1
00:00:01 --> 00:00:03
The following content is
provided under a Creative

2
00:00:03 --> 00:00:05
Commons license.
Your support will help MIT

3
00:00:05 --> 00:00:08
OpenCourseWare continue to offer
high quality educational

4
00:00:08 --> 00:00:13
resources for free.
To make a donation or to view

5
00:00:13 --> 00:00:18
additional materials from
hundreds of MIT courses,

6
00:00:18 --> 00:00:23
visit MIT OpenCourseWare at
ocw.mit.edu.

7
00:00:23 --> 00:00:33
Today I am going to tell you
about flux of a vector field for

8
00:00:33 --> 00:00:35
a curve.
In case you have seen flux in

9
00:00:35 --> 00:00:37
physics,
probably you have seen flux in

10
00:00:37 --> 00:00:39
space,
and we are going to come to

11
00:00:39 --> 00:00:41
that in a couple of weeks,
but for now we are still doing

12
00:00:41 --> 00:00:44
everything in the plane.
So bear with me if you have

13
00:00:44 --> 00:00:47
seen a more complicated version
of flux.

14
00:00:47 --> 00:00:50
We are going to do the easy one
first.

15
00:00:50 --> 00:00:59
What is flux?
Well, flux is actually another

16
00:00:59 --> 00:01:10
kind of line integral.
Let's say that I have a plane

17
00:01:10 --> 00:01:18
curve and a vector field in the
plane.

18
00:01:18 --> 00:01:27
Then the flux of F across a
curve C is, by definition,

19
00:01:27 --> 00:01:35
a line integral,
but I will use notation F dot n

20
00:01:35 --> 00:01:38
ds.
I have to explain to you what

21
00:01:38 --> 00:01:43
it means, but let me first box
that because that is the

22
00:01:43 --> 00:01:50
important formula to remember.
That is the definition.

23
00:01:50 --> 00:01:55
What does that mean?
First, mostly I have to tell

24
00:01:55 --> 00:02:01
you what this little n is.
The notation suggests it is a

25
00:02:01 --> 00:02:05
normal vector,
so what does that mean?

26
00:02:05 --> 00:02:16
I have a curve in the plane and
I have a vector field.

27
00:02:16 --> 00:02:24
Let's see.
The vector field will be yellow

28
00:02:24 --> 00:02:28
today.
And I will want to integrate

29
00:02:28 --> 00:02:33
along the curve the dot product
of F with the normal vector to

30
00:02:33 --> 00:02:37
the curve, a unit normal vector
to the curve.

31
00:02:37 --> 00:02:44
That means a vector that is at
every point of the curve

32
00:02:44 --> 00:02:51
perpendicular to the curve and
has length one.

33
00:02:51 --> 00:03:06
N everywhere will be the unit
normal vector to the curve C

34
00:03:06 --> 00:03:17
pointing 90 degrees clockwise
from T.

35
00:03:17 --> 00:03:19
What does that mean?
That means I have two normal

36
00:03:19 --> 00:03:22
vectors, one that is pointing
this way, one that is pointing

37
00:03:22 --> 00:03:24
that way.
I have to choose a convention.

38
00:03:24 --> 00:03:28
And the convention is that the
normal vector that I take goes

39
00:03:28 --> 00:03:32
to the right of the curve as I
am traveling along the curve.

40
00:03:32 --> 00:03:36
You mentioned that you were
walking along this curve,

41
00:03:36 --> 00:03:40
then you look to your right,
that is that direction.

42
00:03:40 --> 00:03:44
What we will do is just,
at every point along the curve,

43
00:03:44 --> 00:03:48
the dot product between the
vector field and the normal

44
00:03:48 --> 00:03:51
vector.
And we will sum that along the

45
00:03:51 --> 00:03:58
various pieces of the curve.
What this notation means is

46
00:03:58 --> 00:04:09
that if we actually break C into
small pieces of length delta(s)

47
00:04:09 --> 00:04:17
then the flux will be the limit,
as the pieces become smaller

48
00:04:17 --> 00:04:25
and smaller,
of the sum of F dot n delta S.

49
00:04:25 --> 00:04:29
I take each small piece of my
curve, I do the dot product

50
00:04:29 --> 00:04:32
between F and n and I multiply
by the length of a piece.

51
00:04:32 --> 00:04:35
And then I add these together.
That is what the line integral

52
00:04:35 --> 00:04:38
means.
Of course that is,

53
00:04:38 --> 00:04:41
again, not how I will compute
it.

54
00:04:41 --> 00:04:49
Just to compare this with work, 
conceptually it is similar to

55
00:04:49 --> 00:05:00
the line integral we did for
work except the line integral

56
00:05:00 --> 00:05:10
for work -- Work is the line
integral of F dot dr,

57
00:05:10 --> 00:05:15
which is also the line integral
of F dot T ds.

58
00:05:15 --> 00:05:20
That is how we reformulated it.
That means we take our curve

59
00:05:20 --> 00:05:26
and we figure out at each point
how big the tangent component --

60
00:05:26 --> 00:05:33
I guess I should probably take
the same vector field as before.

61
00:05:33 --> 00:05:37
Let's see.
My field was pointing more like

62
00:05:37 --> 00:05:40
that way.
What I do at any point is

63
00:05:40 --> 00:05:45
project F to the tangent
direction, I figure out how much

64
00:05:45 --> 00:05:50
F is going along my curve and
then I sum these things

65
00:05:50 --> 00:06:02
together.
I am actually summing -- -- the

66
00:06:02 --> 00:06:14
tangential component of my field
F.

67
00:06:14 --> 00:06:17
Roughly-speaking the work
measures, you know,

68
00:06:17 --> 00:06:21
when I move along my curve,
how much I am going with or

69
00:06:21 --> 00:06:23
against F.
Flux, on the other hand,

70
00:06:23 --> 00:06:26
measures, when I go along the
curve, roughly how much the

71
00:06:26 --> 00:06:28
field is going to across the
curve.

72
00:06:28 --> 00:06:31
Counting positively what goes
to the right,

73
00:06:31 --> 00:06:34
negatively what goes to the
left.

74
00:06:34 --> 00:06:45
Flux is integral F dot n ds,
and that one corresponds to

75
00:06:45 --> 00:06:54
summing the normal component of
a vector field.

76
00:06:54 --> 00:06:57
But apart from that
conceptually it is the same kind

77
00:06:57 --> 00:06:59
of thing.
Just the physical

78
00:06:59 --> 00:07:01
interpretations will be very
different,

79
00:07:01 --> 00:07:10
but for a mathematician these
are two line integrals that you

80
00:07:10 --> 00:07:17
set up and compute in pretty
much the same way.

81
00:07:17 --> 00:07:21
Let's see.
I should probably tell you what

82
00:07:21 --> 00:07:22
it means.
Why do we make this definition?

83
00:07:22 --> 00:07:27
What does it correspond to?
Well, the interpretation for

84
00:07:27 --> 00:07:31
work made a lot of sense when F
was representing a force.

85
00:07:31 --> 00:07:36
The line integral was actually
the work done by the force.

86
00:07:36 --> 00:07:40
The interpretation for flux
makes more sense if you think of

87
00:07:40 --> 00:07:50
F as a velocity field.
What is the interpretation?

88
00:07:50 --> 00:07:55
Let's say that for F is a
velocity field.

89
00:07:55 --> 00:08:00
That means I am thinking of
some fluid that is moving,

90
00:08:00 --> 00:08:04
maybe water or something,
and it is moving at a certain

91
00:08:04 --> 00:08:08
speed.
And my vector field represents

92
00:08:08 --> 00:08:14
how things are moving at every
point of the plane.

93
00:08:14 --> 00:08:31
I claim that flux measures how
much fluid passes through -- --

94
00:08:31 --> 00:08:41
the curve C per unit time.
If you imagine that maybe you

95
00:08:41 --> 00:08:45
have a river and you are somehow
building a damn here,

96
00:08:45 --> 00:08:48
a damn with holes in it so that
the water still passes through,

97
00:08:48 --> 00:08:53
then this measures how much
water passes through your

98
00:08:53 --> 00:08:57
membrane per unit time.
Let's try to figure out why

99
00:08:57 --> 00:09:00
this is true.
Why does this make sense?

100
00:09:00 --> 00:09:07
Let's look at what happens on a
small portion of our curve C.

101
00:09:07 --> 00:09:21
I am zooming in on my curve C.
I guess I need to zoom further.

102
00:09:21 --> 00:09:26
That is a little piece of my
curve, of length delta S,

103
00:09:26 --> 00:09:30
and there is a fluid flow.
On my picture things are

104
00:09:30 --> 00:09:33
flowing to the right.
Here I am drawing a constant

105
00:09:33 --> 00:09:38
vector field because if you zoom
in enough then your vectors will

106
00:09:38 --> 00:09:40
pretty much be the same
everywhere.

107
00:09:40 --> 00:09:44
If you enlarge the picture
enough then things will be

108
00:09:44 --> 00:09:48
pretty much a uniform flow.
Now, how much stuff goes

109
00:09:48 --> 00:09:51
through this little piece of
curve per unit time?

110
00:09:51 --> 00:09:57
Well, what happens over time is
the fluid is moving while my

111
00:09:57 --> 00:10:04
curve is staying the same place
so it corresponds to something

112
00:10:04 --> 00:10:09
like this.
I claim that what goes through

113
00:10:09 --> 00:10:16
C in unit time is actually going
to be a parallelogram.

114
00:10:16 --> 00:10:21
Here is a better picture.
I claim that what will be going

115
00:10:21 --> 00:10:24
through C is this shaded
parallelogram to the left of C.

116
00:10:24 --> 00:10:32
Let's see.
If I move for unit time it

117
00:10:32 --> 00:10:35
works.
That is the stuff that goes

118
00:10:35 --> 00:10:38
through my curve,
for a small portion of curve in

119
00:10:38 --> 00:10:40
unit time.
And, of course,

120
00:10:40 --> 00:10:43
I would need to add all of
these together to get the entire

121
00:10:43 --> 00:10:47
curve.
Let's try to understand how big

122
00:10:47 --> 00:10:50
this parallelogram is.
To know how big this

123
00:10:50 --> 00:10:53
parallelogram is I would like to
use base times height or

124
00:10:53 --> 00:10:56
something like that.
And maybe I want to actually

125
00:10:56 --> 00:10:59
flip my picture so that the base
and the height make more sense

126
00:10:59 --> 00:11:04
to me.
Let me actually turn it this

127
00:11:04 --> 00:11:10
way.
And, in case you have trouble

128
00:11:10 --> 00:11:21
reading the rotated picture,
let me redo it on the board.

129
00:11:21 --> 00:11:31
What passes through a portion
of C in unit time is the

130
00:11:31 --> 00:11:40
contents of a parallelogram
whose base is on C.

131
00:11:40 --> 00:11:49
So it has length delta s.
That is a piece of C.

132
00:11:49 --> 00:12:06
And the other side is going to
be given by my velocity vector

133
00:12:06 --> 00:12:11
F.
And to find the height of this

134
00:12:11 --> 00:12:17
thing, I need to know what
actually the normal component of

135
00:12:17 --> 00:12:24
this vector is.
If I call n the unit normal

136
00:12:24 --> 00:12:35
vector to the curve then the
area is base times height.

137
00:12:35 --> 00:12:42
The base is delta S and the
height is the normal component

138
00:12:42 --> 00:12:48
of F, so it is F dot n.
And so you see that when you

139
00:12:48 --> 00:12:54
sum these things together you
get, what I said,

140
00:12:54 --> 00:12:56
flux.
Now, if you are worried about

141
00:12:56 --> 00:13:00
the fact that actually -- If
your unit time is too long then

142
00:13:00 --> 00:13:03
of course things might start
changing as it flows.

143
00:13:03 --> 00:13:07
You have to take the time unit
and the length unit that are

144
00:13:07 --> 00:13:11
sufficiently small so that
really this approximation where

145
00:13:11 --> 00:13:15
C is a straight line and where
flow is at constant speed are

146
00:13:15 --> 00:13:17
valid.
You want to take maybe a

147
00:13:17 --> 00:13:20
segment here that is a few
micrometers.

148
00:13:20 --> 00:13:24
And the time unit might be a
few nanoseconds or whatever,

149
00:13:24 --> 00:13:28
and then it is a good
approximation.

150
00:13:28 --> 00:13:31
What I mean by per unit time
is, well, actually,

151
00:13:31 --> 00:13:35
that works, but you want to
think of a really,

152
00:13:35 --> 00:13:39
really small time.
And then the amount of matter

153
00:13:39 --> 00:13:44
that passes in that really,
really small time is the flux

154
00:13:44 --> 00:13:48
times the amount of time.
Let's be a tiny bit more

155
00:13:48 --> 00:13:50
careful.
And what I am saying is the

156
00:13:50 --> 00:13:53
amount of stuff that passes
through C depends actually on

157
00:13:53 --> 00:13:56
whether n is going this way or
the opposite way.

158
00:13:56 --> 00:14:00
Actually, 
what is implicit in this

159
00:14:00 --> 00:14:05
explanation is that I am
counting positively all the

160
00:14:05 --> 00:14:11
stuff that flows across C in the
direction of n and negatively

161
00:14:11 --> 00:14:15
what flows in the opposite
direction.

162
00:14:15 --> 00:14:32
What flows to the right of C,
well, across C from left to

163
00:14:32 --> 00:14:47
right is counted positively.
While what flows right to left

164
00:14:47 --> 00:14:53
is counted negatively.
So, in fact,

165
00:14:53 --> 00:15:00
it is the net flow through C
per unit time.

166
00:15:00 --> 00:15:06
Any questions about the
definition or the interpretation

167
00:15:06 --> 00:15:16
or things like that?
Yes?

168
00:15:16 --> 00:15:19
Well, you can have both not in
the same small segment.

169
00:15:19 --> 00:15:24
But it could be that, 
well, imagine that my vector

170
00:15:24 --> 00:15:28
field accidentally goes in the
opposite direction then this

171
00:15:28 --> 00:15:32
part of the curve,
while things are flowing to the

172
00:15:32 --> 00:15:35
left,
contributes negatively to flux.

173
00:15:35 --> 00:15:39
And here maybe the field is
tangent so the normal component

174
00:15:39 --> 00:15:42
becomes zero.
And then it becomes positive

175
00:15:42 --> 00:15:47
and this part of the curve
contributes positively.

176
00:15:47 --> 00:15:51
For example, 
if you imagine that you have a

177
00:15:51 --> 00:15:54
round tank in which the fluid is
rotating and you put your dam

178
00:15:54 --> 00:15:57
just on a diameter across then
things are going one way on one

179
00:15:57 --> 00:15:59
side,
the other way on the other

180
00:15:59 --> 00:16:04
side,
and actually it just evens out.

181
00:16:04 --> 00:16:06
We don't have complete
information.

182
00:16:06 --> 00:16:13
It is just the total net flux.
OK.

183
00:16:13 --> 00:16:19
If there are no other questions
then I guess we will need to

184
00:16:19 --> 00:16:25
figure out how to compute this
guy and how to actually do this

185
00:16:25 --> 00:16:33
line integral.
Well, let's start with a couple

186
00:16:33 --> 00:16:43
of easy examples.
Let's say that C is a circle of

187
00:16:43 --> 00:16:55
radius (a) centered at the
origin going counterclockwise.

188
00:16:55 --> 00:17:02
And let's say that our vector
field is xi yj.

189
00:17:02 --> 00:17:09
What does that look like?
Remember, xi plus yj is a

190
00:17:09 --> 00:17:15
vector field that is pointing
radially away from the origin.

191
00:17:15 --> 00:17:19
Because at every point it is
equal to the vector from the

192
00:17:19 --> 00:17:25
origin to that point.
Now, if we have a circle and

193
00:17:25 --> 00:17:30
let's say we are going
counterclockwise.

194
00:17:30 --> 00:17:32
Actually, I have a nicer
picture.

195
00:17:32 --> 00:17:48
Let me do it here.
That is my curve and my vector

196
00:17:48 --> 00:17:55
field.
And the normal vector, see, 

197
00:17:55 --> 00:17:57
when you go counterclockwise in
a closed curve,

198
00:17:57 --> 00:18:01
this convention that a normal
vector points to the right of

199
00:18:01 --> 00:18:04
curve makes it point out.
The usual convention,

200
00:18:04 --> 00:18:08
when you take flux for a closed
curve, is that you are counting

201
00:18:08 --> 00:18:11
the flux going out of the region
enclosed by the curve.

202
00:18:11 --> 00:18:13
And, of course,
if you went clockwise it would

203
00:18:13 --> 00:18:18
be the other way around.
You choose to do it the way you

204
00:18:18 --> 00:18:27
want, but the most common one is
to count flux going out of the

205
00:18:27 --> 00:18:31
region.
Let's see what happens.

206
00:18:31 --> 00:18:35
Well, if I am anywhere on my
circle, see, the normal vector

207
00:18:35 --> 00:18:38
is sticking straight out of the
circle.

208
00:18:38 --> 00:18:43
That is a property of the
circle that the radial direction

209
00:18:43 --> 00:18:49
is perpendicular to the circle.
Actually, let me complete this

210
00:18:49 --> 00:18:52
picture.
If I take a point on the

211
00:18:52 --> 00:18:59
circle, I have my normal vector
that is pointing straight out so

212
00:18:59 --> 00:19:05
it is parallel to F.
Along C we know that F is

213
00:19:05 --> 00:19:10
parallel to n,
so F dot n will be equal to the

214
00:19:10 --> 00:19:16
magnitude of F times,
well, the magnitude of n,

215
00:19:16 --> 00:19:20
but that is one.
Let me put it anywhere,

216
00:19:20 --> 00:19:23
but that is the unit normal
vector.

217
00:19:23 --> 00:19:27
Now, what is the magnitude of
this vector field if I am at a

218
00:19:27 --> 00:19:29
point x, y?
Well, it is square root of x

219
00:19:29 --> 00:19:32
squared plus y squared,
which is the same as the

220
00:19:32 --> 00:19:36
distance from the origin.
So if this distance,

221
00:19:36 --> 00:19:46
if this radius is a then the
magnitude of F will just be a.

222
00:19:46 --> 00:19:51
In fact, F dot n is constant,
always equal to a.

223
00:19:51 --> 00:19:57
So the line integral will be
pretty easy because all I have

224
00:19:57 --> 00:20:04
to do is the integral of F dot n
ds becomes the integral of a ds.

225
00:20:04 --> 00:20:07
(a) is a constant so I can take
it out.

226
00:20:07 --> 00:20:16
And integral ds is just a
length of C which is 2pi a,

227
00:20:16 --> 00:20:24
so I will get 2pi a squared.
And that is positive,

228
00:20:24 --> 00:20:28
as we expected,
because stuff is flowing out of

229
00:20:28 --> 00:20:36
the circle.
Any questions about that?

230
00:20:36 --> 00:20:41
No.
OK.

231
00:20:41 --> 00:20:45
Just out of curiosity,
let's say that we had taken our

232
00:20:45 --> 00:20:52
other favorite vector field.
Let's say that we had the same

233
00:20:52 --> 00:20:57
C, but now the vector field
.

234
00:20:57 --> 00:21:05
Remember, that one goes
counterclockwise around the

235
00:21:05 --> 00:21:09
origin.
If you remember what we did

236
00:21:09 --> 00:21:12
several times,
well, along the circle that

237
00:21:12 --> 00:21:16
vector field now is tangent to
the circle.

238
00:21:16 --> 00:21:19
If it is tangent to the circle
it doesn't have any normal

239
00:21:19 --> 00:21:22
component.
The normal component is zero.

240
00:21:22 --> 00:21:25
Things are not flowing into the
circle or out of it.

241
00:21:25 --> 00:21:30
They are just flowing along the
circle around and around so the

242
00:21:30 --> 00:21:38
flux will be zero.
F now is tangent to C.

243
00:21:38 --> 00:21:51
F dot n is zero and,
therefore, the flux will be

244
00:21:51 --> 00:21:55
zero.
These are examples where you

245
00:21:55 --> 00:21:57
can compute things
geometrically.

246
00:21:57 --> 00:22:00
And I would say,
generally speaking,

247
00:22:00 --> 00:22:03
with flux, well,
if it is a very complicated

248
00:22:03 --> 00:22:06
field then you cannot.
But, if a field is fairly

249
00:22:06 --> 00:22:08
simple,
you should be able to get some

250
00:22:08 --> 00:22:11
general feeling for whether your
answer should be positive,

251
00:22:11 --> 00:22:15
negative or zero just by
thinking about which way is my

252
00:22:15 --> 00:22:20
flow going.
Is it going across the curve

253
00:22:20 --> 00:22:32
one way or the other way?
Still no questions about these

254
00:22:32 --> 00:22:36
examples?
The next thing we need to know

255
00:22:36 --> 00:22:40
is how we will actually compute
these things because here,

256
00:22:40 --> 00:22:43
yeah, it works pretty well, 
but what if you don't have a

257
00:22:43 --> 00:22:47
simple geometric interpretation.
What if I give you a really

258
00:22:47 --> 00:22:50
complicated curve and then you
have trouble finding the normal

259
00:22:50 --> 00:22:53
vector?
It is going to be annoying to

260
00:22:53 --> 00:22:56
set up things this way.
Actually, there is a better way

261
00:22:56 --> 00:22:59
to do it in coordinates.
Just as we do work,

262
00:22:59 --> 00:23:04
when we compute this line
integral, usually we don't do it

263
00:23:04 --> 00:23:08
geometrically like this.
Most of the time we just

264
00:23:08 --> 00:23:12
integrate M dx plus N dy in
coordinates.

265
00:23:12 --> 00:23:16
That is a similar way to do it
because it is,

266
00:23:16 --> 00:23:20
again, a line integral so it
should work the same way.

267
00:23:20 --> 00:23:21
Let's try to figure that out.

268
00:23:21 --> 00:24:05

269
00:24:05 --> 00:24:22
How do we do the calculation in
coordinates, or I should say

270
00:24:22 --> 00:24:29
using components?
That is the general method of

271
00:24:29 --> 00:24:33
calculation when we don't have
something geometric to do.

272
00:24:33 --> 00:24:41
Remember, when we were doing
things for work we said this

273
00:24:41 --> 00:24:49
vector dr, or if you prefer T
ds, we said just becomes

274
00:24:49 --> 00:24:56
symbolically dx and dy.
When you do the line integral

275
00:24:56 --> 00:25:01
of F dot dr you get line
integral of n dx plus n dy.

276
00:25:01 --> 00:25:07
Now let's think for a second
about how we would express n ds.

277
00:25:07 --> 00:25:11
Well, what is n ds compared to
T ds?

278
00:25:11 --> 00:25:15
Well, M is just T rotated by 90
degrees, so n ds is T ds rotated

279
00:25:15 --> 00:25:19
by 90 degrees.
That might sound a little bit

280
00:25:19 --> 00:25:23
outrageous because these are
really symbolic notations but it

281
00:25:23 --> 00:25:25
works.
I am not going to spend too

282
00:25:25 --> 00:25:28
much time trying to convince you
carefully.

283
00:25:28 --> 00:25:33
But if you go back to where we
wrote this and how we tried to

284
00:25:33 --> 00:25:36
justify this and you work your
way through it,

285
00:25:36 --> 00:25:42
you will see that n ds can be
analyzed the same way.

286
00:25:42 --> 00:25:51
N is T rotated 90 degrees
clockwise.

287
00:25:51 --> 00:25:57
That tells us that n ds is --
How do we rotate a vector by 90

288
00:25:57 --> 00:26:00
degrees?
Well, we swept the two

289
00:26:00 --> 00:26:05
components and we put a minus
sign.

290
00:26:05 --> 00:26:07
You have dy and dx.
And you have to be careful

291
00:26:07 --> 00:26:11
where to put the minus sign.
Well, if you are doing it

292
00:26:11 --> 00:26:13
clockwise, it is in front of dx.

293
00:26:13 --> 00:26:26

294
00:26:26 --> 00:26:29
Well, actually,
let me just convince you

295
00:26:29 --> 00:26:32
quickly.
Let's say we have a small piece

296
00:26:32 --> 00:26:36
of C.
If we do T delta S,

297
00:26:36 --> 00:26:44
that is also vector delta r.
That is going to be just the

298
00:26:44 --> 00:26:48
vector that goes along the curve
given by this.

299
00:26:48 --> 00:26:54
Its components will be indeed
the change in x,

300
00:26:54 --> 00:27:00
delta x, and the change in y,
delta y.

301
00:27:00 --> 00:27:07
And now, if I want to get n
delta S, well,

302
00:27:07 --> 00:27:15
I claim now that it is
perfectly valid and rigorous to

303
00:27:15 --> 00:27:24
just rotate that by 90 degrees.
If I want to rotate this by 90

304
00:27:24 --> 00:27:31
degrees clockwise then the x
component will become the same

305
00:27:31 --> 00:27:36
as the old y component.
And the y component will be

306
00:27:36 --> 00:27:40
minus delta x.
Then you take the limit when

307
00:27:40 --> 00:27:44
the segment becomes shorter and
shorter, and that is how you can

308
00:27:44 --> 00:27:47
justify this.
That is the key to computing

309
00:27:47 --> 00:27:50
things in practice.
It means, actually,

310
00:27:50 --> 00:27:55
you already know how to compute
line integrals for flux.

311
00:27:55 --> 00:28:05
Let me just write it explicitly.
Let's say that our vector field

312
00:28:05 --> 00:28:08
has two components.
And let me just confuse you a

313
00:28:08 --> 00:28:12
little bit and not call them M
and N for this time just to

314
00:28:12 --> 00:28:16
stress the fact that we are
doing a different line integral.

315
00:28:16 --> 00:28:22
Let me call them P and Q for
now.

316
00:28:22 --> 00:28:31
Then the line integral of F dot
n ds will be the line integral

317
00:28:31 --> 00:28:39
of 
dot product .

318
00:28:39 --> 00:28:46
That will be the integral of -
Q dx P dy.

319
00:28:46 --> 00:28:50
Well, I am running out of space
here.

320
00:28:50 --> 00:29:01
It is integral along C of
negative Q dx plus P dy.

321
00:29:01 --> 00:29:04
And from that point onwards you
just do it the usual way.

322
00:29:04 --> 00:29:10
Remember, here you have two
variables x and y but you are

323
00:29:10 --> 00:29:14
integrating along a curve.
If you are integrating along a

324
00:29:14 --> 00:29:18
curve x and y are related.
They depend on each other or

325
00:29:18 --> 00:29:21
maybe on some other parameter
like T or theta or whatever.

326
00:29:21 --> 00:29:28
You express everything in terms
of a single variable and then

327
00:29:28 --> 00:29:36
you do a usual single integral.
Any questions about that?

328
00:29:36 --> 00:29:39
I see a lot of confused faces
so maybe I shouldn't have called

329
00:29:39 --> 00:29:41
my component P and Q.

330
00:29:41 --> 00:30:04

331
00:30:04 --> 00:30:14
If you prefer,
if you are really sentimentally

332
00:30:14 --> 00:30:27
attached to M and N then this
new line integral becomes the

333
00:30:27 --> 00:30:35
integral of - N dx M dy.
If a problem tells you compute

334
00:30:35 --> 00:30:37
flux instead of saying compute
work,

335
00:30:37 --> 00:30:41
the only thing you change is
instead of doing M dx plus N dy

336
00:30:41 --> 00:30:45
you do minus N dx plus M dy.
And I am sorry to say that I

337
00:30:45 --> 00:30:49
don't have any good way of
helping you remember which one

338
00:30:49 --> 00:30:52
of the two gets the minus sign,
so you just have to remember

339
00:30:52 --> 00:30:58
this formula by heart.
That is the only way I know.

340
00:30:58 --> 00:31:04
Well, you can try to go through
this argument again,

341
00:31:04 --> 00:31:10
but it is really best if you
just remember that formula.

342
00:31:10 --> 00:31:15
I am not going to do an example
because we already know how to

343
00:31:15 --> 00:31:19
do line integrals.
Hopefully you will get to see

344
00:31:19 --> 00:31:23
one at least in recitation on
Monday.

345
00:31:23 --> 00:31:29
That is all pretty good.
Let me tell you now what if I

346
00:31:29 --> 00:31:35
have to compute flux along a
closed curve and I don't want to

347
00:31:35 --> 00:31:39
compute it?
Well, remember in the case of

348
00:31:39 --> 00:31:43
work we had Green's theorem.
We saw yesterday Green's

349
00:31:43 --> 00:31:45
theorem.
Let's us replace a line

350
00:31:45 --> 00:31:48
integral along a closed curve by
a double integral.

351
00:31:48 --> 00:31:51
Well, here it is the same.
We have a line integral along a

352
00:31:51 --> 00:31:53
curve.
If it is a closed curve,

353
00:31:53 --> 00:31:57
we should be able to replace it
by a double integral.

354
00:31:57 --> 00:32:09
There is a version of Green's
theorem for flux.

355
00:32:09 --> 00:32:13
And you will see it is not
scarier than the other one.

356
00:32:13 --> 00:32:18
It is perhaps less scarier or
perhaps just as scary or just

357
00:32:18 --> 00:32:22
not as scary,
depending on how you feel about

358
00:32:22 --> 00:32:26
it, but it works pretty much the
same way.

359
00:32:26 --> 00:32:30
What does Green's theorem for
flux say?

360
00:32:30 --> 00:32:39
It says if C is a curve that
encloses a region R

361
00:32:39 --> 00:32:51
counterclockwise and if I have a
vector field that is defined

362
00:32:51 --> 00:32:56
everywhere,
not just on C but also inside, 

363
00:32:56 --> 00:33:11
so also on R.
Well, maybe I should give names

364
00:33:11 --> 00:33:14
to the components.
If you will forgive me for a

365
00:33:14 --> 00:33:16
second, I will still use P and Q
for now.

366
00:33:16 --> 00:33:23
You will see why.
It is defined and

367
00:33:23 --> 00:33:30
differentiable in R.
Then I can actually -- --

368
00:33:30 --> 00:33:40
replace the line integral for
flux by a double integral over R

369
00:33:40 --> 00:33:47
of some function.
And that function is called the

370
00:33:47 --> 00:33:58
divergence of F dA.
This is the divergence of F.

371
00:33:58 --> 00:34:08
And I have to define for you
what this guy is.

372
00:34:08 --> 00:34:15
The divergence of a vector
field with components P and Q is

373
00:34:15 --> 00:34:20
just P sub x Q sub y.
This one is actually easier to

374
00:34:20 --> 00:34:23
remember than curl because you
just take the x component,

375
00:34:23 --> 00:34:26
take its partial with respect
to x,

376
00:34:26 --> 00:34:29
take the y component, 
take its partial with respect

377
00:34:29 --> 00:34:31
to y and add them together.
No signs.

378
00:34:31 --> 00:34:36
No switching things around.
This one is pretty

379
00:34:36 --> 00:34:43
straightforward.
The picture again is if I have

380
00:34:43 --> 00:34:50
my curve C going
counterclockwise around a region

381
00:34:50 --> 00:34:58
R and I want to find the flux of
some vector field F that is

382
00:34:58 --> 00:35:03
everywhere in here.
Maybe some parts of C will

383
00:35:03 --> 00:35:06
contribute positively and some
parts will contribute

384
00:35:06 --> 00:35:10
negatively.
Just to reiterate what I said,

385
00:35:10 --> 00:35:14
positively here means,
because we are going

386
00:35:14 --> 00:35:19
counterclockwise,
the normal vector points out of

387
00:35:19 --> 00:35:31
the region.
This guy here is the flux out

388
00:35:31 --> 00:35:39
of R through C.
That is the formula.

389
00:35:39 --> 00:35:45
Any questions about what the
statement says or how to use it

390
00:35:45 --> 00:35:48
concretely?
No.

391
00:35:48 --> 00:35:51
OK.
It is pretty similar to Green's

392
00:35:51 --> 00:35:58
theorem for work.
Actually, I should say -- This

393
00:35:58 --> 00:36:07
is called Green's theorem in
normal form also.

394
00:36:07 --> 00:36:17
Not that the other one is
abnormal, but just that the old

395
00:36:17 --> 00:36:23
one for work was,
you could say,

396
00:36:23 --> 00:36:28
in tangential form.
That just means,

397
00:36:28 --> 00:36:32
well, Green's theorem,
as seen yesterday was for the

398
00:36:32 --> 00:36:36
line integral F dot T ds,
integrating the tangent

399
00:36:36 --> 00:36:39
component of F.
The one today is for

400
00:36:39 --> 00:36:43
integrating the normal component
of F.

401
00:36:43 --> 00:36:47
OK. Let's prove this.
Good news.

402
00:36:47 --> 00:36:50
It is much easier to prove than
the one we did yesterday because

403
00:36:50 --> 00:36:53
we are just going to show that
it is the same thing just using

404
00:36:53 --> 00:36:54
different notations.

405
00:36:54 --> 00:37:23

406
00:37:23 --> 00:37:29
How do I prove it?
Well, maybe actually it would

407
00:37:29 --> 00:37:33
help if first,
before proving it,

408
00:37:33 --> 00:37:38
I actually rewrite what it
means in components.

409
00:37:38 --> 00:37:46
We said the line integral of F
dot n ds is actually the line

410
00:37:46 --> 00:37:54
integral of - Q dx P dy.
And we want to show that this

411
00:37:54 --> 00:38:03
is equal to the double integral
of P sub x Q sub y dA.

412
00:38:03 --> 00:38:09
This is really one of the
features of Green's theorem.

413
00:38:09 --> 00:38:12
No matter which form it is,
it relates a line integral to a

414
00:38:12 --> 00:38:16
double integral.
Let's just try to see if we can

415
00:38:16 --> 00:38:19
reduce it to the one we had
yesterday.

416
00:38:19 --> 00:38:24
Let me forget what these things
mean physically and just focus

417
00:38:24 --> 00:38:26
on the math.
On the math it is a line

418
00:38:26 --> 00:38:29
integral of something dx plus
something dy.

419
00:38:29 --> 00:38:36
Let's call this guy M and let's
call this guy N.

420
00:38:36 --> 00:38:42
Let M equal negative Q and N
equal P.

421
00:38:42 --> 00:38:53
Then this guy here becomes
integral of M dx plus N dy.

422
00:38:53 --> 00:38:57
And I know from yesterday what
this is equal to,

423
00:38:57 --> 00:39:01
namely using the tangential
form of Green's theorem.

424
00:39:01 --> 00:39:05
Green for work.
This is the double integral of

425
00:39:05 --> 00:39:11
curl of this guy.
That is Nx minus My dA.

426
00:39:11 --> 00:39:15
But now let's think about what
this is in terms of M and N.

427
00:39:15 --> 00:39:24
Well, we said that M is
negative Q so this is negative

428
00:39:24 --> 00:39:29
My.
And we said P is the same as N,

429
00:39:29 --> 00:39:33
so this is Nx.
Just by renaming the

430
00:39:33 --> 00:39:37
components, I go from one form
to the other one.

431
00:39:37 --> 00:39:38
So it is really the same
theorem.

432
00:39:38 --> 00:39:41
That's why it is also called
Green's theorem.

433
00:39:41 --> 00:39:45
But the way we think about it
when we use it is different,

434
00:39:45 --> 00:39:48
because one of them computes
the work done by a force along a

435
00:39:48 --> 00:39:53
closed curve,
the other one computes the flux

436
00:39:53 --> 00:39:59
maybe of a velocity field out of
region.

437
00:39:59 --> 00:40:10
Questions?
Yes?

438
00:40:10 --> 00:40:14
That is correct.
If you are trying to compute a

439
00:40:14 --> 00:40:18
line integral for flux,
wait, where did I put it?

440
00:40:18 --> 00:40:20
A line integral for flux just
becomes this.

441
00:40:20 --> 00:40:25
And once you are here you know
how to compute that kind of

442
00:40:25 --> 00:40:27
thing.
The double integral side does

443
00:40:27 --> 00:40:29
not even have any kind of
renaming to do.

444
00:40:29 --> 00:40:31
You know how to compute a
double integral of a function.

445
00:40:31 --> 00:40:35
This is just a particular kind
of function that you get out of

446
00:40:35 --> 00:40:38
a vector field,
but it is like any function.

447
00:40:38 --> 00:40:41
The way you would evaluate
these double integrals is just

448
00:40:41 --> 00:40:46
the usual way.
Namely, you have a function of

449
00:40:46 --> 00:40:54
x and y, you have a region and
you set up the bounds for the

450
00:40:54 --> 00:40:57
isolated integral.
The way you would evaluate the

451
00:40:57 --> 00:40:59
double integrals is really the
usual way,

452
00:40:59 --> 00:41:02
by slicing the region and
setting up the bounds for

453
00:41:02 --> 00:41:06
iterated integrals in dx,
dy or dydx or maybe rd, 

454
00:41:06 --> 00:41:12
rd theta or whatever you want.
In fact, in terms of computing

455
00:41:12 --> 00:41:14
integrals, we just have two sets
of skills.

456
00:41:14 --> 00:41:18
One is setting up and
evaluating double integrals.

457
00:41:18 --> 00:41:21
The other one is setting up and
evaluating line integrals.

458
00:41:21 --> 00:41:25
And whether these line
integrals or double integrals

459
00:41:25 --> 00:41:29
are representing work,
flux, integral of a curve,

460
00:41:29 --> 00:41:34
whatever, 
the way that we actually

461
00:41:34 --> 00:41:40
compute them is the same.
Let's do an example.

462
00:41:40 --> 00:41:47
Oh, first. Sorry.
This renaming here, see, 

463
00:41:47 --> 00:41:51
that is why actually I call my
components P and Q because the

464
00:41:51 --> 00:41:54
argument would have gotten very
messy if I had told you now I

465
00:41:54 --> 00:41:57
call M ,N and I call N minus M
and so on.

466
00:41:57 --> 00:42:00
But, now that we are through
with this,

467
00:42:00 --> 00:42:03
if you still like M and N
better,

468
00:42:03 --> 00:42:15
you know, what this says -- The
formulation of Green's theorem

469
00:42:15 --> 00:42:27
in this language is just
integral of minus N dx plus M dy

470
00:42:27 --> 00:42:37
is the double integral over R of
Mx plus Ny dA.

471
00:42:37 --> 00:42:42
Now let's do an example.
Let's look at this picture

472
00:42:42 --> 00:42:51
again, the flux of xi plus yj
out of the circle of radius A.

473
00:42:51 --> 00:42:53
We did the calculation directly
using geometry,

474
00:42:53 --> 00:42:57
and it wasn't all that bad.
But let's see what Green's

475
00:42:57 --> 00:42:58
theorem does for us here.

476
00:42:58 --> 00:43:19

477
00:43:19 --> 00:43:22
Example.
Let's take the same example as

478
00:43:22 --> 00:43:28
last time.
F equals xi yj.

479
00:43:28 --> 00:43:43
C equals circle of radius a
counterclockwise.

480
00:43:43 --> 00:43:46
How do we set up Green's
theorem.

481
00:43:46 --> 00:43:57
Well, let's first figure out
the divergence of F.

482
00:43:57 --> 00:44:00
The divergence of this field,
I take the x component,

483
00:44:00 --> 00:44:03
which is x, and I take its
partial respect to x.

484
00:44:03 --> 00:44:08
And then I do the same with the
y component, and I will get one

485
00:44:08 --> 00:44:12
plus one equals two.
So, the divergence of this

486
00:44:12 --> 00:44:17
field is two.
Now, Green's theorem tells us

487
00:44:17 --> 00:44:25
that the flux out of this region
is going to be the double

488
00:44:25 --> 00:44:29
integral of 2 dA.
What is R now?

489
00:44:29 --> 00:44:31
Well, R is the region enclosed
by C.

490
00:44:31 --> 00:44:38
So if C is the circle,
R is the disk of radius A.

491
00:44:38 --> 00:44:42
Of course, we can compute it, 
but we don't have to because

492
00:44:42 --> 00:44:46
double integral of 2dA is just
twice the double integral of dA

493
00:44:46 --> 00:44:51
so it is twice the area of R.
And we know the area of a

494
00:44:51 --> 00:44:54
circle of radius A.
That is piA2.

495
00:44:54 --> 00:45:01
So, it is 2piA2.
That is the same answer that we

496
00:45:01 --> 00:45:04
got directly,
which is good news.

497
00:45:04 --> 00:45:08
Now we can even do better.
Let's say that my circle is not

498
00:45:08 --> 00:45:12
at the origin.
Let's say that it is out here.

499
00:45:12 --> 00:45:17
Well, then it becomes harder to
calculate the flux directly.

500
00:45:17 --> 00:45:21
And it is harder even to guess
exactly what will happen because

501
00:45:21 --> 00:45:24
on this side here the vector
field will go into the region so

502
00:45:24 --> 00:45:27
the contribution to flux will be
negative here.

503
00:45:27 --> 00:45:31
Here it will be positive
because it is going out of the

504
00:45:31 --> 00:45:33
region.
There are positive and negative

505
00:45:33 --> 00:45:35
terms.
Well, it looks like positive

506
00:45:35 --> 00:45:38
should win because here the
vector field is much larger than

507
00:45:38 --> 00:45:41
over there.
But, short of computing it,

508
00:45:41 --> 00:45:45
we won't actually know what it
is.

509
00:45:45 --> 00:45:48
If you want to do it by direct
calculation then you have to

510
00:45:48 --> 00:45:51
parametize this circle and
figure out what the line

511
00:45:51 --> 00:45:55
integral will be.
But if you use Green's theorem,

512
00:45:55 --> 00:46:00
well, we never used the fact
that it is the circle of radius

513
00:46:00 --> 00:46:03
A at the origin.
It is true actually for any

514
00:46:03 --> 00:46:08
closed curve that the flux out
of it is going to be twice the

515
00:46:08 --> 00:46:12
area of the region inside.
It still will be 2piA2 even if

516
00:46:12 --> 00:46:16
my circle is anywhere else in
the plane.

517
00:46:16 --> 00:46:18
If I had asked you a trick
question where do you want to

518
00:46:18 --> 00:46:21
place this circle so that that
the flux is the largest?

519
00:46:21 --> 00:46:28
Well, the answer is it doesn't
matter.

520
00:46:28 --> 00:46:33
Now, let's just finish quickly
by answering a question that

521
00:46:33 --> 00:46:36
some of you,
I am sure, must have, 

522
00:46:36 --> 00:46:40
which is what does divergence
mean and what does it measure?

523
00:46:40 --> 00:46:44
I mean, we said for curl,
curl measures how much things

524
00:46:44 --> 00:46:48
are rotating somehow.
What does divergence mean?

525
00:46:48 --> 00:46:53
Well, the answer is divergence
measures how much things are

526
00:46:53 --> 00:47:08
diverging.
Let's be more explicit.

527
00:47:08 --> 00:47:20
Interpretation of divergence.
You can think of it,

528
00:47:20 --> 00:47:23
you know, what do I want to say
first?

529
00:47:23 --> 00:47:28
If you take a vector field that
is a constant vector field where

530
00:47:28 --> 00:47:32
everything just translates then
there is no divergence involved

531
00:47:32 --> 00:47:34
because the derivatives will be
zero.

532
00:47:34 --> 00:47:37
If you take the guy that
rotates things around you will

533
00:47:37 --> 00:47:40
also compute and find zero for
divergence.

534
00:47:40 --> 00:47:43
This is not sensitive to
translation motions where

535
00:47:43 --> 00:47:46
everything moves together or to
rotation motions,

536
00:47:46 --> 00:47:51
but instead it is sensitive to
explaining motions.

537
00:47:51 --> 00:48:04
A possible answer is that it
measures how much the flow is

538
00:48:04 --> 00:48:10
expanding areas.
If you imagine this flow that

539
00:48:10 --> 00:48:14
we have here on the picture,
things are moving away from the

540
00:48:14 --> 00:48:16
origin and they fill out the
plane.

541
00:48:16 --> 00:48:19
If we mention this fluid
flowing out there,

542
00:48:19 --> 00:48:21
it is occupying more and more
space.

543
00:48:21 --> 00:48:24
And so that is what it means to
have positive divergence.

544
00:48:24 --> 00:48:28
If you took the opposite vector
field that contracts everything

545
00:48:28 --> 00:48:31
to the origin that will have
negative divergence.

546
00:48:31 --> 00:48:34
That is a good way to think
about it if you are thinking of

547
00:48:34 --> 00:48:37
a gas maybe that can expand to
fill out more volume.

548
00:48:37 --> 00:48:41
If you thinking of water,
well, water doesn't really

549
00:48:41 --> 00:48:43
shrink or expand.
The fact that it is taking more

550
00:48:43 --> 00:48:46
and more space actually means
that there is more and more

551
00:48:46 --> 00:48:51
water.
The other way to think about it

552
00:48:51 --> 00:48:56
is divergence is the source
rate,

553
00:48:56 --> 00:49:00
it is the amount of fluid that
is being inserted into the

554
00:49:00 --> 00:49:12
system,
that is being pumped into the

555
00:49:12 --> 00:49:27
system per unit time per unit
area.

556
00:49:27 --> 00:49:31
What div F equals two here
means is that here you actually

557
00:49:31 --> 00:49:35
have matter being created or
being pumped into the system so

558
00:49:35 --> 00:49:39
that you have more and more
water filling more and more

559
00:49:39 --> 00:49:41
space as it flows.
But, actually,

560
00:49:41 --> 00:49:43
divergence is not two just at
the origin.

561
00:49:43 --> 00:49:46
It is two everywhere.
So, in fact,

562
00:49:46 --> 00:49:49
to have this you need to have a
system of pumps that actually is

563
00:49:49 --> 00:49:52
in something water absolutely
everywhere uniformly.

564
00:49:52 --> 00:49:55
That is the only way to do this.
I mean if you imagine that you

565
00:49:55 --> 00:49:57
just have one spring at the
origin then,

566
00:49:57 --> 00:50:00
sure, water will flow out, 
but as you go further and

567
00:50:00 --> 00:50:02
further away it will do so more
and more slowly.

568
00:50:02 --> 00:50:04
Well, here it is flowing away
faster and faster.

569
00:50:04 --> 00:50:09
And that means everywhere you
are still pumping more water

570
00:50:09 --> 00:50:11
into it.
So, that is what divergence

571
00:50:11 --> 00:50:13
measures.
 

572
00:50:13 --> 00:50:18