that in space is Stokes' theorem. 534 00:42:10 --> 00:42:15 Stokes says if I have a closed curve in space, 535 00:42:15 --> 00:42:20 now I have to decide what kind of thing it bounds. 536 00:42:20 --> 00:42:23 And the answer is it will have to bound some surface, 537 00:42:23 --> 00:42:28 but I have a choice of surface. I choose my favorite surface 538 00:42:28 --> 00:42:31 bounded by C. I guess I will just draw it 539 00:42:31 --> 00:42:33 like that. And I have to choose a 540 00:42:33 --> 00:42:36 compatible orientation. Remember, we have seen this 541 00:42:36 --> 00:42:39 right hand rule for choosing how to orient the surface. 542 00:42:39 --> 00:42:43 I believe, in this case, if I take C like that then the 543 00:42:43 --> 00:42:49 normal vector has to go up. And then it tells me how to 544 00:42:49 --> 00:42:53 compute the work done by F along C. 545 00:42:53 --> 00:43:01 Namely, that becomes the double integral over that surface S of 546 00:43:01 --> 00:43:08 curl F, which I will write as dell cross F dot n dS. 547 00:43:08 --> 00:43:10 This line integral is a usual line integral, 548 00:43:10 --> 00:43:14 but if for some reason we don't want to compute it directly we 549 00:43:14 --> 00:43:18 can actually replace it by a surface integral over any 550 00:43:18 --> 00:43:22 surface bounded by the curve. It might be that a problem will 551 00:43:22 --> 00:43:23 tell you which surface you have to consider. 552 00:43:23 --> 00:43:28 It might be that you will be left to choose the simplest 553 00:43:28 --> 00:43:33 possible surface you can think of that is somehow having this 554 00:43:33 --> 00:43:36 curve as its boundary. And so now, remember, 555 00:43:36 --> 00:43:40 curl of a vector field in space is going to be another vector 556 00:43:40 --> 00:43:42 expression. It has three components. 557 00:43:42 --> 00:43:45 And the way you compute it is not by remembering the actual 558 00:43:45 --> 00:43:48 formula, which is really complicated by, 559 00:43:48 --> 00:43:54 but by instead computing the cross-product between dell and 560 00:43:54 --> 00:44:01 F. You set up the cross-product. 561 00:44:01 --> 00:44:03 And, of course, it is a highly symbolic 562 00:44:03 --> 00:44:07 cross-product. I mean it is not an 563 00:44:07 --> 00:44:16 cross-product of actual vectors but it works the same way. 564 00:44:16 --> 00:44:20 Both of these formulas basically relate work on a curve 565 00:44:20 --> 00:44:24 with what happens to the curl on the surface that is enclosed by 566 00:44:24 --> 00:44:27 this curve, that is bounded by this curve. 567 00:44:27 --> 00:44:30 And in this one you have less freedom of choice because you 568 00:44:30 --> 00:44:34 don't have somehow a z direction in which you could move your 569 00:44:34 --> 00:44:36 surface. There is only possible choice 570 00:44:36 --> 00:44:39 of surface. There is only one thing that is 571 00:44:39 --> 00:44:42 enclosed by this curve in the plane. 572 00:44:42 --> 00:44:45 In both cases, these things tell you that you 573 00:44:45 --> 00:44:48 can think of curl as measuring how much the field fails to be 574 00:44:48 --> 00:44:51 conservative. See, if your field was 575 00:44:51 --> 00:44:55 conservative -- If a curl was zero then the right-hand side 576 00:44:55 --> 00:44:57 would just be zero. And that would be fortunate 577 00:44:57 --> 00:45:01 because if a curl is zero then your field is less conservative. 578 00:45:01 --> 00:45:02 That means it comes from a potential. 579 00:45:02 --> 00:45:05 That means when you go along a closed curve, 580 00:45:05 --> 00:45:09 well, the change of value of a potential should be zero. 581 00:45:09 --> 00:45:13 Another way to say it is path independence tells you no work. 582 00:45:13 --> 00:45:16 And, of course, if you have a vector field that 583 00:45:16 --> 00:45:20 is not a gradient field then the curl is not necessarily zero and 584 00:45:20 --> 00:45:23 then you get a more interesting answer. 585 00:45:23 --> 00:45:27 Finally, let's move onto the theorems about flux. 586 00:45:27 --> 00:45:43 That is Green for flux and that is the divergence theorem. 587 00:45:43 --> 00:45:54 Flux theorems. Here I say that will be 588 00:45:54 --> 00:46:01 divergence. And here it will be Green again. 589 00:46:01 --> 00:46:07 Green's theorem for flux says I have a closed curve that goes 590 00:46:07 --> 00:46:10 counterclockwise around some region. 591 00:46:10 --> 00:46:13 In particular, counterclockwise means that the 592 00:46:13 --> 00:46:16 normal vector will be going out of the region. 593 00:46:16 --> 00:46:20 And then it tells us that the flux out of the region, 594 00:46:20 --> 00:46:24 through the curve C, so that will be the line 595 00:46:24 --> 00:46:30 integral of F dot n ds is equal to the integral of a region 596 00:46:30 --> 00:46:37 inside of div F dA. And remember the divergence of 597 00:46:37 --> 00:46:42 M, N is just Mx plus Ny. This one here, 598 00:46:42 --> 00:46:46 the divergence theorem, tells you something similar but 599 00:46:46 --> 00:46:49 now for a region of space bounded by a closed surface. 600 00:46:49 --> 00:46:55 So if you have some region of space and you call its boundary 601 00:46:55 --> 00:47:00 surface S and you let n be the normal vector that goes out of 602 00:47:00 --> 00:47:07 the region R. You orient S outwards. 603 00:47:07 --> 00:47:15 Then the flux out of the region through S is going to be the 604 00:47:15 --> 00:47:23 same as the triple integral over the region of divergence F dV. 605 00:47:23 --> 00:47:31 Remember, the divergence of a vector field with components P, 606 00:47:31 --> 00:47:36 Q, R is Px plus Qy plus Rz. What do these two theorems say? 607 00:47:36 --> 00:47:39 Well, they say essentially the same thing. 608 00:47:39 --> 00:47:43 They say the total flux out of a region is equal to the 609 00:47:43 --> 00:47:46 integral of divergence over whatever is inside. 610 00:47:46 --> 00:47:49 And the reason for that is, again, we have seen for a 611 00:47:49 --> 00:47:52 velocity field that divergence measures how much things are 612 00:47:52 --> 00:47:55 expanding or how much stuff is being created. 613 00:47:55 --> 00:47:59 It tells you the amount of sources per unit portion of the 614 00:47:59 --> 00:48:01 region. When you sum that over 615 00:48:01 --> 00:48:03 everything, you get the amount of fluid 616 00:48:03 --> 00:48:05 that is being, you know, the total amount of 617 00:48:05 --> 00:48:08 sources inside here, and that tells us how much 618 00:48:08 --> 00:48:10 stuff has to go out per unit time. 619 00:48:10 --> 00:48:12 That is basically the interpretation. 620 00:48:12 --> 00:48:16 In a way, I would be tempted to say that this table of four 621 00:48:16 --> 00:48:20 theorems is somehow the crucial point of 18.02. 622 00:48:20 --> 00:48:23 And you would do well to remember them. 623 00:48:23 --> 00:48:26 However, I would like also to point out that these theorems 624 00:48:26 --> 00:48:28 are completely useless if you don't know how to compute any of 625 00:48:28 --> 00:48:31 the integrals that are in there. So all the stuff that was 626 00:48:31 --> 00:48:35 around there before is actually somehow more fundamental. 627 00:48:35 --> 00:48:38 And if you don't know how to compute the double or triple 628 00:48:38 --> 00:48:41 integrals then this is of little use to you. 629 00:48:41 --> 00:48:48 That is the end. I guess I have to wish you 630 00:48:48 --> 00:48:52 happy holidays. 631 00:48:52 --> 00:48:57