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OK, so last time we've seen the
curl of the vector field with

8
00:00:32 --> 00:00:39
components M and N.
We defined that to be N sub x

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00:00:39 --> 00:00:43
minus M sub y.
And, we said this measures how

10
00:00:43 --> 00:00:47
far that vector field is from
being conservative.

11
00:00:47 --> 00:00:50
If the curl is zero,
and if the field is defined

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00:00:50 --> 00:00:53
everywhere, then it's going to
be conservative.

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00:00:53 --> 00:00:55
And so, when I take the line
integral along a closed curve,

14
00:00:55 --> 00:00:59
I don't have to compute it.
I notes going to be zero.

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00:00:59 --> 00:01:04
But now, let's say that I have
a general vector field.

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00:01:04 --> 00:01:07
So, the curl will not be zero.
And, I still want to compute

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00:01:07 --> 00:01:10
the line integral along a closed
curve.

18
00:01:10 --> 00:01:14
Well, I could compute it
directly or there's another way.

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00:01:14 --> 00:01:17
And that's what we are going to
see today.

20
00:01:17 --> 00:01:26
So, say that I have a closed
curve, C, and I want to find the

21
00:01:26 --> 00:01:30
work.
So, there's two options.

22
00:01:30 --> 00:01:40
One is direct calculation,
and the other one is Green's

23
00:01:40 --> 00:01:47
theorem.
So, Green's theorem is another

24
00:01:47 --> 00:01:56
way to avoid calculating line
integrals if we don't want to.

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00:01:56 --> 00:02:04
OK, so what does it say?
It says if C is a closed curve

26
00:02:04 --> 00:02:16
enclosing a region R in the
plane, and I have to insist C

27
00:02:16 --> 00:02:28
should go counterclockwise.
And, if I have a vector field

28
00:02:28 --> 00:02:37
that's defined and
differentiable everywhere not

29
00:02:37 --> 00:02:42
only on the curve,
C, which is what I need to

30
00:02:42 --> 00:02:48
define the line integral,
but also on the region inside.

31
00:02:48 --> 00:02:57
Then -- -- the line integral
for the work done along C is

32
00:02:57 --> 00:03:07
actually equal to a double
integral over the region inside

33
00:03:07 --> 00:03:15
of curl F dA.
OK, so that's the conclusion.

34
00:03:15 --> 00:03:19
And, if you want me to write it
in coordinates,

35
00:03:19 --> 00:03:24
maybe I should do that.
So, the line integral in terms

36
00:03:24 --> 00:03:28
of the components,
that's the integral of M dx

37
00:03:28 --> 00:03:36
plus N dy.
And, the curl is (Nx-My)dA.

38
00:03:36 --> 00:03:42
OK, so that's the other way to
state it.

39
00:03:42 --> 00:03:47
So, that's a really strange
statement if you think about it

40
00:03:47 --> 00:03:51
because the left-hand side is a
line integral.

41
00:03:51 --> 00:03:57
OK, so the way we compute it is
we take this expression Mdx Ndy

42
00:03:57 --> 00:04:01
and we parameterize the curve.
We express x and y in terms of

43
00:04:01 --> 00:04:04
some variable,
t, maybe, or whatever you want

44
00:04:04 --> 00:04:07
to call it.
And then, you'll do a one

45
00:04:07 --> 00:04:11
variable integral over t.
This right-hand side here,

46
00:04:11 --> 00:04:13
it's a double integral,
dA.

47
00:04:13 --> 00:04:16
So, we do it the way that we
learn how to couple of weeks

48
00:04:16 --> 00:04:18
ago.
You take your region, 

49
00:04:18 --> 00:04:22
you slice it in the x direction
or in the y direction,

50
00:04:22 --> 00:04:26
and you integrate dx dy after
setting up the bounds carefully,

51
00:04:26 --> 00:04:29
or maybe in polar coordinates r
dr d theta.

52
00:04:29 --> 00:04:32
But, see, the way you compute
these things is completely

53
00:04:32 --> 00:04:36
different.
This one on the left-hand side

54
00:04:36 --> 00:04:42
lives only on the curve,
while the right-hand side lives

55
00:04:42 --> 00:04:46
everywhere in this region
inside.

56
00:04:46 --> 00:04:49
So, here, x and y are related,
they live on the curve.

57
00:04:49 --> 00:04:53
Here, x and y are independent.
There just are some bounds

58
00:04:53 --> 00:04:54
between them.
And, of course,

59
00:04:54 --> 00:04:56
what you're integrating is
different.

60
00:04:56 --> 00:05:01
It's a line integral for work.
Here, it's a double integral of

61
00:05:01 --> 00:05:08
some function of x and y.
So, it's a very perplexing

62
00:05:08 --> 00:05:14
statement at first.
But, it's a very powerful tool.

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00:05:14 --> 00:05:18
So, we're going to try to see
how it works concretely,

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00:05:18 --> 00:05:20
what it says, 
what are the consequences, 

65
00:05:20 --> 00:05:23
how we could convince ourselves
that, yes,

66
00:05:23 --> 00:05:26
this works, and so on.
That's going to be the topic

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00:05:26 --> 00:05:32
for today.
Any questions about the

68
00:05:32 --> 00:05:38
statement first?
No?

69
00:05:38 --> 00:05:43
OK, yeah, one remark, sorry.
So, here, it stays

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00:05:43 --> 00:05:46
counterclockwise.
What if I have a curve that

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00:05:46 --> 00:05:49
goes clockwise?
Well, you could just take the

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00:05:49 --> 00:05:52
negative, and integrate
counterclockwise.

73
00:05:52 --> 00:05:57
Why does the theorem choose
counterclockwise over clockwise?

74
00:05:57 --> 00:06:00
How doesn't know that it's
counterclockwise rather than

75
00:06:00 --> 00:06:02
clockwise?
Well, the answer is basically

76
00:06:02 --> 00:06:05
in our convention for curl.
See, we've said curl is Nx

77
00:06:05 --> 00:06:08
minus My, and not the other way
around.

78
00:06:08 --> 00:06:10
And, that's a convention as
well.

79
00:06:10 --> 00:06:13
So, somehow,
the two conventions match with

80
00:06:13 --> 00:06:15
each other.
That's the best answer I can

81
00:06:15 --> 00:06:18
give you.
So, if you met somebody from a

82
00:06:18 --> 00:06:20
different planet,
they might have Green's theorem

83
00:06:20 --> 00:06:23
with the opposite conventions,
with curves going clockwise, 

84
00:06:23 --> 00:06:27
and the curl defined the other
way around.

85
00:06:27 --> 00:06:35
Probably if you met an alien,
I'm not sure if you would be

86
00:06:35 --> 00:06:43
discussing Green's theorem
first, but just in case.

87
00:06:43 --> 00:06:53
OK, so that being said,
there is a warning here which

88
00:06:53 --> 00:07:00
is that this is only for closed
curves.

89
00:07:00 --> 00:07:03
OK, so if I give you a curve
that's not closed,

90
00:07:03 --> 00:07:05
and I tell you,
well, compute the line

91
00:07:05 --> 00:07:08
integral, then you have to do it
by hand.

92
00:07:08 --> 00:07:10
You have to parameterize the
curve.

93
00:07:10 --> 00:07:12
Or, if you really don't like
that line integral,

94
00:07:12 --> 00:07:16
you could close the path by
adding some other line integral

95
00:07:16 --> 00:07:19
to it,
and then compute using Green's

96
00:07:19 --> 00:07:23
theorem.
But, you can't use Green's

97
00:07:23 --> 00:07:30
theorem directly if the curve is
not closed.

98
00:07:30 --> 00:07:42
OK, so let's do a quick example.
So, let's say that I give you

99
00:07:42 --> 00:07:52
C, the circle of radius one,
centered at the point (2,0).

100
00:07:52 --> 00:08:00
So, it's out here.
That's my curve, C.

101
00:08:00 --> 00:08:09
And, let's say that I do it
counterclockwise so that it will

102
00:08:09 --> 00:08:16
match with the statement of the
theorem.

103
00:08:16 --> 00:08:24
And, let's say that I want you
to compute the line integral

104
00:08:24 --> 00:08:34
along C of ye^(-x) dx plus (one
half of x squared minus e^(-x))

105
00:08:34 --> 00:08:37
dy.
And, that's a kind of sadistic

106
00:08:37 --> 00:08:41
example, but maybe I'll ask you
to do that.

107
00:08:41 --> 00:08:44
So, how would you do it
directly?

108
00:08:44 --> 00:08:48
Well, to do it directly you
would have to parameterize this

109
00:08:48 --> 00:08:53
curve.
So that would probably involve

110
00:08:53 --> 00:09:03
setting x equals two plus cosine
theta y equals sine theta.

111
00:09:03 --> 00:09:06
But, I'm using as parameter of
the angle around the circle,

112
00:09:06 --> 00:09:09
it's like the unit circle,
the usual ones that shifted by

113
00:09:09 --> 00:09:15
two in the x direction.
And then, I would set dx equals

114
00:09:15 --> 00:09:21
minus sine theta d theta.
I would set dy equals cosine

115
00:09:21 --> 00:09:24
theta d theta.
And, I will substitute,

116
00:09:24 --> 00:09:26
and I will integrate from zero
to 2pi.

117
00:09:26 --> 00:09:29
And, I would probably run into
a bit of trouble because I would

118
00:09:29 --> 00:09:32
have these e to the minus x,
which would give me something

119
00:09:32 --> 00:09:35
that I really don't want to
integrate.

120
00:09:35 --> 00:09:44
So, instead of doing that,
which looks pretty much doomed,

121
00:09:44 --> 00:09:51
instead, I'm going to use
Green's theorem.

122
00:09:51 --> 00:09:58
So, using Green's theorem,
the way we'll do it is I will,

123
00:09:58 --> 00:10:03
instead, compute a double
integral.

124
00:10:03 --> 00:10:16
So, I will -- -- compute the
double integral over the region

125
00:10:16 --> 00:10:27
inside of curl F dA.
So, I should say probably what

126
00:10:27 --> 00:10:30
F was.
So, let's call this M.

127
00:10:30 --> 00:10:37
Let's call this N.
And, then I will actually just

128
00:10:37 --> 00:10:45
choose the form coordinates,
(Nx minus My) dA.

129
00:10:45 --> 00:10:53
And, what is R here?
Well, R is the disk in here.

130
00:10:53 --> 00:10:56
OK, so, of course,
it might not be that pleasant

131
00:10:56 --> 00:10:58
because we'll also have to set
up this double integral.

132
00:10:58 --> 00:11:02
And, for that,
we'll have to figure out a way

133
00:11:02 --> 00:11:05
to slice this region nicely.
We could do it dx dy.

134
00:11:05 --> 00:11:08
We could do it dy dx.
Or, maybe we will want to

135
00:11:08 --> 00:11:11
actually make a change of
variables to first shift this to

136
00:11:11 --> 00:11:14
the origin,
you know, change x to x minus

137
00:11:14 --> 00:11:17
two and then switch to polar
coordinates.

138
00:11:17 --> 00:11:22
Well, let's see what happens
later.

139
00:11:22 --> 00:11:33
OK, so what is, so this is R.
So, what is N sub x?

140
00:11:33 --> 00:11:42
Well, N sub x is x plus e to
the minus x minus,

141
00:11:42 --> 00:11:48
what is M sub y,
e to the minus x,

142
00:11:48 --> 00:11:52
OK?
This is Nx.

143
00:11:52 --> 00:11:58
This is My dA.
Well, it seems to simplify a

144
00:11:58 --> 00:12:02
bit.
I will just get double integral

145
00:12:02 --> 00:12:06
over R of x dA,
which looks certainly a lot

146
00:12:06 --> 00:12:10
more pleasant.
Of course, I made up the

147
00:12:10 --> 00:12:14
example in that way so that it
simplifies when you use Green's

148
00:12:14 --> 00:12:16
theorem.
But, you know,

149
00:12:16 --> 00:12:20
it gives you an example where
you can turn are really hard

150
00:12:20 --> 00:12:23
line integral into an easier
double integral.

151
00:12:23 --> 00:12:28
Now, how do we compute that
double integral?

152
00:12:28 --> 00:12:32
Well, so one way would be to
set it up.

153
00:12:32 --> 00:12:41
Or, let's actually be a bit
smarter and observe that this is

154
00:12:41 --> 00:12:50
actually the area of the region
R, times the x coordinate of its

155
00:12:50 --> 00:12:55
center of mass.
If I look at the definition of

156
00:12:55 --> 00:12:59
the center of mass,
it's the average value of x.

157
00:12:59 --> 00:13:03
So, it's one over the area
times the double integral of x

158
00:13:03 --> 00:13:07
dA, well, possibly with the
density, but here I'm thinking

159
00:13:07 --> 00:13:11
uniform density one.
And, now, I think I know just

160
00:13:11 --> 00:13:15
by looking at the picture where
the center of mass of this

161
00:13:15 --> 00:13:16
circle will be,
right?

162
00:13:16 --> 00:13:19
I mean, it would be right in
the middle.

163
00:13:19 --> 00:13:24
So, that is two,
if you want,

164
00:13:24 --> 00:13:29
by symmetry.
And, the area of the guy is

165
00:13:29 --> 00:13:33
just pi because it's a disk of
radius one.

166
00:13:33 --> 00:13:37
So, I will just get 2pi.
I mean, of course,

167
00:13:37 --> 00:13:40
if you didn't see that,
then you can also compute that

168
00:13:40 --> 00:13:43
double integral directly.
It's a nice exercise.

169
00:13:43 --> 00:13:47
But see, here,
using geometry helps you to

170
00:13:47 --> 00:13:50
actually streamline the
calculation.

171
00:13:50 --> 00:14:01
OK, any questions?
Yes?

172
00:14:01 --> 00:14:04
OK, yes, let me just repeat the
last part.

173
00:14:04 --> 00:14:10
So, I said we had to compute
the double integral of x dA over

174
00:14:10 --> 00:14:14
this region here,
which is a disk of radius one,

175
00:14:14 --> 00:14:18
centered at,
this point is (2,0).

176
00:14:18 --> 00:14:22
So, instead of setting up the
integral with bounds and

177
00:14:22 --> 00:14:26
integrating dx dy or dy dx or in
polar coordinates,

178
00:14:26 --> 00:14:30
I'm just going to say, well, 
let's remember the definition

179
00:14:30 --> 00:14:32
of a center of mass.
It's the average value of a

180
00:14:32 --> 00:14:37
function, x in the region.
So, it's one over the area of

181
00:14:37 --> 00:14:42
origin times the double integral
of x dA.

182
00:14:42 --> 00:14:46
If you look,
again, at the definition of x

183
00:14:46 --> 00:14:51
bar, it's one over area of
double integral x dA.

184
00:14:51 --> 00:14:54
Well, maybe if there's a
density, then it's one over mass

185
00:14:54 --> 00:14:57
times double integral of x
density dA.

186
00:14:57 --> 00:15:02
But, if density is one,
then it just becomes this.

187
00:15:02 --> 00:15:06
So, switching the area, 
moving the area to the other

188
00:15:06 --> 00:15:08
side,
I'll get double integral of x

189
00:15:08 --> 00:15:12
dA is the area of origin times
the x coordinate of the center

190
00:15:12 --> 00:15:14
of mass.
The area of origin is pi

191
00:15:14 --> 00:15:18
because it's a unit disk.
And, the center of mass is the

192
00:15:18 --> 00:15:23
center of a disk.
So, its x bar is two,

193
00:15:23 --> 00:15:27
and I get 2 pi.
OK, that I didn't actually have

194
00:15:27 --> 00:15:30
to do this in my example today,
but of course that would be

195
00:15:30 --> 00:15:36
good review.
It will remind you of center of

196
00:15:36 --> 00:15:47
mass and all that.
OK, any other questions?

197
00:15:47 --> 00:15:50
No?
OK, so let's see,

198
00:15:50 --> 00:15:54
now that we've seen how to use
it practice, how to avoid

199
00:15:54 --> 00:15:58
calculating the line integral if
we don't want to.

200
00:15:58 --> 00:16:04
Let's try to convince ourselves
that this theorem makes sense.

201
00:16:04 --> 00:16:09
OK, so, well,
let's start with an easy case

202
00:16:09 --> 00:16:15
where we should be able to know
the answer to both sides.

203
00:16:15 --> 00:16:23
So let's look at the special
case.

204
00:16:23 --> 00:16:32
Let's look at the case where
curl F is zero.

205
00:16:32 --> 00:16:45
Then, well, we'd like to
conclude that F is conservative.

206
00:16:45 --> 00:16:53
That's what we said.
Well let's see what happens.

207
00:16:53 --> 00:16:59
So, Green's theorem says that
if I have a closed curve,

208
00:16:59 --> 00:17:06
then the line integral of F is
equal to the double integral of

209
00:17:06 --> 00:17:12
curl on the region inside.
And, if the curl is zero,

210
00:17:12 --> 00:17:15
then I will be integrating
zero.

211
00:17:15 --> 00:17:22
I will get zero.
OK, so this is actually how you

212
00:17:22 --> 00:17:26
prove that if your vector field
has curve zero,

213
00:17:26 --> 00:17:27
then it's conservative.

214
00:17:27 --> 00:17:54

215
00:17:54 --> 00:17:57
OK, so in particular,
if you have a vector field

216
00:17:57 --> 00:18:01
that's defined everywhere the
plane, then you take any closed

217
00:18:01 --> 00:18:04
curve.
Well, you will get that the

218
00:18:04 --> 00:18:06
line integral will be zero.
Straightly speaking,

219
00:18:06 --> 00:18:10
that will only work here if the
curve goes counterclockwise.

220
00:18:10 --> 00:18:13
But otherwise,
just look at the various loops

221
00:18:13 --> 00:18:16
that it makes,
and orient each of them

222
00:18:16 --> 00:18:19
counterclockwise and sum things
together.

223
00:18:19 --> 00:18:22
So let me state that again.

224
00:18:22 --> 00:18:45

225
00:18:45 --> 00:18:51
So, 
OK, 

226
00:18:51 --> 00:19:02
so a consequence of Green's
theorem is that if F is defined

227
00:19:02 --> 00:19:12
everywhere in the plane -- --
and the curl of F is zero

228
00:19:12 --> 00:19:24
everywhere,
then F is conservative.

229
00:19:24 --> 00:19:29
And so, this actually is the
input we needed to justify our

230
00:19:29 --> 00:19:33
criterion.
The test that we saw last time

231
00:19:33 --> 00:19:35
saying,
well, to check if something is

232
00:19:35 --> 00:19:37
a gradient field if it's
conservative,

233
00:19:37 --> 00:19:40
we just have to compute the
curl and check whether it's

234
00:19:40 --> 00:19:45
zero.
OK, so how do we prove that now

235
00:19:45 --> 00:19:49
carefully?
Well, you just take a closed

236
00:19:49 --> 00:19:52
curve in the plane.
You switch the orientation if

237
00:19:52 --> 00:19:55
needed so it becomes
counterclockwise.

238
00:19:55 --> 00:19:59
And then you look at the region
inside.

239
00:19:59 --> 00:20:05
And then you know that the line
integral inside will be equal to

240
00:20:05 --> 00:20:12
the double integral of curl,
which is the double integral of

241
00:20:12 --> 00:20:16
zero.
Therefore, that's zero.

242
00:20:16 --> 00:20:19
But see, OK,
so now let's say that we try to

243
00:20:19 --> 00:20:23
do that for the vector field
that was on your problems that

244
00:20:23 --> 00:20:27
was not defined at the origin.
So if you've done the problem

245
00:20:27 --> 00:20:30
sets and found the same answers
that I did, then you will have

246
00:20:30 --> 00:20:33
found that this vector field had
curve zero everywhere.

247
00:20:33 --> 00:20:36
But still it wasn't
conservative because if you went

248
00:20:36 --> 00:20:39
around the unit circle,
then you got a line integral

249
00:20:39 --> 00:20:42
that was 2pi.
Or, if you compared the two

250
00:20:42 --> 00:20:45
halves, you got different
answers for two parts that go

251
00:20:45 --> 00:20:47
from the same point to the same
point.

252
00:20:47 --> 00:20:51
So, it fails this property but
that's because it's not defined

253
00:20:51 --> 00:20:54
everywhere.
So, what goes wrong with this

254
00:20:54 --> 00:20:57
argument?
Well, if I take the vector

255
00:20:57 --> 00:21:03
field that was in the problem
set, and if I do things,

256
00:21:03 --> 00:21:07
say that I look at the unit
circle.

257
00:21:07 --> 00:21:10
That's a closed curve.
So, I would like to use Green's

258
00:21:10 --> 00:21:13
theorem.
Green's theorem would tell me

259
00:21:13 --> 00:21:17
the line integral along this
loop is equal to the double

260
00:21:17 --> 00:21:21
integral of curl over this
region here, the unit disk.

261
00:21:21 --> 00:21:25
And, of course the curl is
zero, well, except at the

262
00:21:25 --> 00:21:27
origin.
At the origin,

263
00:21:27 --> 00:21:29
the vector field is not
defined.

264
00:21:29 --> 00:21:32
You cannot take the
derivatives, and the curl is not

265
00:21:32 --> 00:21:34
defined.
And somehow that messes things

266
00:21:34 --> 00:21:38
up.
You cannot apply Green's

267
00:21:38 --> 00:21:49
theorem to the vector field.
So, you cannot apply Green's

268
00:21:49 --> 00:22:02
theorem to the vector field on
problem set eight problem two

269
00:22:02 --> 00:22:12
when C encloses the origin.
And so, that's why this guy,

270
00:22:12 --> 00:22:16
even though it has curl zero,
is not conservative.

271
00:22:16 --> 00:22:20
There's no contradiction.
And somehow,

272
00:22:20 --> 00:22:23
you have to imagine that,
well, the curl here is really

273
00:22:23 --> 00:22:26
not defined.
But somehow it becomes infinite

274
00:22:26 --> 00:22:30
so that when you do the double
integral, you actually get 2 pi

275
00:22:30 --> 00:22:37
instead of zero.
I mean, that doesn't make any

276
00:22:37 --> 00:22:46
sense, of course,
but that's one way to think

277
00:22:46 --> 00:22:51
about it.
OK, any questions?

278
00:22:51 --> 00:23:02
Yes?
Well, though actually it's not

279
00:23:02 --> 00:23:06
defined because the curl is zero
everywhere else.

280
00:23:06 --> 00:23:08
So, if a curl was well defined
at the origin,

281
00:23:08 --> 00:23:11
you would try to,
then, take the double integral.

282
00:23:11 --> 00:23:12
no matter what value you put
for a function,

283
00:23:12 --> 00:23:15
if you have a function that's
zero everywhere except at the

284
00:23:15 --> 00:23:17
origin,
and some other value at the

285
00:23:17 --> 00:23:20
origin,
the integral is still zero.

286
00:23:20 --> 00:23:24
So, it's worse than that.
It's not only that you can't

287
00:23:24 --> 00:23:29
compute it, it's that is not
defined.

288
00:23:29 --> 00:23:36
OK, anyway, that's like a
slightly pathological example.

289
00:23:36 --> 00:23:44
Yes?
Well, we wouldn't be able to

290
00:23:44 --> 00:23:46
because the curl is not defined
at the origin.

291
00:23:46 --> 00:23:49
So, you can actually integrate
it.

292
00:23:49 --> 00:23:52
OK, so that's the problem.
I mean, if you try to

293
00:23:52 --> 00:23:55
integrate, we've said everywhere
where it's defined,

294
00:23:55 --> 00:23:57
the curl is zero.
So, what you would be

295
00:23:57 --> 00:24:01
integrating would be zero.
But, that doesn't work because

296
00:24:01 --> 00:24:09
at the origin it's not defined.
Yes?

297
00:24:09 --> 00:24:11
Ah, so if you take a curve that
makes a figure 8,

298
00:24:11 --> 00:24:14
then indeed my proof over there
is false.

299
00:24:14 --> 00:24:19
So, I kind of tricked you.
It's not actually correct.

300
00:24:19 --> 00:24:24
So, if the curve does a figure
8, then what you do is you would

301
00:24:24 --> 00:24:27
actually cut it into its two
halves.

302
00:24:27 --> 00:24:30
And for each of them,
you will apply Green's theorem.

303
00:24:30 --> 00:24:32
And then, you'd still get,
if a curl is zero then this

304
00:24:32 --> 00:24:35
line integral is zero.
That one is also zero.

305
00:24:35 --> 00:24:38
So this one is zero.
OK, small details that you

306
00:24:38 --> 00:24:41
don't really need to worry too
much about,

307
00:24:41 --> 00:24:47
but indeed if you want to be
careful with details then my

308
00:24:47 --> 00:24:54
proof is not quite complete.
But the computation is still

309
00:24:54 --> 00:24:58
true.
Let's move on.

310
00:24:58 --> 00:25:06
So, I want to tell you how to
prove Green's theorem because

311
00:25:06 --> 00:25:15
it's such a strange formula that
where can it come from possibly?

312
00:25:15 --> 00:25:21
I mean, 
so let me remind you first of

313
00:25:21 --> 00:25:26
all the statement we want to
prove is that the line integral

314
00:25:26 --> 00:25:31
along a closed curve of Mdx plus
Ndy is equal to the double

315
00:25:31 --> 00:25:36
integral over the region inside
of (Nx minus My)dA.

316
00:25:36 --> 00:25:40
And, let's simplify our lives a
bit by proving easier

317
00:25:40 --> 00:25:43
statements.
So actually, 

318
00:25:43 --> 00:25:53
the first observation will
actually prove something easier,

319
00:25:53 --> 00:25:58
namely, that the line integral, 
let's see, 

320
00:25:58 --> 00:26:03
of Mdx along a closed curve is
equal to the double integral

321
00:26:03 --> 00:26:08
over the region inside of minus
M sub y dA.

322
00:26:08 --> 00:26:13
OK, so that's the special case
where N is zero,

323
00:26:13 --> 00:26:19
where you have only an x
component for your vector field.

324
00:26:19 --> 00:26:23
Now, why is that good enough?
Well, the claim is if I can

325
00:26:23 --> 00:26:28
prove this, I claim you will be
able to do the same thing to

326
00:26:28 --> 00:26:33
prove the other case where there
is only the y component.

327
00:26:33 --> 00:26:38
And then, if the other
together, you will get the

328
00:26:38 --> 00:26:40
general case.
So, let me explain.

329
00:26:40 --> 00:27:00
 
 

330
00:27:00 --> 00:27:06
OK, so a similar argument which
I will not do,

331
00:27:06 --> 00:27:11
to save time,
will show, so actually it's

332
00:27:11 --> 00:27:15
just the same thing but
switching the roles of x and y,

333
00:27:15 --> 00:27:20
that if I integrate along a
closed curve N dy,

334
00:27:20 --> 00:27:29
then I'll get the double
integral of N sub x dA.

335
00:27:29 --> 00:27:36
And so, now if I have proved
these two formulas separately,

336
00:27:36 --> 00:27:44
then if you sum them together
will get the correct statement.

337
00:27:44 --> 00:27:52
Let me write it.
We get Green's theorem.

338
00:27:52 --> 00:27:55
OK, so we've simplified our
task a little bit.

339
00:27:55 --> 00:28:00
We'll just be trying to prove
the case where there's only an x

340
00:28:00 --> 00:28:04
component.
So, let's do it.

341
00:28:04 --> 00:28:07
Well, we have another problem
which is the region that we are

342
00:28:07 --> 00:28:10
looking at, the curve that we're
looking at might be very

343
00:28:10 --> 00:28:12
complicated.
If I give you,

344
00:28:12 --> 00:28:17
let's say I give you,
I don't know,

345
00:28:17 --> 00:28:22
a curve that does something
like this.

346
00:28:22 --> 00:28:26
Well, it will be kind of tricky
to set up a double integral over

347
00:28:26 --> 00:28:29
the region inside.
So maybe we first want to look

348
00:28:29 --> 00:28:33
at curves that are simpler,
that will actually allow us to

349
00:28:33 --> 00:28:36
set up the double integral
easily.

350
00:28:36 --> 00:28:42
So, the second observation,
so that was the first

351
00:28:42 --> 00:28:51
observation.
The second observation is that

352
00:28:51 --> 00:29:02
we can decompose R into simpler
regions.

353
00:29:02 --> 00:29:10
So what do I mean by that?
Well, let's say that I have a

354
00:29:10 --> 00:29:13
region and I'm going to cut it
into two.

355
00:29:13 --> 00:29:18
So, I'll have R1 and R2.
And then, of course,

356
00:29:18 --> 00:29:22
I need to have the curves that
go around them.

357
00:29:22 --> 00:29:29
So, I had my initial curve,
C, was going around everybody.

358
00:29:29 --> 00:29:41
They have curves C1 that goes
around R1, and C2 goes around

359
00:29:41 --> 00:29:46
R2.
OK, so, 

360
00:29:46 --> 00:29:55
what I would like to say is if
we can prove that the statement

361
00:29:55 --> 00:30:07
is true, so let's see,
for C1 and also for C2 -- --

362
00:30:07 --> 00:30:23
then I claim we can prove the
statement for C.

363
00:30:23 --> 00:30:26
How do we do that?
Well, we just add these two

364
00:30:26 --> 00:30:28
equalities together.
OK, why does that work?

365
00:30:28 --> 00:30:31
There's something fishy going
on because C1 and C2 have this

366
00:30:31 --> 00:30:35
piece here in the middle.
That's not there in C.

367
00:30:35 --> 00:30:39
So, if you add the line
integral along C1 and C2,

368
00:30:39 --> 00:30:44
you get these unwanted pieces.
But, the good news is actually

369
00:30:44 --> 00:30:47
you go twice through that edge
in the middle.

370
00:30:47 --> 00:30:51
See, it appears once in C1
going up, and once in C2 going

371
00:30:51 --> 00:30:52
down.
So, in fact,

372
00:30:52 --> 00:30:55
when you will do the work,
when you will sum the work,

373
00:30:55 --> 00:30:57
you will add these two guys
together.

374
00:30:57 --> 00:31:06
They will cancel.
OK, so the line integral along

375
00:31:06 --> 00:31:14
C will be, then,
it will be the sum of the line

376
00:31:14 --> 00:31:21
integrals on C1 and C2.
And, that will equal,

377
00:31:21 --> 00:31:29
therefore, the double integral
over R1 plus the double integral

378
00:31:29 --> 00:31:36
over R2, which is the double
integral over R of negative My.

379
00:31:36 --> 00:31:47
OK and the reason for this
equality here is because we go

380
00:31:47 --> 00:31:56
twice through the inner part.
What do I want to say?

381
00:31:56 --> 00:32:15
Along the boundary between R1
and R2 -- -- with opposite

382
00:32:15 --> 00:32:25
orientations.
So, the extra things cancel out.

383
00:32:25 --> 00:32:29
OK, so that means I just need
to look at smaller pieces if

384
00:32:29 --> 00:32:34
that makes my life easier.
So, now, will make my life easy?

385
00:32:34 --> 00:32:41
Well, let's say that I have a
curve like that.

386
00:32:41 --> 00:32:45
Well, I guess I should really
draw a pumpkin or something like

387
00:32:45 --> 00:32:48
that because it would be more
seasonal.

388
00:32:48 --> 00:32:53
But, well, I don't really know
how to draw a pumpkin.

389
00:32:53 --> 00:32:57
OK, so what I will do is I will
cut this into smaller regions

390
00:32:57 --> 00:33:01
for which I have a well-defined
lower and upper boundary so that

391
00:33:01 --> 00:33:05
I will be able to set up a
double integral,

392
00:33:05 --> 00:33:10
dy dx, easily.
So, a region like this I will

393
00:33:10 --> 00:33:17
actually cut it here and here
into five smaller pieces so that

394
00:33:17 --> 00:33:23
each small piece will let me set
up the double integral,

395
00:33:23 --> 00:33:31
dy dx.
OK, so we'll cut R in to what I

396
00:33:31 --> 00:33:41
will call vertically simple --
-- regions.

397
00:33:41 --> 00:33:43
So, what's a vertically simple
region?

398
00:33:43 --> 00:33:48
That's a region that's given by
looking at x between a and b for

399
00:33:48 --> 00:33:53
some values of a and b.
And, for each value of x,

400
00:33:53 --> 00:34:00
y is between some function of x
and some other function of x.

401
00:34:00 --> 00:34:03
OK, so for example,
this guy is vertically simple.

402
00:34:03 --> 00:34:07
See, x runs from this value of
x to that value of x.

403
00:34:07 --> 00:34:13
And, for each x,
y goes between this value to

404
00:34:13 --> 00:34:16
that value.
And, same with each of these.

405
00:34:16 --> 00:34:39
 
 

406
00:34:39 --> 00:34:49
OK, so now we are down to the
main step that we have to do,

407
00:34:49 --> 00:35:05
which is to prove this identity
if C is, sorry,

408
00:35:05 --> 00:35:23
if -- -- if R is vertically
simple -- -- and C is the

409
00:35:23 --> 00:35:36
boundary of R going
counterclockwise.

410
00:35:36 --> 00:35:40
OK, so let's look at how we
would do it.

411
00:35:40 --> 00:35:46
So, we said vertically simple
region looks like x goes between

412
00:35:46 --> 00:35:52
a and b, and y goes between two
values that are given by

413
00:35:52 --> 00:35:57
functions of x.
OK, so this is y equals f2 of x.

414
00:35:57 --> 00:36:02
This is y equals f1 of x.
This is a.

415
00:36:02 --> 00:36:09
This is b.
Our region is this thing in

416
00:36:09 --> 00:36:13
here.
So, let's compute both sides.

417
00:36:13 --> 00:36:15
And, when I say compute,
of course we will not get

418
00:36:15 --> 00:36:17
numbers because we don't know
what M is.

419
00:36:17 --> 00:36:19
We don't know what f1 and f2
are.

420
00:36:19 --> 00:36:24
But, I claim we should be able
to simplify things a bit.

421
00:36:24 --> 00:36:28
So, let's start with the line
integral.

422
00:36:28 --> 00:36:35
How do I compute the line
integral along the curve that

423
00:36:35 --> 00:36:40
goes all around here?
Well, it looks like there will

424
00:36:40 --> 00:36:45
be four pieces.
OK, so we actually have four

425
00:36:45 --> 00:36:50
things to compute,
C1, C2, C3, and C4.

426
00:36:50 --> 00:37:01
OK?
Well, let's start with C1.

427
00:37:01 --> 00:37:06
So, if we integrate on C1 Mdx,
how do we do that?

428
00:37:06 --> 00:37:10
Well, we know that on C1,
y is given by a function of x.

429
00:37:10 --> 00:37:15
So, we can just get rid of y
and express everything in terms

430
00:37:15 --> 00:37:21
of x.
OK, so, we know y is f1 of x,

431
00:37:21 --> 00:37:27
and x goes from a to b.
So, that will be the integral

432
00:37:27 --> 00:37:30
from a to b of,
well, I have to take the

433
00:37:30 --> 00:37:33
function, M.
And so, M depends normally on x

434
00:37:33 --> 00:37:38
and y.
Maybe I should put x and y here.

435
00:37:38 --> 00:37:46
And then, I will plug y equals
f1 of x dx.

436
00:37:46 --> 00:37:49
And, then I have a single
variable integral.

437
00:37:49 --> 00:37:51
And that's what I have to
compute.

438
00:37:51 --> 00:37:54
Of course, I cannot compute it
here because I don't know what

439
00:37:54 --> 00:37:59
this is.
So, it has to stay this way.

440
00:37:59 --> 00:38:06
OK, next one.
The integral along C2,

441
00:38:06 --> 00:38:13
well, let's think for a second.
On C2, x equals b.

442
00:38:13 --> 00:38:16
It's constant.
So, dx is zero,

443
00:38:16 --> 00:38:20
and you would integrate,
actually, above a variable,

444
00:38:20 --> 00:38:23
y.
But, well, we don't have a y

445
00:38:23 --> 00:38:26
component.
See, this is the reason why we

446
00:38:26 --> 00:38:30
made the first observation.
We got rid of the other term

447
00:38:30 --> 00:38:33
because it's simplifies our life
here.

448
00:38:33 --> 00:38:38
So, we just get zero.
OK, just looking quickly ahead,

449
00:38:38 --> 00:38:40
there's another one that would
be zero as well,

450
00:38:40 --> 00:38:42
right?
Which one?

451
00:38:42 --> 00:38:52
Yeah, C4.
This one gives me zero.

452
00:38:52 --> 00:38:55
What about C3?
Well, C3 will look a lot like

453
00:38:55 --> 00:38:57
C1.
So, we're going to use the same

454
00:38:57 --> 00:38:59
kind of thing that we did with
C.

455
00:38:59 --> 00:39:22
 
 

456
00:39:22 --> 00:39:27
OK, so along C3,
well, let's see,

457
00:39:27 --> 00:39:34
so on C3, y is a function of x,
again.

458
00:39:34 --> 00:39:40
And so we are using as our
variable x, but now x goes down

459
00:39:40 --> 00:39:45
from b to a.
So, it will be the integral

460
00:39:45 --> 00:39:51
from b to a of M of (x and f2 of
x) dx.

461
00:39:51 --> 00:39:57
Or, if you prefer,
that's negative integral from a

462
00:39:57 --> 00:40:04
to b of M of (x and f2 of x) dx.
OK, so now if I sum all these

463
00:40:04 --> 00:40:10
pieces together,
I get that the line integral

464
00:40:10 --> 00:40:20
along the closed curve is the
integral from a to b of M(x1f1

465
00:40:20 --> 00:40:30
of x) dx minus the integral from
a to b of M(x1f2 of x) dx.

466
00:40:30 --> 00:40:39
So, that's the left hand side.
Next, I should try to look at

467
00:40:39 --> 00:40:47
my double integral and see if I
can make it equal to that.

468
00:40:47 --> 00:40:58
So, let's look at the other
guy, double integral over R of

469
00:40:58 --> 00:41:02
negative MydA.
Well, first,

470
00:41:02 --> 00:41:05
I'll take the minus sign out.
It will make my life a little

471
00:41:05 --> 00:41:09
bit easier.
And second, so I said I will

472
00:41:09 --> 00:41:14
try to set this up in the way
that's the most efficient.

473
00:41:14 --> 00:41:20
And, my choice of this kind of
region means that it's easier to

474
00:41:20 --> 00:41:22
set up dy dx,
right?

475
00:41:22 --> 00:41:30
So, if I set it up dy dx,
then I know for a given value

476
00:41:30 --> 00:41:36
of x, y goes from f1 of x to f2
of x.

477
00:41:36 --> 00:41:49
And, x goes from a to b, right?
Is that OK with everyone?

478
00:41:49 --> 00:41:53
OK, so now if I compute the
inner integral,

479
00:41:53 --> 00:41:58
well, what do I get if I get
partial M partial y with respect

480
00:41:58 --> 00:42:02
to y?
I'll get M back, OK?

481
00:42:02 --> 00:42:19
So -- So, I will get M at the
point x f2 of x minus M at the

482
00:42:19 --> 00:42:27
point x f1 of x.
And so, this becomes the

483
00:42:27 --> 00:42:35
integral from a to b.
I guess that was a minus sign,

484
00:42:35 --> 00:42:45
of M of (x1f2 of x) minus M of
(x1f1 of x) dx.

485
00:42:45 --> 00:42:50
And so, that's the same as up
there.

486
00:42:50 --> 00:42:54
And so, that's the end of the
proof because we've checked that

487
00:42:54 --> 00:42:58
for this special case,
when we have only an x

488
00:42:58 --> 00:43:01
component and a vertically
simple region,

489
00:43:01 --> 00:43:04
things work.
Then, we can remove the

490
00:43:04 --> 00:43:07
assumption that things are
vertically simple using this

491
00:43:07 --> 00:43:10
second observation.
We can just glue the various

492
00:43:10 --> 00:43:13
pieces together,
and prove it for any region.

493
00:43:13 --> 00:43:17
Then, we do same thing with the
y component.

494
00:43:17 --> 00:43:22
That's the first observation.
When we add things together,

495
00:43:22 --> 00:43:29
we get Green's theorem in its
full generality.

496
00:43:29 --> 00:43:39
OK, so let me finish with a
cool example.

497
00:43:39 --> 00:43:47
So, there's one place in real
life where Green's theorem used

498
00:43:47 --> 00:43:51
to be extremely useful.
I say used to because computers

499
00:43:51 --> 00:43:53
have actually made that
obsolete.

500
00:43:53 --> 00:44:02
But, so let me show you a
picture of this device.

501
00:44:02 --> 00:44:12
This is called a planimeter.
And what it does is it measures

502
00:44:12 --> 00:44:17
areas.
So, it used to be that when you

503
00:44:17 --> 00:44:23
were an experimental scientist,
you would run your chemical or

504
00:44:23 --> 00:44:27
biological experiment or
whatever.

505
00:44:27 --> 00:44:29
And, you would have all of
these recording devices.

506
00:44:29 --> 00:44:32
And, the data would go,
well, not onto a floppy disk or

507
00:44:32 --> 00:44:35
hard disk or whatever because
you didn't have those at the

508
00:44:35 --> 00:44:37
time.
You didn't have a computer in

509
00:44:37 --> 00:44:39
your lab.
They would go onto a piece of

510
00:44:39 --> 00:44:42
graph paper.
So, you would have your graph

511
00:44:42 --> 00:44:46
paper, and you would have some
curve on it.

512
00:44:46 --> 00:44:48
And, very often,
you wanted to know,

513
00:44:48 --> 00:44:51
what's the total amount of
product that you have

514
00:44:51 --> 00:44:54
synthesized, or whatever the
question might be.

515
00:44:54 --> 00:44:58
It might relate with the area
under your curve.

516
00:44:58 --> 00:45:01
So, you'd say, oh, it's easy.
Let's just integrate,

517
00:45:01 --> 00:45:02
except you don't have a
function.

518
00:45:02 --> 00:45:05
You can put that into
calculator.

519
00:45:05 --> 00:45:07
The next thing you could do is,
well, let's count the little

520
00:45:07 --> 00:45:09
squares.
But, if you've seen a piece of

521
00:45:09 --> 00:45:12
graph paper, that's kind of
time-consuming.

522
00:45:12 --> 00:45:14
So, people invented these
things called planimeters.

523
00:45:14 --> 00:45:19
It's something where there is a
really heavy thing based at one

524
00:45:19 --> 00:45:23
corner, and there's a lot of
dials and gauges and everything.

525
00:45:23 --> 00:45:25
And, there's one arm that you
move.

526
00:45:25 --> 00:45:30
And so, what you do is you take
the moving arm and you just

527
00:45:30 --> 00:45:35
slide it all around your curve.
And, you look at one of the

528
00:45:35 --> 00:45:37
dials.
And, suddenly what comes,

529
00:45:37 --> 00:45:41
as you go around,
it gives you complete garbage.

530
00:45:41 --> 00:45:45
But when you come back here,
that dial suddenly gives you

531
00:45:45 --> 00:45:48
the value of the area of this
region.

532
00:45:48 --> 00:45:51
So, how does it work?
This gadget never knows about

533
00:45:51 --> 00:45:55
the region inside because you
don't take it all over here.

534
00:45:55 --> 00:45:57
You only take it along the
curve.

535
00:45:57 --> 00:46:00
So, what it does actually is it
computes a line integral.

536
00:46:00 --> 00:46:04
OK, so it has this system of
wheels and everything that

537
00:46:04 --> 00:46:08
compute for you the line
integral along C of,

538
00:46:08 --> 00:46:11
well, it depends on the model.
But some of them compute the

539
00:46:11 --> 00:46:14
line integral of x dy.
Some of them compute different

540
00:46:14 --> 00:46:17
line integrals.
But, they compute some line

541
00:46:17 --> 00:46:21
integral, OK?
And, now, if you apply Green's

542
00:46:21 --> 00:46:26
theorem, you see that when you
have a counterclockwise curve,

543
00:46:26 --> 00:46:31
this will be just the area of
the region inside.

544
00:46:31 --> 00:46:34
And so, that's how it works.
I mean, of course,

545
00:46:34 --> 00:46:36
now you use a computer and it
does the sums.

546
00:46:36 --> 00:46:39
Yes?
That costs several thousand

547
00:46:39 --> 00:46:43
dollars, possibly more.
So, that's why I didn't bring

548
00:46:43 --> 00:46:44
one.
 

549
00:46:44 --> 00:46:49