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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
9
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
12
00:00:50 --> 00:00:53
everywhere, then it's going to
be conservative.
13
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.
15
00:00:59 --> 00:01:04
But now, let's say that I have
a general vector field.
16
00:01:04 --> 00:01:07
So, the curl will not be zero.
And, I still want to compute
17
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.
19
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.
25
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.
63
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
70
00:05:43 --> 00:05:46
counterclockwise.
What if I have a curve that
71
00:05:46 --> 00:05:49
goes clockwise?
Well, you could just take the
72
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