1 00:00:07,034 --> 00:00:08,950 CHRISTINE BREINER: Welcome back to recitation. 2 00:00:08,950 --> 00:00:12,650 In this video, I'd like us to see how geometric methods can 3 00:00:12,650 --> 00:00:16,220 help us understand the flux of a vector field across a curve. 4 00:00:16,220 --> 00:00:18,120 So in particular, what we're going to do 5 00:00:18,120 --> 00:00:20,155 is try and use geometric methods to compute 6 00:00:20,155 --> 00:00:24,940 the flux of four different vector fields F across curve C. 7 00:00:24,940 --> 00:00:27,870 So I've labeled them for later purposes. 8 00:00:27,870 --> 00:00:31,010 I've labeled them F_1, F_2, F_3, and F_4, 9 00:00:31,010 --> 00:00:32,960 and they are as follows. 10 00:00:32,960 --> 00:00:35,720 F sub 1 is the vector field that is 11 00:00:35,720 --> 00:00:39,760 a scalar function of only the radius, 12 00:00:39,760 --> 00:00:42,817 times the vector x comma y. 13 00:00:42,817 --> 00:00:44,650 And the curve I'm interested in in this part 14 00:00:44,650 --> 00:00:46,440 is the unit circle. 15 00:00:46,440 --> 00:00:51,390 The second one, part b, is the vector field that is g of r-- 16 00:00:51,390 --> 00:00:53,530 again, where g is a scalar function 17 00:00:53,530 --> 00:00:56,020 and r is the radius, so it depends only 18 00:00:56,020 --> 00:00:59,410 on the radius-- times the vector minus y, x. 19 00:00:59,410 --> 00:01:02,470 And again the C will be the unit circle. 20 00:01:02,470 --> 00:01:05,300 The third and fourth ones, we use a different C, 21 00:01:05,300 --> 00:01:06,720 but they will be the same there. 22 00:01:06,720 --> 00:01:08,479 I'll point that out. 23 00:01:08,479 --> 00:01:10,020 So in the third one, the vector field 24 00:01:10,020 --> 00:01:12,970 is 3 times the vector [1, 1]. 25 00:01:12,970 --> 00:01:15,870 And C in this case will be the segment connecting (0, 0) 26 00:01:15,870 --> 00:01:16,470 to (1, 1). 27 00:01:16,470 --> 00:01:20,740 So it's a piece of the line y equals x. 28 00:01:20,740 --> 00:01:25,500 And then in part d, F will be 3 times the vector [-1, 1]. 29 00:01:25,500 --> 00:01:29,089 And C is, again, this segment from (0, 0) to (1, 1). 30 00:01:29,089 --> 00:01:31,380 So again, what I'd like you to do is rather than trying 31 00:01:31,380 --> 00:01:34,550 to parametrize the curve and do the entire calculation, 32 00:01:34,550 --> 00:01:36,740 I'd like to see you try and understand 33 00:01:36,740 --> 00:01:39,050 the relationship between each of these vector fields 34 00:01:39,050 --> 00:01:43,540 F and the normal to the curve that they're on, 35 00:01:43,540 --> 00:01:46,030 and see if you can figure out the flux based 36 00:01:46,030 --> 00:01:47,590 on that relationship. 37 00:01:47,590 --> 00:01:49,780 So why don't you pause the video, give that a shot, 38 00:01:49,780 --> 00:01:51,240 and then when you're ready to see how I did it, 39 00:01:51,240 --> 00:01:52,573 you can bring the video back up. 40 00:02:01,030 --> 00:02:03,160 OK, welcome back. 41 00:02:03,160 --> 00:02:06,110 Again, I'm going to try and use my geometric intuition 42 00:02:06,110 --> 00:02:09,140 to understand what the flux is for each of these four vector 43 00:02:09,140 --> 00:02:11,020 fields along these four curves. 44 00:02:11,020 --> 00:02:14,000 So what I'd like to do is, if I'm going to try and understand 45 00:02:14,000 --> 00:02:15,590 geometrically what's happening, it's 46 00:02:15,590 --> 00:02:17,086 always good to draw a picture. 47 00:02:17,086 --> 00:02:19,460 So I'm going to draw a picture that I'll use for a and b, 48 00:02:19,460 --> 00:02:21,140 and then I'll draw another picture later 49 00:02:21,140 --> 00:02:22,690 that I'll use for c and d. 50 00:02:22,690 --> 00:02:25,970 So notice again in a, my curve is the unit circle. 51 00:02:25,970 --> 00:02:29,190 And I have a somewhat explicit understanding 52 00:02:29,190 --> 00:02:30,440 of what the vector fields are. 53 00:02:30,440 --> 00:02:31,940 So I'm going to draw the unit circle 54 00:02:31,940 --> 00:02:34,750 and see if I can figure out where F_1 and F_2 are. 55 00:02:41,330 --> 00:02:45,280 It's not a perfect unit circle, but it looks sort of 56 00:02:45,280 --> 00:02:46,340 like the unit circle. 57 00:02:46,340 --> 00:02:48,540 So this is going to be my unit circle. 58 00:02:48,540 --> 00:02:50,440 And what I want to point out first 59 00:02:50,440 --> 00:02:53,090 is that I was not trying to trick you, 60 00:02:53,090 --> 00:02:55,700 but just to help you notice, that in a and b, 61 00:02:55,700 --> 00:02:59,130 they both depended on this radial function. 62 00:02:59,130 --> 00:03:02,270 But the radius on the unit circle is fixed at 1. 63 00:03:02,270 --> 00:03:04,120 So in both of these vector fields, 64 00:03:04,120 --> 00:03:06,500 it will simply be g of 1. 65 00:03:06,500 --> 00:03:08,340 So that will be a constant value. 66 00:03:08,340 --> 00:03:11,997 So all that's giving me is some scalar multiple of whatever 67 00:03:11,997 --> 00:03:13,080 the length of this one is. 68 00:03:13,080 --> 00:03:18,470 So it's this direction times this scalar g of 1. 69 00:03:18,470 --> 00:03:19,940 One other thing I want to point out 70 00:03:19,940 --> 00:03:23,320 is that both of these vectors, F_1 and F_2, 71 00:03:23,320 --> 00:03:27,340 if I ignore the g part, and I look at just the x comma 72 00:03:27,340 --> 00:03:31,027 y and the minus y comma x, is these parts have unit length. 73 00:03:31,027 --> 00:03:33,360 And you should be able to see that right away, because x 74 00:03:33,360 --> 00:03:35,780 and y are on the unit circle. 75 00:03:35,780 --> 00:03:40,032 And so the length of the vector whose tail 76 00:03:40,032 --> 00:03:42,490 is at the origin and head is on the edge of the unit circle 77 00:03:42,490 --> 00:03:43,550 has length 1. 78 00:03:43,550 --> 00:03:46,070 And then you can easily see that this vector and this vector 79 00:03:46,070 --> 00:03:49,370 have the same length, because their individual components are 80 00:03:49,370 --> 00:03:51,920 the same absolute value. 81 00:03:51,920 --> 00:03:54,940 We already understand a few things about F_1 and F_2 82 00:03:54,940 --> 00:03:55,910 right away. 83 00:03:55,910 --> 00:03:58,720 That all the length is coming from this g, 84 00:03:58,720 --> 00:04:00,810 and g is fixed all the way around the unit circle 85 00:04:00,810 --> 00:04:02,430 at g of 1. 86 00:04:02,430 --> 00:04:06,330 Now let's figure out what the flux is for these two things. 87 00:04:06,330 --> 00:04:10,220 So if you notice first, F sub 1 is the vector [x, y] 88 00:04:10,220 --> 00:04:12,190 times the scalar g of 1. 89 00:04:12,190 --> 00:04:14,714 So if I come over here, I want to point out-- 90 00:04:14,714 --> 00:04:16,130 I'll do this part in white first-- 91 00:04:16,130 --> 00:04:21,080 that if I'm at this point-- this is the point (x, y)-- 92 00:04:21,080 --> 00:04:28,210 the vector [x, y] is equal to this vector. 93 00:04:28,210 --> 00:04:31,190 So if I think about putting that at this point-- 94 00:04:31,190 --> 00:04:32,860 I'm going to draw it here-- I get 95 00:04:32,860 --> 00:04:35,120 something that looks like this. 96 00:04:35,120 --> 00:04:38,820 So this is F sub 1, probably. 97 00:04:38,820 --> 00:04:41,180 I'm going to have to make one comment about that. 98 00:04:41,180 --> 00:04:44,460 But notice, this should look like it's all in one direction. 99 00:04:44,460 --> 00:04:47,600 So this is the vector [x, y]. 100 00:04:47,600 --> 00:04:49,820 If I slide it so its tail is here, 101 00:04:49,820 --> 00:04:51,460 it's again in the same direction, 102 00:04:51,460 --> 00:04:54,800 and now I've just scaled it by g of 1. 103 00:04:54,800 --> 00:04:56,920 Now, this is assuming, obviously, that g of 1 104 00:04:56,920 --> 00:04:57,471 is positive. 105 00:04:57,471 --> 00:04:59,720 So we're going to assume that throughout this problem. 106 00:04:59,720 --> 00:05:02,740 I'll mention what happens when g is negative at the end. 107 00:05:02,740 --> 00:05:05,060 So this is my vector F sub 1. 108 00:05:05,060 --> 00:05:07,500 Let's think about what is the normal to this curve. 109 00:05:07,500 --> 00:05:10,210 The normal to this curve, actually 110 00:05:10,210 --> 00:05:12,430 at each point on the circle, points 111 00:05:12,430 --> 00:05:15,627 in exactly the same direction as F sub 1. 112 00:05:15,627 --> 00:05:17,585 Because if I'm parametrizing in this direction, 113 00:05:17,585 --> 00:05:19,476 the normal-- I'll draw one down here 114 00:05:19,476 --> 00:05:21,850 so we can see what it looks like-- the normal-- actually, 115 00:05:21,850 --> 00:05:24,290 let me come from there-- the normal is going 116 00:05:24,290 --> 00:05:26,331 to look something like this. 117 00:05:26,331 --> 00:05:26,830 OK. 118 00:05:26,830 --> 00:05:29,309 So it would be connecting from the origin 119 00:05:29,309 --> 00:05:30,850 to that point on the circle, and keep 120 00:05:30,850 --> 00:05:32,016 going out in that direction. 121 00:05:32,016 --> 00:05:33,600 That's the normal direction. 122 00:05:33,600 --> 00:05:36,160 So F sub 1, if g is a positive function, 123 00:05:36,160 --> 00:05:38,410 it points in exactly the same direction as the normal. 124 00:05:38,410 --> 00:05:41,330 If g is a negative function, it points in exactly 125 00:05:41,330 --> 00:05:42,460 the opposite direction. 126 00:05:42,460 --> 00:05:44,590 So F_1 would be flipped exactly around 127 00:05:44,590 --> 00:05:46,350 if g was a negative function. 128 00:05:46,350 --> 00:05:48,960 So if I want to compute the flux for part a-- I'll do it 129 00:05:48,960 --> 00:05:52,110 down here-- if I want to compute the flux, remember, 130 00:05:52,110 --> 00:05:54,110 I'm taking the integral along the curve 131 00:05:54,110 --> 00:05:58,020 of F dotted with n ds. 132 00:05:58,020 --> 00:06:00,400 Well, F dotted with n is constant. 133 00:06:00,400 --> 00:06:03,110 And that's the main point that's going to make this easier. 134 00:06:03,110 --> 00:06:07,530 At each point-- I guess I should say F_1-- F_1 dotted with n 135 00:06:07,530 --> 00:06:12,740 is always equal to the length of F_1 times the length of n times 136 00:06:12,740 --> 00:06:15,090 cosine of the angle between them. 137 00:06:15,090 --> 00:06:16,520 With a very quick calculation, you 138 00:06:16,520 --> 00:06:19,990 can see that winds up being g of 1. 139 00:06:19,990 --> 00:06:22,980 So the only reason I don't have to worry about absolute value, 140 00:06:22,980 --> 00:06:25,575 is if g is positive, I'm pointing in the same direction. 141 00:06:25,575 --> 00:06:27,950 If g is negative, I'm pointing in the opposite direction. 142 00:06:27,950 --> 00:06:32,421 And so the cosine theta is minus 1 instead of plus 1. 143 00:06:32,421 --> 00:06:34,170 You might want to check that for yourself, 144 00:06:34,170 --> 00:06:36,320 but this is just g of one. 145 00:06:36,320 --> 00:06:37,380 So that's a constant. 146 00:06:37,380 --> 00:06:39,970 So this is actually equal to g of 1 times 147 00:06:39,970 --> 00:06:42,820 the integral over C of ds. 148 00:06:42,820 --> 00:06:43,960 Now, what is this? 149 00:06:43,960 --> 00:06:46,590 If I integrate this, I should pick up exactly the length 150 00:06:46,590 --> 00:06:47,800 of the curve. 151 00:06:47,800 --> 00:06:48,300 OK. 152 00:06:48,300 --> 00:06:50,174 Because this is the derivative of arc length, 153 00:06:50,174 --> 00:06:52,090 so when I integrate this, I get arc length. 154 00:06:52,090 --> 00:06:55,070 But it's a unit circle, so the arc length 155 00:06:55,070 --> 00:06:56,960 is just the circumference of the unit circle. 156 00:06:56,960 --> 00:07:03,500 So that's 2*pi times g of 1. 157 00:07:03,500 --> 00:07:04,312 OK. 158 00:07:04,312 --> 00:07:06,060 And that's all you get. 159 00:07:06,060 --> 00:07:06,560 That's it. 160 00:07:06,560 --> 00:07:08,685 So we didn't actually have to parametrize anything. 161 00:07:08,685 --> 00:07:11,250 We just had to understand F_1 relating to the normal. 162 00:07:11,250 --> 00:07:14,609 So I didn't draw the normal here, but if I take this normal 163 00:07:14,609 --> 00:07:16,150 and I spin around to here, the normal 164 00:07:16,150 --> 00:07:18,880 is in the same direction as F sub 1. 165 00:07:18,880 --> 00:07:21,260 So now let's look at F sub 2. 166 00:07:21,260 --> 00:07:23,970 And let's do this first by pointing something out 167 00:07:23,970 --> 00:07:27,750 about the relationship between F sub 2 and F sub 1. 168 00:07:27,750 --> 00:07:31,230 Notice that if I take F sub 1 and I dot it with F sub 2, 169 00:07:31,230 --> 00:07:32,401 I get 0. 170 00:07:32,401 --> 00:07:32,900 Right? 171 00:07:32,900 --> 00:07:38,150 Because ignoring even the scalar part, I get x times minus y, 172 00:07:38,150 --> 00:07:41,590 plus y times x, which gives me 0. 173 00:07:41,590 --> 00:07:44,640 If the scalars come along for the ride, I still get 0. 174 00:07:44,640 --> 00:07:47,810 So F sub 1 and F sub 2 are orthogonal. 175 00:07:47,810 --> 00:07:51,144 And in fact-- you can do this for yourself, 176 00:07:51,144 --> 00:07:52,810 but if I come over and draw the picture, 177 00:07:52,810 --> 00:07:55,660 F sub 2 is going to be F sub 1 rotated by 90 degrees. 178 00:07:58,560 --> 00:07:59,620 Something like this. 179 00:07:59,620 --> 00:08:01,590 This is my F sub 2. 180 00:08:01,590 --> 00:08:06,270 Again, if g of 1 was negative, F sub 1 would be this direction, 181 00:08:06,270 --> 00:08:08,960 and F sub 2 would then be around here. 182 00:08:08,960 --> 00:08:11,670 But ultimately, it's not going to matter in this case 183 00:08:11,670 --> 00:08:13,750 whether g is positive or negative, because notice 184 00:08:13,750 --> 00:08:14,860 what happens. 185 00:08:14,860 --> 00:08:19,130 If I want to integrate F sub 2 dotted with the normal, 186 00:08:19,130 --> 00:08:21,930 notice the normal is in the same direction as F sub 1, 187 00:08:21,930 --> 00:08:24,400 so F sub 2 dotted with the normal is 0. 188 00:08:24,400 --> 00:08:27,500 So if I'm going to integrate the function 0 all along the curve, 189 00:08:27,500 --> 00:08:33,410 I shouldn't be surprised that my answer to part b is 0. 190 00:08:33,410 --> 00:08:35,190 So there was even less work in part b. 191 00:08:35,190 --> 00:08:37,450 Because I immediately had that F sub 192 00:08:37,450 --> 00:08:40,199 2 is really in the direction of the tangent to the curve. 193 00:08:40,199 --> 00:08:42,490 And so I have something in the direction of the tangent 194 00:08:42,490 --> 00:08:44,370 dotted with something in the direction 195 00:08:44,370 --> 00:08:47,930 of the normal-- in fact, the normal-- so I get 0. 196 00:08:47,930 --> 00:08:48,430 All right. 197 00:08:48,430 --> 00:08:51,239 So now I'm going to draw a picture for c and d, 198 00:08:51,239 --> 00:08:52,780 and then we're going to use that one. 199 00:08:55,960 --> 00:08:58,070 And I have to make sure I come on this side. 200 00:08:58,070 --> 00:08:59,630 Sorry about that. 201 00:08:59,630 --> 00:09:00,795 So here is (0, 0). 202 00:09:04,110 --> 00:09:06,930 And here is (1, 1). 203 00:09:06,930 --> 00:09:07,930 Actually, you know what? 204 00:09:07,930 --> 00:09:09,020 I'm going to make it a little longer. 205 00:09:09,020 --> 00:09:10,160 I might need more room. 206 00:09:10,160 --> 00:09:14,970 So (1, 1) I'll make a little further up. 207 00:09:14,970 --> 00:09:15,960 OK? 208 00:09:15,960 --> 00:09:18,192 (0, 0) and (1, 1), and that's my curve. 209 00:09:18,192 --> 00:09:19,260 There's (1, 1). 210 00:09:19,260 --> 00:09:20,240 OK. 211 00:09:20,240 --> 00:09:23,190 Now, if I parametrize it in this direction, 212 00:09:23,190 --> 00:09:26,360 then I can draw my normal. 213 00:09:26,360 --> 00:09:28,760 Because it's a line segment, my normal 214 00:09:28,760 --> 00:09:31,030 is constant in its length and direction. 215 00:09:31,030 --> 00:09:39,040 So at any given point, it's exactly equal to this vector, 216 00:09:39,040 --> 00:09:40,350 up to the right scaling. 217 00:09:40,350 --> 00:09:45,740 And it should be something like [1, -1] 218 00:09:45,740 --> 00:09:47,722 divided by square root 2. 219 00:09:47,722 --> 00:09:48,430 That's my normal. 220 00:09:48,430 --> 00:09:50,930 If you want it precisely, that's what it is. 221 00:09:50,930 --> 00:09:53,520 You don't actually need it to solve this problem, though. 222 00:09:53,520 --> 00:09:56,030 But that's what it is if you want it precisely. 223 00:09:56,030 --> 00:09:59,440 Now let's look at what F_3 is, and then we'll 224 00:09:59,440 --> 00:10:00,780 look at what F_4 actually is. 225 00:10:00,780 --> 00:10:03,450 So let's look at F sub 3. 226 00:10:03,450 --> 00:10:08,960 F sub 3 was the vector 3 times [1, 1]. 227 00:10:08,960 --> 00:10:11,390 So if we come back to our picture, 228 00:10:11,390 --> 00:10:16,990 at any point on this curve, if I go in the [1, 1] direction, 229 00:10:16,990 --> 00:10:19,860 I stay parallel to this curve. 230 00:10:22,329 --> 00:10:23,870 I don't want to draw the whole thing, 231 00:10:23,870 --> 00:10:25,620 because it would take up the entire curve. 232 00:10:25,620 --> 00:10:27,420 It's longer than the curve itself. 233 00:10:27,420 --> 00:10:32,520 But F sub 3, at any given point, points in this direction. 234 00:10:32,520 --> 00:10:37,920 So this is F sub 3, but not as long as it actually is. 235 00:10:37,920 --> 00:10:40,500 But only the direction is going to matter in this case. 236 00:10:40,500 --> 00:10:43,000 So F sub 3 is pointing in this direction. 237 00:10:43,000 --> 00:10:45,809 So if I want to compute the flux, I dot it with the normal. 238 00:10:45,809 --> 00:10:46,850 But look at what happens. 239 00:10:46,850 --> 00:10:50,580 The normal is orthogonal to F sub 3 at every point. 240 00:10:50,580 --> 00:10:52,760 And so F sub 3 dotted with the normal is 0. 241 00:10:52,760 --> 00:10:57,150 And so again, for exactly the same reason as part b, 242 00:10:57,150 --> 00:10:59,960 in part c, the flux is 0. 243 00:10:59,960 --> 00:11:03,260 So again, it's exactly the same, that the vector field 244 00:11:03,260 --> 00:11:05,700 I was looking at and I wanted to compute the flux for, 245 00:11:05,700 --> 00:11:08,460 is actually tangent to the curve at every point. 246 00:11:08,460 --> 00:11:11,190 And so when I dot it to the normal 247 00:11:11,190 --> 00:11:13,870 to the curve at every point, I get 0. 248 00:11:13,870 --> 00:11:16,420 And so computing the flux is 0. 249 00:11:16,420 --> 00:11:20,460 So now, I have one more to do, and that one is part d. 250 00:11:20,460 --> 00:11:23,010 And in this case, F sub 4-- let me just 251 00:11:23,010 --> 00:11:28,600 remind you-- is 3 times the vector [-1, 1]. 252 00:11:28,600 --> 00:11:30,760 And so if we go back to our picture here, 253 00:11:30,760 --> 00:11:32,830 F sub 4, if I compare it to the normal, 254 00:11:32,830 --> 00:11:36,226 in fact, what I get is very long. 255 00:11:36,226 --> 00:11:37,850 This is probably not quite long enough. 256 00:11:37,850 --> 00:11:43,400 But that's at least the direction of F sub 4. 257 00:11:43,400 --> 00:11:46,580 And so F sub 4 is exactly the opposite direction 258 00:11:46,580 --> 00:11:48,350 to the normal. 259 00:11:48,350 --> 00:11:54,290 So if I want to compute the flux of F sub 4 along this curve, 260 00:11:54,290 --> 00:11:57,100 all I have to understand is F sub 4 dotted with the normal 261 00:11:57,100 --> 00:11:59,120 and the length of the curve. 262 00:11:59,120 --> 00:12:02,600 This is exactly the same type of solution as part a. 263 00:12:02,600 --> 00:12:03,540 So let's notice this. 264 00:12:03,540 --> 00:12:06,690 First, F sub 4, the length is 3 root 2. 265 00:12:06,690 --> 00:12:08,610 You can compute that pretty quickly. 266 00:12:08,610 --> 00:12:10,870 The length of n is just 1. 267 00:12:10,870 --> 00:12:11,482 It's a normal. 268 00:12:11,482 --> 00:12:12,940 That's why this stuff didn't really 269 00:12:12,940 --> 00:12:14,950 matter what exactly it was. 270 00:12:14,950 --> 00:12:17,710 It's good to know what direction it points in. 271 00:12:17,710 --> 00:12:20,040 So F sub 4 dotted with n is exactly 272 00:12:20,040 --> 00:12:25,340 3 root 2 times cosine of the angle between n and F sub 4. 273 00:12:25,340 --> 00:12:26,350 The angle is pi. 274 00:12:26,350 --> 00:12:29,610 You notice they differ by 180 degrees, right? 275 00:12:29,610 --> 00:12:33,080 So it's cosine pi, which is minus 1. 276 00:12:33,080 --> 00:12:37,160 So what I do in part d, is I'm integrating along the curve 277 00:12:37,160 --> 00:12:42,240 the constant negative 3 square root 2 ds. 278 00:12:42,240 --> 00:12:46,690 This is exactly F sub 4 dotted with the normal. 279 00:12:46,690 --> 00:12:51,030 And so as before, this is going to equal to negative 3 280 00:12:51,030 --> 00:12:52,870 root 2 times the arc length. 281 00:12:52,870 --> 00:12:55,110 Because the integral over the curve 282 00:12:55,110 --> 00:12:57,590 ds is going to be the arc length. 283 00:12:57,590 --> 00:13:00,480 And the arc length is very easy to see. 284 00:13:00,480 --> 00:13:02,470 You've gone over 1 and up 1. 285 00:13:02,470 --> 00:13:02,970 Right? 286 00:13:02,970 --> 00:13:06,050 So Pythagorean theorem, understanding right triangles, 287 00:13:06,050 --> 00:13:10,160 however you want to do it, the length of this curve is root 2. 288 00:13:10,160 --> 00:13:16,380 So this works out to be negative 3 root 2 root 2, which is just 289 00:13:16,380 --> 00:13:19,210 negative 6. 290 00:13:19,210 --> 00:13:22,450 So let me just remind you, there were four problems here. 291 00:13:22,450 --> 00:13:25,720 There are two sets of problems, where in each case, 292 00:13:25,720 --> 00:13:27,560 you have one similar to the other. 293 00:13:27,560 --> 00:13:30,490 So let me point this out one more time, and just 294 00:13:30,490 --> 00:13:31,530 sort of step back. 295 00:13:31,530 --> 00:13:34,060 We had a circle, and then in the next part 296 00:13:34,060 --> 00:13:35,140 we had a line segment. 297 00:13:35,140 --> 00:13:36,990 But in the circle, one of the problems 298 00:13:36,990 --> 00:13:39,927 had the vector in the direction of the normal, 299 00:13:39,927 --> 00:13:41,760 and you wanted to compute the flux for that. 300 00:13:41,760 --> 00:13:44,580 And in the other, the vector was tangent to the curve, 301 00:13:44,580 --> 00:13:46,205 and so it was orthogonal to the normal. 302 00:13:46,205 --> 00:13:48,620 And you wanted to compute the flux for that. 303 00:13:48,620 --> 00:13:51,020 Obviously, when you're tangent to the curve 304 00:13:51,020 --> 00:13:53,800 and then orthogonal to the normal, you get 0. 305 00:13:53,800 --> 00:13:56,370 And that was the case for b and for c. 306 00:13:56,370 --> 00:14:00,390 When you're normal to the curve and of constant length-- which 307 00:14:00,390 --> 00:14:04,490 was the case actually for both a and d-- then all you have to do 308 00:14:04,490 --> 00:14:08,810 is find F dotted with the normal, 309 00:14:08,810 --> 00:14:11,320 and then find the arc length, and multiply them together. 310 00:14:11,320 --> 00:14:16,010 So that was the real strategy we had to use for a and for d. 311 00:14:16,010 --> 00:14:19,390 So hopefully, this helps you see how the geometric quantities 312 00:14:19,390 --> 00:14:22,380 are interacting to understand the flux of a vector field 313 00:14:22,380 --> 00:14:23,200 across a curve. 314 00:14:23,200 --> 00:14:25,032 And that's where I'll stop.