1 00:00:01,000 --> 00:00:03,000 The following content is provided under a Creative 2 00:00:03,000 --> 00:00:05,000 Commons license. Your support will help MIT 3 00:00:05,000 --> 00:00:08,000 OpenCourseWare continue to offer high quality educational 4 00:00:08,000 --> 00:00:13,000 resources for free. To make a donation or to view 5 00:00:13,000 --> 00:00:18,000 additional materials from hundreds of MIT courses, 6 00:00:18,000 --> 00:00:23,000 visit MIT OpenCourseWare at ocw.mit.edu. 7 00:00:23,000 --> 00:00:28,000 Yesterday we learned about flux and we have seen the first few 8 00:00:28,000 --> 00:00:33,000 examples of how to set up and compute integrals for a flux of 9 00:00:33,000 --> 00:00:40,000 a vector field for a surface. Remember the flux of a vector 10 00:00:40,000 --> 00:00:47,000 field F through the surface S is defined by taking the double 11 00:00:47,000 --> 00:00:55,000 integral on the surface of F dot n dS where n is the unit normal 12 00:00:55,000 --> 00:01:03,000 to the surface and dS is the area element on the surface. 13 00:01:03,000 --> 00:01:06,000 As we have seen, for various surfaces, 14 00:01:06,000 --> 00:01:09,000 we have various formulas telling us what the normal 15 00:01:09,000 --> 00:01:12,000 vector is and what the area element becomes. 16 00:01:12,000 --> 00:01:14,000 For example, on spheres we typically 17 00:01:14,000 --> 00:01:18,000 integrate with respect to phi and theta for latitude and 18 00:01:18,000 --> 00:01:21,000 longitude angles. On a horizontal plane, 19 00:01:21,000 --> 00:01:25,000 we would just end up degrading dx, dy and so on. 20 00:01:25,000 --> 00:01:29,000 At the end of lecture we saw a formula. 21 00:01:29,000 --> 00:01:32,000 A lot of you asked me how we got it. 22 00:01:32,000 --> 00:01:36,000 Well, we didn't get it yet. We are going to try to explain 23 00:01:36,000 --> 00:01:39,000 where it comes from and why it works. 24 00:01:39,000 --> 00:01:50,000 The case we want to look at is if S is the graph of a function, 25 00:01:50,000 --> 00:02:01,000 it is given by z equals some function in terms of x and y. 26 00:02:01,000 --> 00:02:08,000 Our surface is out here. Z is a function of x and y. 27 00:02:08,000 --> 00:02:17,000 And x and y will range over some domain in the x, 28 00:02:17,000 --> 00:02:28,000 y plane, namely the region that is the shadow of the surface on 29 00:02:28,000 --> 00:02:34,000 the x, y plane. I said that we will have a 30 00:02:34,000 --> 00:02:40,000 formula for n dS which will end up being plus/minus minus f sub 31 00:02:40,000 --> 00:02:45,000 x, minus f sub y, one dxdy, 32 00:02:45,000 --> 00:02:49,000 so that we will set up and evaluate the integral in terms 33 00:02:49,000 --> 00:02:53,000 of x and y. Every time we see z we will 34 00:02:53,000 --> 00:02:57,000 replace it by f of xy, whatever the formula for f 35 00:02:57,000 --> 00:03:00,000 might be. Actually, if we look at a very 36 00:03:00,000 --> 00:03:04,000 easy case where this is just a horizontal plane, 37 00:03:04,000 --> 00:03:07,000 z equals constant, the function is just a 38 00:03:07,000 --> 00:03:09,000 constant, well, the partial derivatives 39 00:03:09,000 --> 00:03:14,000 become just zero. You get dx, dy. 40 00:03:14,000 --> 00:03:17,000 That is what you would expect for a horizontal plane just from 41 00:03:17,000 --> 00:03:19,000 common sense. This is more interesting, 42 00:03:19,000 --> 00:03:23,000 of course, if a function is more interesting. 43 00:03:23,000 --> 00:03:30,000 How do we get that? Where does this come from? 44 00:03:30,000 --> 00:03:36,000 We need to figure out, for a small piece of our 45 00:03:36,000 --> 00:03:43,000 surface, what will be n delta S. Let's say that we take a small 46 00:03:43,000 --> 00:03:51,000 rectangle in here corresponding to sides delta x and delta y and 47 00:03:51,000 --> 00:03:58,000 we look at the piece of surface that is above that. 48 00:03:58,000 --> 00:04:03,000 Well, the question we have now is what is the area of this 49 00:04:03,000 --> 00:04:08,000 little piece of surface and what is its normal vector? 50 00:04:08,000 --> 00:04:10,000 Observe this little piece up here. 51 00:04:10,000 --> 00:04:12,000 If it is small enough, it will look like a 52 00:04:12,000 --> 00:04:14,000 parallelogram. I mean it might be slightly 53 00:04:14,000 --> 00:04:17,000 curvy, but roughly it looks like a parallelogram in space. 54 00:04:17,000 --> 00:04:22,000 And so we have seen how to find the area of a parallelogram in 55 00:04:22,000 --> 00:04:25,000 space using cross-product. If we can figure out what are 56 00:04:25,000 --> 00:04:29,000 the vectors for this side and that side then taking that 57 00:04:29,000 --> 00:04:33,000 cross-product and taking the magnitude of the cross-product 58 00:04:33,000 --> 00:04:36,000 will give us the area. Moreover, the cross-product 59 00:04:36,000 --> 00:04:38,000 also gives us the normal direction. 60 00:04:38,000 --> 00:04:40,000 In fact, the cross-product gives us two in one. 61 00:04:40,000 --> 00:04:44,000 It gives us the normal direction and the area element. 62 00:04:44,000 --> 00:04:48,000 And that is why I said that we will have an easy formula for n 63 00:04:48,000 --> 00:04:52,000 dS while n and dS taken separately are more complicated 64 00:04:52,000 --> 00:04:55,000 because you would have to actually take the length of a 65 00:04:55,000 --> 00:05:03,000 direction of this guy. Let's carry out this problem. 66 00:05:03,000 --> 00:05:13,000 Let's say I am going to look at a small piece of the x, 67 00:05:13,000 --> 00:05:17,000 y plane. Here I have delta x, 68 00:05:17,000 --> 00:05:22,000 here I have delta y, and I am starting at some point 69 00:05:22,000 --> 00:05:28,000 (x, y). Now, above that I will have a 70 00:05:28,000 --> 00:05:34,000 parallelogram on my surface. This point here, 71 00:05:34,000 --> 00:05:36,000 the point where I start, I know what it is. 72 00:05:36,000 --> 00:05:43,000 It is just (x, y). And, well, z is f(x, y). 73 00:05:43,000 --> 00:05:44,000 Now what I want to find, actually, 74 00:05:44,000 --> 00:05:50,000 is what are these two vectors, let's call them U and V, 75 00:05:50,000 --> 00:05:54,000 that correspond to moving a bit in the x direction or in the y 76 00:05:54,000 --> 00:05:58,000 direction? And then U cross V will be, 77 00:05:58,000 --> 00:06:06,000 well, in terms of the magnitude of this guy will just be the 78 00:06:06,000 --> 00:06:11,000 little piece of surface area, delta S. 79 00:06:11,000 --> 00:06:15,000 And, in terms of direction, it will be normal to the 80 00:06:15,000 --> 00:06:19,000 surface. Actually, I will get just delta 81 00:06:19,000 --> 00:06:23,000 S times my normal vector. Well, up to sign because, 82 00:06:23,000 --> 00:06:26,000 depending on whether I do U V or V U, I might get the normal 83 00:06:26,000 --> 00:06:29,000 vector in the direction I want or in the opposite direction. 84 00:06:29,000 --> 00:06:36,000 But we will take care of that later. 85 00:06:36,000 --> 00:06:41,000 Let's find U and V. And, in case you have trouble 86 00:06:41,000 --> 00:06:47,000 with that small picture, I have a better one here. 87 00:06:47,000 --> 00:06:53,000 Let's keep it just in case this one gets really too cluttered. 88 00:06:53,000 --> 00:06:57,000 It really represents the same thing. 89 00:06:57,000 --> 00:07:05,000 Let's try to figure out these vectors U and V. 90 00:07:05,000 --> 00:07:13,000 Vector U starts at the point x, y, f of x, y and it goes to -- 91 00:07:13,000 --> 00:07:19,000 Whereas, its head, well, I will have moved x by 92 00:07:19,000 --> 00:07:23,000 delta x. So, x plus delta x and y 93 00:07:23,000 --> 00:07:25,000 doesn't change. And, of course, 94 00:07:25,000 --> 00:07:31,000 the z coordinate has to change. It becomes f of x plus delta x 95 00:07:31,000 --> 00:07:34,000 and y. Now, how does f change if I 96 00:07:34,000 --> 00:07:37,000 change x a little bit? Well, we have seen that it is 97 00:07:37,000 --> 00:07:41,000 given by the partial derivative f sub x. 98 00:07:41,000 --> 00:07:47,000 This is approximately equal to f of x, y plus delta x times f 99 00:07:47,000 --> 00:07:51,000 sub x at the given point x, y. 100 00:07:51,000 --> 00:07:56,000 I am not going to add it because the notation is already 101 00:07:56,000 --> 00:07:59,000 long enough. That means my vector U, 102 00:07:59,000 --> 00:08:05,000 well, approximately because I am using this linear 103 00:08:05,000 --> 00:08:08,000 approximation, 00:08:15,000 0, f sub x times delta x>. Is that OK with everyone? 105 00:08:15,000 --> 00:08:19,000 Good. Now, what about V? 106 00:08:19,000 --> 00:08:23,000 Well, V works the same way so I am not going to do all the 107 00:08:23,000 --> 00:08:27,000 details. When I move from here to here x 108 00:08:27,000 --> 00:08:30,000 doesn't change and y changes by delta y. 109 00:08:30,000 --> 00:08:39,000 X component nothing happens. Y component changes by delta y. 110 00:08:39,000 --> 00:08:42,000 What about the z component? Well, f changes by f sub y 111 00:08:42,000 --> 00:08:51,000 times delta y. That is how f changes if I 112 00:08:51,000 --> 00:08:59,000 increase y by delta y. I have my two sides. 113 00:08:59,000 --> 00:09:03,000 Now I can take that cross-product. 114 00:09:03,000 --> 00:09:08,000 Well, maybe I will first factor something out. 115 00:09:08,000 --> 00:09:13,000 See, I can rewrite this as one, zero, f sub x times delta x. 116 00:09:13,000 --> 00:09:20,000 And this one I will rewrite as zero, one, f sub y delta y. 117 00:09:20,000 --> 00:09:26,000 And so now the cross-product, n hat delta S up to sign is 118 00:09:26,000 --> 00:09:31,000 going to be U cross V. We will have to do the 119 00:09:31,000 --> 00:09:35,000 cross-product, and we will have a delta x, 120 00:09:35,000 --> 00:09:41,000 delta y coming out. I am just saving myself the 121 00:09:41,000 --> 00:09:47,000 trouble of writing a lot of delta x's and delta y's, 122 00:09:47,000 --> 00:09:55,000 but if you prefer you can just do directly this cross-product. 123 00:09:55,000 --> 00:09:57,000 Let's compute this cross-product. 124 00:09:57,000 --> 00:10:03,000 Well, the i component is zero minus f sub x. 125 00:10:03,000 --> 00:10:07,000 The y component is going to be, well, f sub y minus zero but 126 00:10:07,000 --> 00:10:12,000 with the minus sign in front of everything, so negative f sub y. 127 00:10:12,000 --> 00:10:20,000 And the z component will be just one times delta x delta y. 128 00:10:20,000 --> 00:10:25,000 Does that make sense? Yes. 129 00:10:25,000 --> 00:10:30,000 Very good. And so now we shrink this 130 00:10:30,000 --> 00:10:32,000 rectangle, we shrink delta x and delta y 131 00:10:32,000 --> 00:10:35,000 to zero, that is how we get this formula 132 00:10:35,000 --> 00:10:38,000 for n dS equals negative fx, negative fy, 133 00:10:38,000 --> 00:10:43,000 one, dxdy. Well, plus/minus because it is 134 00:10:43,000 --> 00:10:47,000 up to us to choose whether we want to take the normal vector 135 00:10:47,000 --> 00:10:50,000 point up or down. See, if you take this 136 00:10:50,000 --> 00:10:55,000 convention then the z component of n dS is positive. 137 00:10:55,000 --> 00:10:59,000 That corresponds to normal vector pointing up. 138 00:10:59,000 --> 00:11:02,000 If you take the opposite signs then the z component will be 139 00:11:02,000 --> 00:11:06,000 negative. That means your normal vector 140 00:11:06,000 --> 00:11:15,000 points down. This one is with n pointing up. 141 00:11:15,000 --> 00:11:17,000 I mean when I say up, of course it is still 142 00:11:17,000 --> 00:11:21,000 perpendicular to the surface. If the surface really has a big 143 00:11:21,000 --> 00:11:26,000 slope then it is not really going to go all that much up, 144 00:11:26,000 --> 00:11:31,000 but more up than down. OK. 145 00:11:31,000 --> 00:11:40,000 That is how we get the formula. Any questions? 146 00:11:40,000 --> 00:11:41,000 No. OK. 147 00:11:41,000 --> 00:11:44,000 That is a really useful formula. 148 00:11:44,000 --> 00:11:49,000 You don't really need to remember all the details of how 149 00:11:49,000 --> 00:11:54,000 we got it, but please remember that formula. 150 00:11:54,000 --> 00:12:06,000 Let's do an example, actually. Let's say we want to find the 151 00:12:06,000 --> 00:12:08,000 flux of the vector field z times k, 152 00:12:08,000 --> 00:12:12,000 so it is a vertical vector field, 153 00:12:12,000 --> 00:12:27,000 through the portion of the paraboloid z equals x^2 y^2 that 154 00:12:27,000 --> 00:12:36,000 lives above the unit disk. What does that mean? 155 00:12:36,000 --> 00:12:39,000 z = x^2 y^2. We have seen it many times. 156 00:12:39,000 --> 00:12:43,000 It is this parabola and is pointing up. 157 00:12:43,000 --> 00:12:47,000 Above the unit disk means I don't care about this infinite 158 00:12:47,000 --> 00:12:52,000 surface. I will actually stop when I hit 159 00:12:52,000 --> 00:12:56,000 a radius of one away from the z-axis. 160 00:12:56,000 --> 00:13:03,000 And so now I have my vector field which is going to point 161 00:13:03,000 --> 00:13:09,000 overall up because, well, it is z times k. 162 00:13:09,000 --> 00:13:13,000 The more z is positive, the more your vector field goes 163 00:13:13,000 --> 00:13:16,000 up. Of course, if z were negative 164 00:13:16,000 --> 00:13:20,000 then it would point down, but it will live above. 165 00:13:20,000 --> 00:13:25,000 Actually, a quick opinion poll. What do you think the flux 166 00:13:25,000 --> 00:13:28,000 should be? Should it be positive, 167 00:13:28,000 --> 00:13:32,000 zero, negative or we don't know? 168 00:13:32,000 --> 00:13:36,000 I see some I don't know, I see some negative and I see 169 00:13:36,000 --> 00:13:39,000 some positive. Of course, I didn't tell you 170 00:13:39,000 --> 00:13:42,000 which way I am orienting my paraboloid. 171 00:13:42,000 --> 00:13:44,000 So far both answers are correct. The only one that is probably 172 00:13:44,000 --> 00:13:47,000 not correct is zero because, no matter which way you choose 173 00:13:47,000 --> 00:13:49,000 to orient it you should get something. 174 00:13:49,000 --> 00:13:51,000 It is not looking like it will be zero. 175 00:13:51,000 --> 00:14:00,000 Let's say that I am going to do it with the normal pointing 176 00:14:00,000 --> 00:14:06,000 upwards. Second chance. 177 00:14:06,000 --> 00:14:11,000 I see some people changing back and forth from one and two. 178 00:14:11,000 --> 00:14:14,000 Let's draw a picture. Which one is pointing upwards? 179 00:14:14,000 --> 00:14:16,000 Well, let's look at the bottom point. 180 00:14:16,000 --> 00:14:19,000 The normal vector pointing up, here we know what it means. 181 00:14:19,000 --> 00:14:22,000 It is this guy. If you continue to follow your 182 00:14:22,000 --> 00:14:26,000 normal vector, see, they are actually pointing 183 00:14:26,000 --> 00:14:30,000 up and into the paraboloid. And I claim that the answer 184 00:14:30,000 --> 00:14:34,000 should be positive because the vector field is crossing our 185 00:14:34,000 --> 00:14:38,000 paraboliod going upwards, going from the outside out and 186 00:14:38,000 --> 00:14:46,000 below to the inside and upside. So, in the direction that we 187 00:14:46,000 --> 00:14:55,000 are counting positively. We will see how it turns out 188 00:14:55,000 --> 00:15:03,000 when we do the calculation. We have to compute the integral 189 00:15:03,000 --> 00:15:08,000 for flux. Double integral over a surface 190 00:15:08,000 --> 00:15:15,000 of F dot n dS is going to be -- What are we going to do? 191 00:15:15,000 --> 00:15:21,000 Well, F we said is <0,0, z>. 192 00:15:21,000 --> 00:15:25,000 What is n dS. Well, let's use our brand new 193 00:15:25,000 --> 00:15:28,000 formula. It says negative f sub x, 194 00:15:28,000 --> 00:15:31,000 negative f sub y, one, dxdy. 195 00:15:31,000 --> 00:15:38,000 What does little f in here? It is x^2 y^2. 196 00:15:38,000 --> 00:15:42,000 When we are using this formula, we need to know what little x 197 00:15:42,000 --> 00:15:47,000 stands for. It is whatever the formula is 198 00:15:47,000 --> 00:15:53,000 for z as a function of x and y. We take x^2 y^2 and we take the 199 00:15:53,000 --> 00:15:57,000 partial derivatives with minus signs. 200 00:15:57,000 --> 00:16:01,000 We get negative 2x, negative 2y and one, 201 00:16:01,000 --> 00:16:03,000 dxdy. Well, of course here it didn't 202 00:16:03,000 --> 00:16:05,000 really matter because we are going to dot them with zero. 203 00:16:05,000 --> 00:16:11,000 Actually, even if we had made a mistake we somehow wouldn't have 204 00:16:11,000 --> 00:16:15,000 had to pay the price. But still. 205 00:16:15,000 --> 00:16:22,000 We will end up with double integral on S of z dxdy. 206 00:16:22,000 --> 00:16:25,000 Now, what do we do with that? Well, we have too many things. 207 00:16:25,000 --> 00:16:43,000 We have to get rid of z. Let's use z equals x^2 y^2 once 208 00:16:43,000 --> 00:16:51,000 more. That becomes double integral of 209 00:16:51,000 --> 00:16:54,000 x^2 y^2 dxdy. And here, see, 210 00:16:54,000 --> 00:16:57,000 we are using the fact that we are only looking at things that 211 00:16:57,000 --> 00:16:59,000 are on the surface. It is not like in a triple 212 00:16:59,000 --> 00:17:01,000 integral. You could never do that because 213 00:17:01,000 --> 00:17:04,000 z, x and y are independent. Here they are related by the 214 00:17:04,000 --> 00:17:09,000 equation of a surface. If I sound like I am ranting, 215 00:17:09,000 --> 00:17:14,000 but I know from experience this is where one of the most sticky 216 00:17:14,000 --> 00:17:18,000 and tricky points is. OK. 217 00:17:18,000 --> 00:17:20,000 How will we actually integrate that? 218 00:17:20,000 --> 00:17:22,000 Well, now that we have just x and y, we should figure out what 219 00:17:22,000 --> 00:17:25,000 is the range for x and y. Well, the range for x and y is 220 00:17:25,000 --> 00:17:27,000 going to be the shadow of our region. 221 00:17:27,000 --> 00:17:34,000 It is going to be this unit disk. 222 00:17:34,000 --> 00:17:43,000 I can just do that for now. And this is finally where I 223 00:17:43,000 --> 00:17:46,000 have left the world of surface integrals to go back to a usual 224 00:17:46,000 --> 00:17:49,000 double integral. And now I have to set it up. 225 00:17:49,000 --> 00:17:52,000 Well, I can do it this way with dxdy, but it looks like there is 226 00:17:52,000 --> 00:17:55,000 a smarter thing to do. I am going to use polar 227 00:17:55,000 --> 00:17:59,000 coordinates. In fact, I am going to say this 228 00:17:59,000 --> 00:18:03,000 is double integral of r^2 times r dr d theta. 229 00:18:03,000 --> 00:18:07,000 I am on the unit disk so r goes zero to one, theta goes zero to 230 00:18:07,000 --> 00:18:12,000 2pi. And, if you do the calculation, 231 00:18:12,000 --> 00:18:19,000 you will find that this is going to be pi over two. 232 00:18:19,000 --> 00:18:26,000 Any questions about the example. Yes? 233 00:18:26,000 --> 00:18:29,000 How did I get this negative 2x and negative 2y? 234 00:18:29,000 --> 00:18:33,000 I want to use my formula for n dS. 235 00:18:33,000 --> 00:18:36,000 My surface is given by the graph of a function. 236 00:18:36,000 --> 00:18:40,000 It is the graph of a function x^2 y^2. 237 00:18:40,000 --> 00:18:43,000 I will use this formula that is up here. 238 00:18:43,000 --> 00:18:47,000 I will take the function x^2 y^2 and I will take its partial 239 00:18:47,000 --> 00:18:50,000 derivatives. If I take the partial of f, 240 00:18:50,000 --> 00:18:55,000 so x^2 y^2 with respect to x, I get 2x, so I put negative 2x. 241 00:18:55,000 --> 00:18:57,000 And then the same thing, negative 2y, 242 00:18:57,000 --> 00:19:01,000 one, dxdy. Yes? 243 00:19:01,000 --> 00:19:04,000 Which k hat? Oh, you mean the vector field. 244 00:19:04,000 --> 00:19:06,000 It is a different part of the story. 245 00:19:06,000 --> 00:19:10,000 Whenever you do a surface integral for flux you have two 246 00:19:10,000 --> 00:19:13,000 parts of the story. One is the vector field whose 247 00:19:13,000 --> 00:19:17,000 flux you are taking. The other one is the surface 248 00:19:17,000 --> 00:19:21,000 for which you will be taking flux. 249 00:19:21,000 --> 00:19:26,000 The vector field only comes as this f in the notation, 250 00:19:26,000 --> 00:19:28,000 and everything else, the bounds in the double 251 00:19:28,000 --> 00:19:32,000 integral and the n dS, all come from the surface that 252 00:19:32,000 --> 00:19:36,000 we are looking at. Basically, in all of this 253 00:19:36,000 --> 00:19:40,000 calculation, this is coming from f equals zk. 254 00:19:40,000 --> 00:19:47,000 Everything else comes from the information paraboloid z = x^2 255 00:19:47,000 --> 00:19:51,000 y^2 above the unit disk. In particular, 256 00:19:51,000 --> 00:19:55,000 if we wanted to now find the flux of any other vector field 257 00:19:55,000 --> 00:19:58,000 for the same paraboloid, well, all we would have to do 258 00:19:58,000 --> 00:20:02,000 is just replace this guy by whatever the new vector field 259 00:20:02,000 --> 00:20:05,000 is. We have learned how to set up 260 00:20:05,000 --> 00:20:09,000 flux integrals for this paraboloid. 261 00:20:09,000 --> 00:20:10,000 Not that you should remember this one by heart. 262 00:20:10,000 --> 00:20:13,000 I mean there are many paraboloids in life and other 263 00:20:13,000 --> 00:20:17,000 surfaces, too. It is better to remember the 264 00:20:17,000 --> 00:20:22,000 general method. Any other questions? 265 00:20:22,000 --> 00:20:25,000 No. OK. 266 00:20:25,000 --> 00:20:31,000 Let's see more ways of taking flux integrals. 267 00:20:31,000 --> 00:20:33,000 But, just to reassure you, at this point we have seen the 268 00:20:33,000 --> 00:20:36,000 most important ones. 90% of the problems that we 269 00:20:36,000 --> 00:20:41,000 will be looking at we can do with what we have seen so far in 270 00:20:41,000 --> 00:20:49,000 less time and this formula. Let's look a little bit at a 271 00:20:49,000 --> 00:20:56,000 more general situation. Let's say that my surface is so 272 00:20:56,000 --> 00:21:00,000 complicated that I cannot actually express z as a function 273 00:21:00,000 --> 00:21:04,000 of x and y, but let's say that I know how to parametize it. 274 00:21:04,000 --> 00:21:06,000 I have a parametric equation for my surface. 275 00:21:06,000 --> 00:21:11,000 That means I can express x, y and z in terms of any two 276 00:21:11,000 --> 00:21:16,000 parameter variables that might be relevant for me. 277 00:21:16,000 --> 00:21:19,000 If you want, this one here is a special case 278 00:21:19,000 --> 00:21:23,000 where you can parameterize things in terms of x and y as 279 00:21:23,000 --> 00:21:26,000 your two variables. How would you do it in the 280 00:21:26,000 --> 00:21:29,000 fully general case? In a way, that will answer your 281 00:21:29,000 --> 00:21:30,000 question that, I think one of you, 282 00:21:30,000 --> 00:21:34,000 I forgot, asked yesterday how would I do it in general? 283 00:21:34,000 --> 00:21:36,000 Is there a formula like M dx plus N dy? 284 00:21:36,000 --> 00:21:39,000 Well, that is going to be the general formula. 285 00:21:39,000 --> 00:21:42,000 And you will see that it is a little bit too complicated, 286 00:21:42,000 --> 00:21:46,000 so the really useful ones are actually the special ones. 287 00:21:46,000 --> 00:21:56,000 Let's say that we are given a parametric description -- -- of 288 00:21:56,000 --> 00:22:01,000 a surface S. That means we can describe S by 289 00:22:01,000 --> 00:22:05,000 formulas saying x is some function of two parameter 290 00:22:05,000 --> 00:22:08,000 variables. I am going to call them u and v. 291 00:22:08,000 --> 00:22:10,000 I hope you don't mind. You can call them t1 and t2. 292 00:22:10,000 --> 00:22:18,000 You can call them whatever you want. 293 00:22:18,000 --> 00:22:21,000 One of the basic properties of a surface is because I have only 294 00:22:21,000 --> 00:22:23,000 two independent directions to move on. 295 00:22:23,000 --> 00:22:26,000 I should be able to express x, y and z in terms of two 296 00:22:26,000 --> 00:22:29,000 variables. Now, let's say that I know how 297 00:22:29,000 --> 00:22:32,000 to do that. Or, maybe I should instead 298 00:22:32,000 --> 00:22:35,000 think of it in terms of a position vector if it helps you. 299 00:22:35,000 --> 00:22:40,000 That is just a vector with components 00:22:46,000 y, z> is given as a function of u and v. 301 00:22:46,000 --> 00:22:50,000 It works like a parametric curve but with two parameters. 302 00:22:50,000 --> 00:22:54,000 Now, how would we actually set up a flux integral on such a 303 00:22:54,000 --> 00:22:57,000 surface. Well, because we are locating 304 00:22:57,000 --> 00:23:01,000 ourselves in terms of u and v, we will end up with an integral 305 00:23:01,000 --> 00:23:06,000 du dv. We need to figure out how to 306 00:23:06,000 --> 00:23:11,000 express n dS in terms of du and dv. 307 00:23:11,000 --> 00:23:19,000 N dS should be something du dv. How do we do that? 308 00:23:19,000 --> 00:23:25,000 Well, we can use the same method that we have actually 309 00:23:25,000 --> 00:23:28,000 used over here. Because, if you think for a 310 00:23:28,000 --> 00:23:30,000 second, here we used, of course, a rectangle in the 311 00:23:30,000 --> 00:23:34,000 x, y plane and we lifted it to a parallelogram and so on. 312 00:23:34,000 --> 00:23:37,000 But more generally you can think what happens if I change u 313 00:23:37,000 --> 00:23:41,000 by delta u keeping v constant or the other way around? 314 00:23:41,000 --> 00:23:45,000 You will get some sort of mesh grid on your surface and you 315 00:23:45,000 --> 00:23:48,000 will look at a little parallelogram that is an 316 00:23:48,000 --> 00:23:52,000 elementary piece of that mesh and figure out what is its area 317 00:23:52,000 --> 00:23:57,000 and normal vector. Well, that will again be given 318 00:23:57,000 --> 00:24:01,000 by the cross-product of the two sides. 319 00:24:01,000 --> 00:24:07,000 Let's think a little bit about what happens when I move a 320 00:24:07,000 --> 00:24:12,000 little bit on my surface. I am taking this grid on my 321 00:24:12,000 --> 00:24:16,000 surface given by the u and v directions. 322 00:24:16,000 --> 00:24:25,000 And, if I take a piece of that corresponding to small changes 323 00:24:25,000 --> 00:24:33,000 delta u and delta v, what is going to be going on 324 00:24:33,000 --> 00:24:36,000 here? Well, I have to deal with two 325 00:24:36,000 --> 00:24:39,000 vectors, one corresponding to changing u, the other one 326 00:24:39,000 --> 00:24:42,000 corresponding to changing v. If I change u, 327 00:24:42,000 --> 00:24:46,000 how does my point change? Well, it is given by the 328 00:24:46,000 --> 00:24:49,000 derivative of this with respect to u. 329 00:24:49,000 --> 00:24:57,000 This vector here I will call, so the sides are given by, 330 00:24:57,000 --> 00:25:05,000 let me say, partial r over partial u times delta u. 331 00:25:05,000 --> 00:25:09,000 If you prefer, maybe I should write it as 332 00:25:09,000 --> 00:25:13,000 partial x over partial u times delta u. 333 00:25:13,000 --> 00:25:17,000 Well, it is just too boring to write. 334 00:25:17,000 --> 00:25:21,000 And so on. It means if I change u a little 335 00:25:21,000 --> 00:25:24,000 bit, keeping v constant, then how x changes is, 336 00:25:24,000 --> 00:25:26,000 given by partial x over partial u times delta u, 337 00:25:26,000 --> 00:25:28,000 same thing with y, same thing with z, 338 00:25:28,000 --> 00:25:33,000 and I am just using vector notation to do it this way. 339 00:25:33,000 --> 00:25:41,000 That is the analog of when I said delta r for line integrals 340 00:25:41,000 --> 00:25:47,000 along a curve, vector delta r is the velocity 341 00:25:47,000 --> 00:25:58,000 vector dr dt times delta t. Now, if I look at the other 342 00:25:58,000 --> 00:26:07,000 side -- Let me start again. I ran out of space. 343 00:26:07,000 --> 00:26:12,000 One side is partial r over partial u times delta u. 344 00:26:12,000 --> 00:26:17,000 And the other one would be partial r over partial v times 345 00:26:17,000 --> 00:26:19,000 delta v. Because that is how the 346 00:26:19,000 --> 00:26:24,000 position of your point changes if you just change u or v and 347 00:26:24,000 --> 00:26:31,000 not the other one. To find the surface element 348 00:26:31,000 --> 00:26:40,000 together with a normal vector, I would just take the 349 00:26:40,000 --> 00:26:46,000 cross-product between these guys. 350 00:26:46,000 --> 00:26:50,000 If you prefer, that is the cross-product of 351 00:26:50,000 --> 00:26:56,000 partial r over partial u with partial r over partial v, 352 00:26:56,000 --> 00:27:02,000 delta u delta v. And so n dS is this 353 00:27:02,000 --> 00:27:11,000 cross-product times du dv up to sign. 354 00:27:11,000 --> 00:27:27,000 It depends on which choice I make for my normal vector, 355 00:27:27,000 --> 00:27:32,000 of course. That, of course, 356 00:27:32,000 --> 00:27:35,000 is a slightly confusing equation to think of. 357 00:27:35,000 --> 00:27:37,000 A good exercise, if you want to really 358 00:27:37,000 --> 00:27:40,000 understand what is going on, try this in two good examples 359 00:27:40,000 --> 00:27:43,000 to look at. One good example to look at is 360 00:27:43,000 --> 00:27:45,000 the previous one. What is it? 361 00:27:45,000 --> 00:27:47,000 It is when u and v are just x and y. 362 00:27:47,000 --> 00:27:51,000 The parametric equations are just x equals x, 363 00:27:51,000 --> 00:27:54,000 y equals y and z is f of x, y. 364 00:27:54,000 --> 00:27:59,000 You should end up with the same formula that we had over there. 365 00:27:59,000 --> 00:28:03,000 And you should see why because both of them are given by a 366 00:28:03,000 --> 00:28:06,000 cross-product. The other case you can look at 367 00:28:06,000 --> 00:28:08,000 just to convince yourselves even further. 368 00:28:08,000 --> 00:28:12,000 We don't need to do that because we have seen the formula 369 00:28:12,000 --> 00:28:17,000 before, but in the case of a sphere we have seen the formula 370 00:28:17,000 --> 00:28:22,000 for n and for dS separately. We know what n dS are in terms 371 00:28:22,000 --> 00:28:26,000 of d phi, d theta. Well, you could parametize a 372 00:28:26,000 --> 00:28:28,000 sphere in terms of phi and theta. 373 00:28:28,000 --> 00:28:33,000 Namely, the formulas would be x equals a sine phi cosine theta, 374 00:28:33,000 --> 00:28:38,000 y equals a sign phi sine theta, z equals a cosine phi. 375 00:28:38,000 --> 00:28:42,000 The formulas for circle coordinates setting Ro equals a 376 00:28:42,000 --> 00:28:44,000 . That is a parametric equation 377 00:28:44,000 --> 00:28:47,000 for the sphere. And then, if you try to use 378 00:28:47,000 --> 00:28:50,000 this formula here, you should end up with the same 379 00:28:50,000 --> 00:28:52,000 things we have already seen for n dS, 380 00:28:52,000 --> 00:28:57,000 just with a lot more pain to actually get there because 381 00:28:57,000 --> 00:29:00,000 cross-product is going to be a bit complicated. 382 00:29:00,000 --> 00:29:03,000 But we are seeing all of these formulas all fitting together. 383 00:29:03,000 --> 00:29:05,000 Somehow it is always the same question. 384 00:29:05,000 --> 00:29:10,000 We just have different angles of attack on this general 385 00:29:10,000 --> 00:29:18,000 problem. Questions? 386 00:29:18,000 --> 00:29:20,000 No. OK. 387 00:29:20,000 --> 00:29:30,000 Let's look at yet another last way of finding n dS. 388 00:29:30,000 --> 00:29:36,000 And then I promise we will switch to something else because 389 00:29:36,000 --> 00:29:42,000 I can feel that you are getting a bit overwhelmed for all these 390 00:29:42,000 --> 00:29:47,000 formulas for n dS. What happens very often is we 391 00:29:47,000 --> 00:29:51,000 don't actually know how to parametize our surface. 392 00:29:51,000 --> 00:29:54,000 Maybe we don't know how to solve for z as a function of x 393 00:29:54,000 --> 00:29:58,000 and y, but our surface is given by some equation. 394 00:29:58,000 --> 00:30:05,000 And so what that means is actually maybe what we know is 395 00:30:05,000 --> 00:30:12,000 not really these kinds of formulas, but maybe we know a 396 00:30:12,000 --> 00:30:17,000 normal vector. And I am going to call this one 397 00:30:17,000 --> 00:30:22,000 capital N because I don't even need it to be a unit vector. 398 00:30:22,000 --> 00:30:27,000 You will see. It can be a normal vector of 399 00:30:27,000 --> 00:30:33,000 any length you want to the surfaces. 400 00:30:33,000 --> 00:30:35,000 Why would we ever know a normal vector? 401 00:30:35,000 --> 00:30:39,000 Well, for example, if our surface is a plane, 402 00:30:39,000 --> 00:30:43,000 a slanted plane given by some equation, ax by cz = d. 403 00:30:43,000 --> 00:30:44,000 Well, you know the normal vector. 404 00:30:44,000 --> 00:30:48,000 It is . Of course, you could solve for 405 00:30:48,000 --> 00:30:52,000 z and then go back to that case, which is why I said that one is 406 00:30:52,000 --> 00:30:55,000 very useful. But you can also just stay with 407 00:30:55,000 --> 00:30:58,000 a normal vector. Why else would you know a 408 00:30:58,000 --> 00:31:01,000 normal vector? Well, let's say that you know 409 00:31:01,000 --> 00:31:05,000 an equation that is of a form g of x, y, z equals zero. 410 00:31:05,000 --> 00:31:08,000 Well, then you know that the gradient of g is perpendicular 411 00:31:08,000 --> 00:31:14,000 to the level surface. Let me just give you two 412 00:31:14,000 --> 00:31:19,000 examples. If you have a plane, 413 00:31:19,000 --> 00:31:24,000 ax by cz = d, then the normal vector would 414 00:31:24,000 --> 00:31:28,000 just be . 415 00:31:28,000 --> 00:31:34,000 If you have a surface S given by an equation, 416 00:31:34,000 --> 00:31:40,000 g(x, y, z) = 0, then you can take a normal 417 00:31:40,000 --> 00:31:46,000 vector to be the gradient of g. We have seen that the gradient 418 00:31:46,000 --> 00:31:49,000 is perpendicular to the level surface. 419 00:31:49,000 --> 00:31:51,000 Now, of course, we don't necessarily have to 420 00:31:51,000 --> 00:31:55,000 follow what is going to come. Because, if we could solve for 421 00:31:55,000 --> 00:31:59,000 z, then we might be better off doing what we did over there. 422 00:31:59,000 --> 00:32:02,000 But let's say that we want to do it this. 423 00:32:02,000 --> 00:32:06,000 What can we do? Well, I am going to give you 424 00:32:06,000 --> 00:32:09,000 another way to think geometrically about n dS. 425 00:32:40,000 --> 00:32:43,000 Let's start by thinking about the slanted plane. 426 00:32:43,000 --> 00:32:47,000 Let's say that my surface is just a slanted plane. 427 00:32:47,000 --> 00:32:52,000 My normal vector would be maybe somewhere here. 428 00:32:52,000 --> 00:32:55,000 And let's say that I am going to try -- I need to get some 429 00:32:55,000 --> 00:32:57,000 handle on how to set up my integrals, 430 00:32:57,000 --> 00:33:00,000 so maybe I am going to express things in terms of x and y. 431 00:33:00,000 --> 00:33:05,000 I have my coordinates, and I will try to use x and y. 432 00:33:05,000 --> 00:33:11,000 Then I would like to relate delta S or dS to the area in the 433 00:33:11,000 --> 00:33:14,000 x y plane. That means I want maybe to look 434 00:33:14,000 --> 00:33:19,000 at the projection of this guy onto a horizontal plane. 435 00:33:19,000 --> 00:33:31,000 Let's squish it horizontally. Then you have here another area. 436 00:33:31,000 --> 00:33:35,000 The guy on the slanted plane, let's call that delta S. 437 00:33:35,000 --> 00:33:38,000 And let's call this guy down here delta A. 438 00:33:38,000 --> 00:33:42,000 And delta A would become ultimately maybe delta x, 439 00:33:42,000 --> 00:33:47,000 delta y or something like that. The question is how do we find 440 00:33:47,000 --> 00:33:51,000 the conversion rate between these two areas? 441 00:33:51,000 --> 00:33:53,000 I mean they are not the same. Visually, I hope it is clear to 442 00:33:53,000 --> 00:33:56,000 you that if my plane is actually horizontal then, 443 00:33:56,000 --> 00:34:00,000 of course, they are the same. But the more slanted it becomes 444 00:34:00,000 --> 00:34:04,000 the more delta A becomes smaller than delta S. 445 00:34:04,000 --> 00:34:09,000 If you buy land and it is on the side of a cliff, 446 00:34:09,000 --> 00:34:12,000 well, whether you look at it on a map or whether you look at it 447 00:34:12,000 --> 00:34:15,000 on the actual cliff, the area is going to be very 448 00:34:15,000 --> 00:34:18,000 different. I am not sure if that is a wise 449 00:34:18,000 --> 00:34:22,000 thing to do if you want to build a house there, 450 00:34:22,000 --> 00:34:26,000 but I bet you can get really cheap land. 451 00:34:26,000 --> 00:34:31,000 Anyway, delta S versus delta A depends on how slanted things 452 00:34:31,000 --> 00:34:34,000 are. And let's try to make that more 453 00:34:34,000 --> 00:34:41,000 precise by looking at the angel that our plane makes with the 454 00:34:41,000 --> 00:34:47,000 horizontal direction. Let's call this angle alpha, 455 00:34:47,000 --> 00:34:53,000 the angle that our plane makes with the horizontal direction. 456 00:34:53,000 --> 00:34:57,000 See, it is all coming together. The first unit about 457 00:34:57,000 --> 00:35:03,000 cross-products, normal vectors and so on is 458 00:35:03,000 --> 00:35:09,000 actually useful now. I claim that the surface 459 00:35:09,000 --> 00:35:16,000 element is related to the area in the plane by delta A equals 460 00:35:16,000 --> 00:35:19,000 delta S times the cosine of alpha. 461 00:35:19,000 --> 00:35:24,000 Why is that? Well, let's look at this small 462 00:35:24,000 --> 00:35:27,000 rectangle with one horizontal side and one slanted side. 463 00:35:27,000 --> 00:35:33,000 When you project this side does not change, but this side gets 464 00:35:33,000 --> 00:35:37,000 shortened by a factor of cosine alpha. 465 00:35:37,000 --> 00:35:41,000 Whatever this length was, this length here is that one 466 00:35:41,000 --> 00:35:44,000 times cosine alpha. That is why the area gets 467 00:35:44,000 --> 00:35:48,000 shrunk by cosine alpha. In one direction nothing 468 00:35:48,000 --> 00:35:52,000 happens. In the other direction you 469 00:35:52,000 --> 00:35:59,000 squish by cosine alpha. What that means is that, 470 00:35:59,000 --> 00:36:05,000 well, we will have to deal with this. 471 00:36:05,000 --> 00:36:08,000 And, of course, the one we will care about 472 00:36:08,000 --> 00:36:11,000 actually is delta S expressed in terms of delta A. 473 00:36:11,000 --> 00:36:13,000 But what are we going to do with this cosine? 474 00:36:13,000 --> 00:36:15,000 It is not very convenient to have a cosine left in here. 475 00:36:15,000 --> 00:36:19,000 Remember, the angle between two planes is the same thing as the 476 00:36:19,000 --> 00:36:21,000 angle between the normal vectors. 477 00:36:21,000 --> 00:36:23,000 If you want to see this angle alpha elsewhere, 478 00:36:23,000 --> 00:36:27,000 what you can do is you can just take the vertical direction. 479 00:36:27,000 --> 00:36:38,000 Let's take k. Then here we have our angle 480 00:36:38,000 --> 00:36:43,000 alpha again. In particular, 481 00:36:43,000 --> 00:36:46,000 cosine of alpha, I can get, well, 482 00:36:46,000 --> 00:36:50,000 we know how to find the angle between two vectors. 483 00:36:50,000 --> 00:36:57,000 If we have our normal vector N, we will do N dot k, 484 00:36:57,000 --> 00:37:02,000 and we will divide by length N, length k. 485 00:37:02,000 --> 00:37:06,000 Well, length k is one. That is one easy guy. 486 00:37:06,000 --> 00:37:12,000 That is how we find the angle. Now I am going to say, 487 00:37:12,000 --> 00:37:21,000 well, delta S is going to be one over cosine alpha delta A. 488 00:37:21,000 --> 00:37:36,000 And I can rewrite that as length of N divided by N dot k 489 00:37:36,000 --> 00:37:42,000 times delta A. Now, let's multiply that by the 490 00:37:42,000 --> 00:37:48,000 unit normal vector. Because what we are about is 491 00:37:48,000 --> 00:37:52,000 not so much dS but actually n dS. 492 00:37:52,000 --> 00:38:04,000 N delta S will be, I am just going to multiply by 493 00:38:04,000 --> 00:38:07,000 N. Well, let's think for a second. 494 00:38:07,000 --> 00:38:11,000 What happens if I take a unit normal N and I multiply it by 495 00:38:11,000 --> 00:38:14,000 the length of my other normal big N? 496 00:38:14,000 --> 00:38:19,000 Well, I get big N again. This is a normal vector of the 497 00:38:19,000 --> 00:38:22,000 same length as N, well, up to sign. 498 00:38:22,000 --> 00:38:27,000 The only thing I don't know is whether this guy will be going 499 00:38:27,000 --> 00:38:32,000 in the same direction as big N or in the opposite direction. 500 00:38:32,000 --> 00:38:35,000 Say that, for example, my capital N has, 501 00:38:35,000 --> 00:38:39,000 I don't know, length three for example. 502 00:38:39,000 --> 00:38:43,000 Then the normal unit vector might be this guy, 503 00:38:43,000 --> 00:38:47,000 in which case indeed three times little n will be big n. 504 00:38:47,000 --> 00:38:52,000 Or it might be this one in which case three times little n 505 00:38:52,000 --> 00:38:58,000 will be negative big N. But up to sign it is N. 506 00:38:58,000 --> 00:39:02,000 And then I will have N over N dot k delta A. 507 00:39:02,000 --> 00:39:07,000 And so the final formula, the one that we care about in 508 00:39:07,000 --> 00:39:11,000 case you don't really like my explanations of how we get 509 00:39:11,000 --> 00:39:19,000 there, is that N dS is plus or minus N 510 00:39:19,000 --> 00:39:27,000 over N dot k dx dy. That one is actually kind of 511 00:39:27,000 --> 00:39:31,000 useful so let's box it. Now, 512 00:39:31,000 --> 00:39:35,000 just in case you are wondering, of course, if you didn't want 513 00:39:35,000 --> 00:39:36,000 to project to x, y, 514 00:39:36,000 --> 00:39:39,000 you would have maybe preferred to project to say the plane of a 515 00:39:39,000 --> 00:39:40,000 blackboard, y, z, 516 00:39:40,000 --> 00:39:45,000 well, you can do the same thing. To express n dS in terms of dy 517 00:39:45,000 --> 00:39:49,000 dz you do the same argument. Simply, the only thing that 518 00:39:49,000 --> 00:39:51,000 changes, instead of using the vertical vector k, 519 00:39:51,000 --> 00:39:56,000 you use the normal vector i. So you would be doing N over N 520 00:39:56,000 --> 00:39:58,000 dot i dy dz. The same thing. 521 00:39:58,000 --> 00:40:04,000 So just keep an open mind that this also works with other 522 00:40:04,000 --> 00:40:09,000 variables. Anyway, that is how you can 523 00:40:09,000 --> 00:40:17,000 basically project the vectors of this area element onto the x, 524 00:40:17,000 --> 00:40:23,000 y plane in a way. Let's look at the special case 525 00:40:23,000 --> 00:40:29,000 just to see how this fits with stuff we have seen before. 526 00:40:29,000 --> 00:40:37,000 Let's do a special example where our surface is given by 527 00:40:37,000 --> 00:40:44,000 the equation z minus f of x, y equals zero. 528 00:40:44,000 --> 00:40:46,000 That is a strange way to write the equation. 529 00:40:46,000 --> 00:40:50,000 z equals f of x, y. That we saw before. 530 00:40:50,000 --> 00:40:53,000 But now it looks like some function of x, 531 00:40:53,000 --> 00:40:57,000 y, z equals zero. Let's try to use this new 532 00:40:57,000 --> 00:41:05,000 method. Let's call this guy g(x, y, z). 533 00:41:05,000 --> 00:41:07,000 Well, now let's look at the normal vector. 534 00:41:07,000 --> 00:41:10,000 The normal vector would be the gradient of g, 535 00:41:10,000 --> 00:41:13,000 you see. What is the gradient of this 536 00:41:13,000 --> 00:41:17,000 function? The gradient of g -- Well, 537 00:41:17,000 --> 00:41:22,000 partial g, partial x, that is just negative partial 538 00:41:22,000 --> 00:41:25,000 f, partial x. The y component, 539 00:41:25,000 --> 00:41:32,000 partial g, partial y is going to be negative f sub y, 540 00:41:32,000 --> 00:41:42,000 and g sub z is just one. Now, if you take N over N dot k 541 00:41:42,000 --> 00:41:46,000 dx dy, well, it looks like it is going 542 00:41:46,000 --> 00:41:50,000 to be negative f sub x, negative f sub y, 543 00:41:50,000 --> 00:41:53,000 one divided by -- Well, what is N dot k? 544 00:41:53,000 --> 00:41:57,000 If you dot that with k you will get just one, 545 00:41:57,000 --> 00:42:01,000 so I am not going to write it, dx dy. 546 00:42:01,000 --> 00:42:05,000 See, that is again our favorite formula. 547 00:42:05,000 --> 00:42:12,000 This one is actually more general because you don't need 548 00:42:12,000 --> 00:42:18,000 to solve for z, but if you cannot solve for z 549 00:42:18,000 --> 00:42:28,000 then it is the same as before. I think that is enough formulas 550 00:42:28,000 --> 00:42:34,000 for n dS. After spending a lot of time 551 00:42:34,000 --> 00:42:41,000 telling you how to compute surface integrals, 552 00:42:41,000 --> 00:42:51,000 now I am going to try to tell you how to avoid computing them. 553 00:42:51,000 --> 00:43:05,000 And that is called the divergence theorem. 554 00:43:05,000 --> 00:43:09,000 And we will see the proof and everything and applications on 555 00:43:09,000 --> 00:43:12,000 Tuesday, but I want to at least the theorem and see how it works 556 00:43:12,000 --> 00:43:15,000 in one example. It is also known as the 557 00:43:15,000 --> 00:43:19,000 Gauss-Green theorem or just the Gauss theorem, 558 00:43:19,000 --> 00:43:24,000 depending in who you talk to. The Green here is the same 559 00:43:24,000 --> 00:43:28,000 Green as in Green's theorem, because somehow that is a space 560 00:43:28,000 --> 00:43:31,000 version of Green's theorem. What does it say? 561 00:43:31,000 --> 00:43:43,000 It is 3D analog of Green for flux. 562 00:43:43,000 --> 00:43:47,000 What it says is if S is a closed surface -- Remember, 563 00:43:47,000 --> 00:43:52,000 it is the same as with Green's theorem, we need to have 564 00:43:52,000 --> 00:43:56,000 something that is completely enclosed. 565 00:43:56,000 --> 00:44:00,000 You have a surface and there is somehow no gaps in it. 566 00:44:00,000 --> 00:44:06,000 There is no boundary to it. It is really completely 567 00:44:06,000 --> 00:44:17,000 enclosing a region in space that I will call D. 568 00:44:17,000 --> 00:44:19,000 And I need to choose my orientation. 569 00:44:19,000 --> 00:44:29,000 The orientation that will work for this theorem is choosing the 570 00:44:29,000 --> 00:44:43,000 normal vector to point outwards. N needs to be outwards. 571 00:44:43,000 --> 00:44:46,000 That is one part of the puzzle. The other part is a vector 572 00:44:46,000 --> 00:44:51,000 field. I need to have a vector field 573 00:44:51,000 --> 00:44:58,000 that is defined and differentiable -- -- everywhere 574 00:44:58,000 --> 00:45:04,000 in D, so same instructions as usual. 575 00:45:04,000 --> 00:45:12,000 Then I don't have actually to compute the flux integral. 576 00:45:12,000 --> 00:45:17,000 Double integral of f dot n dS of a closed surface S. 577 00:45:17,000 --> 00:45:19,000 I am going to put a circle just to remind you it is has got to 578 00:45:19,000 --> 00:45:22,000 be a closed surface. It is just a notation to remind 579 00:45:22,000 --> 00:45:26,000 us closed surface. I can replace that by the 580 00:45:26,000 --> 00:45:32,000 triple integral of a region inside of divergence of F dV. 581 00:45:32,000 --> 00:45:36,000 Now, I need to tell you what the divergence of a 3D vector 582 00:45:36,000 --> 00:45:39,000 field is. Well, you will see that it is 583 00:45:39,000 --> 00:45:41,000 not much harder than in the 2D case. 584 00:45:41,000 --> 00:45:58,000 What you do is just -- Say that your vector field has components 585 00:45:58,000 --> 00:46:06,000 P, Q and R. Then you will take P sub x Q 586 00:46:06,000 --> 00:46:10,000 sub y R sub z. That is the definition. 587 00:46:10,000 --> 00:46:13,000 It is pretty easy to remember. You take the x component 588 00:46:13,000 --> 00:46:18,000 partial respect to S plus partial respect to y over y 589 00:46:18,000 --> 00:46:23,000 component plus partial respect to z of the z component. 590 00:46:23,000 --> 00:46:33,000 For example, last time we saw that the flux 591 00:46:33,000 --> 00:46:42,000 of the vector field zk through a sphere of radius a was 592 00:46:42,000 --> 00:46:53,000 four-thirds pi a cubed by computing the surface integral. 593 00:46:53,000 --> 00:46:56,000 Well, if we do it more efficiently now by Green's 594 00:46:56,000 --> 00:46:59,000 theorem, we are going to use Green's theorem for this sphere 595 00:46:59,000 --> 00:47:02,000 because we are doing the whole sphere. 596 00:47:02,000 --> 00:47:04,000 It is fine. It is a closed surface. 597 00:47:04,000 --> 00:47:05,000 We couldn't do it for, say, the hemisphere or 598 00:47:05,000 --> 00:47:09,000 something like that. Well, for a hemisphere we would 599 00:47:09,000 --> 00:47:15,000 need to add maybe the flat face of a bottom or something like 600 00:47:15,000 --> 00:47:20,000 that. Green's theorem says that our 601 00:47:20,000 --> 00:47:28,000 flux integral can actually be replaced by the triple integral 602 00:47:28,000 --> 00:47:36,000 over the solid bowl of radius a of the divergence of zk dV. 603 00:47:36,000 --> 00:47:40,000 But now what is the divergence of this field? 604 00:47:40,000 --> 00:47:48,000 Well, you have zero, zero, z so you get zero plus 605 00:47:48,000 --> 00:47:52,000 zero plus one. It looks like it will be one. 606 00:47:52,000 --> 00:47:59,000 If you do the triple integral of 1dV, you will get just the 607 00:47:59,000 --> 00:48:06,000 volume -- -- of the region inside, which is four-thirds by 608 00:48:06,000 --> 00:48:09,000 a cubed. And so it was no accident. 609 00:48:09,000 --> 00:48:13,000 In fact, before that we looked at also xi yj zk and we found 610 00:48:13,000 --> 00:48:17,000 three times the volume. That is because the divergence 611 00:48:17,000 --> 00:48:19,000 of that field was actually three. 612 00:48:19,000 --> 00:48:22,000 Very quickly, let me just say what this means 613 00:48:22,000 --> 00:48:24,000 physically. Physically, see, 614 00:48:24,000 --> 00:48:29,000 this guy on the left is the total amount of stuff that goes 615 00:48:29,000 --> 00:48:34,000 out of the region per unit time. I want to figure out how much 616 00:48:34,000 --> 00:48:37,000 stuff comes out of there. What does the divergence mean? 617 00:48:37,000 --> 00:48:41,000 The divergence means it measures how much the flow is 618 00:48:41,000 --> 00:48:43,000 expanding things. It measures how much, 619 00:48:43,000 --> 00:48:46,000 I said that probably when we were trying to understand 2D 620 00:48:46,000 --> 00:48:49,000 divergence. It measures the amount of 621 00:48:49,000 --> 00:48:54,000 sources or sinks that you have inside your fluid. 622 00:48:54,000 --> 00:48:57,000 Now it becomes commonsense. If you take a region of space, 623 00:48:57,000 --> 00:49:01,000 the total amount of water that flows out of it is the total 624 00:49:01,000 --> 00:49:05,000 amount of sources that you have in there minus the sinks. 625 00:49:05,000 --> 00:49:08,000 I mean, in spite of this commonsense explanation, 626 00:49:08,000 --> 00:49:10,000 we are going to see how to prove this. 627 00:49:10,000 --> 00:49:14,000 And we will see how it works and what it says.