1 00:00:00,000 --> 00:00:07,140 2 00:00:07,140 --> 00:00:08,950 CHRISTINE BREINER: Welcome back to recitation. 3 00:00:08,950 --> 00:00:12,650 In this video, I'd like us to see how geometric methods can 4 00:00:12,650 --> 00:00:16,220 help us understand the flux of a vector field across a curve. 5 00:00:16,220 --> 00:00:18,980 So in particular, what we're going to do is try and use 6 00:00:18,980 --> 00:00:22,400 geometric methods to compute the flux of four different 7 00:00:22,400 --> 00:00:27,140 vector fields F across curve C. So I've labeled them for 8 00:00:27,140 --> 00:00:27,870 later purposes. 9 00:00:27,870 --> 00:00:31,010 I've labeled them F1, F2, F3, and F4, 10 00:00:31,010 --> 00:00:32,960 and they are as follows. 11 00:00:32,960 --> 00:00:38,080 F sub 1 is the vector field that is a scalar function of 12 00:00:38,080 --> 00:00:42,940 only the radius times the vector x comma y. 13 00:00:42,940 --> 00:00:44,260 And the curve I'm interested in in this 14 00:00:44,260 --> 00:00:46,440 part 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 --> 00:00:54,990 Again, where g is a scalar function and r is the radius-- 17 00:00:54,990 --> 00:00:57,010 so it depends only on the radius-- 18 00:00:57,010 --> 00:00:59,410 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 --> 00:01:05,550 The third and fourth ones, we use a different C, but they 21 00:01:05,550 --> 00:01:06,720 will be the same there. 22 00:01:06,720 --> 00:01:08,590 I'll point that out. 23 00:01:08,590 --> 00:01:11,660 So in the third one, the vector field is 3 times the 24 00:01:11,660 --> 00:01:12,970 vector (1, 1). 25 00:01:12,970 --> 00:01:15,350 And C in this case will be the segment connecting 26 00:01:15,350 --> 00:01:16,470 (0, 0) 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 --> 00:01:29,510 And C is, again, this segment from (0, 0) to (1, 1). 30 00:01:29,510 --> 00:01:31,490 So again, what I'd like you to do is rather than trying to 31 00:01:31,490 --> 00:01:34,740 parametrize the curve and do the entire calculation, I'd 32 00:01:34,740 --> 00:01:37,480 like to see you try and understand the relationship 33 00:01:37,480 --> 00:01:41,950 between each of these factor fields F and the normal to the 34 00:01:41,950 --> 00:01:43,540 curve that they're on. 35 00:01:43,540 --> 00:01:46,310 And see if you can figure out the flux based on that 36 00:01:46,310 --> 00:01:47,590 relationship. 37 00:01:47,590 --> 00:01:49,930 So why don't you pause the video, give that a shot, and 38 00:01:49,930 --> 00:01:51,610 then when you're ready to see how I did it, you can bring 39 00:01:51,610 --> 00:01:52,860 the video back up. 40 00:01:52,860 --> 00:02:01,030 41 00:02:01,030 --> 00:02:03,160 OK, welcome back. 42 00:02:03,160 --> 00:02:06,240 Again, I'm going to try and use my geometric intuition to 43 00:02:06,240 --> 00:02:09,140 understand what the flux is for each of these four vector 44 00:02:09,140 --> 00:02:11,020 fields along these four curves. 45 00:02:11,020 --> 00:02:13,530 So what I'd like to do is if I'm going to try and 46 00:02:13,530 --> 00:02:15,840 understand geometrically what's happening, it's always 47 00:02:15,840 --> 00:02:17,130 good to draw a picture. 48 00:02:17,130 --> 00:02:19,630 So I'm going to draw a picture that I'll use for a and b, and 49 00:02:19,630 --> 00:02:21,350 then I'll draw another picture later that I'll 50 00:02:21,350 --> 00:02:22,690 use for c and d. 51 00:02:22,690 --> 00:02:25,970 So notice again in a, my curve is the unit circle. 52 00:02:25,970 --> 00:02:29,670 And I have a somewhat explicit understanding of what the 53 00:02:29,670 --> 00:02:30,440 vector fields are. 54 00:02:30,440 --> 00:02:32,880 So I'm going to draw the unit circle and see if I can figure 55 00:02:32,880 --> 00:02:34,750 out where F1 and F2 are. 56 00:02:34,750 --> 00:02:41,330 57 00:02:41,330 --> 00:02:45,280 It's not a perfect unit circle, but it looks sort of 58 00:02:45,280 --> 00:02:46,340 like the unit circle. 59 00:02:46,340 --> 00:02:48,540 So this is going to be my unit circle. 60 00:02:48,540 --> 00:02:52,600 And what I want to point out first is that I was not trying 61 00:02:52,600 --> 00:02:55,700 to trick you, but just to help you notice, that in a and b, 62 00:02:55,700 --> 00:02:59,130 they both depended on this radial function. 63 00:02:59,130 --> 00:03:02,270 But the radius on the unit circle is fixed at 1. 64 00:03:02,270 --> 00:03:04,390 So in both of these vector fields, it will 65 00:03:04,390 --> 00:03:06,500 simply be g of 1. 66 00:03:06,500 --> 00:03:08,340 So that will be a constant value. 67 00:03:08,340 --> 00:03:11,760 So all that's giving me is some scalar multiple of 68 00:03:11,760 --> 00:03:13,080 whatever the length of this one is. 69 00:03:13,080 --> 00:03:18,470 So it's this direction times this scalar g of 1. 70 00:03:18,470 --> 00:03:21,250 One other thing I want to point out is that both of 71 00:03:21,250 --> 00:03:26,040 these vectors, F1 and F2, if I ignore the g part, and I look 72 00:03:26,040 --> 00:03:29,570 at just the x comma y and the minus y comma x, is these 73 00:03:29,570 --> 00:03:31,080 parts have unit length. 74 00:03:31,080 --> 00:03:33,500 And you should be able to see that right away, because x and 75 00:03:33,500 --> 00:03:35,780 y are on the unit circle. 76 00:03:35,780 --> 00:03:40,650 And so the length of the vector whose tail is at the 77 00:03:40,650 --> 00:03:42,190 origin and head is on the edge of the unit 78 00:03:42,190 --> 00:03:43,550 circle has length 1. 79 00:03:43,550 --> 00:03:45,790 And then you can easily see that this vector and this 80 00:03:45,790 --> 00:03:48,320 vector have the same length, because their individual 81 00:03:48,320 --> 00:03:51,920 components are the same absolute value. 82 00:03:51,920 --> 00:03:54,520 We already understand a few things about F1 83 00:03:54,520 --> 00:03:55,910 and F2 right away. 84 00:03:55,910 --> 00:03:59,770 That all the length is coming from this g, and g is fixed 85 00:03:59,770 --> 00:04:02,430 all the way around the unit circle 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 --> 00:04:10,520 So if you notice first, F sub 1 is the vector (x, y) times 88 00:04:10,520 --> 00:04:12,190 the scalar g of 1. 89 00:04:12,190 --> 00:04:14,770 So if I come over here, I want to point out-- 90 00:04:14,770 --> 00:04:16,250 I'll do this part in white first-- that 91 00:04:16,250 --> 00:04:19,360 if I'm at this point-- 92 00:04:19,360 --> 00:04:21,080 this is the point (x, y)-- 93 00:04:21,080 --> 00:04:28,210 the vector (x, y) is equal to this vector. 94 00:04:28,210 --> 00:04:31,190 So if I think about putting that at this point-- 95 00:04:31,190 --> 00:04:32,520 I'm going to draw it here-- 96 00:04:32,520 --> 00:04:35,120 I get something that looks like this. 97 00:04:35,120 --> 00:04:38,820 So this is F sub 1, probably. 98 00:04:38,820 --> 00:04:41,180 I'm going to have to make one comment about that. 99 00:04:41,180 --> 00:04:42,700 But notice, this should look like 100 00:04:42,700 --> 00:04:44,460 it's all in one direction. 101 00:04:44,460 --> 00:04:47,600 So this is the vector (x, y). 102 00:04:47,600 --> 00:04:50,900 If I slide it so its tail is here, it's again in the same 103 00:04:50,900 --> 00:04:54,800 direction, and now I've just scaled it by g of 1. 104 00:04:54,800 --> 00:04:56,200 Now, this is assuming, obviously, 105 00:04:56,200 --> 00:04:57,560 that g of 1 is positive. 106 00:04:57,560 --> 00:04:59,720 So we're going to assume that throughout this problem. 107 00:04:59,720 --> 00:05:02,740 I'll mention what happens when g is negative at the end. 108 00:05:02,740 --> 00:05:05,060 So this is my vector F sub 1. 109 00:05:05,060 --> 00:05:07,500 Let's think about what is the normal to this curve. 110 00:05:07,500 --> 00:05:11,600 The normal to this curve, actually at each point on the 111 00:05:11,600 --> 00:05:15,840 circle, points in exactly the same direction as F sub 1. 112 00:05:15,840 --> 00:05:18,480 Because if I'm parametrizing in this direction-- 113 00:05:18,480 --> 00:05:20,620 I'll draw one down here so we can see what it looks like-- 114 00:05:20,620 --> 00:05:23,490 the normal-- actually, let me come from there-- 115 00:05:23,490 --> 00:05:26,500 the normal is going to look something like this. 116 00:05:26,500 --> 00:05:26,830 OK. 117 00:05:26,830 --> 00:05:30,130 So it would be connecting from the origin to that point on 118 00:05:30,130 --> 00:05:31,790 the circle, and keep going out in that direction. 119 00:05:31,790 --> 00:05:33,600 That's the normal direction. 120 00:05:33,600 --> 00:05:37,150 So F sub 1, if g is a positive function, it points in exactly 121 00:05:37,150 --> 00:05:38,410 the same direction as the normal. 122 00:05:38,410 --> 00:05:41,440 If g is a negative function, it points in exactly the 123 00:05:41,440 --> 00:05:42,460 opposite direction. 124 00:05:42,460 --> 00:05:44,860 So F1 would be flipped exactly around if g 125 00:05:44,860 --> 00:05:46,350 was a negative function. 126 00:05:46,350 --> 00:05:48,710 So if I want to compute the flux for part a-- 127 00:05:48,710 --> 00:05:51,800 I'll do it down here-- 128 00:05:51,800 --> 00:05:52,595 if I want to compute the flux, remember, I'm taking the 129 00:05:52,595 --> 00:05:58,020 integral along the curve of F dotted with n ds. 130 00:05:58,020 --> 00:06:00,400 Well, F dotted with n is constant. 131 00:06:00,400 --> 00:06:01,620 And that's the main point that's going 132 00:06:01,620 --> 00:06:03,110 to make this easier. 133 00:06:03,110 --> 00:06:04,550 At each point-- 134 00:06:04,550 --> 00:06:06,090 I guess I should say F1-- 135 00:06:06,090 --> 00:06:11,160 F1 dotted with n is always equal to the length of F1 136 00:06:11,160 --> 00:06:13,230 times the length of n times cosine of the 137 00:06:13,230 --> 00:06:15,090 angle between them. 138 00:06:15,090 --> 00:06:17,560 With a very quick calculation, you can see that winds up 139 00:06:17,560 --> 00:06:19,990 being g of 1. 140 00:06:19,990 --> 00:06:22,720 So the only reason I don't have to worry about absolute 141 00:06:22,720 --> 00:06:24,930 value, is if g is positive, I'm 142 00:06:24,930 --> 00:06:25,950 pointing in the same direction. 143 00:06:25,950 --> 00:06:27,950 If g is negative, I'm pointing in the opposite direction. 144 00:06:27,950 --> 00:06:32,550 And so the cosine theta is minus 1 instead of plus 1. 145 00:06:32,550 --> 00:06:34,960 You might want to check that for yourself, but this 146 00:06:34,960 --> 00:06:36,320 is just g of one. 147 00:06:36,320 --> 00:06:37,380 So that's a constant. 148 00:06:37,380 --> 00:06:40,150 So this is actually equal to g of 1 times the 149 00:06:40,150 --> 00:06:42,820 integral over C of ds. 150 00:06:42,820 --> 00:06:43,960 Now, what is this? 151 00:06:43,960 --> 00:06:46,250 If I integrate this, I should pick up exactly the 152 00:06:46,250 --> 00:06:48,120 length of the curve. 153 00:06:48,120 --> 00:06:48,300 OK. 154 00:06:48,300 --> 00:06:50,200 Because this is the derivative of arc length, so when I 155 00:06:50,200 --> 00:06:52,090 integrate this, I get arc length. 156 00:06:52,090 --> 00:06:55,490 But it's a unit circle, so the arc length is just the 157 00:06:55,490 --> 00:06:56,960 circumference of the unit circle. 158 00:06:56,960 --> 00:07:03,500 So that's 2 pi times g of 1. 159 00:07:03,500 --> 00:07:04,312 OK. 160 00:07:04,312 --> 00:07:06,150 And that's all you get. 161 00:07:06,150 --> 00:07:06,560 That's it. 162 00:07:06,560 --> 00:07:08,310 So we didn't actually have to parametrize anything. 163 00:07:08,310 --> 00:07:11,250 We just had to understand F1 relating to the normal. 164 00:07:11,250 --> 00:07:14,360 So I didn't draw the normal here, but if I take this 165 00:07:14,360 --> 00:07:16,380 normal and I spin around to here, the normal is in the 166 00:07:16,380 --> 00:07:18,880 same direction as F sub 1. 167 00:07:18,880 --> 00:07:21,260 So now let's look at F sub 2. 168 00:07:21,260 --> 00:07:24,200 And let's do this first by pointing something out about 169 00:07:24,200 --> 00:07:27,750 the relationship between F sub 2 and F sub 1. 170 00:07:27,750 --> 00:07:30,975 Notice that if I take F sub 1 and I dot it with F 171 00:07:30,975 --> 00:07:32,630 sub 2, I get 0. 172 00:07:32,630 --> 00:07:32,900 Right? 173 00:07:32,900 --> 00:07:36,070 Because ignoring even the scalar part, I get x times 174 00:07:36,070 --> 00:07:41,590 minus y, plus y times x, which gives me 0. 175 00:07:41,590 --> 00:07:44,640 If the scalars come along for the ride, I still get 0. 176 00:07:44,640 --> 00:07:47,810 So F sub 1 and F sub 2 are orthogonal. 177 00:07:47,810 --> 00:07:49,970 And in fact-- 178 00:07:49,970 --> 00:07:52,460 you can do this for yourself-- but if I come over and draw 179 00:07:52,460 --> 00:07:54,280 the picture, F sub 2 is going to be F sub 1 180 00:07:54,280 --> 00:07:58,560 rotated by 90 degrees. 181 00:07:58,560 --> 00:07:59,620 Something like this. 182 00:07:59,620 --> 00:08:01,590 This is my F sub 2. 183 00:08:01,590 --> 00:08:05,810 Again, if g of 1 was negative, F sub 1 would be this 184 00:08:05,810 --> 00:08:08,960 direction, and F sub 2 would then be around here. 185 00:08:08,960 --> 00:08:11,940 But ultimately, it's not going to matter in this case whether 186 00:08:11,940 --> 00:08:14,860 g is positive or negative, because notice what happens. 187 00:08:14,860 --> 00:08:19,470 If I want to integrate F sub 2 dotted with the normal, notice 188 00:08:19,470 --> 00:08:22,460 the normal is in the same direction as F sub 1, so F sub 189 00:08:22,460 --> 00:08:24,400 2 dotted with the normal is 0. 190 00:08:24,400 --> 00:08:26,810 So if I'm going to integrate the function 0 all along the 191 00:08:26,810 --> 00:08:30,000 curve, I shouldn't be surprised that my answer to 192 00:08:30,000 --> 00:08:33,410 part b is 0. 193 00:08:33,410 --> 00:08:35,190 So there was even less work in part b. 194 00:08:35,190 --> 00:08:38,260 Because I immediately had that F sub 2 is really in the 195 00:08:38,260 --> 00:08:40,740 direction of the tangent to the curve. 196 00:08:40,740 --> 00:08:42,490 And so I have something in the direction of the tangent 197 00:08:42,490 --> 00:08:45,150 dotted with something in the direction of the normal-- in 198 00:08:45,150 --> 00:08:46,410 fact, the normal-- 199 00:08:46,410 --> 00:08:48,190 so I get 0. 200 00:08:48,190 --> 00:08:48,430 All right. 201 00:08:48,430 --> 00:08:51,750 So now I'm going to draw a picture for c and d, and then 202 00:08:51,750 --> 00:08:53,000 we're going to use that one. 203 00:08:53,000 --> 00:08:55,960 204 00:08:55,960 --> 00:08:58,070 And I have to make sure I come on this side. 205 00:08:58,070 --> 00:08:59,630 Sorry about that. 206 00:08:59,630 --> 00:09:00,880 So here is (0, 0). 207 00:09:00,880 --> 00:09:04,110 208 00:09:04,110 --> 00:09:07,100 And here is (1, 1). 209 00:09:07,100 --> 00:09:07,930 Actually, you know what? 210 00:09:07,930 --> 00:09:09,020 I'm going to make it a little longer. 211 00:09:09,020 --> 00:09:10,160 I might need more room. 212 00:09:10,160 --> 00:09:14,970 So (1, 1) I'll make a little further up. 213 00:09:14,970 --> 00:09:15,960 OK? 214 00:09:15,960 --> 00:09:18,192 (0, 0) and (1, 1) and that's my curve. 215 00:09:18,192 --> 00:09:19,260 There's (1, 1). 216 00:09:19,260 --> 00:09:20,240 OK. 217 00:09:20,240 --> 00:09:23,710 Now, if I parametrize it in this direction, then I can 218 00:09:23,710 --> 00:09:26,360 draw my normal. 219 00:09:26,360 --> 00:09:30,100 Because it's a line segment, my normal is constant in its 220 00:09:30,100 --> 00:09:31,030 length and direction. 221 00:09:31,030 --> 00:09:39,040 So at any given point, it's exactly equal to this vector, 222 00:09:39,040 --> 00:09:40,350 up to the right scaling. 223 00:09:40,350 --> 00:09:46,170 And it should be something like (1, -1) divided 224 00:09:46,170 --> 00:09:47,750 by square root 2. 225 00:09:47,750 --> 00:09:48,430 That's my normal. 226 00:09:48,430 --> 00:09:50,930 If you want it precisely, that's what it is. 227 00:09:50,930 --> 00:09:53,520 You don't actually need it to solve this problem, though. 228 00:09:53,520 --> 00:09:56,030 But that's what it is if you want it precisely. 229 00:09:56,030 --> 00:10:00,260 Now let's look at what F3 is, and then we'll look at what F4 230 00:10:00,260 --> 00:10:00,780 actually is. 231 00:10:00,780 --> 00:10:03,450 So let's look at F sub 3. 232 00:10:03,450 --> 00:10:08,960 F sub 3 was the vector 3 times (1, 1). 233 00:10:08,960 --> 00:10:13,560 So if we come back to our picture, at any point on this 234 00:10:13,560 --> 00:10:19,350 curve, if I go in the (1, 1) direction, I stay parallel to 235 00:10:19,350 --> 00:10:20,600 this curve. 236 00:10:20,600 --> 00:10:22,600 237 00:10:22,600 --> 00:10:24,420 I don't want to draw the whole thing, because it would take 238 00:10:24,420 --> 00:10:25,340 up the entire curve. 239 00:10:25,340 --> 00:10:27,420 It's longer than the curve itself. 240 00:10:27,420 --> 00:10:29,340 But F sub 3, at any given point, 241 00:10:29,340 --> 00:10:32,520 points in this direction. 242 00:10:32,520 --> 00:10:37,920 So this is F sub 3, but not as long as it actually is. 243 00:10:37,920 --> 00:10:40,500 But only the direction is going to matter in this case. 244 00:10:40,500 --> 00:10:43,000 So F sub 3 is pointing in this direction. 245 00:10:43,000 --> 00:10:44,500 So if I want to compute the flux, I 246 00:10:44,500 --> 00:10:45,870 dot it with the normal. 247 00:10:45,870 --> 00:10:46,850 But look at what happens. 248 00:10:46,850 --> 00:10:50,580 The normal is orthogonal to F sub 3 at every point. 249 00:10:50,580 --> 00:10:52,760 And so F sub 3 dotted with the normal is 0. 250 00:10:52,760 --> 00:10:57,850 And so again, for exactly the same reason as part b, in part 251 00:10:57,850 --> 00:10:59,960 c, the flux is 0. 252 00:10:59,960 --> 00:11:03,290 So again, it's exactly the same that the vector field I 253 00:11:03,290 --> 00:11:05,930 was looking at and I wanted to compute the flux for, is 254 00:11:05,930 --> 00:11:08,460 actually tangent to the curve at every point. 255 00:11:08,460 --> 00:11:12,050 And so when I dot it to the normal to the curve at every 256 00:11:12,050 --> 00:11:13,870 point, I get 0. 257 00:11:13,870 --> 00:11:16,420 And so computing the flux is 0. 258 00:11:16,420 --> 00:11:20,460 So now, I have one more to do, and that one is part d. 259 00:11:20,460 --> 00:11:22,630 And in this case, F sub 4-- 260 00:11:22,630 --> 00:11:24,750 let me just remind you-- 261 00:11:24,750 --> 00:11:28,600 is 3 times the vector (-1, 1). 262 00:11:28,600 --> 00:11:31,770 And so if we go back to our picture here, F sub 4, if I 263 00:11:31,770 --> 00:11:36,320 compare it to the normal, in fact, what I get is very long. 264 00:11:36,320 --> 00:11:37,850 This is probably not quite long enough. 265 00:11:37,850 --> 00:11:43,400 But that's at least the direction of F sub 4. 266 00:11:43,400 --> 00:11:46,120 And so F sub 4 is exactly the opposite 267 00:11:46,120 --> 00:11:48,350 direction to the normal. 268 00:11:48,350 --> 00:11:52,580 So if I want to compute the flux of F sub 4 along this 269 00:11:52,580 --> 00:11:56,440 curve, all I have to understand is F sub 4 dotted 270 00:11:56,440 --> 00:11:59,120 with the normal and the length of the curve. 271 00:11:59,120 --> 00:12:02,600 This is exactly the same type of solution as part a. 272 00:12:02,600 --> 00:12:03,540 So let's notice this. 273 00:12:03,540 --> 00:12:06,690 First, F sub 4, the length is 3 root 2. 274 00:12:06,690 --> 00:12:08,610 You can compute that pretty quickly. 275 00:12:08,610 --> 00:12:10,870 The length of n is just 1. 276 00:12:10,870 --> 00:12:11,510 It's a normal. 277 00:12:11,510 --> 00:12:13,830 That's why this stuff didn't really matter 278 00:12:13,830 --> 00:12:14,950 what exactly it was. 279 00:12:14,950 --> 00:12:17,710 It's good to know what direction it points in. 280 00:12:17,710 --> 00:12:23,090 So F sub 4 dotted with n is exactly 3 root 2 times cosine 281 00:12:23,090 --> 00:12:25,340 of the angle between n and F sub 4. 282 00:12:25,340 --> 00:12:26,350 The angle is pi. 283 00:12:26,350 --> 00:12:29,610 You notice they differ by 180 degrees, right? 284 00:12:29,610 --> 00:12:33,080 So it's cosine pi, which is minus 1. 285 00:12:33,080 --> 00:12:37,160 So what I do in part d, is I'm integrating along the curve 286 00:12:37,160 --> 00:12:42,240 the constant negative 3 square root 2 ds. 287 00:12:42,240 --> 00:12:46,690 This is exactly F sub 4 dotted with the normal. 288 00:12:46,690 --> 00:12:51,030 And so as before, this is going to equal to negative 3 289 00:12:51,030 --> 00:12:52,870 root 2 times the arc length. 290 00:12:52,870 --> 00:12:56,130 Because the integral over the curve ds is going to be the 291 00:12:56,130 --> 00:12:57,590 arc length. 292 00:12:57,590 --> 00:13:00,480 And the arc length is very easy to see. 293 00:13:00,480 --> 00:13:02,590 You've gone over 1 and up 1. 294 00:13:02,590 --> 00:13:02,970 Right? 295 00:13:02,970 --> 00:13:06,050 So Pythagorean theorem, understanding right triangles, 296 00:13:06,050 --> 00:13:08,290 however you want to do it, the length of this 297 00:13:08,290 --> 00:13:10,160 curve is root 2. 298 00:13:10,160 --> 00:13:15,380 So this works out to be negative 3 root 2 root 2, 299 00:13:15,380 --> 00:13:19,210 which is just negative 6. 300 00:13:19,210 --> 00:13:22,450 So let me just remind you, there were four problems here. 301 00:13:22,450 --> 00:13:25,720 There are two sets of problems, where in each case, 302 00:13:25,720 --> 00:13:27,560 you have one similar to the other. 303 00:13:27,560 --> 00:13:30,490 So let me point this out one more time, and just 304 00:13:30,490 --> 00:13:31,530 sort of step back. 305 00:13:31,530 --> 00:13:34,060 We had a circle, and then in the next part 306 00:13:34,060 --> 00:13:35,140 we had a line segment. 307 00:13:35,140 --> 00:13:38,410 But in the circle, one of the problems had the vector in the 308 00:13:38,410 --> 00:13:40,930 direction of the normal, and you wanted to compute 309 00:13:40,930 --> 00:13:41,760 the flux for that. 310 00:13:41,760 --> 00:13:44,740 And in the other, the vector was tangent to the curve, and 311 00:13:44,740 --> 00:13:46,205 so it was orthogonal to the normal. 312 00:13:46,205 --> 00:13:48,620 And you wanted to compute the flux for that. 313 00:13:48,620 --> 00:13:51,310 Obviously, when you're tangent to the curve and then 314 00:13:51,310 --> 00:13:53,800 orthogonal to the normal, you get 0. 315 00:13:53,800 --> 00:13:56,370 And that was the case for b and for c. 316 00:13:56,370 --> 00:14:00,220 When you're normal to the curve and of constant length-- 317 00:14:00,220 --> 00:14:03,580 which was the case actually for both a and d-- 318 00:14:03,580 --> 00:14:08,810 then all you have to do is find F dotted with the normal, 319 00:14:08,810 --> 00:14:11,320 and then find the arc length, and multiply them together. 320 00:14:11,320 --> 00:14:16,010 So that was the real strategy we had to use for a and for d. 321 00:14:16,010 --> 00:14:18,900 So hopefully, this helps you see how the geometric 322 00:14:18,900 --> 00:14:22,065 quantities are interacting to understand the flux of a 323 00:14:22,065 --> 00:14:23,200 vector field across a curve. 324 00:14:23,200 --> 00:14:25,340 And that's where I'll stop. 325 00:14:25,340 --> 00:14:25,532