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Yesterday we learned about flux
and we have seen the first few

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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

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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,

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00:01:06 --> 00:01:09
we have various formulas
telling us what the normal

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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.

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00:01:25 --> 00:01:29
At the end of lecture we saw a
formula.

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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

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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,
<delta x,

104
00:08:08 --> 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 <x,

300
00:22:40 --> 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 <a,
b, c>.

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