1
00:00:01,000 --> 00:00:03,000
The following content is
provided under a Creative
2
00:00:03,000 --> 00:00:05,000
Commons license.
Your support will help MIT
3
00:00:05,000 --> 00:00:08,000
OpenCourseWare continue to offer
high quality educational
4
00:00:08,000 --> 00:00:13,000
resources for free.
To make a donation or to view
5
00:00:13,000 --> 00:00:18,000
additional materials from
hundreds of MIT courses,
6
00:00:18,000 --> 00:00:23,000
visit MIT OpenCourseWare at
ocw.mit.edu.
7
00:00:23,000 --> 00:00:32,000
OK, so last time we've seen the
curl of the vector field with
8
00:00:32,000 --> 00:00:39,000
components M and N.
We defined that to be N sub x
9
00:00:39,000 --> 00:00:43,000
minus M sub y.
And, we said this measures how
10
00:00:43,000 --> 00:00:47,000
far that vector field is from
being conservative.
11
00:00:47,000 --> 00:00:50,000
If the curl is zero,
and if the field is defined
12
00:00:50,000 --> 00:00:53,000
everywhere, then it's going to
be conservative.
13
00:00:53,000 --> 00:00:55,000
And so, when I take the line
integral along a closed curve,
14
00:00:55,000 --> 00:00:59,000
I don't have to compute it.
I notes going to be zero.
15
00:00:59,000 --> 00:01:04,000
But now, let's say that I have
a general vector field.
16
00:01:04,000 --> 00:01:07,000
So, the curl will not be zero.
And, I still want to compute
17
00:01:07,000 --> 00:01:10,000
the line integral along a closed
curve.
18
00:01:10,000 --> 00:01:14,000
Well, I could compute it
directly or there's another way.
19
00:01:14,000 --> 00:01:17,000
And that's what we are going to
see today.
20
00:01:17,000 --> 00:01:26,000
So, say that I have a closed
curve, C, and I want to find the
21
00:01:26,000 --> 00:01:30,000
work.
So, there's two options.
22
00:01:30,000 --> 00:01:40,000
One is direct calculation,
and the other one is Green's
23
00:01:40,000 --> 00:01:47,000
theorem.
So, Green's theorem is another
24
00:01:47,000 --> 00:01:56,000
way to avoid calculating line
integrals if we don't want to.
25
00:01:56,000 --> 00:02:04,000
OK, so what does it say?
It says if C is a closed curve
26
00:02:04,000 --> 00:02:16,000
enclosing a region R in the
plane, and I have to insist C
27
00:02:16,000 --> 00:02:28,000
should go counterclockwise.
And, if I have a vector field
28
00:02:28,000 --> 00:02:37,000
that's defined and
differentiable everywhere not
29
00:02:37,000 --> 00:02:42,000
only on the curve,
C, which is what I need to
30
00:02:42,000 --> 00:02:48,000
define the line integral,
but also on the region inside.
31
00:02:48,000 --> 00:02:57,000
Then -- -- the line integral
for the work done along C is
32
00:02:57,000 --> 00:03:07,000
actually equal to a double
integral over the region inside
33
00:03:07,000 --> 00:03:15,000
of curl F dA.
OK, so that's the conclusion.
34
00:03:15,000 --> 00:03:19,000
And, if you want me to write it
in coordinates,
35
00:03:19,000 --> 00:03:24,000
maybe I should do that.
So, the line integral in terms
36
00:03:24,000 --> 00:03:28,000
of the components,
that's the integral of M dx
37
00:03:28,000 --> 00:03:36,000
plus N dy.
And, the curl is (Nx-My)dA.
38
00:03:36,000 --> 00:03:42,000
OK, so that's the other way to
state it.
39
00:03:42,000 --> 00:03:47,000
So, that's a really strange
statement if you think about it
40
00:03:47,000 --> 00:03:51,000
because the left-hand side is a
line integral.
41
00:03:51,000 --> 00:03:57,000
OK, so the way we compute it is
we take this expression Mdx Ndy
42
00:03:57,000 --> 00:04:01,000
and we parameterize the curve.
We express x and y in terms of
43
00:04:01,000 --> 00:04:04,000
some variable,
t, maybe, or whatever you want
44
00:04:04,000 --> 00:04:07,000
to call it.
And then, you'll do a one
45
00:04:07,000 --> 00:04:11,000
variable integral over t.
This right-hand side here,
46
00:04:11,000 --> 00:04:13,000
it's a double integral,
dA.
47
00:04:13,000 --> 00:04:16,000
So, we do it the way that we
learn how to couple of weeks
48
00:04:16,000 --> 00:04:18,000
ago.
You take your region,
49
00:04:18,000 --> 00:04:22,000
you slice it in the x direction
or in the y direction,
50
00:04:22,000 --> 00:04:26,000
and you integrate dx dy after
setting up the bounds carefully,
51
00:04:26,000 --> 00:04:29,000
or maybe in polar coordinates r
dr d theta.
52
00:04:29,000 --> 00:04:32,000
But, see, the way you compute
these things is completely
53
00:04:32,000 --> 00:04:36,000
different.
This one on the left-hand side
54
00:04:36,000 --> 00:04:42,000
lives only on the curve,
while the right-hand side lives
55
00:04:42,000 --> 00:04:46,000
everywhere in this region
inside.
56
00:04:46,000 --> 00:04:49,000
So, here, x and y are related,
they live on the curve.
57
00:04:49,000 --> 00:04:53,000
Here, x and y are independent.
There just are some bounds
58
00:04:53,000 --> 00:04:54,000
between them.
And, of course,
59
00:04:54,000 --> 00:04:56,000
what you're integrating is
different.
60
00:04:56,000 --> 00:05:01,000
It's a line integral for work.
Here, it's a double integral of
61
00:05:01,000 --> 00:05:08,000
some function of x and y.
So, it's a very perplexing
62
00:05:08,000 --> 00:05:14,000
statement at first.
But, it's a very powerful tool.
63
00:05:14,000 --> 00:05:18,000
So, we're going to try to see
how it works concretely,
64
00:05:18,000 --> 00:05:20,000
what it says,
what are the consequences,
65
00:05:20,000 --> 00:05:23,000
how we could convince ourselves
that, yes,
66
00:05:23,000 --> 00:05:26,000
this works, and so on.
That's going to be the topic
67
00:05:26,000 --> 00:05:32,000
for today.
Any questions about the
68
00:05:32,000 --> 00:05:38,000
statement first?
No?
69
00:05:38,000 --> 00:05:43,000
OK, yeah, one remark, sorry.
So, here, it stays
70
00:05:43,000 --> 00:05:46,000
counterclockwise.
What if I have a curve that
71
00:05:46,000 --> 00:05:49,000
goes clockwise?
Well, you could just take the
72
00:05:49,000 --> 00:05:52,000
negative, and integrate
counterclockwise.
73
00:05:52,000 --> 00:05:57,000
Why does the theorem choose
counterclockwise over clockwise?
74
00:05:57,000 --> 00:06:00,000
How doesn't know that it's
counterclockwise rather than
75
00:06:00,000 --> 00:06:02,000
clockwise?
Well, the answer is basically
76
00:06:02,000 --> 00:06:05,000
in our convention for curl.
See, we've said curl is Nx
77
00:06:05,000 --> 00:06:08,000
minus My, and not the other way
around.
78
00:06:08,000 --> 00:06:10,000
And, that's a convention as
well.
79
00:06:10,000 --> 00:06:13,000
So, somehow,
the two conventions match with
80
00:06:13,000 --> 00:06:15,000
each other.
That's the best answer I can
81
00:06:15,000 --> 00:06:18,000
give you.
So, if you met somebody from a
82
00:06:18,000 --> 00:06:20,000
different planet,
they might have Green's theorem
83
00:06:20,000 --> 00:06:23,000
with the opposite conventions,
with curves going clockwise,
84
00:06:23,000 --> 00:06:27,000
and the curl defined the other
way around.
85
00:06:27,000 --> 00:06:35,000
Probably if you met an alien,
I'm not sure if you would be
86
00:06:35,000 --> 00:06:43,000
discussing Green's theorem
first, but just in case.
87
00:06:43,000 --> 00:06:53,000
OK, so that being said,
there is a warning here which
88
00:06:53,000 --> 00:07:00,000
is that this is only for closed
curves.
89
00:07:00,000 --> 00:07:03,000
OK, so if I give you a curve
that's not closed,
90
00:07:03,000 --> 00:07:05,000
and I tell you,
well, compute the line
91
00:07:05,000 --> 00:07:08,000
integral, then you have to do it
by hand.
92
00:07:08,000 --> 00:07:10,000
You have to parameterize the
curve.
93
00:07:10,000 --> 00:07:12,000
Or, if you really don't like
that line integral,
94
00:07:12,000 --> 00:07:16,000
you could close the path by
adding some other line integral
95
00:07:16,000 --> 00:07:19,000
to it,
and then compute using Green's
96
00:07:19,000 --> 00:07:23,000
theorem.
But, you can't use Green's
97
00:07:23,000 --> 00:07:30,000
theorem directly if the curve is
not closed.
98
00:07:30,000 --> 00:07:42,000
OK, so let's do a quick example.
So, let's say that I give you
99
00:07:42,000 --> 00:07:52,000
C, the circle of radius one,
centered at the point (2,0).
100
00:07:52,000 --> 00:08:00,000
So, it's out here.
That's my curve, C.
101
00:08:00,000 --> 00:08:09,000
And, let's say that I do it
counterclockwise so that it will
102
00:08:09,000 --> 00:08:16,000
match with the statement of the
theorem.
103
00:08:16,000 --> 00:08:24,000
And, let's say that I want you
to compute the line integral
104
00:08:24,000 --> 00:08:34,000
along C of ye^(-x) dx plus (one
half of x squared minus e^(-x))
105
00:08:34,000 --> 00:08:37,000
dy.
And, that's a kind of sadistic
106
00:08:37,000 --> 00:08:41,000
example, but maybe I'll ask you
to do that.
107
00:08:41,000 --> 00:08:44,000
So, how would you do it
directly?
108
00:08:44,000 --> 00:08:48,000
Well, to do it directly you
would have to parameterize this
109
00:08:48,000 --> 00:08:53,000
curve.
So that would probably involve
110
00:08:53,000 --> 00:09:03,000
setting x equals two plus cosine
theta y equals sine theta.
111
00:09:03,000 --> 00:09:06,000
But, I'm using as parameter of
the angle around the circle,
112
00:09:06,000 --> 00:09:09,000
it's like the unit circle,
the usual ones that shifted by
113
00:09:09,000 --> 00:09:15,000
two in the x direction.
And then, I would set dx equals
114
00:09:15,000 --> 00:09:21,000
minus sine theta d theta.
I would set dy equals cosine
115
00:09:21,000 --> 00:09:24,000
theta d theta.
And, I will substitute,
116
00:09:24,000 --> 00:09:26,000
and I will integrate from zero
to 2pi.
117
00:09:26,000 --> 00:09:29,000
And, I would probably run into
a bit of trouble because I would
118
00:09:29,000 --> 00:09:32,000
have these e to the minus x,
which would give me something
119
00:09:32,000 --> 00:09:35,000
that I really don't want to
integrate.
120
00:09:35,000 --> 00:09:44,000
So, instead of doing that,
which looks pretty much doomed,
121
00:09:44,000 --> 00:09:51,000
instead, I'm going to use
Green's theorem.
122
00:09:51,000 --> 00:09:58,000
So, using Green's theorem,
the way we'll do it is I will,
123
00:09:58,000 --> 00:10:03,000
instead, compute a double
integral.
124
00:10:03,000 --> 00:10:16,000
So, I will -- -- compute the
double integral over the region
125
00:10:16,000 --> 00:10:27,000
inside of curl F dA.
So, I should say probably what
126
00:10:27,000 --> 00:10:30,000
F was.
So, let's call this M.
127
00:10:30,000 --> 00:10:37,000
Let's call this N.
And, then I will actually just
128
00:10:37,000 --> 00:10:45,000
choose the form coordinates,
(Nx minus My) dA.
129
00:10:45,000 --> 00:10:53,000
And, what is R here?
Well, R is the disk in here.
130
00:10:53,000 --> 00:10:56,000
OK, so, of course,
it might not be that pleasant
131
00:10:56,000 --> 00:10:58,000
because we'll also have to set
up this double integral.
132
00:10:58,000 --> 00:11:02,000
And, for that,
we'll have to figure out a way
133
00:11:02,000 --> 00:11:05,000
to slice this region nicely.
We could do it dx dy.
134
00:11:05,000 --> 00:11:08,000
We could do it dy dx.
Or, maybe we will want to
135
00:11:08,000 --> 00:11:11,000
actually make a change of
variables to first shift this to
136
00:11:11,000 --> 00:11:14,000
the origin,
you know, change x to x minus
137
00:11:14,000 --> 00:11:17,000
two and then switch to polar
coordinates.
138
00:11:17,000 --> 00:11:22,000
Well, let's see what happens
later.
139
00:11:22,000 --> 00:11:33,000
OK, so what is, so this is R.
So, what is N sub x?
140
00:11:33,000 --> 00:11:42,000
Well, N sub x is x plus e to
the minus x minus,
141
00:11:42,000 --> 00:11:48,000
what is M sub y,
e to the minus x,
142
00:11:48,000 --> 00:11:52,000
OK?
This is Nx.
143
00:11:52,000 --> 00:11:58,000
This is My dA.
Well, it seems to simplify a
144
00:11:58,000 --> 00:12:02,000
bit.
I will just get double integral
145
00:12:02,000 --> 00:12:06,000
over R of x dA,
which looks certainly a lot
146
00:12:06,000 --> 00:12:10,000
more pleasant.
Of course, I made up the
147
00:12:10,000 --> 00:12:14,000
example in that way so that it
simplifies when you use Green's
148
00:12:14,000 --> 00:12:16,000
theorem.
But, you know,
149
00:12:16,000 --> 00:12:20,000
it gives you an example where
you can turn are really hard
150
00:12:20,000 --> 00:12:23,000
line integral into an easier
double integral.
151
00:12:23,000 --> 00:12:28,000
Now, how do we compute that
double integral?
152
00:12:28,000 --> 00:12:32,000
Well, so one way would be to
set it up.
153
00:12:32,000 --> 00:12:41,000
Or, let's actually be a bit
smarter and observe that this is
154
00:12:41,000 --> 00:12:50,000
actually the area of the region
R, times the x coordinate of its
155
00:12:50,000 --> 00:12:55,000
center of mass.
If I look at the definition of
156
00:12:55,000 --> 00:12:59,000
the center of mass,
it's the average value of x.
157
00:12:59,000 --> 00:13:03,000
So, it's one over the area
times the double integral of x
158
00:13:03,000 --> 00:13:07,000
dA, well, possibly with the
density, but here I'm thinking
159
00:13:07,000 --> 00:13:11,000
uniform density one.
And, now, I think I know just
160
00:13:11,000 --> 00:13:15,000
by looking at the picture where
the center of mass of this
161
00:13:15,000 --> 00:13:16,000
circle will be,
right?
162
00:13:16,000 --> 00:13:19,000
I mean, it would be right in
the middle.
163
00:13:19,000 --> 00:13:24,000
So, that is two,
if you want,
164
00:13:24,000 --> 00:13:29,000
by symmetry.
And, the area of the guy is
165
00:13:29,000 --> 00:13:33,000
just pi because it's a disk of
radius one.
166
00:13:33,000 --> 00:13:37,000
So, I will just get 2pi.
I mean, of course,
167
00:13:37,000 --> 00:13:40,000
if you didn't see that,
then you can also compute that
168
00:13:40,000 --> 00:13:43,000
double integral directly.
It's a nice exercise.
169
00:13:43,000 --> 00:13:47,000
But see, here,
using geometry helps you to
170
00:13:47,000 --> 00:13:50,000
actually streamline the
calculation.
171
00:13:50,000 --> 00:14:01,000
OK, any questions?
Yes?
172
00:14:01,000 --> 00:14:04,000
OK, yes, let me just repeat the
last part.
173
00:14:04,000 --> 00:14:10,000
So, I said we had to compute
the double integral of x dA over
174
00:14:10,000 --> 00:14:14,000
this region here,
which is a disk of radius one,
175
00:14:14,000 --> 00:14:18,000
centered at,
this point is (2,0).
176
00:14:18,000 --> 00:14:22,000
So, instead of setting up the
integral with bounds and
177
00:14:22,000 --> 00:14:26,000
integrating dx dy or dy dx or in
polar coordinates,
178
00:14:26,000 --> 00:14:30,000
I'm just going to say, well,
let's remember the definition
179
00:14:30,000 --> 00:14:32,000
of a center of mass.
It's the average value of a
180
00:14:32,000 --> 00:14:37,000
function, x in the region.
So, it's one over the area of
181
00:14:37,000 --> 00:14:42,000
origin times the double integral
of x dA.
182
00:14:42,000 --> 00:14:46,000
If you look,
again, at the definition of x
183
00:14:46,000 --> 00:14:51,000
bar, it's one over area of
double integral x dA.
184
00:14:51,000 --> 00:14:54,000
Well, maybe if there's a
density, then it's one over mass
185
00:14:54,000 --> 00:14:57,000
times double integral of x
density dA.
186
00:14:57,000 --> 00:15:02,000
But, if density is one,
then it just becomes this.
187
00:15:02,000 --> 00:15:06,000
So, switching the area,
moving the area to the other
188
00:15:06,000 --> 00:15:08,000
side,
I'll get double integral of x
189
00:15:08,000 --> 00:15:12,000
dA is the area of origin times
the x coordinate of the center
190
00:15:12,000 --> 00:15:14,000
of mass.
The area of origin is pi
191
00:15:14,000 --> 00:15:18,000
because it's a unit disk.
And, the center of mass is the
192
00:15:18,000 --> 00:15:23,000
center of a disk.
So, its x bar is two,
193
00:15:23,000 --> 00:15:27,000
and I get 2 pi.
OK, that I didn't actually have
194
00:15:27,000 --> 00:15:30,000
to do this in my example today,
but of course that would be
195
00:15:30,000 --> 00:15:36,000
good review.
It will remind you of center of
196
00:15:36,000 --> 00:15:47,000
mass and all that.
OK, any other questions?
197
00:15:47,000 --> 00:15:50,000
No?
OK, so let's see,
198
00:15:50,000 --> 00:15:54,000
now that we've seen how to use
it practice, how to avoid
199
00:15:54,000 --> 00:15:58,000
calculating the line integral if
we don't want to.
200
00:15:58,000 --> 00:16:04,000
Let's try to convince ourselves
that this theorem makes sense.
201
00:16:04,000 --> 00:16:09,000
OK, so, well,
let's start with an easy case
202
00:16:09,000 --> 00:16:15,000
where we should be able to know
the answer to both sides.
203
00:16:15,000 --> 00:16:23,000
So let's look at the special
case.
204
00:16:23,000 --> 00:16:32,000
Let's look at the case where
curl F is zero.
205
00:16:32,000 --> 00:16:45,000
Then, well, we'd like to
conclude that F is conservative.
206
00:16:45,000 --> 00:16:53,000
That's what we said.
Well let's see what happens.
207
00:16:53,000 --> 00:16:59,000
So, Green's theorem says that
if I have a closed curve,
208
00:16:59,000 --> 00:17:06,000
then the line integral of F is
equal to the double integral of
209
00:17:06,000 --> 00:17:12,000
curl on the region inside.
And, if the curl is zero,
210
00:17:12,000 --> 00:17:15,000
then I will be integrating
zero.
211
00:17:15,000 --> 00:17:22,000
I will get zero.
OK, so this is actually how you
212
00:17:22,000 --> 00:17:26,000
prove that if your vector field
has curve zero,
213
00:17:26,000 --> 00:17:27,000
then it's conservative.
214
00:17:54,000 --> 00:17:57,000
OK, so in particular,
if you have a vector field
215
00:17:57,000 --> 00:18:01,000
that's defined everywhere the
plane, then you take any closed
216
00:18:01,000 --> 00:18:04,000
curve.
Well, you will get that the
217
00:18:04,000 --> 00:18:06,000
line integral will be zero.
Straightly speaking,
218
00:18:06,000 --> 00:18:10,000
that will only work here if the
curve goes counterclockwise.
219
00:18:10,000 --> 00:18:13,000
But otherwise,
just look at the various loops
220
00:18:13,000 --> 00:18:16,000
that it makes,
and orient each of them
221
00:18:16,000 --> 00:18:19,000
counterclockwise and sum things
together.
222
00:18:19,000 --> 00:18:22,000
So let me state that again.
223
00:18:45,000 --> 00:18:51,000
So,
OK,
224
00:18:51,000 --> 00:19:02,000
so a consequence of Green's
theorem is that if F is defined
225
00:19:02,000 --> 00:19:12,000
everywhere in the plane -- --
and the curl of F is zero
226
00:19:12,000 --> 00:19:24,000
everywhere,
then F is conservative.
227
00:19:24,000 --> 00:19:29,000
And so, this actually is the
input we needed to justify our
228
00:19:29,000 --> 00:19:33,000
criterion.
The test that we saw last time
229
00:19:33,000 --> 00:19:35,000
saying,
well, to check if something is
230
00:19:35,000 --> 00:19:37,000
a gradient field if it's
conservative,
231
00:19:37,000 --> 00:19:40,000
we just have to compute the
curl and check whether it's
232
00:19:40,000 --> 00:19:45,000
zero.
OK, so how do we prove that now
233
00:19:45,000 --> 00:19:49,000
carefully?
Well, you just take a closed
234
00:19:49,000 --> 00:19:52,000
curve in the plane.
You switch the orientation if
235
00:19:52,000 --> 00:19:55,000
needed so it becomes
counterclockwise.
236
00:19:55,000 --> 00:19:59,000
And then you look at the region
inside.
237
00:19:59,000 --> 00:20:05,000
And then you know that the line
integral inside will be equal to
238
00:20:05,000 --> 00:20:12,000
the double integral of curl,
which is the double integral of
239
00:20:12,000 --> 00:20:16,000
zero.
Therefore, that's zero.
240
00:20:16,000 --> 00:20:19,000
But see, OK,
so now let's say that we try to
241
00:20:19,000 --> 00:20:23,000
do that for the vector field
that was on your problems that
242
00:20:23,000 --> 00:20:27,000
was not defined at the origin.
So if you've done the problem
243
00:20:27,000 --> 00:20:30,000
sets and found the same answers
that I did, then you will have
244
00:20:30,000 --> 00:20:33,000
found that this vector field had
curve zero everywhere.
245
00:20:33,000 --> 00:20:36,000
But still it wasn't
conservative because if you went
246
00:20:36,000 --> 00:20:39,000
around the unit circle,
then you got a line integral
247
00:20:39,000 --> 00:20:42,000
that was 2pi.
Or, if you compared the two
248
00:20:42,000 --> 00:20:45,000
halves, you got different
answers for two parts that go
249
00:20:45,000 --> 00:20:47,000
from the same point to the same
point.
250
00:20:47,000 --> 00:20:51,000
So, it fails this property but
that's because it's not defined
251
00:20:51,000 --> 00:20:54,000
everywhere.
So, what goes wrong with this
252
00:20:54,000 --> 00:20:57,000
argument?
Well, if I take the vector
253
00:20:57,000 --> 00:21:03,000
field that was in the problem
set, and if I do things,
254
00:21:03,000 --> 00:21:07,000
say that I look at the unit
circle.
255
00:21:07,000 --> 00:21:10,000
That's a closed curve.
So, I would like to use Green's
256
00:21:10,000 --> 00:21:13,000
theorem.
Green's theorem would tell me
257
00:21:13,000 --> 00:21:17,000
the line integral along this
loop is equal to the double
258
00:21:17,000 --> 00:21:21,000
integral of curl over this
region here, the unit disk.
259
00:21:21,000 --> 00:21:25,000
And, of course the curl is
zero, well, except at the
260
00:21:25,000 --> 00:21:27,000
origin.
At the origin,
261
00:21:27,000 --> 00:21:29,000
the vector field is not
defined.
262
00:21:29,000 --> 00:21:32,000
You cannot take the
derivatives, and the curl is not
263
00:21:32,000 --> 00:21:34,000
defined.
And somehow that messes things
264
00:21:34,000 --> 00:21:38,000
up.
You cannot apply Green's
265
00:21:38,000 --> 00:21:49,000
theorem to the vector field.
So, you cannot apply Green's
266
00:21:49,000 --> 00:22:02,000
theorem to the vector field on
problem set eight problem two
267
00:22:02,000 --> 00:22:12,000
when C encloses the origin.
And so, that's why this guy,
268
00:22:12,000 --> 00:22:16,000
even though it has curl zero,
is not conservative.
269
00:22:16,000 --> 00:22:20,000
There's no contradiction.
And somehow,
270
00:22:20,000 --> 00:22:23,000
you have to imagine that,
well, the curl here is really
271
00:22:23,000 --> 00:22:26,000
not defined.
But somehow it becomes infinite
272
00:22:26,000 --> 00:22:30,000
so that when you do the double
integral, you actually get 2 pi
273
00:22:30,000 --> 00:22:37,000
instead of zero.
I mean, that doesn't make any
274
00:22:37,000 --> 00:22:46,000
sense, of course,
but that's one way to think
275
00:22:46,000 --> 00:22:51,000
about it.
OK, any questions?
276
00:22:51,000 --> 00:23:02,000
Yes?
Well, though actually it's not
277
00:23:02,000 --> 00:23:06,000
defined because the curl is zero
everywhere else.
278
00:23:06,000 --> 00:23:08,000
So, if a curl was well defined
at the origin,
279
00:23:08,000 --> 00:23:11,000
you would try to,
then, take the double integral.
280
00:23:11,000 --> 00:23:12,000
no matter what value you put
for a function,
281
00:23:12,000 --> 00:23:15,000
if you have a function that's
zero everywhere except at the
282
00:23:15,000 --> 00:23:17,000
origin,
and some other value at the
283
00:23:17,000 --> 00:23:20,000
origin,
the integral is still zero.
284
00:23:20,000 --> 00:23:24,000
So, it's worse than that.
It's not only that you can't
285
00:23:24,000 --> 00:23:29,000
compute it, it's that is not
defined.
286
00:23:29,000 --> 00:23:36,000
OK, anyway, that's like a
slightly pathological example.
287
00:23:36,000 --> 00:23:44,000
Yes?
Well, we wouldn't be able to
288
00:23:44,000 --> 00:23:46,000
because the curl is not defined
at the origin.
289
00:23:46,000 --> 00:23:49,000
So, you can actually integrate
it.
290
00:23:49,000 --> 00:23:52,000
OK, so that's the problem.
I mean, if you try to
291
00:23:52,000 --> 00:23:55,000
integrate, we've said everywhere
where it's defined,
292
00:23:55,000 --> 00:23:57,000
the curl is zero.
So, what you would be
293
00:23:57,000 --> 00:24:01,000
integrating would be zero.
But, that doesn't work because
294
00:24:01,000 --> 00:24:09,000
at the origin it's not defined.
Yes?
295
00:24:09,000 --> 00:24:11,000
Ah, so if you take a curve that
makes a figure 8,
296
00:24:11,000 --> 00:24:14,000
then indeed my proof over there
is false.
297
00:24:14,000 --> 00:24:19,000
So, I kind of tricked you.
It's not actually correct.
298
00:24:19,000 --> 00:24:24,000
So, if the curve does a figure
8, then what you do is you would
299
00:24:24,000 --> 00:24:27,000
actually cut it into its two
halves.
300
00:24:27,000 --> 00:24:30,000
And for each of them,
you will apply Green's theorem.
301
00:24:30,000 --> 00:24:32,000
And then, you'd still get,
if a curl is zero then this
302
00:24:32,000 --> 00:24:35,000
line integral is zero.
That one is also zero.
303
00:24:35,000 --> 00:24:38,000
So this one is zero.
OK, small details that you
304
00:24:38,000 --> 00:24:41,000
don't really need to worry too
much about,
305
00:24:41,000 --> 00:24:47,000
but indeed if you want to be
careful with details then my
306
00:24:47,000 --> 00:24:54,000
proof is not quite complete.
But the computation is still
307
00:24:54,000 --> 00:24:58,000
true.
Let's move on.
308
00:24:58,000 --> 00:25:06,000
So, I want to tell you how to
prove Green's theorem because
309
00:25:06,000 --> 00:25:15,000
it's such a strange formula that
where can it come from possibly?
310
00:25:15,000 --> 00:25:21,000
I mean,
so let me remind you first of
311
00:25:21,000 --> 00:25:26,000
all the statement we want to
prove is that the line integral
312
00:25:26,000 --> 00:25:31,000
along a closed curve of Mdx plus
Ndy is equal to the double
313
00:25:31,000 --> 00:25:36,000
integral over the region inside
of (Nx minus My)dA.
314
00:25:36,000 --> 00:25:40,000
And, let's simplify our lives a
bit by proving easier
315
00:25:40,000 --> 00:25:43,000
statements.
So actually,
316
00:25:43,000 --> 00:25:53,000
the first observation will
actually prove something easier,
317
00:25:53,000 --> 00:25:58,000
namely, that the line integral,
let's see,
318
00:25:58,000 --> 00:26:03,000
of Mdx along a closed curve is
equal to the double integral
319
00:26:03,000 --> 00:26:08,000
over the region inside of minus
M sub y dA.
320
00:26:08,000 --> 00:26:13,000
OK, so that's the special case
where N is zero,
321
00:26:13,000 --> 00:26:19,000
where you have only an x
component for your vector field.
322
00:26:19,000 --> 00:26:23,000
Now, why is that good enough?
Well, the claim is if I can
323
00:26:23,000 --> 00:26:28,000
prove this, I claim you will be
able to do the same thing to
324
00:26:28,000 --> 00:26:33,000
prove the other case where there
is only the y component.
325
00:26:33,000 --> 00:26:38,000
And then, if the other
together, you will get the
326
00:26:38,000 --> 00:26:40,000
general case.
So, let me explain.
327
00:27:00,000 --> 00:27:06,000
OK, so a similar argument which
I will not do,
328
00:27:06,000 --> 00:27:11,000
to save time,
will show, so actually it's
329
00:27:11,000 --> 00:27:15,000
just the same thing but
switching the roles of x and y,
330
00:27:15,000 --> 00:27:20,000
that if I integrate along a
closed curve N dy,
331
00:27:20,000 --> 00:27:29,000
then I'll get the double
integral of N sub x dA.
332
00:27:29,000 --> 00:27:36,000
And so, now if I have proved
these two formulas separately,
333
00:27:36,000 --> 00:27:44,000
then if you sum them together
will get the correct statement.
334
00:27:44,000 --> 00:27:52,000
Let me write it.
We get Green's theorem.
335
00:27:52,000 --> 00:27:55,000
OK, so we've simplified our
task a little bit.
336
00:27:55,000 --> 00:28:00,000
We'll just be trying to prove
the case where there's only an x
337
00:28:00,000 --> 00:28:04,000
component.
So, let's do it.
338
00:28:04,000 --> 00:28:07,000
Well, we have another problem
which is the region that we are
339
00:28:07,000 --> 00:28:10,000
looking at, the curve that we're
looking at might be very
340
00:28:10,000 --> 00:28:12,000
complicated.
If I give you,
341
00:28:12,000 --> 00:28:17,000
let's say I give you,
I don't know,
342
00:28:17,000 --> 00:28:22,000
a curve that does something
like this.
343
00:28:22,000 --> 00:28:26,000
Well, it will be kind of tricky
to set up a double integral over
344
00:28:26,000 --> 00:28:29,000
the region inside.
So maybe we first want to look
345
00:28:29,000 --> 00:28:33,000
at curves that are simpler,
that will actually allow us to
346
00:28:33,000 --> 00:28:36,000
set up the double integral
easily.
347
00:28:36,000 --> 00:28:42,000
So, the second observation,
so that was the first
348
00:28:42,000 --> 00:28:51,000
observation.
The second observation is that
349
00:28:51,000 --> 00:29:02,000
we can decompose R into simpler
regions.
350
00:29:02,000 --> 00:29:10,000
So what do I mean by that?
Well, let's say that I have a
351
00:29:10,000 --> 00:29:13,000
region and I'm going to cut it
into two.
352
00:29:13,000 --> 00:29:18,000
So, I'll have R1 and R2.
And then, of course,
353
00:29:18,000 --> 00:29:22,000
I need to have the curves that
go around them.
354
00:29:22,000 --> 00:29:29,000
So, I had my initial curve,
C, was going around everybody.
355
00:29:29,000 --> 00:29:41,000
They have curves C1 that goes
around R1, and C2 goes around
356
00:29:41,000 --> 00:29:46,000
R2.
OK, so,
357
00:29:46,000 --> 00:29:55,000
what I would like to say is if
we can prove that the statement
358
00:29:55,000 --> 00:30:07,000
is true, so let's see,
for C1 and also for C2 -- --
359
00:30:07,000 --> 00:30:23,000
then I claim we can prove the
statement for C.
360
00:30:23,000 --> 00:30:26,000
How do we do that?
Well, we just add these two
361
00:30:26,000 --> 00:30:28,000
equalities together.
OK, why does that work?
362
00:30:28,000 --> 00:30:31,000
There's something fishy going
on because C1 and C2 have this
363
00:30:31,000 --> 00:30:35,000
piece here in the middle.
That's not there in C.
364
00:30:35,000 --> 00:30:39,000
So, if you add the line
integral along C1 and C2,
365
00:30:39,000 --> 00:30:44,000
you get these unwanted pieces.
But, the good news is actually
366
00:30:44,000 --> 00:30:47,000
you go twice through that edge
in the middle.
367
00:30:47,000 --> 00:30:51,000
See, it appears once in C1
going up, and once in C2 going
368
00:30:51,000 --> 00:30:52,000
down.
So, in fact,
369
00:30:52,000 --> 00:30:55,000
when you will do the work,
when you will sum the work,
370
00:30:55,000 --> 00:30:57,000
you will add these two guys
together.
371
00:30:57,000 --> 00:31:06,000
They will cancel.
OK, so the line integral along
372
00:31:06,000 --> 00:31:14,000
C will be, then,
it will be the sum of the line
373
00:31:14,000 --> 00:31:21,000
integrals on C1 and C2.
And, that will equal,
374
00:31:21,000 --> 00:31:29,000
therefore, the double integral
over R1 plus the double integral
375
00:31:29,000 --> 00:31:36,000
over R2, which is the double
integral over R of negative My.
376
00:31:36,000 --> 00:31:47,000
OK and the reason for this
equality here is because we go
377
00:31:47,000 --> 00:31:56,000
twice through the inner part.
What do I want to say?
378
00:31:56,000 --> 00:32:15,000
Along the boundary between R1
and R2 -- -- with opposite
379
00:32:15,000 --> 00:32:25,000
orientations.
So, the extra things cancel out.
380
00:32:25,000 --> 00:32:29,000
OK, so that means I just need
to look at smaller pieces if
381
00:32:29,000 --> 00:32:34,000
that makes my life easier.
So, now, will make my life easy?
382
00:32:34,000 --> 00:32:41,000
Well, let's say that I have a
curve like that.
383
00:32:41,000 --> 00:32:45,000
Well, I guess I should really
draw a pumpkin or something like
384
00:32:45,000 --> 00:32:48,000
that because it would be more
seasonal.
385
00:32:48,000 --> 00:32:53,000
But, well, I don't really know
how to draw a pumpkin.
386
00:32:53,000 --> 00:32:57,000
OK, so what I will do is I will
cut this into smaller regions
387
00:32:57,000 --> 00:33:01,000
for which I have a well-defined
lower and upper boundary so that
388
00:33:01,000 --> 00:33:05,000
I will be able to set up a
double integral,
389
00:33:05,000 --> 00:33:10,000
dy dx, easily.
So, a region like this I will
390
00:33:10,000 --> 00:33:17,000
actually cut it here and here
into five smaller pieces so that
391
00:33:17,000 --> 00:33:23,000
each small piece will let me set
up the double integral,
392
00:33:23,000 --> 00:33:31,000
dy dx.
OK, so we'll cut R in to what I
393
00:33:31,000 --> 00:33:41,000
will call vertically simple --
-- regions.
394
00:33:41,000 --> 00:33:43,000
So, what's a vertically simple
region?
395
00:33:43,000 --> 00:33:48,000
That's a region that's given by
looking at x between a and b for
396
00:33:48,000 --> 00:33:53,000
some values of a and b.
And, for each value of x,
397
00:33:53,000 --> 00:34:00,000
y is between some function of x
and some other function of x.
398
00:34:00,000 --> 00:34:03,000
OK, so for example,
this guy is vertically simple.
399
00:34:03,000 --> 00:34:07,000
See, x runs from this value of
x to that value of x.
400
00:34:07,000 --> 00:34:13,000
And, for each x,
y goes between this value to
401
00:34:13,000 --> 00:34:16,000
that value.
And, same with each of these.
402
00:34:39,000 --> 00:34:49,000
OK, so now we are down to the
main step that we have to do,
403
00:34:49,000 --> 00:35:05,000
which is to prove this identity
if C is, sorry,
404
00:35:05,000 --> 00:35:23,000
if -- -- if R is vertically
simple -- -- and C is the
405
00:35:23,000 --> 00:35:36,000
boundary of R going
counterclockwise.
406
00:35:36,000 --> 00:35:40,000
OK, so let's look at how we
would do it.
407
00:35:40,000 --> 00:35:46,000
So, we said vertically simple
region looks like x goes between
408
00:35:46,000 --> 00:35:52,000
a and b, and y goes between two
values that are given by
409
00:35:52,000 --> 00:35:57,000
functions of x.
OK, so this is y equals f2 of x.
410
00:35:57,000 --> 00:36:02,000
This is y equals f1 of x.
This is a.
411
00:36:02,000 --> 00:36:09,000
This is b.
Our region is this thing in
412
00:36:09,000 --> 00:36:13,000
here.
So, let's compute both sides.
413
00:36:13,000 --> 00:36:15,000
And, when I say compute,
of course we will not get
414
00:36:15,000 --> 00:36:17,000
numbers because we don't know
what M is.
415
00:36:17,000 --> 00:36:19,000
We don't know what f1 and f2
are.
416
00:36:19,000 --> 00:36:24,000
But, I claim we should be able
to simplify things a bit.
417
00:36:24,000 --> 00:36:28,000
So, let's start with the line
integral.
418
00:36:28,000 --> 00:36:35,000
How do I compute the line
integral along the curve that
419
00:36:35,000 --> 00:36:40,000
goes all around here?
Well, it looks like there will
420
00:36:40,000 --> 00:36:45,000
be four pieces.
OK, so we actually have four
421
00:36:45,000 --> 00:36:50,000
things to compute,
C1, C2, C3, and C4.
422
00:36:50,000 --> 00:37:01,000
OK?
Well, let's start with C1.
423
00:37:01,000 --> 00:37:06,000
So, if we integrate on C1 Mdx,
how do we do that?
424
00:37:06,000 --> 00:37:10,000
Well, we know that on C1,
y is given by a function of x.
425
00:37:10,000 --> 00:37:15,000
So, we can just get rid of y
and express everything in terms
426
00:37:15,000 --> 00:37:21,000
of x.
OK, so, we know y is f1 of x,
427
00:37:21,000 --> 00:37:27,000
and x goes from a to b.
So, that will be the integral
428
00:37:27,000 --> 00:37:30,000
from a to b of,
well, I have to take the
429
00:37:30,000 --> 00:37:33,000
function, M.
And so, M depends normally on x
430
00:37:33,000 --> 00:37:38,000
and y.
Maybe I should put x and y here.
431
00:37:38,000 --> 00:37:46,000
And then, I will plug y equals
f1 of x dx.
432
00:37:46,000 --> 00:37:49,000
And, then I have a single
variable integral.
433
00:37:49,000 --> 00:37:51,000
And that's what I have to
compute.
434
00:37:51,000 --> 00:37:54,000
Of course, I cannot compute it
here because I don't know what
435
00:37:54,000 --> 00:37:59,000
this is.
So, it has to stay this way.
436
00:37:59,000 --> 00:38:06,000
OK, next one.
The integral along C2,
437
00:38:06,000 --> 00:38:13,000
well, let's think for a second.
On C2, x equals b.
438
00:38:13,000 --> 00:38:16,000
It's constant.
So, dx is zero,
439
00:38:16,000 --> 00:38:20,000
and you would integrate,
actually, above a variable,
440
00:38:20,000 --> 00:38:23,000
y.
But, well, we don't have a y
441
00:38:23,000 --> 00:38:26,000
component.
See, this is the reason why we
442
00:38:26,000 --> 00:38:30,000
made the first observation.
We got rid of the other term
443
00:38:30,000 --> 00:38:33,000
because it's simplifies our life
here.
444
00:38:33,000 --> 00:38:38,000
So, we just get zero.
OK, just looking quickly ahead,
445
00:38:38,000 --> 00:38:40,000
there's another one that would
be zero as well,
446
00:38:40,000 --> 00:38:42,000
right?
Which one?
447
00:38:42,000 --> 00:38:52,000
Yeah, C4.
This one gives me zero.
448
00:38:52,000 --> 00:38:55,000
What about C3?
Well, C3 will look a lot like
449
00:38:55,000 --> 00:38:57,000
C1.
So, we're going to use the same
450
00:38:57,000 --> 00:38:59,000
kind of thing that we did with
C.
451
00:39:22,000 --> 00:39:27,000
OK, so along C3,
well, let's see,
452
00:39:27,000 --> 00:39:34,000
so on C3, y is a function of x,
again.
453
00:39:34,000 --> 00:39:40,000
And so we are using as our
variable x, but now x goes down
454
00:39:40,000 --> 00:39:45,000
from b to a.
So, it will be the integral
455
00:39:45,000 --> 00:39:51,000
from b to a of M of (x and f2 of
x) dx.
456
00:39:51,000 --> 00:39:57,000
Or, if you prefer,
that's negative integral from a
457
00:39:57,000 --> 00:40:04,000
to b of M of (x and f2 of x) dx.
OK, so now if I sum all these
458
00:40:04,000 --> 00:40:10,000
pieces together,
I get that the line integral
459
00:40:10,000 --> 00:40:20,000
along the closed curve is the
integral from a to b of M(x1f1
460
00:40:20,000 --> 00:40:30,000
of x) dx minus the integral from
a to b of M(x1f2 of x) dx.
461
00:40:30,000 --> 00:40:39,000
So, that's the left hand side.
Next, I should try to look at
462
00:40:39,000 --> 00:40:47,000
my double integral and see if I
can make it equal to that.
463
00:40:47,000 --> 00:40:58,000
So, let's look at the other
guy, double integral over R of
464
00:40:58,000 --> 00:41:02,000
negative MydA.
Well, first,
465
00:41:02,000 --> 00:41:05,000
I'll take the minus sign out.
It will make my life a little
466
00:41:05,000 --> 00:41:09,000
bit easier.
And second, so I said I will
467
00:41:09,000 --> 00:41:14,000
try to set this up in the way
that's the most efficient.
468
00:41:14,000 --> 00:41:20,000
And, my choice of this kind of
region means that it's easier to
469
00:41:20,000 --> 00:41:22,000
set up dy dx,
right?
470
00:41:22,000 --> 00:41:30,000
So, if I set it up dy dx,
then I know for a given value
471
00:41:30,000 --> 00:41:36,000
of x, y goes from f1 of x to f2
of x.
472
00:41:36,000 --> 00:41:49,000
And, x goes from a to b, right?
Is that OK with everyone?
473
00:41:49,000 --> 00:41:53,000
OK, so now if I compute the
inner integral,
474
00:41:53,000 --> 00:41:58,000
well, what do I get if I get
partial M partial y with respect
475
00:41:58,000 --> 00:42:02,000
to y?
I'll get M back, OK?
476
00:42:02,000 --> 00:42:19,000
So -- So, I will get M at the
point x f2 of x minus M at the
477
00:42:19,000 --> 00:42:27,000
point x f1 of x.
And so, this becomes the
478
00:42:27,000 --> 00:42:35,000
integral from a to b.
I guess that was a minus sign,
479
00:42:35,000 --> 00:42:45,000
of M of (x1f2 of x) minus M of
(x1f1 of x) dx.
480
00:42:45,000 --> 00:42:50,000
And so, that's the same as up
there.
481
00:42:50,000 --> 00:42:54,000
And so, that's the end of the
proof because we've checked that
482
00:42:54,000 --> 00:42:58,000
for this special case,
when we have only an x
483
00:42:58,000 --> 00:43:01,000
component and a vertically
simple region,
484
00:43:01,000 --> 00:43:04,000
things work.
Then, we can remove the
485
00:43:04,000 --> 00:43:07,000
assumption that things are
vertically simple using this
486
00:43:07,000 --> 00:43:10,000
second observation.
We can just glue the various
487
00:43:10,000 --> 00:43:13,000
pieces together,
and prove it for any region.
488
00:43:13,000 --> 00:43:17,000
Then, we do same thing with the
y component.
489
00:43:17,000 --> 00:43:22,000
That's the first observation.
When we add things together,
490
00:43:22,000 --> 00:43:29,000
we get Green's theorem in its
full generality.
491
00:43:29,000 --> 00:43:39,000
OK, so let me finish with a
cool example.
492
00:43:39,000 --> 00:43:47,000
So, there's one place in real
life where Green's theorem used
493
00:43:47,000 --> 00:43:51,000
to be extremely useful.
I say used to because computers
494
00:43:51,000 --> 00:43:53,000
have actually made that
obsolete.
495
00:43:53,000 --> 00:44:02,000
But, so let me show you a
picture of this device.
496
00:44:02,000 --> 00:44:12,000
This is called a planimeter.
And what it does is it measures
497
00:44:12,000 --> 00:44:17,000
areas.
So, it used to be that when you
498
00:44:17,000 --> 00:44:23,000
were an experimental scientist,
you would run your chemical or
499
00:44:23,000 --> 00:44:27,000
biological experiment or
whatever.
500
00:44:27,000 --> 00:44:29,000
And, you would have all of
these recording devices.
501
00:44:29,000 --> 00:44:32,000
And, the data would go,
well, not onto a floppy disk or
502
00:44:32,000 --> 00:44:35,000
hard disk or whatever because
you didn't have those at the
503
00:44:35,000 --> 00:44:37,000
time.
You didn't have a computer in
504
00:44:37,000 --> 00:44:39,000
your lab.
They would go onto a piece of
505
00:44:39,000 --> 00:44:42,000
graph paper.
So, you would have your graph
506
00:44:42,000 --> 00:44:46,000
paper, and you would have some
curve on it.
507
00:44:46,000 --> 00:44:48,000
And, very often,
you wanted to know,
508
00:44:48,000 --> 00:44:51,000
what's the total amount of
product that you have
509
00:44:51,000 --> 00:44:54,000
synthesized, or whatever the
question might be.
510
00:44:54,000 --> 00:44:58,000
It might relate with the area
under your curve.
511
00:44:58,000 --> 00:45:01,000
So, you'd say, oh, it's easy.
Let's just integrate,
512
00:45:01,000 --> 00:45:02,000
except you don't have a
function.
513
00:45:02,000 --> 00:45:05,000
You can put that into
calculator.
514
00:45:05,000 --> 00:45:07,000
The next thing you could do is,
well, let's count the little
515
00:45:07,000 --> 00:45:09,000
squares.
But, if you've seen a piece of
516
00:45:09,000 --> 00:45:12,000
graph paper, that's kind of
time-consuming.
517
00:45:12,000 --> 00:45:14,000
So, people invented these
things called planimeters.
518
00:45:14,000 --> 00:45:19,000
It's something where there is a
really heavy thing based at one
519
00:45:19,000 --> 00:45:23,000
corner, and there's a lot of
dials and gauges and everything.
520
00:45:23,000 --> 00:45:25,000
And, there's one arm that you
move.
521
00:45:25,000 --> 00:45:30,000
And so, what you do is you take
the moving arm and you just
522
00:45:30,000 --> 00:45:35,000
slide it all around your curve.
And, you look at one of the
523
00:45:35,000 --> 00:45:37,000
dials.
And, suddenly what comes,
524
00:45:37,000 --> 00:45:41,000
as you go around,
it gives you complete garbage.
525
00:45:41,000 --> 00:45:45,000
But when you come back here,
that dial suddenly gives you
526
00:45:45,000 --> 00:45:48,000
the value of the area of this
region.
527
00:45:48,000 --> 00:45:51,000
So, how does it work?
This gadget never knows about
528
00:45:51,000 --> 00:45:55,000
the region inside because you
don't take it all over here.
529
00:45:55,000 --> 00:45:57,000
You only take it along the
curve.
530
00:45:57,000 --> 00:46:00,000
So, what it does actually is it
computes a line integral.
531
00:46:00,000 --> 00:46:04,000
OK, so it has this system of
wheels and everything that
532
00:46:04,000 --> 00:46:08,000
compute for you the line
integral along C of,
533
00:46:08,000 --> 00:46:11,000
well, it depends on the model.
But some of them compute the
534
00:46:11,000 --> 00:46:14,000
line integral of x dy.
Some of them compute different
535
00:46:14,000 --> 00:46:17,000
line integrals.
But, they compute some line
536
00:46:17,000 --> 00:46:21,000
integral, OK?
And, now, if you apply Green's
537
00:46:21,000 --> 00:46:26,000
theorem, you see that when you
have a counterclockwise curve,
538
00:46:26,000 --> 00:46:31,000
this will be just the area of
the region inside.
539
00:46:31,000 --> 00:46:34,000
And so, that's how it works.
I mean, of course,
540
00:46:34,000 --> 00:46:36,000
now you use a computer and it
does the sums.
541
00:46:36,000 --> 00:46:39,000
Yes?
That costs several thousand
542
00:46:39,000 --> 00:46:43,000
dollars, possibly more.
So, that's why I didn't bring
543
00:46:43,000 --> 00:46:44,000
one.