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Yesterday we learned about flux
and we have seen the first few
8
00:00:28 --> 00:00:33
examples of how to set up and
compute integrals for a flux of
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00:00:33 --> 00:00:40
a vector field for a surface.
Remember the flux of a vector
10
00:00:40 --> 00:00:47
field F through the surface S is
defined by taking the double
11
00:00:47 --> 00:00:55
integral on the surface of F dot
n dS where n is the unit normal
12
00:00:55 --> 00:01:03
to the surface and dS is the
area element on the surface.
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00:01:03 --> 00:01:06
As we have seen,
for various surfaces,
14
00:01:06 --> 00:01:09
we have various formulas
telling us what the normal
15
00:01:09 --> 00:01:12
vector is and what the area
element becomes.
16
00:01:12 --> 00:01:14
For example,
on spheres we typically
17
00:01:14 --> 00:01:18
integrate with respect to phi
and theta for latitude and
18
00:01:18 --> 00:01:21
longitude angles.
On a horizontal plane,
19
00:01:21 --> 00:01:25
we would just end up degrading
dx, dy and so on.
20
00:01:25 --> 00:01:29
At the end of lecture we saw a
formula.
21
00:01:29 --> 00:01:32
A lot of you asked me how we
got it.
22
00:01:32 --> 00:01:36
Well, we didn't get it yet.
We are going to try to explain
23
00:01:36 --> 00:01:39
where it comes from and why it
works.
24
00:01:39 --> 00:01:50
The case we want to look at is
if S is the graph of a function,
25
00:01:50 --> 00:02:01
it is given by z equals some
function in terms of x and y.
26
00:02:01 --> 00:02:08
Our surface is out here.
Z is a function of x and y.
27
00:02:08 --> 00:02:17
And x and y will range over
some domain in the x,
28
00:02:17 --> 00:02:28
y plane, namely the region that
is the shadow of the surface on
29
00:02:28 --> 00:02:34
the x, y plane.
I said that we will have a
30
00:02:34 --> 00:02:40
formula for n dS which will end
up being plus/minus minus f sub
31
00:02:40 --> 00:02:45
x, minus f sub y,
one dxdy,
32
00:02:45 --> 00:02:49
so that we will set up and
evaluate the integral in terms
33
00:02:49 --> 00:02:53
of x and y.
Every time we see z we will
34
00:02:53 --> 00:02:57
replace it by f of xy,
whatever the formula for f
35
00:02:57 --> 00:03:00
might be.
Actually, if we look at a very
36
00:03:00 --> 00:03:04
easy case where this is just a
horizontal plane,
37
00:03:04 --> 00:03:07
z equals constant,
the function is just a
38
00:03:07 --> 00:03:09
constant,
well, the partial derivatives
39
00:03:09 --> 00:03:14
become just zero.
You get dx, dy.
40
00:03:14 --> 00:03:17
That is what you would expect
for a horizontal plane just from
41
00:03:17 --> 00:03:19
common sense.
This is more interesting,
42
00:03:19 --> 00:03:23
of course, if a function is
more interesting.
43
00:03:23 --> 00:03:30
How do we get that?
Where does this come from?
44
00:03:30 --> 00:03:36
We need to figure out,
for a small piece of our
45
00:03:36 --> 00:03:43
surface, what will be n delta S.
Let's say that we take a small
46
00:03:43 --> 00:03:51
rectangle in here corresponding
to sides delta x and delta y and
47
00:03:51 --> 00:03:58
we look at the piece of surface
that is above that.
48
00:03:58 --> 00:04:03
Well, the question we have now
is what is the area of this
49
00:04:03 --> 00:04:08
little piece of surface and what
is its normal vector?
50
00:04:08 --> 00:04:10
Observe this little piece up
here.
51
00:04:10 --> 00:04:12
If it is small enough,
it will look like a
52
00:04:12 --> 00:04:14
parallelogram.
I mean it might be slightly
53
00:04:14 --> 00:04:17
curvy, but roughly it looks like
a parallelogram in space.
54
00:04:17 --> 00:04:22
And so we have seen how to find
the area of a parallelogram in
55
00:04:22 --> 00:04:25
space using cross-product.
If we can figure out what are
56
00:04:25 --> 00:04:29
the vectors for this side and
that side then taking that
57
00:04:29 --> 00:04:33
cross-product and taking the
magnitude of the cross-product
58
00:04:33 --> 00:04:36
will give us the area.
Moreover, the cross-product
59
00:04:36 --> 00:04:38
also gives us the normal
direction.
60
00:04:38 --> 00:04:40
In fact, the cross-product
gives us two in one.
61
00:04:40 --> 00:04:44
It gives us the normal
direction and the area element.
62
00:04:44 --> 00:04:48
And that is why I said that we
will have an easy formula for n
63
00:04:48 --> 00:04:52
dS while n and dS taken
separately are more complicated
64
00:04:52 --> 00:04:55
because you would have to
actually take the length of a
65
00:04:55 --> 00:05:03
direction of this guy.
Let's carry out this problem.
66
00:05:03 --> 00:05:13
Let's say I am going to look at
a small piece of the x,
67
00:05:13 --> 00:05:17
y plane.
Here I have delta x,
68
00:05:17 --> 00:05:22
here I have delta y,
and I am starting at some point
69
00:05:22 --> 00:05:28
(x, y).
Now, above that I will have a
70
00:05:28 --> 00:05:34
parallelogram on my surface.
This point here,
71
00:05:34 --> 00:05:36
the point where I start,
I know what it is.
72
00:05:36 --> 00:05:43
It is just (x, y).
And, well, z is f(x, y).
73
00:05:43 --> 00:05:44
Now what I want to find,
actually,
74
00:05:44 --> 00:05:50
is what are these two vectors,
let's call them U and V,
75
00:05:50 --> 00:05:54
that correspond to moving a bit
in the x direction or in the y
76
00:05:54 --> 00:05:58
direction?
And then U cross V will be,
77
00:05:58 --> 00:06:06
well, in terms of the magnitude
of this guy will just be the
78
00:06:06 --> 00:06:11
little piece of surface area,
delta S.
79
00:06:11 --> 00:06:15
And, in terms of direction,
it will be normal to the
80
00:06:15 --> 00:06:19
surface.
Actually, I will get just delta
81
00:06:19 --> 00:06:23
S times my normal vector.
Well, up to sign because,
82
00:06:23 --> 00:06:26
depending on whether I do U V
or V U, I might get the normal
83
00:06:26 --> 00:06:29
vector in the direction I want
or in the opposite direction.
84
00:06:29 --> 00:06:36
But we will take care of that
later.
85
00:06:36 --> 00:06:41
Let's find U and V.
And, in case you have trouble
86
00:06:41 --> 00:06:47
with that small picture,
I have a better one here.
87
00:06:47 --> 00:06:53
Let's keep it just in case this
one gets really too cluttered.
88
00:06:53 --> 00:06:57
It really represents the same
thing.
89
00:06:57 --> 00:07:05
Let's try to figure out these
vectors U and V.
90
00:07:05 --> 00:07:13
Vector U starts at the point x,
y, f of x, y and it goes to --
91
00:07:13 --> 00:07:19
Whereas, its head,
well, I will have moved x by
92
00:07:19 --> 00:07:23
delta x.
So, x plus delta x and y
93
00:07:23 --> 00:07:25
doesn't change.
And, of course,
94
00:07:25 --> 00:07:31
the z coordinate has to change.
It becomes f of x plus delta x
95
00:07:31 --> 00:07:34
and y.
Now, how does f change if I
96
00:07:34 --> 00:07:37
change x a little bit?
Well, we have seen that it is
97
00:07:37 --> 00:07:41
given by the partial derivative
f sub x.
98
00:07:41 --> 00:07:47
This is approximately equal to
f of x, y plus delta x times f
99
00:07:47 --> 00:07:51
sub x at the given point x,
y.
100
00:07:51 --> 00:07:56
I am not going to add it
because the notation is already
101
00:07:56 --> 00:07:59
long enough.
That means my vector U,
102
00:07:59 --> 00:08:05
well, approximately because I
am using this linear
103
00:08:05 --> 00:08:08
approximation,
00:08:15
0, f sub x times delta x>.
Is that OK with everyone?
105
00:08:15 --> 00:08:19
Good.
Now, what about V?
106
00:08:19 --> 00:08:23
Well, V works the same way so I
am not going to do all the
107
00:08:23 --> 00:08:27
details.
When I move from here to here x
108
00:08:27 --> 00:08:30
doesn't change and y changes by
delta y.
109
00:08:30 --> 00:08:39
X component nothing happens.
Y component changes by delta y.
110
00:08:39 --> 00:08:42
What about the z component?
Well, f changes by f sub y
111
00:08:42 --> 00:08:51
times delta y.
That is how f changes if I
112
00:08:51 --> 00:08:59
increase y by delta y.
I have my two sides.
113
00:08:59 --> 00:09:03
Now I can take that
cross-product.
114
00:09:03 --> 00:09:08
Well, maybe I will first factor
something out.
115
00:09:08 --> 00:09:13
See, I can rewrite this as one,
zero, f sub x times delta x.
116
00:09:13 --> 00:09:20
And this one I will rewrite as
zero, one, f sub y delta y.
117
00:09:20 --> 00:09:26
And so now the cross-product,
n hat delta S up to sign is
118
00:09:26 --> 00:09:31
going to be U cross V.
We will have to do the
119
00:09:31 --> 00:09:35
cross-product,
and we will have a delta x,
120
00:09:35 --> 00:09:41
delta y coming out.
I am just saving myself the
121
00:09:41 --> 00:09:47
trouble of writing a lot of
delta x's and delta y's,
122
00:09:47 --> 00:09:55
but if you prefer you can just
do directly this cross-product.
123
00:09:55 --> 00:09:57
Let's compute this
cross-product.
124
00:09:57 --> 00:10:03
Well, the i component is zero
minus f sub x.
125
00:10:03 --> 00:10:07
The y component is going to be,
well, f sub y minus zero but
126
00:10:07 --> 00:10:12
with the minus sign in front of
everything, so negative f sub y.
127
00:10:12 --> 00:10:20
And the z component will be
just one times delta x delta y.
128
00:10:20 --> 00:10:25
Does that make sense?
Yes.
129
00:10:25 --> 00:10:30
Very good.
And so now we shrink this
130
00:10:30 --> 00:10:32
rectangle,
we shrink delta x and delta y
131
00:10:32 --> 00:10:35
to zero,
that is how we get this formula
132
00:10:35 --> 00:10:38
for n dS equals negative fx,
negative fy,
133
00:10:38 --> 00:10:43
one, dxdy.
Well, plus/minus because it is
134
00:10:43 --> 00:10:47
up to us to choose whether we
want to take the normal vector
135
00:10:47 --> 00:10:50
point up or down.
See, if you take this
136
00:10:50 --> 00:10:55
convention then the z component
of n dS is positive.
137
00:10:55 --> 00:10:59
That corresponds to normal
vector pointing up.
138
00:10:59 --> 00:11:02
If you take the opposite signs
then the z component will be
139
00:11:02 --> 00:11:06
negative.
That means your normal vector
140
00:11:06 --> 00:11:15
points down.
This one is with n pointing up.
141
00:11:15 --> 00:11:17
I mean when I say up,
of course it is still
142
00:11:17 --> 00:11:21
perpendicular to the surface.
If the surface really has a big
143
00:11:21 --> 00:11:26
slope then it is not really
going to go all that much up,
144
00:11:26 --> 00:11:31
but more up than down.
OK.
145
00:11:31 --> 00:11:40
That is how we get the formula.
Any questions?
146
00:11:40 --> 00:11:41
No.
OK.
147
00:11:41 --> 00:11:44
That is a really useful
formula.
148
00:11:44 --> 00:11:49
You don't really need to
remember all the details of how
149
00:11:49 --> 00:11:54
we got it, but please remember
that formula.
150
00:11:54 --> 00:12:06
Let's do an example, actually.
Let's say we want to find the
151
00:12:06 --> 00:12:08
flux of the vector field z times
k,
152
00:12:08 --> 00:12:12
so it is a vertical vector
field,
153
00:12:12 --> 00:12:27
through the portion of the
paraboloid z equals x^2 y^2 that
154
00:12:27 --> 00:12:36
lives above the unit disk.
What does that mean?
155
00:12:36 --> 00:12:39
z = x^2 y^2.
We have seen it many times.
156
00:12:39 --> 00:12:43
It is this parabola and is
pointing up.
157
00:12:43 --> 00:12:47
Above the unit disk means I
don't care about this infinite
158
00:12:47 --> 00:12:52
surface.
I will actually stop when I hit
159
00:12:52 --> 00:12:56
a radius of one away from the
z-axis.
160
00:12:56 --> 00:13:03
And so now I have my vector
field which is going to point
161
00:13:03 --> 00:13:09
overall up because,
well, it is z times k.
162
00:13:09 --> 00:13:13
The more z is positive,
the more your vector field goes
163
00:13:13 --> 00:13:16
up.
Of course, if z were negative
164
00:13:16 --> 00:13:20
then it would point down,
but it will live above.
165
00:13:20 --> 00:13:25
Actually, a quick opinion poll.
What do you think the flux
166
00:13:25 --> 00:13:28
should be?
Should it be positive,
167
00:13:28 --> 00:13:32
zero, negative or we don't
know?
168
00:13:32 --> 00:13:36
I see some I don't know,
I see some negative and I see
169
00:13:36 --> 00:13:39
some positive.
Of course, I didn't tell you
170
00:13:39 --> 00:13:42
which way I am orienting my
paraboloid.
171
00:13:42 --> 00:13:44
So far both answers are correct.
The only one that is probably
172
00:13:44 --> 00:13:47
not correct is zero because,
no matter which way you choose
173
00:13:47 --> 00:13:49
to orient it you should get
something.
174
00:13:49 --> 00:13:51
It is not looking like it will
be zero.
175
00:13:51 --> 00:14:00
Let's say that I am going to do
it with the normal pointing
176
00:14:00 --> 00:14:06
upwards.
Second chance.
177
00:14:06 --> 00:14:11
I see some people changing back
and forth from one and two.
178
00:14:11 --> 00:14:14
Let's draw a picture.
Which one is pointing upwards?
179
00:14:14 --> 00:14:16
Well, let's look at the bottom
point.
180
00:14:16 --> 00:14:19
The normal vector pointing up,
here we know what it means.
181
00:14:19 --> 00:14:22
It is this guy.
If you continue to follow your
182
00:14:22 --> 00:14:26
normal vector,
see, they are actually pointing
183
00:14:26 --> 00:14:30
up and into the paraboloid.
And I claim that the answer
184
00:14:30 --> 00:14:34
should be positive because the
vector field is crossing our
185
00:14:34 --> 00:14:38
paraboliod going upwards,
going from the outside out and
186
00:14:38 --> 00:14:46
below to the inside and upside.
So, in the direction that we
187
00:14:46 --> 00:14:55
are counting positively.
We will see how it turns out
188
00:14:55 --> 00:15:03
when we do the calculation.
We have to compute the integral
189
00:15:03 --> 00:15:08
for flux.
Double integral over a surface
190
00:15:08 --> 00:15:15
of F dot n dS is going to be --
What are we going to do?
191
00:15:15 --> 00:15:21
Well, F we said is <0,0,
z>.
192
00:15:21 --> 00:15:25
What is n dS.
Well, let's use our brand new
193
00:15:25 --> 00:15:28
formula.
It says negative f sub x,
194
00:15:28 --> 00:15:31
negative f sub y,
one, dxdy.
195
00:15:31 --> 00:15:38
What does little f in here?
It is x^2 y^2.
196
00:15:38 --> 00:15:42
When we are using this formula,
we need to know what little x
197
00:15:42 --> 00:15:47
stands for.
It is whatever the formula is
198
00:15:47 --> 00:15:53
for z as a function of x and y.
We take x^2 y^2 and we take the
199
00:15:53 --> 00:15:57
partial derivatives with minus
signs.
200
00:15:57 --> 00:16:01
We get negative 2x,
negative 2y and one,
201
00:16:01 --> 00:16:03
dxdy.
Well, of course here it didn't
202
00:16:03 --> 00:16:05
really matter because we are
going to dot them with zero.
203
00:16:05 --> 00:16:11
Actually, even if we had made a
mistake we somehow wouldn't have
204
00:16:11 --> 00:16:15
had to pay the price.
But still.
205
00:16:15 --> 00:16:22
We will end up with double
integral on S of z dxdy.
206
00:16:22 --> 00:16:25
Now, what do we do with that?
Well, we have too many things.
207
00:16:25 --> 00:16:43
We have to get rid of z.
Let's use z equals x^2 y^2 once
208
00:16:43 --> 00:16:51
more.
That becomes double integral of
209
00:16:51 --> 00:16:54
x^2 y^2 dxdy.
And here, see,
210
00:16:54 --> 00:16:57
we are using the fact that we
are only looking at things that
211
00:16:57 --> 00:16:59
are on the surface.
It is not like in a triple
212
00:16:59 --> 00:17:01
integral.
You could never do that because
213
00:17:01 --> 00:17:04
z, x and y are independent.
Here they are related by the
214
00:17:04 --> 00:17:09
equation of a surface.
If I sound like I am ranting,
215
00:17:09 --> 00:17:14
but I know from experience this
is where one of the most sticky
216
00:17:14 --> 00:17:18
and tricky points is.
OK.
217
00:17:18 --> 00:17:20
How will we actually integrate
that?
218
00:17:20 --> 00:17:22
Well, now that we have just x
and y, we should figure out what
219
00:17:22 --> 00:17:25
is the range for x and y.
Well, the range for x and y is
220
00:17:25 --> 00:17:27
going to be the shadow of our
region.
221
00:17:27 --> 00:17:34
It is going to be this unit
disk.
222
00:17:34 --> 00:17:43
I can just do that for now.
And this is finally where I
223
00:17:43 --> 00:17:46
have left the world of surface
integrals to go back to a usual
224
00:17:46 --> 00:17:49
double integral.
And now I have to set it up.
225
00:17:49 --> 00:17:52
Well, I can do it this way with
dxdy, but it looks like there is
226
00:17:52 --> 00:17:55
a smarter thing to do.
I am going to use polar
227
00:17:55 --> 00:17:59
coordinates.
In fact, I am going to say this
228
00:17:59 --> 00:18:03
is double integral of r^2 times
r dr d theta.
229
00:18:03 --> 00:18:07
I am on the unit disk so r goes
zero to one, theta goes zero to
230
00:18:07 --> 00:18:12
2pi.
And, if you do the calculation,
231
00:18:12 --> 00:18:19
you will find that this is
going to be pi over two.
232
00:18:19 --> 00:18:26
Any questions about the example.
Yes?
233
00:18:26 --> 00:18:29
How did I get this negative 2x
and negative 2y?
234
00:18:29 --> 00:18:33
I want to use my formula for n
dS.
235
00:18:33 --> 00:18:36
My surface is given by the
graph of a function.
236
00:18:36 --> 00:18:40
It is the graph of a function
x^2 y^2.
237
00:18:40 --> 00:18:43
I will use this formula that is
up here.
238
00:18:43 --> 00:18:47
I will take the function x^2
y^2 and I will take its partial
239
00:18:47 --> 00:18:50
derivatives.
If I take the partial of f,
240
00:18:50 --> 00:18:55
so x^2 y^2 with respect to x,
I get 2x, so I put negative 2x.
241
00:18:55 --> 00:18:57
And then the same thing,
negative 2y,
242
00:18:57 --> 00:19:01
one, dxdy.
Yes?
243
00:19:01 --> 00:19:04
Which k hat?
Oh, you mean the vector field.
244
00:19:04 --> 00:19:06
It is a different part of the
story.
245
00:19:06 --> 00:19:10
Whenever you do a surface
integral for flux you have two
246
00:19:10 --> 00:19:13
parts of the story.
One is the vector field whose
247
00:19:13 --> 00:19:17
flux you are taking.
The other one is the surface
248
00:19:17 --> 00:19:21
for which you will be taking
flux.
249
00:19:21 --> 00:19:26
The vector field only comes as
this f in the notation,
250
00:19:26 --> 00:19:28
and everything else,
the bounds in the double
251
00:19:28 --> 00:19:32
integral and the n dS,
all come from the surface that
252
00:19:32 --> 00:19:36
we are looking at.
Basically, in all of this
253
00:19:36 --> 00:19:40
calculation, this is coming from
f equals zk.
254
00:19:40 --> 00:19:47
Everything else comes from the
information paraboloid z = x^2
255
00:19:47 --> 00:19:51
y^2 above the unit disk.
In particular,
256
00:19:51 --> 00:19:55
if we wanted to now find the
flux of any other vector field
257
00:19:55 --> 00:19:58
for the same paraboloid,
well, all we would have to do
258
00:19:58 --> 00:20:02
is just replace this guy by
whatever the new vector field
259
00:20:02 --> 00:20:05
is.
We have learned how to set up
260
00:20:05 --> 00:20:09
flux integrals for this
paraboloid.
261
00:20:09 --> 00:20:10
Not that you should remember
this one by heart.
262
00:20:10 --> 00:20:13
I mean there are many
paraboloids in life and other
263
00:20:13 --> 00:20:17
surfaces, too.
It is better to remember the
264
00:20:17 --> 00:20:22
general method.
Any other questions?
265
00:20:22 --> 00:20:25
No.
OK.
266
00:20:25 --> 00:20:31
Let's see more ways of taking
flux integrals.
267
00:20:31 --> 00:20:33
But, just to reassure you,
at this point we have seen the
268
00:20:33 --> 00:20:36
most important ones.
90% of the problems that we
269
00:20:36 --> 00:20:41
will be looking at we can do
with what we have seen so far in
270
00:20:41 --> 00:20:49
less time and this formula.
Let's look a little bit at a
271
00:20:49 --> 00:20:56
more general situation.
Let's say that my surface is so
272
00:20:56 --> 00:21:00
complicated that I cannot
actually express z as a function
273
00:21:00 --> 00:21:04
of x and y, but let's say that I
know how to parametize it.
274
00:21:04 --> 00:21:06
I have a parametric equation
for my surface.
275
00:21:06 --> 00:21:11
That means I can express x,
y and z in terms of any two
276
00:21:11 --> 00:21:16
parameter variables that might
be relevant for me.
277
00:21:16 --> 00:21:19
If you want,
this one here is a special case
278
00:21:19 --> 00:21:23
where you can parameterize
things in terms of x and y as
279
00:21:23 --> 00:21:26
your two variables.
How would you do it in the
280
00:21:26 --> 00:21:29
fully general case?
In a way, that will answer your
281
00:21:29 --> 00:21:30
question that,
I think one of you,
282
00:21:30 --> 00:21:34
I forgot, asked yesterday how
would I do it in general?
283
00:21:34 --> 00:21:36
Is there a formula like M dx
plus N dy?
284
00:21:36 --> 00:21:39
Well, that is going to be the
general formula.
285
00:21:39 --> 00:21:42
And you will see that it is a
little bit too complicated,
286
00:21:42 --> 00:21:46
so the really useful ones are
actually the special ones.
287
00:21:46 --> 00:21:56
Let's say that we are given a
parametric description -- -- of
288
00:21:56 --> 00:22:01
a surface S.
That means we can describe S by
289
00:22:01 --> 00:22:05
formulas saying x is some
function of two parameter
290
00:22:05 --> 00:22:08
variables.
I am going to call them u and v.
291
00:22:08 --> 00:22:10
I hope you don't mind.
You can call them t1 and t2.
292
00:22:10 --> 00:22:18
You can call them whatever you
want.
293
00:22:18 --> 00:22:21
One of the basic properties of
a surface is because I have only
294
00:22:21 --> 00:22:23
two independent directions to
move on.
295
00:22:23 --> 00:22:26
I should be able to express x,
y and z in terms of two
296
00:22:26 --> 00:22:29
variables.
Now, let's say that I know how
297
00:22:29 --> 00:22:32
to do that.
Or, maybe I should instead
298
00:22:32 --> 00:22:35
think of it in terms of a
position vector if it helps you.
299
00:22:35 --> 00:22:40
That is just a vector with
components 00:22:46
y, z> is given as a function
of u and v.
301
00:22:46 --> 00:22:50
It works like a parametric
curve but with two parameters.
302
00:22:50 --> 00:22:54
Now, how would we actually set
up a flux integral on such a
303
00:22:54 --> 00:22:57
surface.
Well, because we are locating
304
00:22:57 --> 00:23:01
ourselves in terms of u and v,
we will end up with an integral
305
00:23:01 --> 00:23:06
du dv.
We need to figure out how to
306
00:23:06 --> 00:23:11
express n dS in terms of du and
dv.
307
00:23:11 --> 00:23:19
N dS should be something du dv.
How do we do that?
308
00:23:19 --> 00:23:25
Well, we can use the same
method that we have actually
309
00:23:25 --> 00:23:28
used over here.
Because, if you think for a
310
00:23:28 --> 00:23:30
second, here we used,
of course, a rectangle in the
311
00:23:30 --> 00:23:34
x, y plane and we lifted it to a
parallelogram and so on.
312
00:23:34 --> 00:23:37
But more generally you can
think what happens if I change u
313
00:23:37 --> 00:23:41
by delta u keeping v constant or
the other way around?
314
00:23:41 --> 00:23:45
You will get some sort of mesh
grid on your surface and you
315
00:23:45 --> 00:23:48
will look at a little
parallelogram that is an
316
00:23:48 --> 00:23:52
elementary piece of that mesh
and figure out what is its area
317
00:23:52 --> 00:23:57
and normal vector.
Well, that will again be given
318
00:23:57 --> 00:24:01
by the cross-product of the two
sides.
319
00:24:01 --> 00:24:07
Let's think a little bit about
what happens when I move a
320
00:24:07 --> 00:24:12
little bit on my surface.
I am taking this grid on my
321
00:24:12 --> 00:24:16
surface given by the u and v
directions.
322
00:24:16 --> 00:24:25
And, if I take a piece of that
corresponding to small changes
323
00:24:25 --> 00:24:33
delta u and delta v,
what is going to be going on
324
00:24:33 --> 00:24:36
here?
Well, I have to deal with two
325
00:24:36 --> 00:24:39
vectors, one corresponding to
changing u, the other one
326
00:24:39 --> 00:24:42
corresponding to changing v.
If I change u,
327
00:24:42 --> 00:24:46
how does my point change?
Well, it is given by the
328
00:24:46 --> 00:24:49
derivative of this with respect
to u.
329
00:24:49 --> 00:24:57
This vector here I will call,
so the sides are given by,
330
00:24:57 --> 00:25:05
let me say, partial r over
partial u times delta u.
331
00:25:05 --> 00:25:09
If you prefer,
maybe I should write it as
332
00:25:09 --> 00:25:13
partial x over partial u times
delta u.
333
00:25:13 --> 00:25:17
Well, it is just too boring to
write.
334
00:25:17 --> 00:25:21
And so on.
It means if I change u a little
335
00:25:21 --> 00:25:24
bit, keeping v constant,
then how x changes is,
336
00:25:24 --> 00:25:26
given by partial x over partial
u times delta u,
337
00:25:26 --> 00:25:28
same thing with y,
same thing with z,
338
00:25:28 --> 00:25:33
and I am just using vector
notation to do it this way.
339
00:25:33 --> 00:25:41
That is the analog of when I
said delta r for line integrals
340
00:25:41 --> 00:25:47
along a curve,
vector delta r is the velocity
341
00:25:47 --> 00:25:58
vector dr dt times delta t.
Now, if I look at the other
342
00:25:58 --> 00:26:07
side -- Let me start again.
I ran out of space.
343
00:26:07 --> 00:26:12
One side is partial r over
partial u times delta u.
344
00:26:12 --> 00:26:17
And the other one would be
partial r over partial v times
345
00:26:17 --> 00:26:19
delta v.
Because that is how the
346
00:26:19 --> 00:26:24
position of your point changes
if you just change u or v and
347
00:26:24 --> 00:26:31
not the other one.
To find the surface element
348
00:26:31 --> 00:26:40
together with a normal vector,
I would just take the
349
00:26:40 --> 00:26:46
cross-product between these
guys.
350
00:26:46 --> 00:26:50
If you prefer,
that is the cross-product of
351
00:26:50 --> 00:26:56
partial r over partial u with
partial r over partial v,
352
00:26:56 --> 00:27:02
delta u delta v.
And so n dS is this
353
00:27:02 --> 00:27:11
cross-product times du dv up to
sign.
354
00:27:11 --> 00:27:27
It depends on which choice I
make for my normal vector,
355
00:27:27 --> 00:27:32
of course.
That, of course,
356
00:27:32 --> 00:27:35
is a slightly confusing
equation to think of.
357
00:27:35 --> 00:27:37
A good exercise,
if you want to really
358
00:27:37 --> 00:27:40
understand what is going on,
try this in two good examples
359
00:27:40 --> 00:27:43
to look at.
One good example to look at is
360
00:27:43 --> 00:27:45
the previous one.
What is it?
361
00:27:45 --> 00:27:47
It is when u and v are just x
and y.
362
00:27:47 --> 00:27:51
The parametric equations are
just x equals x,
363
00:27:51 --> 00:27:54
y equals y and z is f of x,
y.
364
00:27:54 --> 00:27:59
You should end up with the same
formula that we had over there.
365
00:27:59 --> 00:28:03
And you should see why because
both of them are given by a
366
00:28:03 --> 00:28:06
cross-product.
The other case you can look at
367
00:28:06 --> 00:28:08
just to convince yourselves even
further.
368
00:28:08 --> 00:28:12
We don't need to do that
because we have seen the formula
369
00:28:12 --> 00:28:17
before, but in the case of a
sphere we have seen the formula
370
00:28:17 --> 00:28:22
for n and for dS separately.
We know what n dS are in terms
371
00:28:22 --> 00:28:26
of d phi, d theta.
Well, you could parametize a
372
00:28:26 --> 00:28:28
sphere in terms of phi and
theta.
373
00:28:28 --> 00:28:33
Namely, the formulas would be x
equals a sine phi cosine theta,
374
00:28:33 --> 00:28:38
y equals a sign phi sine theta,
z equals a cosine phi.
375
00:28:38 --> 00:28:42
The formulas for circle
coordinates setting Ro equals a
376
00:28:42 --> 00:28:44
.
That is a parametric equation
377
00:28:44 --> 00:28:47
for the sphere.
And then, if you try to use
378
00:28:47 --> 00:28:50
this formula here,
you should end up with the same
379
00:28:50 --> 00:28:52
things we have already seen for
n dS,
380
00:28:52 --> 00:28:57
just with a lot more pain to
actually get there because
381
00:28:57 --> 00:29:00
cross-product is going to be a
bit complicated.
382
00:29:00 --> 00:29:03
But we are seeing all of these
formulas all fitting together.
383
00:29:03 --> 00:29:05
Somehow it is always the same
question.
384
00:29:05 --> 00:29:10
We just have different angles
of attack on this general
385
00:29:10 --> 00:29:18
problem.
Questions?
386
00:29:18 --> 00:29:20
No.
OK.
387
00:29:20 --> 00:29:30
Let's look at yet another last
way of finding n dS.
388
00:29:30 --> 00:29:36
And then I promise we will
switch to something else because
389
00:29:36 --> 00:29:42
I can feel that you are getting
a bit overwhelmed for all these
390
00:29:42 --> 00:29:47
formulas for n dS.
What happens very often is we
391
00:29:47 --> 00:29:51
don't actually know how to
parametize our surface.
392
00:29:51 --> 00:29:54
Maybe we don't know how to
solve for z as a function of x
393
00:29:54 --> 00:29:58
and y, but our surface is given
by some equation.
394
00:29:58 --> 00:30:05
And so what that means is
actually maybe what we know is
395
00:30:05 --> 00:30:12
not really these kinds of
formulas, but maybe we know a
396
00:30:12 --> 00:30:17
normal vector.
And I am going to call this one
397
00:30:17 --> 00:30:22
capital N because I don't even
need it to be a unit vector.
398
00:30:22 --> 00:30:27
You will see.
It can be a normal vector of
399
00:30:27 --> 00:30:33
any length you want to the
surfaces.
400
00:30:33 --> 00:30:35
Why would we ever know a normal
vector?
401
00:30:35 --> 00:30:39
Well, for example,
if our surface is a plane,
402
00:30:39 --> 00:30:43
a slanted plane given by some
equation, ax by cz = d.
403
00:30:43 --> 00:30:44
Well, you know the normal
vector.
404
00:30:44 --> 00:30:48
It is .
Of course, you could solve for
405
00:30:48 --> 00:30:52
z and then go back to that case,
which is why I said that one is
406
00:30:52 --> 00:30:55
very useful.
But you can also just stay with
407
00:30:55 --> 00:30:58
a normal vector.
Why else would you know a
408
00:30:58 --> 00:31:01
normal vector?
Well, let's say that you know
409
00:31:01 --> 00:31:05
an equation that is of a form g
of x, y, z equals zero.
410
00:31:05 --> 00:31:08
Well, then you know that the
gradient of g is perpendicular
411
00:31:08 --> 00:31:14
to the level surface.
Let me just give you two
412
00:31:14 --> 00:31:19
examples.
If you have a plane,
413
00:31:19 --> 00:31:24
ax by cz = d,
then the normal vector would
414
00:31:24 --> 00:31:28
just be .
415
00:31:28 --> 00:31:34
If you have a surface S given
by an equation,
416
00:31:34 --> 00:31:40
g(x, y, z) = 0,
then you can take a normal
417
00:31:40 --> 00:31:46
vector to be the gradient of g.
We have seen that the gradient
418
00:31:46 --> 00:31:49
is perpendicular to the level
surface.
419
00:31:49 --> 00:31:51
Now, of course,
we don't necessarily have to
420
00:31:51 --> 00:31:55
follow what is going to come.
Because, if we could solve for
421
00:31:55 --> 00:31:59
z, then we might be better off
doing what we did over there.
422
00:31:59 --> 00:32:02
But let's say that we want to
do it this.
423
00:32:02 --> 00:32:06
What can we do?
Well, I am going to give you
424
00:32:06 --> 00:32:09
another way to think
geometrically about n dS.
425
00:32:09 --> 00:32:40
426
00:32:40 --> 00:32:43
Let's start by thinking about
the slanted plane.
427
00:32:43 --> 00:32:47
Let's say that my surface is
just a slanted plane.
428
00:32:47 --> 00:32:52
My normal vector would be maybe
somewhere here.
429
00:32:52 --> 00:32:55
And let's say that I am going
to try -- I need to get some
430
00:32:55 --> 00:32:57
handle on how to set up my
integrals,
431
00:32:57 --> 00:33:00
so maybe I am going to express
things in terms of x and y.
432
00:33:00 --> 00:33:05
I have my coordinates,
and I will try to use x and y.
433
00:33:05 --> 00:33:11
Then I would like to relate
delta S or dS to the area in the
434
00:33:11 --> 00:33:14
x y plane.
That means I want maybe to look
435
00:33:14 --> 00:33:19
at the projection of this guy
onto a horizontal plane.
436
00:33:19 --> 00:33:31
Let's squish it horizontally.
Then you have here another area.
437
00:33:31 --> 00:33:35
The guy on the slanted plane,
let's call that delta S.
438
00:33:35 --> 00:33:38
And let's call this guy down
here delta A.
439
00:33:38 --> 00:33:42
And delta A would become
ultimately maybe delta x,
440
00:33:42 --> 00:33:47
delta y or something like that.
The question is how do we find
441
00:33:47 --> 00:33:51
the conversion rate between
these two areas?
442
00:33:51 --> 00:33:53
I mean they are not the same.
Visually, I hope it is clear to
443
00:33:53 --> 00:33:56
you that if my plane is actually
horizontal then,
444
00:33:56 --> 00:34:00
of course, they are the same.
But the more slanted it becomes
445
00:34:00 --> 00:34:04
the more delta A becomes smaller
than delta S.
446
00:34:04 --> 00:34:09
If you buy land and it is on
the side of a cliff,
447
00:34:09 --> 00:34:12
well, whether you look at it on
a map or whether you look at it
448
00:34:12 --> 00:34:15
on the actual cliff,
the area is going to be very
449
00:34:15 --> 00:34:18
different.
I am not sure if that is a wise
450
00:34:18 --> 00:34:22
thing to do if you want to build
a house there,
451
00:34:22 --> 00:34:26
but I bet you can get really
cheap land.
452
00:34:26 --> 00:34:31
Anyway, delta S versus delta A
depends on how slanted things
453
00:34:31 --> 00:34:34
are.
And let's try to make that more
454
00:34:34 --> 00:34:41
precise by looking at the angel
that our plane makes with the
455
00:34:41 --> 00:34:47
horizontal direction.
Let's call this angle alpha,
456
00:34:47 --> 00:34:53
the angle that our plane makes
with the horizontal direction.
457
00:34:53 --> 00:34:57
See, it is all coming together.
The first unit about
458
00:34:57 --> 00:35:03
cross-products,
normal vectors and so on is
459
00:35:03 --> 00:35:09
actually useful now.
I claim that the surface
460
00:35:09 --> 00:35:16
element is related to the area
in the plane by delta A equals
461
00:35:16 --> 00:35:19
delta S times the cosine of
alpha.
462
00:35:19 --> 00:35:24
Why is that?
Well, let's look at this small
463
00:35:24 --> 00:35:27
rectangle with one horizontal
side and one slanted side.
464
00:35:27 --> 00:35:33
When you project this side does
not change, but this side gets
465
00:35:33 --> 00:35:37
shortened by a factor of cosine
alpha.
466
00:35:37 --> 00:35:41
Whatever this length was,
this length here is that one
467
00:35:41 --> 00:35:44
times cosine alpha.
That is why the area gets
468
00:35:44 --> 00:35:48
shrunk by cosine alpha.
In one direction nothing
469
00:35:48 --> 00:35:52
happens.
In the other direction you
470
00:35:52 --> 00:35:59
squish by cosine alpha.
What that means is that,
471
00:35:59 --> 00:36:05
well, we will have to deal with
this.
472
00:36:05 --> 00:36:08
And, of course,
the one we will care about
473
00:36:08 --> 00:36:11
actually is delta S expressed in
terms of delta A.
474
00:36:11 --> 00:36:13
But what are we going to do
with this cosine?
475
00:36:13 --> 00:36:15
It is not very convenient to
have a cosine left in here.
476
00:36:15 --> 00:36:19
Remember, the angle between two
planes is the same thing as the
477
00:36:19 --> 00:36:21
angle between the normal
vectors.
478
00:36:21 --> 00:36:23
If you want to see this angle
alpha elsewhere,
479
00:36:23 --> 00:36:27
what you can do is you can just
take the vertical direction.
480
00:36:27 --> 00:36:38
Let's take k.
Then here we have our angle
481
00:36:38 --> 00:36:43
alpha again.
In particular,
482
00:36:43 --> 00:36:46
cosine of alpha,
I can get, well,
483
00:36:46 --> 00:36:50
we know how to find the angle
between two vectors.
484
00:36:50 --> 00:36:57
If we have our normal vector N,
we will do N dot k,
485
00:36:57 --> 00:37:02
and we will divide by length N,
length k.
486
00:37:02 --> 00:37:06
Well, length k is one.
That is one easy guy.
487
00:37:06 --> 00:37:12
That is how we find the angle.
Now I am going to say,
488
00:37:12 --> 00:37:21
well, delta S is going to be
one over cosine alpha delta A.
489
00:37:21 --> 00:37:36
And I can rewrite that as
length of N divided by N dot k
490
00:37:36 --> 00:37:42
times delta A.
Now, let's multiply that by the
491
00:37:42 --> 00:37:48
unit normal vector.
Because what we are about is
492
00:37:48 --> 00:37:52
not so much dS but actually n
dS.
493
00:37:52 --> 00:38:04
N delta S will be,
I am just going to multiply by
494
00:38:04 --> 00:38:07
N.
Well, let's think for a second.
495
00:38:07 --> 00:38:11
What happens if I take a unit
normal N and I multiply it by
496
00:38:11 --> 00:38:14
the length of my other normal
big N?
497
00:38:14 --> 00:38:19
Well, I get big N again.
This is a normal vector of the
498
00:38:19 --> 00:38:22
same length as N,
well, up to sign.
499
00:38:22 --> 00:38:27
The only thing I don't know is
whether this guy will be going
500
00:38:27 --> 00:38:32
in the same direction as big N
or in the opposite direction.
501
00:38:32 --> 00:38:35
Say that, for example,
my capital N has,
502
00:38:35 --> 00:38:39
I don't know,
length three for example.
503
00:38:39 --> 00:38:43
Then the normal unit vector
might be this guy,
504
00:38:43 --> 00:38:47
in which case indeed three
times little n will be big n.
505
00:38:47 --> 00:38:52
Or it might be this one in
which case three times little n
506
00:38:52 --> 00:38:58
will be negative big N.
But up to sign it is N.
507
00:38:58 --> 00:39:02
And then I will have N over N
dot k delta A.
508
00:39:02 --> 00:39:07
And so the final formula,
the one that we care about in
509
00:39:07 --> 00:39:11
case you don't really like my
explanations of how we get
510
00:39:11 --> 00:39:19
there,
is that N dS is plus or minus N
511
00:39:19 --> 00:39:27
over N dot k dx dy.
That one is actually kind of
512
00:39:27 --> 00:39:31
useful so let's box it.
Now,
513
00:39:31 --> 00:39:35
just in case you are wondering,
of course, if you didn't want
514
00:39:35 --> 00:39:36
to project to x,
y,
515
00:39:36 --> 00:39:39
you would have maybe preferred
to project to say the plane of a
516
00:39:39 --> 00:39:40
blackboard, y,
z,
517
00:39:40 --> 00:39:45
well, you can do the same thing.
To express n dS in terms of dy
518
00:39:45 --> 00:39:49
dz you do the same argument.
Simply, the only thing that
519
00:39:49 --> 00:39:51
changes, instead of using the
vertical vector k,
520
00:39:51 --> 00:39:56
you use the normal vector i.
So you would be doing N over N
521
00:39:56 --> 00:39:58
dot i dy dz.
The same thing.
522
00:39:58 --> 00:40:04
So just keep an open mind that
this also works with other
523
00:40:04 --> 00:40:09
variables.
Anyway, that is how you can
524
00:40:09 --> 00:40:17
basically project the vectors of
this area element onto the x,
525
00:40:17 --> 00:40:23
y plane in a way.
Let's look at the special case
526
00:40:23 --> 00:40:29
just to see how this fits with
stuff we have seen before.
527
00:40:29 --> 00:40:37
Let's do a special example
where our surface is given by
528
00:40:37 --> 00:40:44
the equation z minus f of x,
y equals zero.
529
00:40:44 --> 00:40:46
That is a strange way to write
the equation.
530
00:40:46 --> 00:40:50
z equals f of x, y.
That we saw before.
531
00:40:50 --> 00:40:53
But now it looks like some
function of x,
532
00:40:53 --> 00:40:57
y, z equals zero.
Let's try to use this new
533
00:40:57 --> 00:41:05
method.
Let's call this guy g(x, y, z).
534
00:41:05 --> 00:41:07
Well, now let's look at the
normal vector.
535
00:41:07 --> 00:41:10
The normal vector would be the
gradient of g,
536
00:41:10 --> 00:41:13
you see.
What is the gradient of this
537
00:41:13 --> 00:41:17
function?
The gradient of g -- Well,
538
00:41:17 --> 00:41:22
partial g, partial x,
that is just negative partial
539
00:41:22 --> 00:41:25
f, partial x.
The y component,
540
00:41:25 --> 00:41:32
partial g, partial y is going
to be negative f sub y,
541
00:41:32 --> 00:41:42
and g sub z is just one.
Now, if you take N over N dot k
542
00:41:42 --> 00:41:46
dx dy,
well, it looks like it is going
543
00:41:46 --> 00:41:50
to be negative f sub x,
negative f sub y,
544
00:41:50 --> 00:41:53
one divided by -- Well,
what is N dot k?
545
00:41:53 --> 00:41:57
If you dot that with k you will
get just one,
546
00:41:57 --> 00:42:01
so I am not going to write it,
dx dy.
547
00:42:01 --> 00:42:05
See, that is again our favorite
formula.
548
00:42:05 --> 00:42:12
This one is actually more
general because you don't need
549
00:42:12 --> 00:42:18
to solve for z,
but if you cannot solve for z
550
00:42:18 --> 00:42:28
then it is the same as before.
I think that is enough formulas
551
00:42:28 --> 00:42:34
for n dS.
After spending a lot of time
552
00:42:34 --> 00:42:41
telling you how to compute
surface integrals,
553
00:42:41 --> 00:42:51
now I am going to try to tell
you how to avoid computing them.
554
00:42:51 --> 00:43:05
And that is called the
divergence theorem.
555
00:43:05 --> 00:43:09
And we will see the proof and
everything and applications on
556
00:43:09 --> 00:43:12
Tuesday, but I want to at least
the theorem and see how it works
557
00:43:12 --> 00:43:15
in one example.
It is also known as the
558
00:43:15 --> 00:43:19
Gauss-Green theorem or just the
Gauss theorem,
559
00:43:19 --> 00:43:24
depending in who you talk to.
The Green here is the same
560
00:43:24 --> 00:43:28
Green as in Green's theorem,
because somehow that is a space
561
00:43:28 --> 00:43:31
version of Green's theorem.
What does it say?
562
00:43:31 --> 00:43:43
It is 3D analog of Green for
flux.
563
00:43:43 --> 00:43:47
What it says is if S is a
closed surface -- Remember,
564
00:43:47 --> 00:43:52
it is the same as with Green's
theorem, we need to have
565
00:43:52 --> 00:43:56
something that is completely
enclosed.
566
00:43:56 --> 00:44:00
You have a surface and there is
somehow no gaps in it.
567
00:44:00 --> 00:44:06
There is no boundary to it.
It is really completely
568
00:44:06 --> 00:44:17
enclosing a region in space that
I will call D.
569
00:44:17 --> 00:44:19
And I need to choose my
orientation.
570
00:44:19 --> 00:44:29
The orientation that will work
for this theorem is choosing the
571
00:44:29 --> 00:44:43
normal vector to point outwards.
N needs to be outwards.
572
00:44:43 --> 00:44:46
That is one part of the puzzle.
The other part is a vector
573
00:44:46 --> 00:44:51
field.
I need to have a vector field
574
00:44:51 --> 00:44:58
that is defined and
differentiable -- -- everywhere
575
00:44:58 --> 00:45:04
in D, so same instructions as
usual.
576
00:45:04 --> 00:45:12
Then I don't have actually to
compute the flux integral.
577
00:45:12 --> 00:45:17
Double integral of f dot n dS
of a closed surface S.
578
00:45:17 --> 00:45:19
I am going to put a circle just
to remind you it is has got to
579
00:45:19 --> 00:45:22
be a closed surface.
It is just a notation to remind
580
00:45:22 --> 00:45:26
us closed surface.
I can replace that by the
581
00:45:26 --> 00:45:32
triple integral of a region
inside of divergence of F dV.
582
00:45:32 --> 00:45:36
Now, I need to tell you what
the divergence of a 3D vector
583
00:45:36 --> 00:45:39
field is.
Well, you will see that it is
584
00:45:39 --> 00:45:41
not much harder than in the 2D
case.
585
00:45:41 --> 00:45:58
What you do is just -- Say that
your vector field has components
586
00:45:58 --> 00:46:06
P, Q and R.
Then you will take P sub x Q
587
00:46:06 --> 00:46:10
sub y R sub z.
That is the definition.
588
00:46:10 --> 00:46:13
It is pretty easy to remember.
You take the x component
589
00:46:13 --> 00:46:18
partial respect to S plus
partial respect to y over y
590
00:46:18 --> 00:46:23
component plus partial respect
to z of the z component.
591
00:46:23 --> 00:46:33
For example,
last time we saw that the flux
592
00:46:33 --> 00:46:42
of the vector field zk through a
sphere of radius a was
593
00:46:42 --> 00:46:53
four-thirds pi a cubed by
computing the surface integral.
594
00:46:53 --> 00:46:56
Well, if we do it more
efficiently now by Green's
595
00:46:56 --> 00:46:59
theorem, we are going to use
Green's theorem for this sphere
596
00:46:59 --> 00:47:02
because we are doing the whole
sphere.
597
00:47:02 --> 00:47:04
It is fine.
It is a closed surface.
598
00:47:04 --> 00:47:05
We couldn't do it for,
say, the hemisphere or
599
00:47:05 --> 00:47:09
something like that.
Well, for a hemisphere we would
600
00:47:09 --> 00:47:15
need to add maybe the flat face
of a bottom or something like
601
00:47:15 --> 00:47:20
that.
Green's theorem says that our
602
00:47:20 --> 00:47:28
flux integral can actually be
replaced by the triple integral
603
00:47:28 --> 00:47:36
over the solid bowl of radius a
of the divergence of zk dV.
604
00:47:36 --> 00:47:40
But now what is the divergence
of this field?
605
00:47:40 --> 00:47:48
Well, you have zero,
zero, z so you get zero plus
606
00:47:48 --> 00:47:52
zero plus one.
It looks like it will be one.
607
00:47:52 --> 00:47:59
If you do the triple integral
of 1dV, you will get just the
608
00:47:59 --> 00:48:06
volume -- -- of the region
inside, which is four-thirds by
609
00:48:06 --> 00:48:09
a cubed.
And so it was no accident.
610
00:48:09 --> 00:48:13
In fact, before that we looked
at also xi yj zk and we found
611
00:48:13 --> 00:48:17
three times the volume.
That is because the divergence
612
00:48:17 --> 00:48:19
of that field was actually
three.
613
00:48:19 --> 00:48:22
Very quickly,
let me just say what this means
614
00:48:22 --> 00:48:24
physically.
Physically, see,
615
00:48:24 --> 00:48:29
this guy on the left is the
total amount of stuff that goes
616
00:48:29 --> 00:48:34
out of the region per unit time.
I want to figure out how much
617
00:48:34 --> 00:48:37
stuff comes out of there.
What does the divergence mean?
618
00:48:37 --> 00:48:41
The divergence means it
measures how much the flow is
619
00:48:41 --> 00:48:43
expanding things.
It measures how much,
620
00:48:43 --> 00:48:46
I said that probably when we
were trying to understand 2D
621
00:48:46 --> 00:48:49
divergence.
It measures the amount of
622
00:48:49 --> 00:48:54
sources or sinks that you have
inside your fluid.
623
00:48:54 --> 00:48:57
Now it becomes commonsense.
If you take a region of space,
624
00:48:57 --> 00:49:01
the total amount of water that
flows out of it is the total
625
00:49:01 --> 00:49:05
amount of sources that you have
in there minus the sinks.
626
00:49:05 --> 00:49:08
I mean, in spite of this
commonsense explanation,
627
00:49:08 --> 00:49:10
we are going to see how to
prove this.
628
00:49:10 --> 00:49:14
And we will see how it works
and what it says.
629
00:49:14 --> 00:49:15