1
00:00:00,000 --> 00:00:08,070
JOEL LEWIS: Hi.

2
00:00:08,070 --> 00:00:09,570
Welcome back to recitation.

3
00:00:09,570 --> 00:00:12,510
In lecture, you've been learning
about two-dimensional flux

4
00:00:12,510 --> 00:00:14,990
and the normal form of
Green's theorem-- so normal

5
00:00:14,990 --> 00:00:19,750
here meaning perpendicular--
so I want to give you

6
00:00:19,750 --> 00:00:20,710
a problem about that.

7
00:00:20,710 --> 00:00:23,830
So what I'd like you
to do is to verify

8
00:00:23,830 --> 00:00:27,140
Green's theorem in normal form
for this particular field,

9
00:00:27,140 --> 00:00:32,310
the field F equals x i hat
plus y j hat and the curve C

10
00:00:32,310 --> 00:00:35,260
that consists of the upper
half of the unit circle

11
00:00:35,260 --> 00:00:37,760
and the x-axis
interval minus 1 to 1.

12
00:00:37,760 --> 00:00:40,900
So first of all, let me
say what I mean by this C.

13
00:00:40,900 --> 00:00:45,500
So C, it's the usual unit
circle, circle of radius 1

14
00:00:45,500 --> 00:00:47,550
centered at the origin,
so just its top half,

15
00:00:47,550 --> 00:00:49,400
And the x-axis
interval minus 1 to 1,

16
00:00:49,400 --> 00:00:53,590
I mean the line segment that
connects the points (-1, 0)

17
00:00:53,590 --> 00:00:57,460
and (1, 0), so the diameter
of that semicircle.

18
00:00:57,460 --> 00:01:01,340
So that's the curve
C and the field F.

19
00:01:01,340 --> 00:01:03,780
What I mean by verify
Green's theorem

20
00:01:03,780 --> 00:01:08,520
is I'd like you to compute
both the double integral that

21
00:01:08,520 --> 00:01:11,672
appears in Green's theorem and
the line integral that appears

22
00:01:11,672 --> 00:01:13,380
in Green's theorem
and check that they're

23
00:01:13,380 --> 00:01:14,700
really equal to each other.

24
00:01:14,700 --> 00:01:17,160
So that'll confirm
Green's theorem

25
00:01:17,160 --> 00:01:19,060
in this particular
instance and hopefully

26
00:01:19,060 --> 00:01:22,880
help give us a feel for
how it works a little bit.

27
00:01:22,880 --> 00:01:25,920
So why don't you pause
the video, have a go,

28
00:01:25,920 --> 00:01:28,730
compute both of those
integrals, come back

29
00:01:28,730 --> 00:01:30,670
and we can work
them out together.

30
00:01:38,447 --> 00:01:40,030
Hopefully you had
some luck with this.

31
00:01:40,030 --> 00:01:41,390
Let's have a go at it.

32
00:01:41,390 --> 00:01:44,520
So what Green's
theorem tells you

33
00:01:44,520 --> 00:01:47,240
is that the flux across
a curve, which we usually

34
00:01:47,240 --> 00:01:49,470
compute as a line
integral, is also

35
00:01:49,470 --> 00:01:53,844
equal to an integral of the
divergence over the region

36
00:01:53,844 --> 00:01:54,760
bounded by that curve.

37
00:01:54,760 --> 00:01:57,950
So here the curve has to be
a closed curve so that it

38
00:01:57,950 --> 00:01:59,960
bounds a region of the plane.

39
00:01:59,960 --> 00:02:04,650
So in particular, let's
draw the picture here

40
00:02:04,650 --> 00:02:06,630
so we know what
we're talking about.

41
00:02:06,630 --> 00:02:14,000
So the curve C, so we've got
the segment from minus 1 to 1

42
00:02:14,000 --> 00:02:16,419
along the x-axis, and
then we have the top half

43
00:02:16,419 --> 00:02:17,210
of the unit circle.

44
00:02:23,180 --> 00:02:26,820
That's the curve C. And I
didn't specify an orientation

45
00:02:26,820 --> 00:02:29,302
but in any context
like this, when

46
00:02:29,302 --> 00:02:31,052
you don't specify an
orientation, what you

47
00:02:31,052 --> 00:02:32,650
mean is positively oriented.

48
00:02:32,650 --> 00:02:37,880
So it's a positively
oriented curve.

49
00:02:37,880 --> 00:02:39,320
So that's our curve C.

50
00:02:39,320 --> 00:02:44,310
And so the region that
it bounds is this half

51
00:02:44,310 --> 00:02:48,210
of the circle, the semicircle.

52
00:02:48,210 --> 00:02:50,730
I keep saying half
of the semicircle.

53
00:02:50,730 --> 00:02:52,450
Sorry about that.

54
00:02:52,450 --> 00:02:57,160
This upper half of the
circle, of the disc, in fact.

55
00:02:57,160 --> 00:02:58,830
That region of the plane.

56
00:02:58,830 --> 00:03:01,370
OK, so what Green's
theorem tells

57
00:03:01,370 --> 00:03:06,410
us is that when we compute
the surface integral-- sorry,

58
00:03:06,410 --> 00:03:11,380
the double integral--
of the divergence of F

59
00:03:11,380 --> 00:03:14,060
over this region, that
should be the same as what

60
00:03:14,060 --> 00:03:16,920
we do if we compute F
dot n around the boundary

61
00:03:16,920 --> 00:03:18,860
of the curve.

62
00:03:18,860 --> 00:03:20,540
And now we're going to check it.

63
00:03:20,540 --> 00:03:23,635
So let's do the
double integral first.

64
00:03:31,310 --> 00:03:32,980
The double integral
in this case,

65
00:03:32,980 --> 00:03:36,130
so it's the double
integral over--

66
00:03:36,130 --> 00:03:41,650
let me call that region R-- so
it's the double integral over R

67
00:03:41,650 --> 00:03:48,790
of the divergence of F dA.

68
00:03:48,790 --> 00:03:50,380
So what is the divergence of F?

69
00:03:50,380 --> 00:03:53,650
Well, here's F. It's
x i hat plus y j hat.

70
00:03:53,650 --> 00:03:56,400
So its divergence
is the partial of x

71
00:03:56,400 --> 00:03:59,760
with respect to x, plus the
partial of y with respect to y.

72
00:03:59,760 --> 00:04:00,800
So that's 1 plus 1.

73
00:04:00,800 --> 00:04:03,990
So the divergence of F
is just 2, in this case.

74
00:04:03,990 --> 00:04:06,120
So it's equal to
the double integral

75
00:04:06,120 --> 00:04:11,570
over the semicircle of 2 dA.

76
00:04:11,570 --> 00:04:15,290
And of course when you integrate
a constant over a region, what

77
00:04:15,290 --> 00:04:18,330
you get is just that constant
times the area of the region.

78
00:04:18,330 --> 00:04:22,640
So dA here is the area
of the semicircle,

79
00:04:22,640 --> 00:04:25,820
it's half of a
circle of radius 1.

80
00:04:25,820 --> 00:04:28,980
So circle of radius
1 has area pi.

81
00:04:28,980 --> 00:04:31,790
So this is 2 times 1/2 pi.

82
00:04:31,790 --> 00:04:34,720
So this is pi.

83
00:04:34,720 --> 00:04:36,010
OK.

84
00:04:36,010 --> 00:04:38,790
So that's the double
integral that we

85
00:04:38,790 --> 00:04:41,880
get from Green's
theorem in normal form.

86
00:04:41,880 --> 00:04:45,100
And what Green's theorem
says is that this

87
00:04:45,100 --> 00:04:47,602
is equal to a particular
line integral.

88
00:04:47,602 --> 00:04:48,810
So what is the line integral?

89
00:04:48,810 --> 00:04:54,110
Well, it's the integral
around C of F dot n.

90
00:04:54,110 --> 00:05:02,770
So let's write down what it
is, the line integral part now.

91
00:05:02,770 --> 00:05:04,620
So it's an integral.

92
00:05:04,620 --> 00:05:05,940
It's a closed curve.

93
00:05:05,940 --> 00:05:20,100
So it's integral
around C of F dot n ds.

94
00:05:20,100 --> 00:05:27,580
So that is the line integral
that we're looking to compute.

95
00:05:27,580 --> 00:05:29,035
So how do we compute this?

96
00:05:29,035 --> 00:05:33,270
Well, usually we compute it
by using the coordinates of F.

97
00:05:33,270 --> 00:05:37,200
So this is equal to, and we know
that this is always equal to,

98
00:05:37,200 --> 00:05:40,982
the integral around
C of-- and let

99
00:05:40,982 --> 00:05:43,440
me make sure I'm getting this
right before I screw anything

100
00:05:43,440 --> 00:05:43,680
up.

101
00:05:43,680 --> 00:05:44,179
Yes.

102
00:05:44,179 --> 00:05:44,690
OK, I am.

103
00:05:44,690 --> 00:05:45,270
Good.

104
00:05:45,270 --> 00:05:52,490
M*dy minus N*dx, where M and
N are the coordinates of F--

105
00:05:52,490 --> 00:05:56,370
or the components of F. M and
N are the components of F.

106
00:05:56,370 --> 00:05:59,640
So M is the first component
and N is the second component.

107
00:05:59,640 --> 00:06:02,530
So in our case, F has
this fairly simple form.

108
00:06:02,530 --> 00:06:05,940
So in our case, this is
equal to the integral

109
00:06:05,940 --> 00:06:13,155
around C of-- so M, the
first component of F is x.

110
00:06:13,155 --> 00:06:20,120
So this is x*dy minus-- the
second component is y*dx.

111
00:06:20,120 --> 00:06:24,730
So this is the line integral
we're interested in computing.

112
00:06:24,730 --> 00:06:27,490
But, of course,
this curve is not

113
00:06:27,490 --> 00:06:29,716
easy to parameterize
as a single go.

114
00:06:29,716 --> 00:06:31,340
So we want to split
it into two pieces.

115
00:06:31,340 --> 00:06:32,710
So let's look.

116
00:06:32,710 --> 00:06:35,320
So the first piece we
want to split it into

117
00:06:35,320 --> 00:06:36,700
is that line segment.

118
00:06:36,700 --> 00:06:40,210
So let's call that maybe--
well, I'm not even going

119
00:06:40,210 --> 00:06:42,890
to bother giving them names.

120
00:06:42,890 --> 00:06:45,090
We want to split it into
the integral over the line

121
00:06:45,090 --> 00:06:48,700
segment plus the integral
over the semicircle.

122
00:06:48,700 --> 00:06:52,040
The boundary of that--
yes, the semicircle.

123
00:06:52,040 --> 00:06:54,960
So this is equal to-- so it's
the integral over the line

124
00:06:54,960 --> 00:06:56,180
segment.

125
00:06:56,180 --> 00:06:57,730
So let's see.

126
00:06:57,730 --> 00:06:59,940
So that integral-- well,
OK, I will give them names.

127
00:06:59,940 --> 00:07:00,898
I will give them names.

128
00:07:00,898 --> 00:07:01,560
I take it back.

129
00:07:01,560 --> 00:07:04,300
We'll call the line
segment C_1 and we'll

130
00:07:04,300 --> 00:07:05,930
call the semicircle C_2.

131
00:07:05,930 --> 00:07:09,310
So it's equal to the
integral over C_1--

132
00:07:09,310 --> 00:07:11,760
OK, well, what is the
integral over C_1?

133
00:07:11,760 --> 00:07:14,260
What are x and dy and
y and dx in this case?

134
00:07:14,260 --> 00:07:18,670
So in this case, well, x
is what we're integrating.

135
00:07:18,670 --> 00:07:22,820
But dy, we're on this line
segment, y isn't changing.

136
00:07:22,820 --> 00:07:24,110
y is constant.

137
00:07:24,110 --> 00:07:26,070
So dy is just 0.

138
00:07:26,070 --> 00:07:32,990
So it's 0 minus-- OK, and now on
this line segment y is 0 also.

139
00:07:32,990 --> 00:07:35,260
So it's 0*dx.

140
00:07:35,260 --> 00:07:37,790
So the first integral,
the integral over C_1,

141
00:07:37,790 --> 00:07:39,870
is the integral of 0.

142
00:07:39,870 --> 00:07:49,998
Plus-- we have to
integrate over C_2 of-- OK,

143
00:07:49,998 --> 00:07:56,870
so x*dy minus y*dx.

144
00:07:56,870 --> 00:07:57,370
All right.

145
00:07:57,370 --> 00:07:59,286
So this one's just going
to be 0, that's easy.

146
00:07:59,286 --> 00:08:01,720
So now we just have to
work with this second one.

147
00:08:01,720 --> 00:08:03,820
OK, so for the second
one, we're integrating

148
00:08:03,820 --> 00:08:06,130
over the semicircle, so we
want to parameterize it.

149
00:08:06,130 --> 00:08:08,213
And we're going to use our
usual parameterization.

150
00:08:08,213 --> 00:08:11,340
x equals cosine t,
y equals sine t.

151
00:08:11,340 --> 00:08:14,710
And in this case, we just
want to do this semicircle.

152
00:08:14,710 --> 00:08:17,660
So we just want to go from
(1, 0) all the way around to

153
00:08:17,660 --> 00:08:18,170
(-1, 0).

154
00:08:18,170 --> 00:08:21,910
So that is from
going from 0 to pi.

155
00:08:21,910 --> 00:08:25,170
So this is equal to-- so this
first integral is just 0.

156
00:08:25,170 --> 00:08:27,430
It's the integral
of 0 and that's 0.

157
00:08:27,430 --> 00:08:31,480
So it's the integral-- OK, so
t is going to go from 0 to pi.

158
00:08:31,480 --> 00:08:37,100
So now x*dy So x is cosine t.

159
00:08:37,100 --> 00:08:39,500
y is sine t.

160
00:08:39,500 --> 00:08:42,890
So dy is sine t dt.

161
00:08:42,890 --> 00:08:44,420
Sorry, it's cosine t dt.

162
00:08:44,420 --> 00:08:45,910
y is sine t.

163
00:08:45,910 --> 00:08:48,840
dy is cosine t dt.

164
00:08:48,840 --> 00:08:54,710
So it's cosine t times cosine
t dt, minus-- all right,

165
00:08:54,710 --> 00:08:57,500
now y is sine t again.

166
00:09:00,390 --> 00:09:02,820
And x is cosine t.

167
00:09:02,820 --> 00:09:07,000
So dx is minus sine t dt.

168
00:09:07,000 --> 00:09:10,777
So this is times
minus sine t dt.

169
00:09:13,950 --> 00:09:15,140
OK, well what happens here?

170
00:09:15,140 --> 00:09:21,930
So this becomes cosine squared
t dt minus minus sine squared t.

171
00:09:21,930 --> 00:09:22,930
Minus minus is plus.

172
00:09:22,930 --> 00:09:27,620
So it's cosine squared t
dt plus sine squared t dt.

173
00:09:27,620 --> 00:09:30,500
But, of course, cosine squared
plus sine squared is just 1.

174
00:09:30,500 --> 00:09:36,253
So we can write this even more
simply as the integral from 0

175
00:09:36,253 --> 00:09:43,660
to pi of 1 dt or just dt, and
the integral dt is just t.

176
00:09:43,660 --> 00:09:48,881
So this is t between
0 and pi, which is pi.

177
00:09:48,881 --> 00:09:49,380
Whew!

178
00:09:49,380 --> 00:09:50,060
OK.

179
00:09:50,060 --> 00:09:50,991
Pi.

180
00:09:50,991 --> 00:09:51,490
Good.

181
00:09:51,490 --> 00:09:52,910
And what did we get before?

182
00:09:52,910 --> 00:09:54,100
We also got pi.

183
00:09:54,100 --> 00:09:54,990
Great.

184
00:09:54,990 --> 00:09:58,550
So we have successfully
verified Green's theorem

185
00:09:58,550 --> 00:10:01,210
in normal form in this
particular instance.

186
00:10:01,210 --> 00:10:03,990
So let's just recap
again what we did.

187
00:10:03,990 --> 00:10:08,340
We had this field
F and this curve

188
00:10:08,340 --> 00:10:11,970
C. This closed curve C
that bounded some region.

189
00:10:11,970 --> 00:10:15,130
And so what we've done is we
computed the double integral

190
00:10:15,130 --> 00:10:18,850
over the region of div F dA.

191
00:10:18,850 --> 00:10:20,540
So that's what we did first.

192
00:10:20,540 --> 00:10:23,140
Double integral over the
region bounded by the curve.

193
00:10:23,140 --> 00:10:25,630
And then second, we
computed the line

194
00:10:25,630 --> 00:10:31,490
integral, around the
boundary, of F dot n ds.

195
00:10:31,490 --> 00:10:34,410
So that was what--
this is always

196
00:10:34,410 --> 00:10:37,950
a useful form in which
to write this F dot n dA.

197
00:10:37,950 --> 00:10:41,620
OK, and then we substituted
and computed it.

198
00:10:41,620 --> 00:10:44,280
And Green's theorem tells
us that the two integrals

199
00:10:44,280 --> 00:10:45,600
have to be equal to each other.

200
00:10:45,600 --> 00:10:48,450
And indeed, for this particular
F and this particular C,

201
00:10:48,450 --> 00:10:52,084
we verified that in this
case they both give us pi.

202
00:10:52,084 --> 00:10:53,750
And, of course, Green's
theorem tells us

203
00:10:53,750 --> 00:10:56,125
that that would have been
true, that they would have come

204
00:10:56,125 --> 00:10:58,160
out the same, regardless
of what the choice of F

205
00:10:58,160 --> 00:11:00,620
and the choice of C
that we made were.

206
00:11:00,620 --> 00:11:02,296
So I'll stop there.