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Recall that yesterday we saw,
no, two days ago we learned
8
00:00:32 --> 00:00:37
about the curl of a vector field
in space.
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00:00:37 --> 00:00:45
And we said the curl of F is
defined by taking a cross
10
00:00:45 --> 00:00:52
product between the symbol dell
and the vector F.
11
00:00:52 --> 00:00:58
Concretely, the way we would
compute this would be by putting
12
00:00:58 --> 00:01:04
the components of F into this
determinant and expanding and
13
00:01:04 --> 00:01:09
then getting a vector with
components Ry minus Qz,
14
00:01:09 --> 00:01:21
Pz minus Rx and Qx minus Py.
I think I also tried to explain
15
00:01:21 --> 00:01:25
very quickly what the
significance of a curl is.
16
00:01:25 --> 00:01:28
Just to tell you again very
quickly,
17
00:01:28 --> 00:01:34
basically curl measures,
if you mention that your vector
18
00:01:34 --> 00:01:40
field measures the velocity in
some fluid then the curl
19
00:01:40 --> 00:01:47
measures how much rotation is
taking place in that fluid.
20
00:01:47 --> 00:02:05
Measures the rotation part of a
velocity field.
21
00:02:05 --> 00:02:13
More precisely the direction
corresponds to the axis of
22
00:02:13 --> 00:02:22
rotation and the magnitude
corresponds to twice the angular
23
00:02:22 --> 00:02:24
velocity.
24
00:02:24 --> 00:02:47
25
00:02:47 --> 00:02:50
Just to give you a few quick
examples.
26
00:02:50 --> 00:02:53
If I take a constant vector
field,
27
00:02:53 --> 00:03:01
so everything translates at the
same speed,
28
00:03:01 --> 00:03:06
then obviously when you take
the partial derivatives you will
29
00:03:06 --> 00:03:10
just get a bunch of zeros so the
curl will be zero.
30
00:03:10 --> 00:03:15
If you take a vector field that
stretches things,
31
00:03:15 --> 00:03:17
let's say, for example,
we are going to stretch things
32
00:03:17 --> 00:03:23
along the x-axis,
that would be a vector field
33
00:03:23 --> 00:03:30
that goes parallel to the x
direction but maybe,
34
00:03:30 --> 00:03:33
say, x times i.
So that when you are in front
35
00:03:33 --> 00:03:35
of a plane of a blackboard you
are moving forward,
36
00:03:35 --> 00:03:36
when you are behind you are
moving backwards,
37
00:03:36 --> 00:03:40
things are getting expanded in
the x direction.
38
00:03:40 --> 00:03:48
If you compute the curl,
you can check each of these.
39
00:03:48 --> 00:03:49
Again, they are going to be
zero.
40
00:03:49 --> 00:03:53
There is no curl.
This is not what curl measures.
41
00:03:53 --> 00:03:58
I mean, actually,
what measures expansion,
42
00:03:58 --> 00:04:03
stretching is actually
divergence.
43
00:04:03 --> 00:04:05
If you take the divergence of
this field,
44
00:04:05 --> 00:04:07
you would get one plus zero
plus zero,
45
00:04:07 --> 00:04:10
it looks like it will be one,
so in case you don't remember,
46
00:04:10 --> 00:04:15
I mean divergence precisely
measures this stretching effect
47
00:04:15 --> 00:04:18
in your field.
And, on the other hand,
48
00:04:18 --> 00:04:22
if you take something that
corresponds to,
49
00:04:22 --> 00:04:26
say,
rotation about the z-axis at
50
00:04:26 --> 00:04:34
unit angular velocity -- That
means they are going to moving
51
00:04:34 --> 00:04:41
in circles around the z-axis.
One way to write down this
52
00:04:41 --> 00:04:46
field, let's see,
the z component is zero because
53
00:04:46 --> 00:04:50
everything is moving
horizontally.
54
00:04:50 --> 00:04:54
And in the x and y directions,
if you look at it from above,
55
00:04:54 --> 00:04:59
well, it is just going to be
our good old friend the vector
56
00:04:59 --> 00:05:02
field that rotates everything
[at unit speed?].
57
00:05:02 --> 00:05:05
And we have seen the formula
for this one many times.
58
00:05:05 --> 00:05:09
The first component is minus y,
the second one is x.
59
00:05:09 --> 00:05:17
Now, if you compute the curl of
this guy, you will get zero,
60
00:05:17 --> 00:05:21
zero, two, two k.
And so k is the axis of
61
00:05:21 --> 00:05:24
rotation, two is twice the
angular velocity.
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00:05:24 --> 00:05:26
And now, of course,
you can imagine much more
63
00:05:26 --> 00:05:29
complicated motions where you
will have -- For example,
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00:05:29 --> 00:05:32
if you look at the Charles
River very carefully then you
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00:05:32 --> 00:05:34
will see that water is flowing,
generally speaking,
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00:05:34 --> 00:05:38
towards the ocean.
But, at the same time,
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00:05:38 --> 00:05:43
there are actually a few eddies
in there and with water
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00:05:43 --> 00:05:47
swirling.
Those are the places where
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00:05:47 --> 00:05:51
there is actually curl in the
flow.
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00:05:51 --> 00:05:56
Yes.
I don't know how to turn out
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00:05:56 --> 00:06:00
the lights a bit,
but I'm sure there is a way.
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00:06:00 --> 00:06:10
Does this do it?
Is it working?
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00:06:10 --> 00:06:23
OK.
You're welcome.
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00:06:23 --> 00:06:31
Hopefully it is easier to see
now.
75
00:06:31 --> 00:06:35
That was about curl.
Now, why do we care about curl
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00:06:35 --> 00:06:40
besides this motivation of
understanding motions?
77
00:06:40 --> 00:06:43
One place where it comes up is
when we try to understand
78
00:06:43 --> 00:06:45
whether a vector field is
conservative.
79
00:06:45 --> 00:06:49
Remember we have seen that a
vector field is conservative if
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00:06:49 --> 00:06:53
and only if its curl is zero.
That is the situation in which
81
00:06:53 --> 00:06:56
we are allowed to try to look
for a potential function and
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00:06:56 --> 00:06:57
then use the fundamental
theorem.
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00:06:57 --> 00:07:00
But another place where this
comes up,
84
00:07:00 --> 00:07:02
if you remember what we did in
the plane,
85
00:07:02 --> 00:07:06
curl also came up when we tried
to convert nine integrals into
86
00:07:06 --> 00:07:10
double integrals.
That was Greene's theorem.
87
00:07:10 --> 00:07:19
Well, it turns out we can do
the same thing in space and that
88
00:07:19 --> 00:07:28
is called Stokes' theorem.
What does Stokes' theorem say?
89
00:07:28 --> 00:07:36
It says that the work done by a
vector field along a closed
90
00:07:36 --> 00:07:44
curve can be replaced by a
double integral of curl F.
91
00:07:44 --> 00:07:47
Let me write it using the dell
notation.
92
00:07:47 --> 00:07:54
That is curl F.
Dot ndS on a suitably chosen
93
00:07:54 --> 00:07:58
surface.
That is a very strange kind of
94
00:07:58 --> 00:08:01
statement.
But actually it is not much
95
00:08:01 --> 00:08:04
more strange than things we have
seen before.
96
00:08:04 --> 00:08:09
I should clarify what this
means.
97
00:08:09 --> 00:08:17
C has to be a closed curve in
space.
98
00:08:17 --> 00:08:32
And S can be any surface
bounded by C.
99
00:08:32 --> 00:08:36
For example,
what Stokes' theorem tells me
100
00:08:36 --> 00:08:41
is that let us say that I have
to compute some line integral on
101
00:08:41 --> 00:08:48
maybe,
say, the unit circle in the x,
102
00:08:48 --> 00:08:52
y plane.
Of course I can set a line
103
00:08:52 --> 00:08:57
integral directly and compute it
by setting x equals cosine T,
104
00:08:57 --> 00:09:00
y equals sine T,
z equals zero.
105
00:09:00 --> 00:09:03
But maybe sometimes I don't
want to do that because my
106
00:09:03 --> 00:09:06
vector field is really
complicated.
107
00:09:06 --> 00:09:11
And instead I will want to
reduce things to a surface
108
00:09:11 --> 00:09:13
integral.
Now, I know that you guys are
109
00:09:13 --> 00:09:16
not necessarily fond of
computing flux of vector fields
110
00:09:16 --> 00:09:19
for surfaces so maybe you don't
really see the point.
111
00:09:19 --> 00:09:22
But sometimes it is useful.
Sometimes it is also useful
112
00:09:22 --> 00:09:25
backwards because,
actually, you have a surface
113
00:09:25 --> 00:09:29
integral that you would like to
turn into a line integral.
114
00:09:29 --> 00:09:34
What Stokes' theorem says is
that I can choose my favorite
115
00:09:34 --> 00:09:38
surface whose boundary is this
circle.
116
00:09:38 --> 00:09:42
I could choose,
for example,
117
00:09:42 --> 00:09:50
a half sphere if I want or I
can choose, let's call that s1,
118
00:09:50 --> 00:09:54
I don't know,
a pointy thing,
119
00:09:54 --> 00:09:57
s2.
Probably the most logical one,
120
00:09:57 --> 00:10:00
actually, would be just to
choose a disk in the x,
121
00:10:00 --> 00:10:02
y plane.
That would probably be the
122
00:10:02 --> 00:10:04
easiest one to set up for
calculating flux.
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00:10:04 --> 00:10:07
Anyway,
what Stokes' theorem tells me
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00:10:07 --> 00:10:09
is I can choose any of these
surfaces,
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00:10:09 --> 00:10:14
whichever one I want,
and I can compute the flux of
126
00:10:14 --> 00:10:18
curl F through this surface.
Curl F is a new vector field
127
00:10:18 --> 00:10:22
when you have this formula that
gives you a vector field you
128
00:10:22 --> 00:10:25
compute its flux through your
favorite surface,
129
00:10:25 --> 00:10:31
and you should get the same
thing as if you had done the
130
00:10:31 --> 00:10:37
line integral for F.
That is the statement.
131
00:10:37 --> 00:10:43
Now, there is a catch here.
What is the catch?
132
00:10:43 --> 00:10:47
Well, the catch is we have to
figure out what conventions to
133
00:10:47 --> 00:10:51
use because remember when we
have a surface there are two
134
00:10:51 --> 00:10:54
possible orientations.
We have to decide which way we
135
00:10:54 --> 00:10:58
will counter flux positively,
which way we will counter flux
136
00:10:58 --> 00:11:01
negatively.
And, if we change our choice,
137
00:11:01 --> 00:11:05
then of course the flux will
become the opposite.
138
00:11:05 --> 00:11:08
Well, similarly to define the
work, I need to choose which way
139
00:11:08 --> 00:11:12
I am going to run my curve.
If I change which way I go
140
00:11:12 --> 00:11:16
around the curve then my work
will become the opposite.
141
00:11:16 --> 00:11:21
What happens is I have to
orient the curve C and the
142
00:11:21 --> 00:11:28
surface S in compatible ways.
We have to figure out what the
143
00:11:28 --> 00:11:36
rule is for how the orientation
of S and that of C relate to
144
00:11:36 --> 00:11:41
each other.
What about orientation?
145
00:11:41 --> 00:11:55
Well, we need the orientations
of C and S to be compatible and
146
00:11:55 --> 00:12:05
they have to explain to you what
the rule is.
147
00:12:05 --> 00:12:15
Let me show you a picture.
The rule is if I walk along C
148
00:12:15 --> 00:12:23
with S to my left then the
normal vector is pointing up for
149
00:12:23 --> 00:12:29
me.
Let me write that.
150
00:12:29 --> 00:12:37
If I walk along C,
I should say in the positive
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00:12:37 --> 00:12:48
direction, in the direction that
I have chosen to orient C.
152
00:12:48 --> 00:13:06
With S to my left then n is
pointing up for me.
153
00:13:06 --> 00:13:10
Here is the example.
If I am walking on this curve,
154
00:13:10 --> 00:13:12
it looks like the surface is to
my left.
155
00:13:12 --> 00:13:19
And so the normal vector is
going towards what is up for me.
156
00:13:19 --> 00:13:26
Any questions about that?
I see some people using their
157
00:13:26 --> 00:13:28
right hands.
That is also right-handable
158
00:13:28 --> 00:13:31
which I am going to say in just
a few moments.
159
00:13:31 --> 00:13:32
That is another way to remember
this.
160
00:13:32 --> 00:13:38
Before I tell you about the
right-handable version,
161
00:13:38 --> 00:13:43
let me just try something.
Actually, I am not happy with
162
00:13:43 --> 00:13:47
this orientation of C and I want
to orient my curve C going
163
00:13:47 --> 00:13:51
clockwise on the picture.
So the other orientation.
164
00:13:51 --> 00:13:55
Then, if I walk on it this way,
the surface would be to my
165
00:13:55 --> 00:13:56
right.
You can remember,
166
00:13:56 --> 00:13:59
if it helps you,
that if a surface is to your
167
00:13:59 --> 00:14:01
right then the normal vector
will go down.
168
00:14:01 --> 00:14:04
The other way to think about
this rule is enough because if
169
00:14:04 --> 00:14:07
you are walking clockwise,
well, you can change that to
170
00:14:07 --> 00:14:10
counterclockwise just by walking
upside down.
171
00:14:10 --> 00:14:14
This guy is walking clockwise
on C.
172
00:14:14 --> 00:14:21
And while for him,
if you look carefully at the
173
00:14:21 --> 00:14:31
picture, the surface is actually
to his left when you flip upside
174
00:14:31 --> 00:14:34
down.
Yeah, it is kind of confusing.
175
00:14:34 --> 00:14:38
But, anyway,
maybe it's easier if you
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00:14:38 --> 00:14:44
actually rotate in the picture.
And now it is getting actually
177
00:14:44 --> 00:14:50
really confusing because his
walking upside up with,
178
00:14:50 --> 00:14:54
actually, the surface is to his
left.
179
00:14:54 --> 00:14:58
I mean where he is at here is
actually at the front and this
180
00:14:58 --> 00:15:01
is the back, but that is kind of
hard to see.
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00:15:01 --> 00:15:05
Anyway, whichever method will
work best for you.
182
00:15:05 --> 00:15:07
Perhaps it is easiest to first
do it with the other
183
00:15:07 --> 00:15:09
orientation,
this one,
184
00:15:09 --> 00:15:13
and this side,
if you want the opposite one,
185
00:15:13 --> 00:15:21
then you will just flip
everything.
186
00:15:21 --> 00:15:25
Now, what is the other way of
remembering this with the
187
00:15:25 --> 00:15:27
right-hand rule?
First of all,
188
00:15:27 --> 00:15:29
take your right hand,
not your left.
189
00:15:29 --> 00:15:32
Even if your right hand is
actually using a pen or
190
00:15:32 --> 00:15:34
something like that in your
right hand do this.
191
00:15:34 --> 00:15:37
And let's take your fingers in
order.
192
00:15:37 --> 00:15:40
First your thumb.
Let's make your thumb go along
193
00:15:40 --> 00:15:42
the object that has only one
dimension in there.
194
00:15:42 --> 00:15:47
That is the curve.
Well, let's look at the top
195
00:15:47 --> 00:15:52
picture up there.
I want my thumb to go along the
196
00:15:52 --> 00:15:56
curve so that is kind of towards
the right.
197
00:15:56 --> 00:16:06
Then I want to make my index
finger point towards the
198
00:16:06 --> 00:16:09
surface.
Towards the surface I mean
199
00:16:09 --> 00:16:12
towards the interior of the
surface from the curve.
200
00:16:12 --> 00:16:15
And when I am on the curve I am
on the boundary of the surface,
201
00:16:15 --> 00:16:18
so there is a direction along
the surface that is the curve
202
00:16:18 --> 00:16:20
and the other one is pointing
into the surface.
203
00:16:20 --> 00:16:24
That one would be pointing kind
of to the back slightly up
204
00:16:24 --> 00:16:27
maybe, so like that.
And now your middle finger is
205
00:16:27 --> 00:16:30
going to point in the direction
of the normal vector.
206
00:16:30 --> 00:16:37
That is up, at least if you
have the same kind of right hand
207
00:16:37 --> 00:16:49
as I do.
The other way of doing it is
208
00:16:49 --> 00:17:08
using the right-hand rule along
C positively.
209
00:17:08 --> 00:17:18
The index finger towards the
interior of S.
210
00:17:18 --> 00:17:27
Sorry, I shouldn't say interior.
I should say tangent to S
211
00:17:27 --> 00:17:34
towards the interior of S.
What I mean by that is really
212
00:17:34 --> 00:17:39
the part of S that is not its
boundary, so the rest of the
213
00:17:39 --> 00:17:49
surface.
Then the middle finger points
214
00:17:49 --> 00:18:00
parallel to n.
Let's practice.
215
00:18:00 --> 00:18:09
Let's say that I gave you this
curve bounding this surface.
216
00:18:09 --> 00:18:13
Which way do you think the
normal vector will be going?
217
00:18:13 --> 00:18:16
Up. Yes. Everyone is voting up.
Imaging that I am walking
218
00:18:16 --> 00:18:18
around C.
That is to my left.
219
00:18:18 --> 00:18:24
Normal vector points up.
Imagine that you put your thumb
220
00:18:24 --> 00:18:32
along C, your index towards S
and then your middle finger
221
00:18:32 --> 00:18:36
points up.
Very good.
222
00:18:36 --> 00:18:43
N points up.
Another one.
223
00:18:43 --> 00:19:05
224
00:19:05 --> 00:19:07
It is interesting to watch you
guys.
225
00:19:07 --> 00:19:13
I think mostly it is going up.
The correct answer is it goes
226
00:19:13 --> 00:19:20
up and into the cone.
How do we see that?
227
00:19:20 --> 00:19:24
Well, one way to think about it
is imagine that you are walking
228
00:19:24 --> 00:19:27
on C, on the rim of this cone.
You have two options.
229
00:19:27 --> 00:19:30
Imagine that you are walking
kind of inside or imagine that
230
00:19:30 --> 00:19:33
you are walking kind of outside.
If you are walking outside then
231
00:19:33 --> 00:19:35
S is to your right,
but it does not sound good.
232
00:19:35 --> 00:19:39
Let's say instead that you are
walking on the inside of a cone
233
00:19:39 --> 00:19:43
following the boundary.
Well, then the surface is to
234
00:19:43 --> 00:19:45
your left.
And so the normal vector will
235
00:19:45 --> 00:19:50
be up for you which means it
will be pointing slightly up and
236
00:19:50 --> 00:19:52
into the cone.
Another way to think about it,
237
00:19:52 --> 00:19:56
through the right-hand rule,
from this way index going kind
238
00:19:56 --> 00:20:00
of down because the surface goes
down and a bit to the back.
239
00:20:00 --> 00:20:04
And then the normal vector
points up and in.
240
00:20:04 --> 00:20:08
Yet another way,
if you deform continuously your
241
00:20:08 --> 00:20:12
surface then the conventions
will not change.
242
00:20:12 --> 00:20:15
See, this is kind of
[UNINTELLIGIBLE]
243
00:20:15 --> 00:20:17
in a way.
You can deform things and
244
00:20:17 --> 00:20:21
nothing will change.
So what if we somehow flatten
245
00:20:21 --> 00:20:27
our cone, push it a bit up so
that it becomes completely flat?
246
00:20:27 --> 00:20:30
Then, if you had a flat disk
with the curve going
247
00:20:30 --> 00:20:33
counterclockwise,
the normal vector would go up.
248
00:20:33 --> 00:20:36
Now take your disk with its
normal vector sticking up.
249
00:20:36 --> 00:20:39
If you want to paint the face a
different color so that you can
250
00:20:39 --> 00:20:43
remember that was beside with a
normal vector and then push it
251
00:20:43 --> 00:20:45
back down to the cone,
you will see that the painted
252
00:20:45 --> 00:20:48
face,
the one with the normal vector
253
00:20:48 --> 00:20:52
on that side is the one that is
inside and up.
254
00:20:52 --> 00:20:59
Does that make sense?
Anyway, I think you have just
255
00:20:59 --> 00:21:03
to play with these examples for
long enough and get it.
256
00:21:03 --> 00:21:07
OK. The last one.
Let's say that I have a
257
00:21:07 --> 00:21:10
cylinder.
So now this guy has actually
258
00:21:10 --> 00:21:12
two boundary curves,
C and C prime.
259
00:21:12 --> 00:21:16
And let's say I want to orient
my cylinder so that the normal
260
00:21:16 --> 00:21:20
vector sticks out.
How should I choose the
261
00:21:20 --> 00:21:30
orientation of my curves?
Let's start with,
262
00:21:30 --> 00:21:40
say, the bottom one.
Would the bottom one be going
263
00:21:40 --> 00:21:44
clockwise or counterclockwise.
Most people seem to say
264
00:21:44 --> 00:21:47
counterclockwise,
and I agree with that.
265
00:21:47 --> 00:21:51
Let me write that down and
claim C prime should go
266
00:21:51 --> 00:21:55
counterclockwise.
One way to think about it,
267
00:21:55 --> 00:21:58
actually, it's quite easy,
you mentioned that you're
268
00:21:58 --> 00:22:02
walking on the outside of the
cylinder along C prime.
269
00:22:02 --> 00:22:06
If you want to walk along C
prime so that the cylinder is to
270
00:22:06 --> 00:22:10
your left, that means you have
to actually go counterclockwise
271
00:22:10 --> 00:22:14
around it.
The other way is use your right
272
00:22:14 --> 00:22:17
hand.
Say when you're at the front of
273
00:22:17 --> 00:22:19
C prime,
your thumb points to the right,
274
00:22:19 --> 00:22:23
your index points up because
that's where the surface is,
275
00:22:23 --> 00:22:28
and then your middle finger
will point out.
276
00:22:28 --> 00:22:39
What about C?
Well, C I claim we should be
277
00:22:39 --> 00:22:43
doing clockwise.
I mean think about just walking
278
00:22:43 --> 00:22:46
again on the surface of the
cylinder along C.
279
00:22:46 --> 00:22:52
If you walk clockwise,
you will see that the surface
280
00:22:52 --> 00:22:57
is to your left or use the
right-hand rule.
281
00:22:57 --> 00:23:00
Now, if a problem gives you
neither the orientation of a
282
00:23:00 --> 00:23:04
curve nor that of the surface
then it's up to you to make them
283
00:23:04 --> 00:23:05
up.
But you have to make them up in
284
00:23:05 --> 00:23:09
a consistent way.
You cannot choose them both at
285
00:23:09 --> 00:23:13
random.
All right.
286
00:23:13 --> 00:23:30
Now we're all set to try to use
Stokes' theorem.
287
00:23:30 --> 00:23:35
Well, let me do an example
first.
288
00:23:35 --> 00:23:43
The first example that I will
do is actually a comparison.
289
00:23:43 --> 00:23:52
Stokes' versus Green.
I want to show you how Green's
290
00:23:52 --> 00:23:55
theorem for work that we saw in
the plane,
291
00:23:55 --> 00:23:58
but also involved work and curl
and so on,
292
00:23:58 --> 00:24:04
is actually a special case of
this.
293
00:24:04 --> 00:24:11
Let's say that we will look at
the special case where our curve
294
00:24:11 --> 00:24:16
C is actually a curve in the x,
y plane.
295
00:24:16 --> 00:24:19
And let's make it go
counterclockwise in the x,
296
00:24:19 --> 00:24:23
y plane because that's what we
did for Green's theorem.
297
00:24:23 --> 00:24:25
Now let's choose a surface
bounded by this curve.
298
00:24:25 --> 00:24:28
Well, as I said,
I could make up any surface
299
00:24:28 --> 00:24:32
that comes to my mind.
But, if I want to relate to
300
00:24:32 --> 00:24:35
this stuff, I should probably
stay in the x,
301
00:24:35 --> 00:24:38
y plane.
So I am just going to take my
302
00:24:38 --> 00:24:43
surface to be the piece of the
x, y plane that is inside my
303
00:24:43 --> 00:24:52
curve.
So let's say S is going to be a
304
00:24:52 --> 00:25:02
portion of x,
y plane bounded by a curve C,
305
00:25:02 --> 00:25:11
and the curve C goes
counterclockwise.
306
00:25:11 --> 00:25:17
Well, then I should look at
[the table?].
307
00:25:17 --> 00:25:24
For work along C of my favorite
vector field F dot dr.
308
00:25:24 --> 00:25:31
So that will be the line
integral of Pdx plus Qdy.
309
00:25:31 --> 00:25:34
Like I said,
if I call the components of my
310
00:25:34 --> 00:25:38
field P, Q and R,
it will be Pdx plus Qdy plus
311
00:25:38 --> 00:25:42
Rdz, but I don't have any Z
here.
312
00:25:42 --> 00:25:49
Dz is zero on C.
If I evaluate for line
313
00:25:49 --> 00:25:53
integral, I don't have any term
involving dz.
314
00:25:53 --> 00:25:59
Z is zero.
Now, let's see what Strokes
315
00:25:59 --> 00:26:05
says.
Stokes says instead I can
316
00:26:05 --> 00:26:13
compute for flux through S of
curve F.
317
00:26:13 --> 00:26:17
But now what's the normal
vector to my surface?
318
00:26:17 --> 00:26:19
Well, it's going to be either k
or negative k.
319
00:26:19 --> 00:26:22
I just have to figure out which
one it is.
320
00:26:22 --> 00:26:25
Well, if you followed what
we've done there,
321
00:26:25 --> 00:26:30
you know that the normal vector
compatible with this choice for
322
00:26:30 --> 00:26:32
the curve C is the one that
points up.
323
00:26:32 --> 00:26:43
My normal vector is just going
to be k hat, so I am going to
324
00:26:43 --> 00:26:49
replace my normal vector by k
hat.
325
00:26:49 --> 00:26:53
That means, actually,
I will be integrating curl dot
326
00:26:53 --> 00:26:56
k.
That means I am integrating the
327
00:26:56 --> 00:27:04
z component of curl.
Let's look at curl F dot k.
328
00:27:04 --> 00:27:14
That's the z component of curl
F.
329
00:27:14 --> 00:27:18
And what's the z component of
curl?
330
00:27:18 --> 00:27:21
Well, I conveniently still have
the values up there.
331
00:27:21 --> 00:27:36
It's Q sub x minus P sub y.
My double integral becomes
332
00:27:36 --> 00:27:42
double integral of Q sub x minus
P sub y.
333
00:27:42 --> 00:27:46
What about dS?
Well, I am in a piece of the x,
334
00:27:46 --> 00:27:52
y plane, so dS is just dxdy or
your favorite combination that
335
00:27:52 --> 00:27:56
does the same thing.
Now, see, if you look at this
336
00:27:56 --> 00:28:02
equality, integral of Pdx plus
Qdy along a closed curve equals
337
00:28:02 --> 00:28:05
double integral of Qx minus Py
dxdy.
338
00:28:05 --> 00:28:11
That is exactly the statement
of Green's theorem.
339
00:28:11 --> 00:28:17
I mean except at that time we
called things m and n,
340
00:28:17 --> 00:28:20
but really that shouldn't
matter.
341
00:28:20 --> 00:28:35
This tells you that,
in fact, Green's theorem is
342
00:28:35 --> 00:28:51
just a special case of Stokes'
in the x, y plane.
343
00:28:51 --> 00:28:55
Now, another small remark I
want to make right away before I
344
00:28:55 --> 00:28:57
forget,
you might think that these
345
00:28:57 --> 00:29:01
rules that we've made up about
compatibility of orientations
346
00:29:01 --> 00:29:05
are completely arbitrary.
Well, they are literally in the
347
00:29:05 --> 00:29:10
same way as our convention for
which we guy curl is arbitrary.
348
00:29:10 --> 00:29:14
We chose to make the curl be
this thing and not the opposite
349
00:29:14 --> 00:29:17
which would have been pretty
much just as sensible.
350
00:29:17 --> 00:29:21
And, ultimately,
that comes from our choice of
351
00:29:21 --> 00:29:25
making the cross-product be what
it is but of the opposite.
352
00:29:25 --> 00:29:30
Ultimately, it all comes from
our preference for right-handed
353
00:29:30 --> 00:29:33
coordinate systems.
If we had been on the planet
354
00:29:33 --> 00:29:36
with left-handed coordinate
systems then actually our
355
00:29:36 --> 00:29:40
conventions would be all the
other way around,
356
00:29:40 --> 00:30:05
but they are this way.
Any other questions?
357
00:30:05 --> 00:30:08
A surface that you use in
Stokes' theorem is usually not
358
00:30:08 --> 00:30:12
going to be closed because its
boundary needs to be the curve
359
00:30:12 --> 00:30:14
C.
So if you had a closed surface
360
00:30:14 --> 00:30:17
you wouldn't know where to put
your curve.
361
00:30:17 --> 00:30:20
I mean of course you could make
a tiny hole in it and get a tiny
362
00:30:20 --> 00:30:22
curve.
Actually, what that would say,
363
00:30:22 --> 00:30:26
and we are going to see more
about that so not very important
364
00:30:26 --> 00:30:28
right now,
but what we would see is that
365
00:30:28 --> 00:30:31
for a close surface we would end
up getting zero for the flux.
366
00:30:31 --> 00:30:34
And that is actually because
divergence of curl is zero,
367
00:30:34 --> 00:30:36
but I am getting ahead of
myself.
368
00:30:36 --> 00:30:45
We are going to see that
probably tomorrow in more
369
00:30:45 --> 00:30:49
detail.
Stokes' theorem only works if
370
00:30:49 --> 00:30:53
you can make sense of this.
That means you need your vector
371
00:30:53 --> 00:30:58
field to be continuous and
differentiable everywhere on the
372
00:30:58 --> 00:31:03
surface S.
Now, why is that relevant?
373
00:31:03 --> 00:31:05
Well, say that your vector
field was not defined at the
374
00:31:05 --> 00:31:07
origin and say that you wanted
to do,
375
00:31:07 --> 00:31:11
you know, the example that I
had first with the unit circling
376
00:31:11 --> 00:31:14
the x, y plane.
Normally, the most sensible
377
00:31:14 --> 00:31:17
choice of surface to apply
Stokes' theorem to would be just
378
00:31:17 --> 00:31:19
the flat disk in the x,
y plane.
379
00:31:19 --> 00:31:23
But that assumes that your
vector field is well-defined
380
00:31:23 --> 00:31:24
there.
If your vector field is not
381
00:31:24 --> 00:31:27
defined at the origin but
defined everywhere else you
382
00:31:27 --> 00:31:29
cannot use this guy,
but maybe you can still use,
383
00:31:29 --> 00:31:31
say, the half-sphere,
for example.
384
00:31:31 --> 00:31:35
Or, you could use a piece of
cylinder plus a flat top or
385
00:31:35 --> 00:31:38
whatever you want but not
pressing for the origin.
386
00:31:38 --> 00:31:41
So you could still use Stokes
but you'd have to be careful
387
00:31:41 --> 00:31:44
about which surface you choose.
Now, if instead your vector
388
00:31:44 --> 00:31:49
field is not defined anywhere on
the z-axis then you're out of
389
00:31:49 --> 00:31:54
luck because there is no way to
find a surface bounded by this
390
00:31:54 --> 00:31:59
unit circle without crossing the
z-axis somewhere.
391
00:31:59 --> 00:32:07
Then you wouldn't be able to
Stokes' theorem at all or at
392
00:32:07 --> 00:32:16
least not directly.
Maybe I should write it F
393
00:32:16 --> 00:32:25
defines a differentiable
everywhere on this.
394
00:32:25 --> 00:32:27
But we don't care about what
happens outside of this.
395
00:32:27 --> 00:32:35
It's really only on the surface
that we need it to be OK.
396
00:32:35 --> 00:32:39
I mean, again,
99% of the vector fields that
397
00:32:39 --> 00:32:44
we see in this class are defined
everywhere so that's not an
398
00:32:44 --> 00:32:47
urgent concern,
but still.
399
00:32:47 --> 00:32:49
OK.
Should we move on?
400
00:32:49 --> 00:33:01
Yes. I have a yes.
Let me explain to you quickly
401
00:33:01 --> 00:33:08
why Stokes is true.
How do we prove a theorem like
402
00:33:08 --> 00:33:10
that?
Well,
403
00:33:10 --> 00:33:12
the strategy,
I mean there are other ways,
404
00:33:12 --> 00:33:16
but the least painful strategy
at this point is to observe what
405
00:33:16 --> 00:33:19
we already know is a special
case of Stokes's theorem.
406
00:33:19 --> 00:33:22
Namely we know the case where
the curve is actually in the x,
407
00:33:22 --> 00:33:24
y plane and the surface is a
flat piece of the x,
408
00:33:24 --> 00:33:34
y plane because that's Green's
theorem which we proved a while
409
00:33:34 --> 00:33:42
ago.
We know it for C and S in the
410
00:33:42 --> 00:33:47
x, y plane.
Now, what if C and S were,
411
00:33:47 --> 00:33:49
say, in the y,
z plane instead of the x,
412
00:33:49 --> 00:33:51
y plane?
Well, then it will not quite
413
00:33:51 --> 00:33:55
give the same picture because
the normal vector would be i hat
414
00:33:55 --> 00:33:58
instead of k hat and they would
be having different notations
415
00:33:58 --> 00:34:01
and it would be integrating with
y and z.
416
00:34:01 --> 00:34:02
But you see that it would
become, again,
417
00:34:02 --> 00:34:05
exactly the same formula.
We'd know it for any of the
418
00:34:05 --> 00:34:08
coordinate planes.
In fact, I claim we know it for
419
00:34:08 --> 00:34:13
absolutely any plane.
And the reason for that is,
420
00:34:13 --> 00:34:15
sure, when we write it in
coordinates,
421
00:34:15 --> 00:34:19
when we write that this line
integral is integral of Pdx plus
422
00:34:19 --> 00:34:24
Qdy plus Rdz or when we write
that the curl is given by this
423
00:34:24 --> 00:34:28
formula we use the x,
y, z coordinate system.
424
00:34:28 --> 00:34:30
But there is something I
haven't quite told you about.
425
00:34:30 --> 00:34:33
Which is if I switch to any
other right-handed coordinate
426
00:34:33 --> 00:34:35
system,
so I do some sort of rotation
427
00:34:35 --> 00:34:40
of my space coordinates,
then somehow the line integral,
428
00:34:40 --> 00:34:44
the flux integral,
the notion of curl makes sense
429
00:34:44 --> 00:34:47
in coordinates.
And the reason is that they all
430
00:34:47 --> 00:34:50
have geometric interpretations.
For example,
431
00:34:50 --> 00:34:52
when I think of this as the
work done by a force,
432
00:34:52 --> 00:34:55
well, the force doesn't care
whether it's being put in x,
433
00:34:55 --> 00:34:56
y coordinates this way or that
way.
434
00:34:56 --> 00:35:00
It still does the same work
because it's the same force.
435
00:35:00 --> 00:35:03
And when I say that the curl
measures the rotation in a
436
00:35:03 --> 00:35:06
motion, well,
that depends on which
437
00:35:06 --> 00:35:09
coordinates you use.
And the same for interpretation
438
00:35:09 --> 00:35:12
of flux.
In fact, if I rotated my
439
00:35:12 --> 00:35:17
coordinates to fit with any
other plane, I could still do
440
00:35:17 --> 00:35:23
the same things.
What I'm trying to say is,
441
00:35:23 --> 00:35:31
in fact, if C and S are in any
plane then we can still claim
442
00:35:31 --> 00:35:37
that it reduces to Green's
theorem.
443
00:35:37 --> 00:35:45
It will be Green's theorem not
in x, y, z coordinates but in
444
00:35:45 --> 00:35:50
some funny rotated coordinate
systems.
445
00:35:50 --> 00:35:56
What I'm saying is that work,
flux and curl makes sense
446
00:35:56 --> 00:35:59
independently of coordinates.
447
00:35:59 --> 00:36:20
448
00:36:20 --> 00:36:23
Now, this has to stop somewhere.
I can start claiming that I can
449
00:36:23 --> 00:36:26
somehow bend my coordinates to a
plane, any surface is flat.
450
00:36:26 --> 00:36:29
That doesn't really work.
But what I can say is if I have
451
00:36:29 --> 00:36:31
any surface I can cut it into
tiny pieces.
452
00:36:31 --> 00:36:35
And these tiny pieces are
basically flat.
453
00:36:35 --> 00:36:39
So that's basically the idea of
a proof.
454
00:36:39 --> 00:36:47
I am going to decompose my
surface into very small flat
455
00:36:47 --> 00:36:56
pieces.
Given any S we are just going
456
00:36:56 --> 00:37:08
to decompose it into tiny almost
flat pieces.
457
00:37:08 --> 00:37:15
For example,
if I have my surface like this,
458
00:37:15 --> 00:37:23
what I will do is I will just
cut it into tiles.
459
00:37:23 --> 00:37:28
I mean a good example of that
is if you look at
460
00:37:28 --> 00:37:31
[UNINTELLIGIBLE],
for example,
461
00:37:31 --> 00:37:36
it's made of all these hexagons
and pentagons.
462
00:37:36 --> 00:37:38
Well, actually,
they're not quite flat in the
463
00:37:38 --> 00:37:41
usual rule, but you could make
them flat and it would still
464
00:37:41 --> 00:37:45
look pretty much like a sphere.
Anyway, you're going to cut
465
00:37:45 --> 00:37:49
your surface into lots of tiny
pieces.
466
00:37:49 --> 00:37:53
And then you can use Stokes'
theorem on each small piece.
467
00:37:53 --> 00:38:00
What it says on each small flat
piece -- It says that the line
468
00:38:00 --> 00:38:04
integral along say,
for example,
469
00:38:04 --> 00:38:08
this curve is equal to the flux
of a curl through this tiny
470
00:38:08 --> 00:38:12
piece of surface.
And now I will add all of these
471
00:38:12 --> 00:38:14
terms together.
If I add all of the small
472
00:38:14 --> 00:38:17
contributions to flux I get the
total flux.
473
00:38:17 --> 00:38:19
What if I add all of the small
line integrals?
474
00:38:19 --> 00:38:23
Well, I get lots of extra junk
because I never asked to compute
475
00:38:23 --> 00:38:26
the line integral along this.
But this guy will come in twice
476
00:38:26 --> 00:38:30
when I do this little plate and
when I do that little plate with
477
00:38:30 --> 00:38:34
opposite orientations.
When I sum all of the little
478
00:38:34 --> 00:38:38
line integrals together,
all of the inner things cancel
479
00:38:38 --> 00:38:40
out,
and the only ones that I go
480
00:38:40 --> 00:38:44
through only once are those that
are at the outer most edges.
481
00:38:44 --> 00:38:50
So, when I sum all of my works
together, I will get the work
482
00:38:50 --> 00:38:54
done just along the outer
boundary C.
483
00:38:54 --> 00:39:12
Sum of work around each little
piece is just actually the work
484
00:39:12 --> 00:39:27
along C, the outer curve.
And the sum of the flux for
485
00:39:27 --> 00:39:39
each piece is going to be the
flux through S.
486
00:39:39 --> 00:39:45
From Stokes' theorem for flat
surfaces, I can get it for any
487
00:39:45 --> 00:39:47
surface.
I am cheating a little bit
488
00:39:47 --> 00:39:50
because you would actually have
to check carefully that this
489
00:39:50 --> 00:39:53
approximately where you flatten
the little pieces that are
490
00:39:53 --> 00:39:56
almost flat is [UNINTELLIGIBLE].
But, trust me,
491
00:39:56 --> 00:39:56
it actually works.
492
00:39:56 --> 00:40:13
493
00:40:13 --> 00:40:15
Let's do an actual example.
I mean I said example,
494
00:40:15 --> 00:40:19
but that was more like getting
us ready for the proof so
495
00:40:19 --> 00:40:22
probably that doesn't count as
an actual example.
496
00:40:22 --> 00:40:25
I should probably keep these
statements for now so I am not
497
00:40:25 --> 00:40:26
going to erase this side.
498
00:40:26 --> 00:41:08
499
00:41:08 --> 00:41:21
Let's do an example.
Let's try to find the work of
500
00:41:21 --> 00:41:40
vector field zi plus xj plus yk
around the unit circle in the x,
501
00:41:40 --> 00:41:58
y plane counterclockwise.
The picture is conveniently
502
00:41:58 --> 00:42:05
already there.
Just as a quick review,
503
00:42:05 --> 00:42:08
let's see how we do that
directly.
504
00:42:08 --> 00:42:14
If we do that directly,
I have to find the integral
505
00:42:14 --> 00:42:21
along C.
So F dot dr becomes zdx plus
506
00:42:21 --> 00:42:28
xdy plus ydz.
But now we actually know that
507
00:42:28 --> 00:42:33
on this circle,
well, z is zero.
508
00:42:33 --> 00:42:39
And we can parameterize x and
y, the unit circle in the x,
509
00:42:39 --> 00:42:44
y plane, so we can take x
equals cosine t,
510
00:42:44 --> 00:42:49
y equals sine t.
That will just become the
511
00:42:49 --> 00:42:54
integral over C.
Well, z times dx,
512
00:42:54 --> 00:43:05
z is zero so we have nothing,
plus x is cosine t times dy is
513
00:43:05 --> 00:43:17
-- Well, if y is sine t then dy
is cosine tdt plus ydz but z is
514
00:43:17 --> 00:43:22
zero.
Now, the range of values for t,
515
00:43:22 --> 00:43:26
well, we are going
counterclockwise around the
516
00:43:26 --> 00:43:31
entire circle so that should go
from zero to 2pi.
517
00:43:31 --> 00:43:39
We will get integral from zero
to 2pi of cosine square tdt
518
00:43:39 --> 00:43:45
which, if you do the
calculation, turns out to be
519
00:43:45 --> 00:43:50
just pi.
Now, let's instead try to use
520
00:43:50 --> 00:43:55
Stokes' theorem to do the
calculation.
521
00:43:55 --> 00:44:00
Now, of course the smart choice
would be to just take the flat
522
00:44:00 --> 00:44:02
unit disk.
I am not going to do that.
523
00:44:02 --> 00:44:06
That would be too boring.
Plus we have already kind of
524
00:44:06 --> 00:44:09
checked it because we already
trust Green's theorem.
525
00:44:09 --> 00:44:11
Instead, just to convince you
that,
526
00:44:11 --> 00:44:14
yes, I can choose really any
surface I want,
527
00:44:14 --> 00:44:23
let's say that I'm going to
choose a piece of paraboloid z
528
00:44:23 --> 00:44:30
equals one minus x squared minus
y squared.
529
00:44:30 --> 00:44:36
Well, to get our conventions
straight, we should take the
530
00:44:36 --> 00:44:43
normal vector pointing up for
compatibility with our choice.
531
00:44:43 --> 00:44:48
Well, we will have to compute
the flux through S.
532
00:44:48 --> 00:44:50
We don't really have to because
we could have chosen the disk,
533
00:44:50 --> 00:44:54
it would be easier,
but if we want to do it this
534
00:44:54 --> 00:45:00
way we will compute the flux of
curl F through our paraboloid.
535
00:45:00 --> 00:45:03
How do we do that?
Well, we need to find the curl
536
00:45:03 --> 00:45:10
and we need to find ndS.
Let's start with the curl.
537
00:45:10 --> 00:45:23
Curl F let's take the
cross-product between dell and F
538
00:45:23 --> 00:45:28
which is zxy.
If we compute this,
539
00:45:28 --> 00:45:31
the i component will be one
minus zero.
540
00:45:31 --> 00:45:37
It looks like it is one i.
Minus the j component is zero
541
00:45:37 --> 00:45:41
minus one.
Plus the k component is one
542
00:45:41 --> 00:45:52
minus zero.
In fact, the curl of the field
543
00:45:52 --> 00:45:59
is one, one, one.
Now, what about ndS?
544
00:45:59 --> 00:46:03
Well, this is a surface for
which we know z is a function of
545
00:46:03 --> 00:46:08
x and y.
ndS we can write as,
546
00:46:08 --> 00:46:14
let's call this F of xy,
then we can use the formula
547
00:46:14 --> 00:46:19
that says ndS equals negative F
sub x, negative F sub y,
548
00:46:19 --> 00:46:26
one dxdy,
which here gives us 2x,
549
00:46:26 --> 00:46:34
2y, one dxdy.
Now, when we want to compute
550
00:46:34 --> 00:46:41
the flux, we will have to do
double integral over S of one,
551
00:46:41 --> 00:46:47
one, one dot product with 2x,
2y, one dxdy.
552
00:46:47 --> 00:46:55
It will become the double
integral of 2x plus 2y plus one
553
00:46:55 --> 00:46:58
dxdy.
And, of course,
554
00:46:58 --> 00:47:01
the region which we are
integrating, the range of values
555
00:47:01 --> 00:47:04
of x and y will be the shadow of
our surface.
556
00:47:04 --> 00:47:07
That is just going to be,
if you look at this paraboloid
557
00:47:07 --> 00:47:11
from above,
all you will see is the unit
558
00:47:11 --> 00:47:17
disk so it will be a double
integral of the unit disk.
559
00:47:17 --> 00:47:23
And the way we will do that,
one way is to switch to polar
560
00:47:23 --> 00:47:28
coordinates and do the
calculation and then you will
561
00:47:28 --> 00:47:31
end up with pi.
The other way is to try to do
562
00:47:31 --> 00:47:34
it by symmetry.
Observe, when you integrate x
563
00:47:34 --> 00:47:37
above this, x is as negative on
the left as it is positive on
564
00:47:37 --> 00:47:40
the right.
So the integral of x will be
565
00:47:40 --> 00:47:42
zero.
The integral of y will be zero
566
00:47:42 --> 00:47:46
also by symmetry.
Then the integral of one dxdy
567
00:47:46 --> 00:47:52
will just be the area of this
unit disk which is pi.
568
00:47:52 --> 00:47:54
That was our first example.
And, of course,
569
00:47:54 --> 00:47:57
if you're actually free to
choose your favorite surface,
570
00:47:57 --> 00:48:01
there is absolutely no reason
why you would actually choose
571
00:48:01 --> 00:48:04
this paraboloid in this example.
I mean it would be much easier
572
00:48:04 --> 00:48:05
to choose a flat disk.
OK.
573
00:48:05 --> 00:48:09
Tomorrow I will tell you a few
more things about curl fits in
574
00:48:09 --> 00:48:13
with conservativeness and with
the divergence theorem,
575
00:48:13 --> 00:48:17
Stokes all together,
and we will look at Practice
576
00:48:17 --> 00:48:20
Exam 4B so please bring the exam
with you.
577
00:48:20 --> 00:48:25
578
00:48:20 --> 00:48:20
with you.
579
00:48:20 --> 00:48:25