1 00:00:00,040 --> 00:00:02,460 The following content is provided under a Creative 2 00:00:02,460 --> 00:00:03,870 Commons license. 3 00:00:03,870 --> 00:00:06,320 Your support will help MIT OpenCourseWare 4 00:00:06,320 --> 00:00:10,560 continue to offer high quality educational resources for free. 5 00:00:10,560 --> 00:00:13,300 To make a donation or view additional materials 6 00:00:13,300 --> 00:00:17,210 from hundreds of MIT courses, visit MIT OpenCourseWare 7 00:00:17,210 --> 00:00:19,500 at ocw.mit.edu. 8 00:00:28,160 --> 00:00:29,000 PROFESSOR: Hi. 9 00:00:29,000 --> 00:00:31,260 In our lesson today, we shall try 10 00:00:31,260 --> 00:00:34,270 to find a relationship between line 11 00:00:34,270 --> 00:00:36,600 integrals and double integrals. 12 00:00:36,600 --> 00:00:41,300 In particular, we shall equate line integrals 13 00:00:41,300 --> 00:00:44,590 along certain types of closed curves 14 00:00:44,590 --> 00:00:48,400 to double integrals over the region enclosed 15 00:00:48,400 --> 00:00:50,620 by the particular closed curve. 16 00:00:50,620 --> 00:00:53,280 In particular, the title of today's lecture 17 00:00:53,280 --> 00:00:55,350 is Green's theorem. 18 00:00:55,350 --> 00:00:58,330 And in terms of some particulars, 19 00:00:58,330 --> 00:01:01,330 I guess the intuitive notion that we'll have to accept 20 00:01:01,330 --> 00:01:04,760 is the idea of what one means by a connected region. 21 00:01:04,760 --> 00:01:07,300 I think this is one of those times in mathematics where 22 00:01:07,300 --> 00:01:10,560 we can say, if you have an intuitive idea of what 23 00:01:10,560 --> 00:01:13,240 the concept means, you have the exact idea 24 00:01:13,240 --> 00:01:14,740 of what the concept means. 25 00:01:14,740 --> 00:01:16,470 Essentially, connected means that you 26 00:01:16,470 --> 00:01:18,920 can get from any place in the region 27 00:01:18,920 --> 00:01:21,770 to any other place in the region without ever 28 00:01:21,770 --> 00:01:23,860 having to leave the region. 29 00:01:23,860 --> 00:01:25,000 It's as simple as that. 30 00:01:25,000 --> 00:01:25,650 OK. 31 00:01:25,650 --> 00:01:29,060 Connected means that you can get from any point in the region 32 00:01:29,060 --> 00:01:31,730 to any other point in the region without ever having 33 00:01:31,730 --> 00:01:33,810 to go outside the region. 34 00:01:33,810 --> 00:01:37,730 The mathematician distinguishes between two different types 35 00:01:37,730 --> 00:01:39,550 of connected regions. 36 00:01:39,550 --> 00:01:42,010 One of which is called simply connected. 37 00:01:42,010 --> 00:01:43,910 The other of which being the opposite 38 00:01:43,910 --> 00:01:46,360 of simply connected, meaning non-simply 39 00:01:46,360 --> 00:01:48,780 connected or multiply connected. 40 00:01:48,780 --> 00:01:52,529 I prefer multiply connected to non-simply connected, 41 00:01:52,529 --> 00:01:55,070 because in non-simply connected you don't know whether you're 42 00:01:55,070 --> 00:01:57,340 modifying simply or connected. 43 00:01:57,340 --> 00:01:59,280 A non-simply connected region, when 44 00:01:59,280 --> 00:02:02,310 you say multiply connected, means that in particular 45 00:02:02,310 --> 00:02:04,240 the region is connected. 46 00:02:04,240 --> 00:02:07,700 The intuitive way of looking at this is that simply connected 47 00:02:07,700 --> 00:02:09,015 means no holes. 48 00:02:11,570 --> 00:02:14,620 For example, this region here is not only connected, 49 00:02:14,620 --> 00:02:16,230 but it's simply connected. 50 00:02:16,230 --> 00:02:17,950 There are no holes. 51 00:02:17,950 --> 00:02:20,630 This region is connected. 52 00:02:20,630 --> 00:02:24,010 You see I can get from any point to any other point 53 00:02:24,010 --> 00:02:26,890 without ever having to leave the region. 54 00:02:26,890 --> 00:02:28,660 But it's called multiply connected, 55 00:02:28,660 --> 00:02:30,870 because there are holes in the region. 56 00:02:30,870 --> 00:02:33,650 By the way, you see, the trouble with using expressions 57 00:02:33,650 --> 00:02:38,110 like holes versus non-holes, is that in that particular case, 58 00:02:38,110 --> 00:02:40,980 where we're dealing with two or three dimensions 59 00:02:40,980 --> 00:02:43,350 there's certainly no problem, because we intuitively 60 00:02:43,350 --> 00:02:45,580 visualize what a hole is. 61 00:02:45,580 --> 00:02:48,180 In n-dimensional space-- and we have 62 00:02:48,180 --> 00:02:51,910 analogs of these results in these n-dimensional space-- 63 00:02:51,910 --> 00:02:54,950 the problem is that how do you describe a hole 64 00:02:54,950 --> 00:02:57,080 without referring to a picture? 65 00:02:57,080 --> 00:03:00,200 And the more rigorous mathematical definition 66 00:03:00,200 --> 00:03:04,870 of a simply connected region is the following: the connected 67 00:03:04,870 --> 00:03:07,070 region-- and it has to be connected before you talk 68 00:03:07,070 --> 00:03:09,320 about simply connected, you see-- the connected region 69 00:03:09,320 --> 00:03:13,550 R is called simply connected if its complement is also 70 00:03:13,550 --> 00:03:14,520 connected. 71 00:03:14,520 --> 00:03:16,550 Remember the complement is the portion 72 00:03:16,550 --> 00:03:19,930 of space left when the particular region is deleted. 73 00:03:19,930 --> 00:03:22,910 For example, here we have a simply connected region. 74 00:03:22,910 --> 00:03:24,930 Notice that the complement of this region 75 00:03:24,930 --> 00:03:27,930 would be, for example, the entire blackboard 76 00:03:27,930 --> 00:03:29,340 with this piece missing. 77 00:03:29,340 --> 00:03:31,530 And the entire blackboard with this piece missing 78 00:03:31,530 --> 00:03:33,660 is still a connected region. 79 00:03:33,660 --> 00:03:37,120 On the other hand, if we were to take this connected region 80 00:03:37,120 --> 00:03:39,910 and look at its complement, its complement 81 00:03:39,910 --> 00:03:44,960 is this little hole together with what's outside here. 82 00:03:44,960 --> 00:03:47,120 And notice that there is no way from getting 83 00:03:47,120 --> 00:03:52,710 from here to here without having to go outside the complement. 84 00:03:52,710 --> 00:03:54,400 You see, in other words, the complement 85 00:03:54,400 --> 00:03:59,640 of the multiply connected region is not connected. 86 00:03:59,640 --> 00:04:03,700 At any rate, whichever way you want to look at this, is fine. 87 00:04:03,700 --> 00:04:05,540 Think of simply connected, at least 88 00:04:05,540 --> 00:04:08,260 in the two-dimensional case, as meaning no holes. 89 00:04:08,260 --> 00:04:12,690 And with this as background, Green's theorem-- 90 00:04:12,690 --> 00:04:17,410 call it the theorem, I guess-- is stated as follows. 91 00:04:17,410 --> 00:04:21,920 Suppose R is simply connected with a boundary C. Notice, 92 00:04:21,920 --> 00:04:22,760 R is the region. 93 00:04:22,760 --> 00:04:24,200 C is its boundary. 94 00:04:24,200 --> 00:04:27,110 And a very interesting thing takes place. 95 00:04:27,110 --> 00:04:31,600 If you form the line integral around the boundary C, 96 00:04:31,600 --> 00:04:33,660 and by the way, with this loop in here, 97 00:04:33,660 --> 00:04:38,400 this is the symbol that we mean around the closed curve. 98 00:04:38,400 --> 00:04:42,430 The arrow head is simply done to invoke the convention-- 99 00:04:42,430 --> 00:04:43,930 and we do have to pick a convention, 100 00:04:43,930 --> 00:04:46,480 just like we did in calculus of a single variable 101 00:04:46,480 --> 00:04:48,160 about positive and negative and how 102 00:04:48,160 --> 00:04:52,080 you move-- we pick the positive direction as being 103 00:04:52,080 --> 00:04:54,900 the direction in which one goes around the boundary 104 00:04:54,900 --> 00:04:58,440 so that the region appears to our left 105 00:04:58,440 --> 00:04:59,920 as we go around the boundary. 106 00:04:59,920 --> 00:05:02,360 As we're going around here, you see, in this direction, 107 00:05:02,360 --> 00:05:05,130 the region appears to our left. 108 00:05:05,130 --> 00:05:08,240 At any rate, I'll mention that in more detail later. 109 00:05:08,240 --> 00:05:11,590 The directed line integral around closed curve C, 110 00:05:11,590 --> 00:05:17,500 M*dx plus N*dy turns out to be a double integral over the region 111 00:05:17,500 --> 00:05:19,100 R enclosed by the curve. 112 00:05:19,100 --> 00:05:22,580 And that double integral has as its integrand 113 00:05:22,580 --> 00:05:26,760 the partial of N with respect to x minus the partial of M 114 00:05:26,760 --> 00:05:28,240 with respect to y. 115 00:05:28,240 --> 00:05:31,400 And I will refer to this again very, very shortly. 116 00:05:31,400 --> 00:05:34,400 But I hope that this rings a familiar bell 117 00:05:34,400 --> 00:05:38,170 in terms of what we talked about exact differentials. 118 00:05:38,170 --> 00:05:41,950 Remember, a differential M*dx plus N*dy was exact 119 00:05:41,950 --> 00:05:45,300 if the partial of N with respect to x equaled the partial of M 120 00:05:45,300 --> 00:05:46,420 with respect to y. 121 00:05:46,420 --> 00:05:48,290 I'll return to that in a moment. 122 00:05:48,290 --> 00:05:50,900 For the time being, all I want you to see 123 00:05:50,900 --> 00:05:52,620 is what Green's theorem says. 124 00:05:52,620 --> 00:05:58,010 What Green's theorem does is it relates a line integral 125 00:05:58,010 --> 00:06:00,920 along the boundary of a simply connected region 126 00:06:00,920 --> 00:06:04,520 to a double integral over the region itself. 127 00:06:04,520 --> 00:06:07,730 The proof of the theorem is in the Thomas text. 128 00:06:07,730 --> 00:06:10,700 It's available in a pretty straightforward form 129 00:06:10,700 --> 00:06:14,940 for anybody who would like to see the proof of the result. 130 00:06:14,940 --> 00:06:17,960 I don't think I want to prove the result in this lecture. 131 00:06:17,960 --> 00:06:23,170 I think I would rather hammer home what the theorem says, 132 00:06:23,170 --> 00:06:26,120 rather than what this rigorous proof is. 133 00:06:26,120 --> 00:06:28,150 The idea is, though, that what? 134 00:06:28,150 --> 00:06:32,070 A line integral is related to a multiple integral. 135 00:06:32,070 --> 00:06:34,060 And all that's required for Green's theorem 136 00:06:34,060 --> 00:06:38,750 to be true is that M, N, the partial of M with respect to y, 137 00:06:38,750 --> 00:06:40,690 the partial of N with respect to x 138 00:06:40,690 --> 00:06:45,840 must exist and be continuous in a region containing R. 139 00:06:45,840 --> 00:06:49,160 And we'll see later that even the condition of simply 140 00:06:49,160 --> 00:06:51,050 connected can be waived. 141 00:06:51,050 --> 00:06:53,410 In fact, in the Thomas demonstration, 142 00:06:53,410 --> 00:06:56,540 we don't even deal with the most general simply connected 143 00:06:56,540 --> 00:06:57,430 regions. 144 00:06:57,430 --> 00:07:01,020 He gives his proof for a very specific type 145 00:07:01,020 --> 00:07:03,080 of simply connected region. 146 00:07:03,080 --> 00:07:06,210 But again, that isn't crucial either. 147 00:07:06,210 --> 00:07:08,760 We'll talk about that in more detail later in the lecture 148 00:07:08,760 --> 00:07:11,270 as well as in the exercises. 149 00:07:11,270 --> 00:07:12,610 The idea again though is what? 150 00:07:12,610 --> 00:07:14,930 We're not relating arbitrary line integrals 151 00:07:14,930 --> 00:07:16,970 to arbitrary multiple integrals. 152 00:07:16,970 --> 00:07:20,000 But Green's theorem relates a line integral 153 00:07:20,000 --> 00:07:23,350 along the boundary of a simply connected region 154 00:07:23,350 --> 00:07:28,230 to a double integral defined on the interior 155 00:07:28,230 --> 00:07:30,800 of the closed curve C, the region R. 156 00:07:30,800 --> 00:07:32,552 And there are two interesting cases. 157 00:07:32,552 --> 00:07:34,760 One of which-- well, there are more than two, but two 158 00:07:34,760 --> 00:07:36,120 that we'd like to talk about. 159 00:07:36,120 --> 00:07:38,530 One of the interesting cases we've already hinted 160 00:07:38,530 --> 00:07:42,730 at and that is if M*dx plus N*dy is exact, 161 00:07:42,730 --> 00:07:47,620 then the integral of M*dx plus N*dy around the closed curve C 162 00:07:47,620 --> 00:07:48,890 is 0. 163 00:07:48,890 --> 00:07:50,630 And the reason for this, you see, 164 00:07:50,630 --> 00:07:53,870 is that the integral around the closed curve, 165 00:07:53,870 --> 00:07:57,050 by Green's theorem, is just a double integral partial N 166 00:07:57,050 --> 00:08:00,310 with respect to x minus partial M with respect to y. 167 00:08:00,310 --> 00:08:05,670 And if M*dx plus N*dy is exact, then these two terms are equal 168 00:08:05,670 --> 00:08:08,110 so that their difference is identically 0. 169 00:08:08,110 --> 00:08:12,660 And to integrate 0*dA over any region R, of course, 170 00:08:12,660 --> 00:08:14,630 is just 0 itself. 171 00:08:14,630 --> 00:08:18,470 In other words, then, if M*dx plus N*dy is exact, 172 00:08:18,470 --> 00:08:22,350 the integral around any closed curve is 0. 173 00:08:22,350 --> 00:08:25,190 And by the way, this ties in with a question 174 00:08:25,190 --> 00:08:28,200 that was raised in our previous lecture. 175 00:08:28,200 --> 00:08:33,150 And that was we saw that if we had a force for example, 176 00:08:33,150 --> 00:08:38,730 and we had a particle move from the point P_0 to the point P_1 177 00:08:38,730 --> 00:08:40,740 under the influence of that force, 178 00:08:40,740 --> 00:08:43,860 that the work done as that particle moved in general 179 00:08:43,860 --> 00:08:48,000 depended on the curve which connected P_0 to P_1. 180 00:08:48,000 --> 00:08:54,950 My claim is that if the force, M*i plus N*j has as its 181 00:08:54,950 --> 00:08:58,950 components-- see M and N-- where the partial of M with respect 182 00:08:58,950 --> 00:09:01,660 to y equals the partial of N with respect x. 183 00:09:01,660 --> 00:09:03,480 My claim is, under those conditions, 184 00:09:03,480 --> 00:09:05,530 the work is independent of the path. 185 00:09:05,530 --> 00:09:10,050 And that's what we mean in turn by a conservative force field. 186 00:09:10,050 --> 00:09:11,490 The proof is quite simple. 187 00:09:11,490 --> 00:09:14,040 What we do is we pick two different paths 188 00:09:14,040 --> 00:09:16,630 that connect P_0 to P_1. 189 00:09:16,630 --> 00:09:21,060 We call one of the paths C_1 and the other path C_2. 190 00:09:21,060 --> 00:09:22,600 And now, what we're saying is, if we 191 00:09:22,600 --> 00:09:25,510 think of C as being the curve that 192 00:09:25,510 --> 00:09:29,450 keeps this region to our left as we go along here, 193 00:09:29,450 --> 00:09:31,890 then the integral around C is 0. 194 00:09:31,890 --> 00:09:33,800 But what is the integral around C? 195 00:09:33,800 --> 00:09:39,440 It's the integral along C_1 minus the integral along C_2. 196 00:09:39,440 --> 00:09:41,650 And I hope that that's clear from the exercises 197 00:09:41,650 --> 00:09:42,780 of last time. 198 00:09:42,780 --> 00:09:46,540 That when you switch the sense of the path, all you do 199 00:09:46,540 --> 00:09:49,290 is change the sign of the integrand. 200 00:09:49,290 --> 00:09:52,540 In fact, by the way, while we're on that topic, let's digress 201 00:09:52,540 --> 00:09:55,950 for a moment and return to our statement of Green's theorem 202 00:09:55,950 --> 00:09:59,950 and observe that if we were to switch the convention, 203 00:09:59,950 --> 00:10:02,440 if we want to switch the convention, 204 00:10:02,440 --> 00:10:05,430 and keep the positive direction as that which 205 00:10:05,430 --> 00:10:09,200 kept the area on our right, notice 206 00:10:09,200 --> 00:10:11,470 that this would change the sense here, which 207 00:10:11,470 --> 00:10:13,130 should give me a minus sign. 208 00:10:13,130 --> 00:10:17,110 But if I interchange, you see, the order of the terms 209 00:10:17,110 --> 00:10:20,140 here, if I reverse the sense, the sign 210 00:10:20,140 --> 00:10:23,560 of the integral, the double integral, will also change. 211 00:10:23,560 --> 00:10:27,740 The important point being that all of these sign conventions 212 00:10:27,740 --> 00:10:30,700 are preserved by things like Green's theorem and the like. 213 00:10:30,700 --> 00:10:33,560 But the important point is, getting back to this now, 214 00:10:33,560 --> 00:10:36,370 the integral around the closed curve C 215 00:10:36,370 --> 00:10:41,290 is the integral along C_1 minus the integral along C_2. 216 00:10:41,290 --> 00:10:44,120 And since the integral along that closed curve is 0, 217 00:10:44,120 --> 00:10:46,410 it says that this minus this is 0. 218 00:10:46,410 --> 00:10:49,490 Consequently, these two integrals are equal. 219 00:10:49,490 --> 00:10:52,150 And since C_1 and C_2 are arbitrary 220 00:10:52,150 --> 00:10:55,160 paths that connect P_0 and P_1, that 221 00:10:55,160 --> 00:10:58,930 shows that the line integral, in this particular case, 222 00:10:58,930 --> 00:11:04,380 does not depend on the path, but only the points P_0 and P_1. 223 00:11:04,380 --> 00:11:08,400 Again, we will emphasize that more in the exercises. 224 00:11:08,400 --> 00:11:12,320 The second interesting application of Green's theorem 225 00:11:12,320 --> 00:11:14,590 is a form which not only shows us 226 00:11:14,590 --> 00:11:18,020 how a line integral and a multiple integral are related, 227 00:11:18,020 --> 00:11:22,630 but it actually shows us a way of computing an area in terms 228 00:11:22,630 --> 00:11:24,110 of a line integral. 229 00:11:24,110 --> 00:11:26,330 In particular, if in Green's theorem 230 00:11:26,330 --> 00:11:30,580 we let N equal x and M equal negative y, 231 00:11:30,580 --> 00:11:33,630 notice that Green's theorem then becomes what? 232 00:11:33,630 --> 00:11:36,070 The integral along a closed curve C, 233 00:11:36,070 --> 00:11:41,020 M*dx plus N*dy is just minus y*dx plus x*dy. 234 00:11:41,020 --> 00:11:43,910 The partial of N with respect to x is 1. 235 00:11:43,910 --> 00:11:47,170 The partial of M with respect to y is minus 1. 236 00:11:47,170 --> 00:11:48,990 Therefore the partial with N with respect 237 00:11:48,990 --> 00:11:52,710 to x minus the partial of M with-- the partial of N 238 00:11:52,710 --> 00:11:55,590 with respect to x minus the partial of M with respect to y 239 00:11:55,590 --> 00:11:58,240 is 1 minus minus 1, which is 2. 240 00:11:58,240 --> 00:12:00,760 In other words, Green's theorem says 241 00:12:00,760 --> 00:12:03,470 that the integral around the closed curve, in this case, 242 00:12:03,470 --> 00:12:07,880 is twice the area of the region R. You see, without the 2 243 00:12:07,880 --> 00:12:11,010 in here, this would just be the area of the region R. 244 00:12:11,010 --> 00:12:13,040 Since 2 is a constant factor, we can take it 245 00:12:13,040 --> 00:12:14,650 outside the integral sign. 246 00:12:14,650 --> 00:12:16,550 In other words, in terms of a picture, 247 00:12:16,550 --> 00:12:20,370 if I want to find the area of the region R, 248 00:12:20,370 --> 00:12:23,030 and I want to use a line integral, 249 00:12:23,030 --> 00:12:26,540 I need only take one half of this integral. 250 00:12:26,540 --> 00:12:30,060 In other words, the area of the region R is one half 251 00:12:30,060 --> 00:12:37,550 the integral around the boundary of R, x*dy minus y*dx. 252 00:12:37,550 --> 00:12:41,140 And again, I will drill on that in the homework. 253 00:12:41,140 --> 00:12:43,610 The one remaining thing I wanted to point out 254 00:12:43,610 --> 00:12:46,350 before we go to a specific example 255 00:12:46,350 --> 00:12:50,750 is how we remove the restriction of the region being 256 00:12:50,750 --> 00:12:52,230 simply connected. 257 00:12:52,230 --> 00:12:54,690 You see, what I claim is that we may replace 258 00:12:54,690 --> 00:12:57,330 simply connected by connected. 259 00:12:57,330 --> 00:13:00,050 Since multiply connected regions are 260 00:13:00,050 --> 00:13:04,250 unions of simply connected ones with cuts, whatever that means. 261 00:13:04,250 --> 00:13:05,890 Let me show you what I'm saying here. 262 00:13:05,890 --> 00:13:08,920 Look at this multiply connected region here. 263 00:13:08,920 --> 00:13:12,890 What I'm saying is, suppose I were to cut this region, 264 00:13:12,890 --> 00:13:18,340 I can then visualize this as the union of two simply connected 265 00:13:18,340 --> 00:13:19,000 regions. 266 00:13:19,000 --> 00:13:21,070 In fact, if I have a piece of paper in my pocket, 267 00:13:21,070 --> 00:13:24,930 I think we might as well-- the entire two parts of calculus 268 00:13:24,930 --> 00:13:27,860 are almost over, and I've never done an experiment for you. 269 00:13:27,860 --> 00:13:29,580 Here's one I think even I can handle. 270 00:13:29,580 --> 00:13:30,900 Here's a connected region. 271 00:13:30,900 --> 00:13:31,990 It's a piece of paper. 272 00:13:31,990 --> 00:13:33,310 I'll fold it. 273 00:13:33,310 --> 00:13:34,550 And I'll now make it. 274 00:13:34,550 --> 00:13:37,410 It's simply connected, because there are no holes in it. 275 00:13:37,410 --> 00:13:39,630 I will now undo that. 276 00:13:39,630 --> 00:13:44,670 I will make the region connected but now multiply connected. 277 00:13:44,670 --> 00:13:46,610 You see, there's a hole in it. 278 00:13:46,610 --> 00:13:49,430 And all I'm saying is, if I cut this region, 279 00:13:49,430 --> 00:13:51,470 not necessarily along a straight line. 280 00:13:51,470 --> 00:13:54,220 All I've got to do is to cut it. 281 00:13:54,220 --> 00:13:59,120 If I cut this region, then what I have 282 00:13:59,120 --> 00:14:04,130 are two simply connected regions that 283 00:14:04,130 --> 00:14:06,680 can be viewed as being fit together this way. 284 00:14:06,680 --> 00:14:08,969 But to see them clearly, we can just separate them. 285 00:14:08,969 --> 00:14:10,760 But the reason I wanted to cut them for you 286 00:14:10,760 --> 00:14:14,010 is to see that I can actually fit these together. 287 00:14:14,010 --> 00:14:15,870 And notice that the cut that joins 288 00:14:15,870 --> 00:14:18,270 these two, from a theoretical point of view, 289 00:14:18,270 --> 00:14:19,160 has no thickness. 290 00:14:19,160 --> 00:14:19,660 Right? 291 00:14:19,660 --> 00:14:22,110 That's just a line that joins these two. 292 00:14:22,110 --> 00:14:24,100 And here's the key point. 293 00:14:24,100 --> 00:14:28,420 You see physically, certainly if I put a cut inside the region, 294 00:14:28,420 --> 00:14:30,825 the resulting picture is different than with the cut 295 00:14:30,825 --> 00:14:31,325 missing. 296 00:14:31,325 --> 00:14:34,305 In other words, I can distinguish between the shape 297 00:14:34,305 --> 00:14:37,760 that I've drawn without the accentuated chalk mark in here 298 00:14:37,760 --> 00:14:40,310 and the shape that I have with the accentuated chalk 299 00:14:40,310 --> 00:14:41,470 mark in here. 300 00:14:41,470 --> 00:14:42,760 But here's the key point. 301 00:14:42,760 --> 00:14:45,500 First of all, as far as computing a double area 302 00:14:45,500 --> 00:14:50,870 is concerned, the cut, having no thickness, has zero area. 303 00:14:50,870 --> 00:14:55,330 So if somehow I were to compute the area of the region R 304 00:14:55,330 --> 00:14:58,880 without the cut, I would expect to get the same area. 305 00:14:58,880 --> 00:15:00,380 Or if I were to compute whatever you 306 00:15:00,380 --> 00:15:02,380 want to call this, the density or what have you, 307 00:15:02,380 --> 00:15:06,220 if I want to compute the mass of the region R with the cut, 308 00:15:06,220 --> 00:15:09,140 it should be the same as the mass without the cut, 309 00:15:09,140 --> 00:15:11,980 because the difference is essentially what? 310 00:15:11,980 --> 00:15:14,651 A set of points which has zero area. 311 00:15:14,651 --> 00:15:16,150 And by the way, for those of you who 312 00:15:16,150 --> 00:15:19,820 may read ahead in mathematics and have heard such expressions 313 00:15:19,820 --> 00:15:23,200 as Lebesgue integrals, the Lebesgue integral and the like. 314 00:15:23,200 --> 00:15:26,860 These are things defined on sets of measure 0. 315 00:15:26,860 --> 00:15:29,377 What these things mean from an intuitive point of view 316 00:15:29,377 --> 00:15:31,210 is essentially what we're talking about now. 317 00:15:31,210 --> 00:15:34,420 You see here are infinitely many points on this cut. 318 00:15:34,420 --> 00:15:37,520 And yet the important thing is that with respect to area 319 00:15:37,520 --> 00:15:39,880 these points, even though they're infinite in number, 320 00:15:39,880 --> 00:15:40,672 contribute no area. 321 00:15:40,672 --> 00:15:42,171 In other words, there are infinitely 322 00:15:42,171 --> 00:15:44,550 many points on a line, but a line having no thickness 323 00:15:44,550 --> 00:15:45,810 has no area. 324 00:15:45,810 --> 00:15:47,890 So certainly, putting the cut in does not 325 00:15:47,890 --> 00:15:50,120 affect the double integral. 326 00:15:50,120 --> 00:15:53,180 On the other hand, since we always 327 00:15:53,180 --> 00:15:57,930 assumed that the region is kept on our left as we go around, 328 00:15:57,930 --> 00:16:01,190 it turns out that from the single integral, the line 329 00:16:01,190 --> 00:16:04,670 integral point of view, each cut cancels 330 00:16:04,670 --> 00:16:06,050 because of opposite sense. 331 00:16:06,050 --> 00:16:09,480 What I'm saying is, if I take this region and slice it, 332 00:16:09,480 --> 00:16:12,400 you see, how do I traverse this piece? 333 00:16:12,400 --> 00:16:16,350 Since I want the area to be enclosed on my left, 334 00:16:16,350 --> 00:16:19,760 I traverse this region in this particular direction. 335 00:16:19,760 --> 00:16:23,320 When it comes to this piece, I traverse this region 336 00:16:23,320 --> 00:16:27,230 in this direction so that the area is enclosed to my left. 337 00:16:27,230 --> 00:16:31,630 When I now superimpose these two, what happens 338 00:16:31,630 --> 00:16:35,560 is, since this line integral and this line integral 339 00:16:35,560 --> 00:16:38,870 go exactly over the same curve, remember I just separated these 340 00:16:38,870 --> 00:16:41,260 so I could see them better, when I push these together, 341 00:16:41,260 --> 00:16:42,660 this is the same line. 342 00:16:42,660 --> 00:16:46,160 Since they have the same function, the same points, 343 00:16:46,160 --> 00:16:50,500 but only opposite sense, these two integrals cancel. 344 00:16:50,500 --> 00:16:53,330 In other words, if I'm given a region that's 345 00:16:53,330 --> 00:16:56,210 multiply connected, I can view it like this, 346 00:16:56,210 --> 00:16:58,720 meaning I go around this piece this way 347 00:16:58,720 --> 00:17:00,310 so that my region is on my left. 348 00:17:00,310 --> 00:17:02,620 I go around the outside border this way 349 00:17:02,620 --> 00:17:04,240 so that the region is on my left. 350 00:17:04,240 --> 00:17:06,240 And when I want to use Green's theorem, 351 00:17:06,240 --> 00:17:08,660 all I do is essentially, I don't even 352 00:17:08,660 --> 00:17:13,560 have to make the two cuts here, I could even make one cut. 353 00:17:13,560 --> 00:17:15,380 And look at this as the union of-- 354 00:17:19,859 --> 00:17:23,795 See, well, the two is-- see, if I make just the one cut 355 00:17:23,795 --> 00:17:27,470 and leave the line out, there are no longer any holes 356 00:17:27,470 --> 00:17:28,462 in the region here. 357 00:17:28,462 --> 00:17:29,920 But that's not the important point. 358 00:17:29,920 --> 00:17:32,750 The important point is that to visualize the integral that I 359 00:17:32,750 --> 00:17:35,510 want, or the regions that I want, all I have to do 360 00:17:35,510 --> 00:17:36,830 is put the cut in. 361 00:17:36,830 --> 00:17:39,780 And whatever the cut does is canceled out 362 00:17:39,780 --> 00:17:41,680 in the statement of Green's theorem. 363 00:17:41,680 --> 00:17:43,990 In other words, again, that quite often 364 00:17:43,990 --> 00:17:45,930 in using Green's theorem, we still 365 00:17:45,930 --> 00:17:49,440 apply it to multiply connected regions. 366 00:17:49,440 --> 00:17:51,360 The reason that we stress simply connected 367 00:17:51,360 --> 00:17:54,440 is that the proofs that are given for Green's theorem 368 00:17:54,440 --> 00:17:57,070 are easiest to handle in the case of simply 369 00:17:57,070 --> 00:17:58,240 connected regions. 370 00:17:58,240 --> 00:18:00,870 And then we simply generalize it to multiply connected 371 00:18:00,870 --> 00:18:04,050 by pointing out what we just did here, that we can always 372 00:18:04,050 --> 00:18:08,070 visualize a multiply connected region as a union of simply 373 00:18:08,070 --> 00:18:11,170 connected regions if the appropriate cuts are made. 374 00:18:11,170 --> 00:18:14,000 And again, that will be emphasized in the exercises. 375 00:18:14,000 --> 00:18:17,330 I thought that what we should do to finish off today's lesson 376 00:18:17,330 --> 00:18:21,200 would be to actually do a trivial exercise. 377 00:18:21,200 --> 00:18:23,290 By trivial, I mean Green's theorem 378 00:18:23,290 --> 00:18:25,700 has profound application. 379 00:18:25,700 --> 00:18:28,310 It's more than just relating line integrals 380 00:18:28,310 --> 00:18:29,250 to multiple integrals. 381 00:18:29,250 --> 00:18:31,870 The applications of these things are enormous. 382 00:18:31,870 --> 00:18:35,150 And we're going to hit a few of these in the exercises. 383 00:18:35,150 --> 00:18:38,600 But what I wanted to do was to pick sort of a straightforward 384 00:18:38,600 --> 00:18:42,490 example just to have you see what Green's theorem means 385 00:18:42,490 --> 00:18:45,650 when we translate into a concrete situation. 386 00:18:45,650 --> 00:18:48,430 And the example I have in mind is simply this. 387 00:18:48,430 --> 00:18:53,982 Let's compute the line integral around the closed curve C, 388 00:18:53,982 --> 00:18:59,910 y cubed dx plus x to the fourth dy, where the curve 389 00:18:59,910 --> 00:19:08,370 C happens to be the following simply closed, or closed curve. 390 00:19:08,370 --> 00:19:11,920 I start at the origin and move to the point (1, 1) 391 00:19:11,920 --> 00:19:14,430 along the parabola y equals x squared. 392 00:19:14,430 --> 00:19:21,710 Then I moved from (1, 1) to (0, 1) along the line y equals 1. 393 00:19:21,710 --> 00:19:25,850 And then I move from (0, 1) down to (0, 0) 394 00:19:25,850 --> 00:19:29,020 along the line x equals 0. 395 00:19:29,020 --> 00:19:32,630 The boundary of this is what I'm calling my region C, my curve 396 00:19:32,630 --> 00:19:38,030 C. Notice that the region R enclosed by C lies on my left 397 00:19:38,030 --> 00:19:42,420 as I traverse the boundary of C. And what I want to do 398 00:19:42,420 --> 00:19:46,160 is to find the value of this line integral 399 00:19:46,160 --> 00:19:49,140 along this particular closed curve C. 400 00:19:49,140 --> 00:19:51,070 And again, if you want to see this 401 00:19:51,070 --> 00:19:54,480 in terms of our lecture of last time, where we talked 402 00:19:54,480 --> 00:19:58,770 about work and force, imagine that my force 403 00:19:58,770 --> 00:20:03,020 is defined in the xy-plane, at any point (x, y), 404 00:20:03,020 --> 00:20:06,420 to be y cubed i plus x to the fourth j, et cetera. 405 00:20:06,420 --> 00:20:11,410 Meaning dR, as usual, will be dx*i plus dy*j. 406 00:20:11,410 --> 00:20:15,400 And what we're saying is that this particular integral 407 00:20:15,400 --> 00:20:19,580 represents the work that would be done if a particle were 408 00:20:19,580 --> 00:20:25,510 starting at the point (0, 0) and being moved along the curve C 409 00:20:25,510 --> 00:20:28,680 so that it returns to the point (0, 0). 410 00:20:28,680 --> 00:20:31,090 Now keep in mind, merely because the particle 411 00:20:31,090 --> 00:20:33,240 is returning to the same point from which it 412 00:20:33,240 --> 00:20:37,580 started is not enough to guarantee that the work done is 413 00:20:37,580 --> 00:20:38,310 0. 414 00:20:38,310 --> 00:20:40,930 In fact, the guarantee that would be necessary 415 00:20:40,930 --> 00:20:44,370 would be that the partial of y cubed with respect to y 416 00:20:44,370 --> 00:20:47,810 would have to be identical to the partial of x to the fourth 417 00:20:47,810 --> 00:20:49,020 with respect x. 418 00:20:49,020 --> 00:20:50,730 And that certainly isn't the case 419 00:20:50,730 --> 00:20:52,160 in this particular example. 420 00:20:52,160 --> 00:20:55,390 We don't have a conservative force field in this case. 421 00:20:55,390 --> 00:20:58,880 In this case, we suspect that the work done 422 00:20:58,880 --> 00:21:01,650 should depend on the path connecting the points. 423 00:21:01,650 --> 00:21:04,460 In particular, if we go around a closed path, 424 00:21:04,460 --> 00:21:07,140 the work done would not have to be 0. 425 00:21:07,140 --> 00:21:09,260 At any rate, that's enough lip service 426 00:21:09,260 --> 00:21:11,600 paid to the physical interpretation of this. 427 00:21:11,600 --> 00:21:14,090 Let's actually try to solve this problem. 428 00:21:14,090 --> 00:21:16,480 And what we're going to do is solve it in two ways. 429 00:21:16,480 --> 00:21:20,150 The first way will be to emphasize Green's theorem. 430 00:21:20,150 --> 00:21:22,390 We'll solve it first by Green's theorem, 431 00:21:22,390 --> 00:21:25,140 because that is the sermon for today. 432 00:21:25,140 --> 00:21:26,960 And the second method will be to show 433 00:21:26,960 --> 00:21:30,520 how we would have solved the same problem if we 434 00:21:30,520 --> 00:21:32,200 didn't have Green's theorem. 435 00:21:32,200 --> 00:21:35,630 The idea being, I won't tell you which of the two ways 436 00:21:35,630 --> 00:21:36,910 one should have done this. 437 00:21:36,910 --> 00:21:39,860 The important point from our point of view 438 00:21:39,860 --> 00:21:42,610 is that the two methods yield the same answer. 439 00:21:42,610 --> 00:21:45,130 At any rate, to use Green's theorem, 440 00:21:45,130 --> 00:21:47,240 notice that M is y cubed. 441 00:21:47,240 --> 00:21:49,830 N is equal to x to the fourth. 442 00:21:49,830 --> 00:21:51,600 So the statement of Green's theorem, 443 00:21:51,600 --> 00:21:54,650 which says that the integral around the closed curve C, 444 00:21:54,650 --> 00:21:58,960 M*dx plus N*dy is the double integral around the region 445 00:21:58,960 --> 00:22:03,340 enclosed by C, the partial of N with respect to x minus 446 00:22:03,340 --> 00:22:05,800 the partial of M with respect to y, dA. 447 00:22:05,800 --> 00:22:08,910 That leads to this. 448 00:22:08,910 --> 00:22:11,620 And now, what is my region A? 449 00:22:11,620 --> 00:22:14,040 I'll think of it as being a dy*dx. 450 00:22:14,040 --> 00:22:16,700 And if I look at my region a, I observe 451 00:22:16,700 --> 00:22:24,170 that for a fixed value of x, y varies from the curve 452 00:22:24,170 --> 00:22:25,810 x squared to 1. 453 00:22:25,810 --> 00:22:29,340 And x, in turn, varies any place from 0 to 1. 454 00:22:29,340 --> 00:22:33,395 So the line integral that I'm trying to evaluate 455 00:22:33,395 --> 00:22:35,620 is this particular double integral. 456 00:22:35,620 --> 00:22:38,120 By the way, we don't really need this, 457 00:22:38,120 --> 00:22:39,860 what comes next, because from here on in, 458 00:22:39,860 --> 00:22:41,010 this should be old hat. 459 00:22:41,010 --> 00:22:43,440 But just by way of review, notice 460 00:22:43,440 --> 00:22:46,630 that I could have written this with the order of integration 461 00:22:46,630 --> 00:22:48,380 reversed, so that I wouldn't have 462 00:22:48,380 --> 00:22:50,870 to worry about an x squared in the lower limit. 463 00:22:50,870 --> 00:22:53,380 I might like all of my lower limits to be zero. 464 00:22:53,380 --> 00:22:56,040 Notice I could have picked a horizontal strip 465 00:22:56,040 --> 00:22:58,300 and shown that for a fixed value of y, 466 00:22:58,300 --> 00:23:01,170 x varies from 0 to the square root of y. 467 00:23:01,170 --> 00:23:03,200 And y varies from 0 to 1. 468 00:23:03,200 --> 00:23:06,910 In any event, this is just garnishing. 469 00:23:06,910 --> 00:23:08,110 This is the easy part. 470 00:23:08,110 --> 00:23:10,280 All I'm saying is, we now evaluate 471 00:23:10,280 --> 00:23:13,360 this iterated integral, integrating with respect to x. 472 00:23:13,360 --> 00:23:16,277 The integrand becomes x to the fourth minus 3x y 473 00:23:16,277 --> 00:23:18,785 squared evaluated between x equals 0, 474 00:23:18,785 --> 00:23:20,940 and x equals the square root of y. 475 00:23:20,940 --> 00:23:25,160 That, of course, leads to y squared minus 3 y to the 5/2. 476 00:23:25,160 --> 00:23:27,580 I integrate that between 0 and 1. 477 00:23:27,580 --> 00:23:32,240 I get 1/3 y cubed minus 6/7 y to the 7/2. 478 00:23:32,240 --> 00:23:36,690 And as y goes from 0 to 1, this becomes minus 11/21. 479 00:23:36,690 --> 00:23:40,050 By the way, again, I hope it's clear from previous exercises 480 00:23:40,050 --> 00:23:43,320 that the minus sign simply has physical significance. 481 00:23:43,320 --> 00:23:45,430 That all we're saying is, is that one 482 00:23:45,430 --> 00:23:48,360 would expect that if the motion takes place 483 00:23:48,360 --> 00:23:50,700 in the direction of the force, the force is 484 00:23:50,700 --> 00:23:52,640 helping us move the particle. 485 00:23:52,640 --> 00:23:55,652 Whereas if the force is taking place opposite 486 00:23:55,652 --> 00:23:57,110 to the direction of motion, we have 487 00:23:57,110 --> 00:23:59,850 to push against the force to get the particle there. 488 00:23:59,850 --> 00:24:02,840 So what we're saying is in one case, the sense is positive. 489 00:24:02,840 --> 00:24:03,780 The other is negative. 490 00:24:03,780 --> 00:24:06,510 I'm not going to worry here physically 491 00:24:06,510 --> 00:24:08,750 about what the meaning of the minus sign is. 492 00:24:08,750 --> 00:24:12,060 But using Green's theorem, we have shown what? 493 00:24:12,060 --> 00:24:14,130 That the integral around the closed curve 494 00:24:14,130 --> 00:24:18,560 C, as we've defined C before, y cubed dx plus x fourth dy, 495 00:24:18,560 --> 00:24:20,920 is minus 11/21. 496 00:24:20,920 --> 00:24:23,950 And what I would like to do now, to finish today's lesson, 497 00:24:23,950 --> 00:24:26,730 is to do the same problem, only doing it 498 00:24:26,730 --> 00:24:28,730 without Green's theorem. 499 00:24:28,730 --> 00:24:31,820 And again, the point is to do this problem 500 00:24:31,820 --> 00:24:33,990 without using Green's theorem. 501 00:24:33,990 --> 00:24:36,020 Remember what my region was over here. 502 00:24:36,020 --> 00:24:38,700 Without having to refer back to the previous work. 503 00:24:38,700 --> 00:24:43,020 What I'm doing is my curve C consists of three separate line 504 00:24:43,020 --> 00:24:43,880 integrals. 505 00:24:43,880 --> 00:24:49,710 C_1 is the curve y equals x squared as x goes from 0 to 1. 506 00:24:49,710 --> 00:24:55,730 C_2 is the curve y equals 1, where x varies from 1 to 0. 507 00:24:55,730 --> 00:25:01,170 And the curve C_3 is x equals 0, and y varying continuously 508 00:25:01,170 --> 00:25:03,250 from 1 to 0. 509 00:25:03,250 --> 00:25:06,270 At any rate, using the fact that, 510 00:25:06,270 --> 00:25:10,580 if a curve is the union of curves, 511 00:25:10,580 --> 00:25:12,580 that the line integral around that curve 512 00:25:12,580 --> 00:25:17,850 is the sum of the integral over the curves making up the union. 513 00:25:17,850 --> 00:25:20,340 I get that the line integral along C, 514 00:25:20,340 --> 00:25:23,070 y cubed dx plus x to the fourth dy, 515 00:25:23,070 --> 00:25:25,470 is the sum of these three line integrals. 516 00:25:25,470 --> 00:25:29,480 In other words, I integrate along C_1, C_2, and C_3, 517 00:25:29,480 --> 00:25:32,470 where C_1 is y equals x squared. 518 00:25:32,470 --> 00:25:35,390 And x varies continuously from 0 to 1. 519 00:25:35,390 --> 00:25:38,220 C_2 is the line y equals 1. 520 00:25:38,220 --> 00:25:40,920 And x varies continuously from 1 to 0. 521 00:25:40,920 --> 00:25:43,680 C_3 is the line x equals 0. 522 00:25:43,680 --> 00:25:46,510 And y varies continuously from 1 to 0, 523 00:25:46,510 --> 00:25:49,930 using the concepts that we talked about 524 00:25:49,930 --> 00:25:51,130 in the last lecture. 525 00:25:51,130 --> 00:25:55,430 For example, to evaluate this integral, this will be what? 526 00:25:55,430 --> 00:26:03,690 I replace y by x squared, dy by 2x*dx and integrate as x goes 527 00:26:03,690 --> 00:26:05,290 from 0 to 1. 528 00:26:05,290 --> 00:26:10,140 So my first integral is simply integral from 0 to 1, 529 00:26:10,140 --> 00:26:15,350 x to sixth dx plus x to the fourth times 2x*dx. 530 00:26:15,350 --> 00:26:20,790 My second integral, coming back to here, y is constantly 1. 531 00:26:20,790 --> 00:26:21,820 All right. 532 00:26:21,820 --> 00:26:23,470 So dy is 0. 533 00:26:23,470 --> 00:26:25,890 That makes this term drop out. 534 00:26:25,890 --> 00:26:28,140 Consequently, this term becomes what? 535 00:26:28,140 --> 00:26:32,710 1*dx, as x varies continuously from 1 to 0. 536 00:26:32,710 --> 00:26:36,680 So that integral becomes integral 1 to 0 dx. 537 00:26:36,680 --> 00:26:40,680 And my final integral, over C_3, notice, 538 00:26:40,680 --> 00:26:43,240 for that integral x is 0. 539 00:26:43,240 --> 00:26:46,440 So both this term and this term drop out. 540 00:26:46,440 --> 00:26:47,590 See x is 0. 541 00:26:47,590 --> 00:26:49,240 So this term drops out. 542 00:26:49,240 --> 00:26:52,690 The fact that x is a constant makes dx equal to 0. 543 00:26:52,690 --> 00:26:54,500 So this term drops out. 544 00:26:54,500 --> 00:26:59,790 So my integrand is just 0*dy as y goes from 1 to 0. 545 00:26:59,790 --> 00:27:01,960 And to make a long story short here, 546 00:27:01,960 --> 00:27:05,020 this integrand is simply x to the sixth plus 2x 547 00:27:05,020 --> 00:27:07,300 to the fifth dx from 0 to 1. 548 00:27:07,300 --> 00:27:10,530 This is simply another way of writing minus 1. 549 00:27:10,530 --> 00:27:12,130 This is 0. 550 00:27:12,130 --> 00:27:16,810 So my line integral along the closed curve C 551 00:27:16,810 --> 00:27:20,500 is 1/7 x to the seventh plus 1/3 x to the sixth, 552 00:27:20,500 --> 00:27:23,710 evaluated between 0 and 1, minus 1. 553 00:27:23,710 --> 00:27:29,340 This comes out to be a 1/7 plus 1/3, which is 10/21 minus 1 554 00:27:29,340 --> 00:27:32,190 makes this minus 11/21. 555 00:27:32,190 --> 00:27:35,390 And hopefully, that checks with the answer 556 00:27:35,390 --> 00:27:37,990 that we got by the previous method. 557 00:27:37,990 --> 00:27:40,440 And at any rate, if nothing else, 558 00:27:40,440 --> 00:27:44,570 that is an application as to how Green's theorem works. 559 00:27:44,570 --> 00:27:47,430 Now what this does as far as this lecture is concerned, 560 00:27:47,430 --> 00:27:49,570 and the exercises will bring this out, 561 00:27:49,570 --> 00:27:53,720 is it ties in line integrals with multiple integrals. 562 00:27:53,720 --> 00:27:57,590 That will be the last lecture of this particular block 563 00:27:57,590 --> 00:27:59,420 of material. 564 00:27:59,420 --> 00:28:03,160 And in the next lesson, what we will do 565 00:28:03,160 --> 00:28:06,740 is introduce a new block of material. 566 00:28:06,740 --> 00:28:10,010 We in particular shall talk about complex numbers. 567 00:28:10,010 --> 00:28:12,550 And that will occupy us for a little while. 568 00:28:12,550 --> 00:28:15,770 And at any rate though, until next time, goodbye. 569 00:28:19,860 --> 00:28:22,250 Funding for the publication of this video 570 00:28:22,250 --> 00:28:27,110 was provided by the Gabriella and Paul Rosenbaum foundation. 571 00:28:27,110 --> 00:28:31,280 Help OCW continue to provide free and open access to MIT 572 00:28:31,280 --> 00:28:38,990 courses by making a donation at ocw.mit.edu/donate.