1
00:00:00,000 --> 00:00:06,866


2
00:00:06,866 --> 00:00:07,490
JOEL LEWIS: Hi.

3
00:00:07,490 --> 00:00:08,970
Welcome back to recitation.

4
00:00:08,970 --> 00:00:11,150
In lecture, you've been
learning about computing

5
00:00:11,150 --> 00:00:13,240
double integrals
and about changing

6
00:00:13,240 --> 00:00:14,360
the order of integration.

7
00:00:14,360 --> 00:00:16,555
And how you can look
at a given region

8
00:00:16,555 --> 00:00:19,470
and you can integrate over
it by integrating dx dy

9
00:00:19,470 --> 00:00:21,380
or by integrating dy dx.

10
00:00:21,380 --> 00:00:23,470
So here I have some examples.

11
00:00:23,470 --> 00:00:25,560
I have two regions.

12
00:00:25,560 --> 00:00:27,930
So one region is
the triangle whose

13
00:00:27,930 --> 00:00:30,840
vertices are the origin,
the point (0, 2),

14
00:00:30,840 --> 00:00:33,120
and the point minus 1, 2.

15
00:00:33,120 --> 00:00:35,530
And the other one is
a sector of a circle.

16
00:00:35,530 --> 00:00:38,690
So the circle has a radius 2
and is centered at the origin.

17
00:00:38,690 --> 00:00:42,290
And I want the part of that
circle that's above the x-axis

18
00:00:42,290 --> 00:00:44,860
and below the line y equals x.

19
00:00:44,860 --> 00:00:46,450
So what I'd like
you to do is I'd

20
00:00:46,450 --> 00:00:48,740
like you to write down
what a double integral

21
00:00:48,740 --> 00:00:51,000
over these regions
looks like, but I'd

22
00:00:51,000 --> 00:00:52,550
like you to do it
two different ways.

23
00:00:52,550 --> 00:00:55,750
I'd like you to do it as an
iterated integral in the order

24
00:00:55,750 --> 00:00:57,610
dx dy.

25
00:00:57,610 --> 00:01:00,590
And I'd also like you to do
it as an iterated integral

26
00:01:00,590 --> 00:01:01,890
in the order dy dx.

27
00:01:01,890 --> 00:01:03,880
So I'd like you to
express the integrals

28
00:01:03,880 --> 00:01:06,930
over these regions in
terms of iterated integrals

29
00:01:06,930 --> 00:01:08,860
in both possible orders.

30
00:01:08,860 --> 00:01:11,371
So why don't you pause the
video, have a go at that,

31
00:01:11,371 --> 00:01:13,120
come back, and we can
work on it together.

32
00:01:13,120 --> 00:01:21,486


33
00:01:21,486 --> 00:01:22,860
So the first thing
to do whenever

34
00:01:22,860 --> 00:01:24,430
you're given a
problem like this--

35
00:01:24,430 --> 00:01:28,620
and in fact, almost anytime you
have to do a double integral--

36
00:01:28,620 --> 00:01:30,620
is to try and understand
the region in question.

37
00:01:30,620 --> 00:01:32,360
It's always a good
idea to understand

38
00:01:32,360 --> 00:01:33,450
the region in question.

39
00:01:33,450 --> 00:01:35,570
And by understand the
region in question,

40
00:01:35,570 --> 00:01:38,011
really the first thing that
I mean is draw a picture.

41
00:01:38,011 --> 00:01:38,510
All right.

42
00:01:38,510 --> 00:01:41,580
So let's do part a first.

43
00:01:41,580 --> 00:01:46,160
So in part a, you
have a triangle,

44
00:01:46,160 --> 00:01:50,820
it has vertices at the
origin, at the point (0, 2),

45
00:01:50,820 --> 00:01:53,070
and at the point minus 1, 2.

46
00:01:53,070 --> 00:01:56,180


47
00:01:56,180 --> 00:02:00,620
So this triangle is
our region in question.

48
00:02:00,620 --> 00:02:03,150
So now that we've got a
picture of it, we can talk

49
00:02:03,150 --> 00:02:06,210
and we can say, what
are the boundaries

50
00:02:06,210 --> 00:02:07,307
of this region, right?

51
00:02:07,307 --> 00:02:09,140
And we want to know
what its boundaries are.

52
00:02:09,140 --> 00:02:14,190
So the top boundary is
the line y equals 2,

53
00:02:14,190 --> 00:02:17,610
the right boundary is
the line x equals 0,

54
00:02:17,610 --> 00:02:21,260
and this sort of lower left
boundary-- the slanted line--

55
00:02:21,260 --> 00:02:25,620
is the line y equals minus 2x.

56
00:02:25,620 --> 00:02:26,120
OK.

57
00:02:26,120 --> 00:02:30,480
So those are the boundary
edges of this triangle.

58
00:02:30,480 --> 00:02:33,017
And so now what we
want to figure out

59
00:02:33,017 --> 00:02:35,350
is we want to figure out, OK,
if you're integrating this

60
00:02:35,350 --> 00:02:37,910
with respect to x and then y,
or if you're integrating this

61
00:02:37,910 --> 00:02:39,930
with respect to y
and then x, what

62
00:02:39,930 --> 00:02:41,890
does that integral look
like when you set it up

63
00:02:41,890 --> 00:02:43,080
as a double integral.

64
00:02:43,080 --> 00:02:46,184
So let's start on one of them.

65
00:02:46,184 --> 00:02:47,350
It doesn't matter which one.

66
00:02:47,350 --> 00:02:53,520
So let's try and write
the double integral

67
00:02:53,520 --> 00:02:56,830
over this region R
in the order dx dy.

68
00:02:56,830 --> 00:03:04,010
OK, so we have
inside bounds dx dy.

69
00:03:04,010 --> 00:03:05,030
So OK.

70
00:03:05,030 --> 00:03:07,700
So we need to find
the bounds on x first,

71
00:03:07,700 --> 00:03:11,540
and those bounds are
going to be in terms of y.

72
00:03:11,540 --> 00:03:12,800
So the bounds on x.

73
00:03:12,800 --> 00:03:14,790
So that means when we
look at this region, what

74
00:03:14,790 --> 00:03:19,250
we want to figure out is we want
to figure out for a given value

75
00:03:19,250 --> 00:03:23,302
y, what is the
leftmost point and what

76
00:03:23,302 --> 00:03:24,260
is the rightmost point?

77
00:03:24,260 --> 00:03:25,520
What are the bounds on x?

78
00:03:25,520 --> 00:03:32,070
So for given value
y, the largest value

79
00:03:32,070 --> 00:03:36,740
x is going to take is
along this line x equals 0.

80
00:03:36,740 --> 00:03:40,430
When you fix some value
of y, the rightmost point

81
00:03:40,430 --> 00:03:45,710
that x can reach in this region
is at this line x equals 0.

82
00:03:45,710 --> 00:03:48,660
So x is going to go up to 0.

83
00:03:48,660 --> 00:03:51,230
That's going to be
its upper bound.

84
00:03:51,230 --> 00:03:54,980
The lower bound is going to be
the left edge of our region.

85
00:03:54,980 --> 00:03:57,890


86
00:03:57,890 --> 00:04:01,860
For a given value of y, what is
that leftmost boundary value?

87
00:04:01,860 --> 00:04:04,030
So what we want to
do is we want to take

88
00:04:04,030 --> 00:04:07,310
that equation for that boundary
and we want to solve it

89
00:04:07,310 --> 00:04:09,780
for x in terms of y.

90
00:04:09,780 --> 00:04:11,750
So that's not hard
to do in this case.

91
00:04:11,750 --> 00:04:13,860
The line y equals
minus 2x is also

92
00:04:13,860 --> 00:04:17,060
the line x equals minus 1/2 y.

93
00:04:17,060 --> 00:04:22,690
So that's that left
boundary: minus 1/2 y.

94
00:04:22,690 --> 00:04:23,320
OK?

95
00:04:23,320 --> 00:04:25,680
So then our outer bounds are dy.

96
00:04:25,680 --> 00:04:28,360
So we want to find the
absolute bounds on y.

97
00:04:28,360 --> 00:04:30,121
What's the smallest
value that y takes,

98
00:04:30,121 --> 00:04:31,870
and what's the largest
value that y takes?

99
00:04:31,870 --> 00:04:34,782
So that means what's the
lowest point of this region

100
00:04:34,782 --> 00:04:35,740
and what's the highest?

101
00:04:35,740 --> 00:04:38,000
And so the lowest point
here is the origin.

102
00:04:38,000 --> 00:04:41,160
So that's when y
takes the value of 0.

103
00:04:41,160 --> 00:04:43,390
And the highest point-- the
very top of this region--

104
00:04:43,390 --> 00:04:45,790
is when y equals 2.

105
00:04:45,790 --> 00:04:46,900
OK.

106
00:04:46,900 --> 00:04:49,390
So this is what
that double integral

107
00:04:49,390 --> 00:04:54,460
is going to become when we
evaluate it in the order dx dy.

108
00:04:54,460 --> 00:04:56,544
So now let's talk
about evaluating it

109
00:04:56,544 --> 00:04:57,460
in the opposite order.

110
00:04:57,460 --> 00:05:01,310
So let's switch our
bounds for dy dx.

111
00:05:01,310 --> 00:05:07,270
So we want the double
integral over R, dy dx.

112
00:05:07,270 --> 00:05:11,410
OK, so this is going to
be an iterated integral.

113
00:05:11,410 --> 00:05:14,150
And this time the
inner bounds are

114
00:05:14,150 --> 00:05:17,000
going to be for y in terms
of x, and the outer bounds

115
00:05:17,000 --> 00:05:19,030
are going to be
absolute bounds on x.

116
00:05:19,030 --> 00:05:21,050
So for y in terms
of x, that means we

117
00:05:21,050 --> 00:05:25,480
look at this region-- we want
to know, for a fixed value of x,

118
00:05:25,480 --> 00:05:27,860
what's the bottom
boundary of this region,

119
00:05:27,860 --> 00:05:29,820
and what's the top boundary?

120
00:05:29,820 --> 00:05:32,830
So here, it's easy to see
that the bottom boundary is

121
00:05:32,830 --> 00:05:35,840
this line y equals minus
2x, and the top boundary

122
00:05:35,840 --> 00:05:37,810
is this line y equals 2.

123
00:05:37,810 --> 00:05:42,310
So y is going from
minus 2x to 2.

124
00:05:42,310 --> 00:05:42,840
Yeah?

125
00:05:42,840 --> 00:05:49,355
So for a fixed value of x,
the values of y that give you

126
00:05:49,355 --> 00:05:51,330
a point in this
region are the values

127
00:05:51,330 --> 00:05:54,205
that y is at least
minus 2x and at most 2.

128
00:05:54,205 --> 00:05:55,840
So OK.

129
00:05:55,840 --> 00:05:57,510
And now we need
the outer bounds.

130
00:05:57,510 --> 00:06:00,900
So the outer bounds have
to be some real numbers,

131
00:06:00,900 --> 00:06:02,740
Those are the
absolute bounds on x.

132
00:06:02,740 --> 00:06:04,830
So we need to know what
the absolute leftmost

133
00:06:04,830 --> 00:06:07,490
point and the absolute rightmost
point in this region are.

134
00:06:07,490 --> 00:06:11,510
And so the absolute leftmost
point is this point minus 1, 2.

135
00:06:11,510 --> 00:06:14,040
So that has an
x-value of minus 1.

136
00:06:14,040 --> 00:06:16,960
And the absolute
rightmost point is along

137
00:06:16,960 --> 00:06:20,280
this right edge at x equals 0.

138
00:06:20,280 --> 00:06:20,970
OK.

139
00:06:20,970 --> 00:06:25,910
So here are the two integrals.

140
00:06:25,910 --> 00:06:28,670
The double integral with
respect to x then y,

141
00:06:28,670 --> 00:06:33,870
and the double integral with
respect to y and then x.

142
00:06:33,870 --> 00:06:34,400
OK.

143
00:06:34,400 --> 00:06:36,450
So that's the answer to part a.

144
00:06:36,450 --> 00:06:39,730
Let's go on to part b.

145
00:06:39,730 --> 00:06:47,540
So for part b, our region is
we take a circle of radius 2,

146
00:06:47,540 --> 00:06:51,690
and we take the line y
equals x, and we take

147
00:06:51,690 --> 00:06:55,980
the line that's the x-axis.

148
00:06:55,980 --> 00:07:00,830
And so we want a circle,
and we want this sector

149
00:07:00,830 --> 00:07:02,852
of the circle in here.

150
00:07:02,852 --> 00:07:09,390
So this region inside the
circle, below the line y

151
00:07:09,390 --> 00:07:11,590
equal x, and above the x-axis.

152
00:07:11,590 --> 00:07:13,311
So this wedge of this circle.

153
00:07:13,311 --> 00:07:13,810
Let's see.

154
00:07:13,810 --> 00:07:19,420
This value is at x equals
2, this is the origin,

155
00:07:19,420 --> 00:07:22,212
and this is the
point square root

156
00:07:22,212 --> 00:07:24,830
of 2 comma square root of 2.

157
00:07:24,830 --> 00:07:28,430
That common point of the line
y equals x and the circle

158
00:07:28,430 --> 00:07:30,832
x squared plus y
squared equals 4.

159
00:07:30,832 --> 00:07:32,290
That's what this
boundary curve is:

160
00:07:32,290 --> 00:07:35,930
x squared plus y
squared equals 4.

161
00:07:35,930 --> 00:07:39,310
And of course, this boundary
curve is the line y equals x.

162
00:07:39,310 --> 00:07:42,450
And this boundary line
is the x-axis, which

163
00:07:42,450 --> 00:07:44,910
has the equation y equals 0.

164
00:07:44,910 --> 00:07:46,910
So those are our boundary
curves for our region.

165
00:07:46,910 --> 00:07:48,740
We've got this nice
picture, so now we

166
00:07:48,740 --> 00:07:56,110
can talk about expressing it
as an iterated integral in two

167
00:07:56,110 --> 00:07:56,900
different orders.

168
00:07:56,900 --> 00:08:01,140
So let's again start off with
this with respect to x first,

169
00:08:01,140 --> 00:08:02,710
and then with respect to y.

170
00:08:02,710 --> 00:08:08,000
So we want the double
integral over R, dx dy.

171
00:08:08,000 --> 00:08:15,360
So this should be an iterated
integral, something dx and then

172
00:08:15,360 --> 00:08:16,280
dy.

173
00:08:16,280 --> 00:08:18,340
OK, so we need
bounds on x, which

174
00:08:18,340 --> 00:08:20,050
means for a fixed
value of y, we need

175
00:08:20,050 --> 00:08:22,040
to know what is the
leftmost boundary

176
00:08:22,040 --> 00:08:23,660
and what's the rightmost bound.

177
00:08:23,660 --> 00:08:27,940
So for a fixed
value of y, we want

178
00:08:27,940 --> 00:08:30,665
to know what the left edge
is and the right edge is.

179
00:08:30,665 --> 00:08:32,960
And it's easy to see because
we've drawn this picture,

180
00:08:32,960 --> 00:08:33,460
right?

181
00:08:33,460 --> 00:08:36,750
Drawing the picture makes
this a much easier process.

182
00:08:36,750 --> 00:08:40,275
The left edge is this line y
equals x and the right edge

183
00:08:40,275 --> 00:08:41,960
is our actual circle.

184
00:08:41,960 --> 00:08:42,460
Yeah?

185
00:08:42,460 --> 00:08:45,550
So those are the left
and right boundaries,

186
00:08:45,550 --> 00:08:48,230
so what we put here are just
the equations of that left edge

187
00:08:48,230 --> 00:08:52,100
and the equation
of that right edge.

188
00:08:52,100 --> 00:08:56,840
But we want their equations in
the form x equals something.

189
00:08:56,840 --> 00:08:58,910
And that's the something
that we put there.

190
00:08:58,910 --> 00:09:04,750
So for this left edge,
it's the line x equals y.

191
00:09:04,750 --> 00:09:08,790
So the left bound is y there.

192
00:09:08,790 --> 00:09:11,810
In this region, x is at least y.

193
00:09:11,810 --> 00:09:13,696
And the upper bound
here, which is

194
00:09:13,696 --> 00:09:15,570
going to be the rightmost
bound-- the largest

195
00:09:15,570 --> 00:09:18,220
value that x takes--
is when x squared

196
00:09:18,220 --> 00:09:19,820
plus y squared equals 4.

197
00:09:19,820 --> 00:09:23,350
So when x is equal to the square
root of 4 minus y squared.

198
00:09:23,350 --> 00:09:27,030


199
00:09:27,030 --> 00:09:28,880
Now you might say
to me, why do I

200
00:09:28,880 --> 00:09:31,310
know that it's the
positive square root here

201
00:09:31,310 --> 00:09:32,810
and not the negative
square root?

202
00:09:32,810 --> 00:09:34,910
And if you said
that to yourself,

203
00:09:34,910 --> 00:09:35,910
that's a great question.

204
00:09:35,910 --> 00:09:38,860
And the answer is that
this part of the circle

205
00:09:38,860 --> 00:09:41,140
is the top half of the
circle and it's also

206
00:09:41,140 --> 00:09:42,600
the right half of the circle.

207
00:09:42,600 --> 00:09:45,010
So here we have
positive values of x.

208
00:09:45,010 --> 00:09:46,960
So it's the right
half of the circle.

209
00:09:46,960 --> 00:09:49,470
We want the positive
values of x, so we

210
00:09:49,470 --> 00:09:51,580
want the positive square root.

211
00:09:51,580 --> 00:09:52,710
OK.

212
00:09:52,710 --> 00:09:53,210
Good.

213
00:09:53,210 --> 00:09:55,520
And so those are
the bounds on x.

214
00:09:55,520 --> 00:09:57,760
Now we need the bounds on y.

215
00:09:57,760 --> 00:09:59,690
So the bounds on y,
well, what are they?

216
00:09:59,690 --> 00:10:01,730
Well, we want the
absolute bounds on y. y

217
00:10:01,730 --> 00:10:04,800
is the outermost variable that
we're integrating with respect

218
00:10:04,800 --> 00:10:09,270
to, so we want the absolute
bounds-- the absolute lowest

219
00:10:09,270 --> 00:10:10,910
value that y takes
in this region,

220
00:10:10,910 --> 00:10:13,090
and the absolute largest
value that y takes.

221
00:10:13,090 --> 00:10:14,710
So the smallest
value that y takes

222
00:10:14,710 --> 00:10:17,430
in this region-- that's
the lowest point-- that's

223
00:10:17,430 --> 00:10:20,160
along this line, and
that's when y equals 0.

224
00:10:20,160 --> 00:10:23,610
And the largest
value that y takes--

225
00:10:23,610 --> 00:10:25,830
that's when y is as
large as possible

226
00:10:25,830 --> 00:10:27,430
as it can get in
this region-- is

227
00:10:27,430 --> 00:10:29,210
up at this point of
intersection there,

228
00:10:29,210 --> 00:10:33,460
so that's when y is equal
to the square root of 2.

229
00:10:33,460 --> 00:10:35,020
OK, three quarters done.

230
00:10:35,020 --> 00:10:36,940
Yeah?

231
00:10:36,940 --> 00:10:39,290
This is that iterated integral.

232
00:10:39,290 --> 00:10:42,902
So now, we want to
do the same thing.

233
00:10:42,902 --> 00:10:49,800
R-- the integral over
this region R-- dy dx.

234
00:10:49,800 --> 00:10:50,599
OK.

235
00:10:50,599 --> 00:10:52,140
So we're going to
look at this region

236
00:10:52,140 --> 00:10:55,536
and we want to say-- dy is
going to be on the inside--

237
00:10:55,536 --> 00:10:57,910
so we're going to say, OK, so
we need to know for a fixed

238
00:10:57,910 --> 00:11:01,395
value of x, what's the
smallest value that y can take

239
00:11:01,395 --> 00:11:03,270
and what's the largest
value that y can take?

240
00:11:03,270 --> 00:11:06,680
So what's the bottom boundary
and what's the top boundary?

241
00:11:06,680 --> 00:11:08,410
But if you look at
this region-- right?--

242
00:11:08,410 --> 00:11:10,290
life is a little
complicated here.

243
00:11:10,290 --> 00:11:12,604
Because if you're in the
left half of this region--

244
00:11:12,604 --> 00:11:13,770
what do I mean by left half?

245
00:11:13,770 --> 00:11:17,285
I mean if you're to the left
of this point of intersection--

246
00:11:17,285 --> 00:11:19,450
if you're at the left
of this line x equals

247
00:11:19,450 --> 00:11:22,430
square root of 2--
when you're over there,

248
00:11:22,430 --> 00:11:26,660
y is going from 0 to x.

249
00:11:26,660 --> 00:11:31,610
But if you're over in the
right part of this region,

250
00:11:31,610 --> 00:11:33,120
there's a different
upper boundary.

251
00:11:33,120 --> 00:11:34,250
Right?

252
00:11:34,250 --> 00:11:36,300
It's a different curve
that it came from.

253
00:11:36,300 --> 00:11:38,200
It has a different equation.

254
00:11:38,200 --> 00:11:44,410
So over here, y is going from
the x-axis up to the circle.

255
00:11:44,410 --> 00:11:47,860
So this is complicated, and what
does this complication mean?

256
00:11:47,860 --> 00:11:49,840
Well, it means that it's
not easy to write this

257
00:11:49,840 --> 00:11:51,770
as a single iterated integral.

258
00:11:51,770 --> 00:11:53,280
If you want to do
this in this way,

259
00:11:53,280 --> 00:11:55,520
you have to break the
region into two pieces,

260
00:11:55,520 --> 00:11:58,600
and write this double
integral as a sum of two

261
00:11:58,600 --> 00:12:00,090
iterated integrals.

262
00:12:00,090 --> 00:12:00,590
OK?

263
00:12:00,590 --> 00:12:03,930
So one iterated integral will
take care of the left part

264
00:12:03,930 --> 00:12:07,020
and one will take care
of the right part.

265
00:12:07,020 --> 00:12:08,665
So let's do the left part first.

266
00:12:08,665 --> 00:12:11,550


267
00:12:11,550 --> 00:12:18,550
So here we're going to have a
iterated integral integrating

268
00:12:18,550 --> 00:12:20,120
with respect to y first.

269
00:12:20,120 --> 00:12:25,290
So to fixed value of x, we want
to know what the bounds on y

270
00:12:25,290 --> 00:12:25,834
are.

271
00:12:25,834 --> 00:12:27,500
And well, we can see
from this picture--

272
00:12:27,500 --> 00:12:30,180
when you're in this
triangle-- that y

273
00:12:30,180 --> 00:12:33,220
is going from the x-axis
up to the line y equals x.

274
00:12:33,220 --> 00:12:35,910
So that means the smallest
value that y can take is 0,

275
00:12:35,910 --> 00:12:38,225
and the largest value
that y can take is x.

276
00:12:38,225 --> 00:12:41,030
So here it's from 0 to x.

277
00:12:41,030 --> 00:12:43,050
And when you're
in this triangle,

278
00:12:43,050 --> 00:12:46,210
we need to know what the
bounds on x are, then.

279
00:12:46,210 --> 00:12:48,220
We need to know
the outer bounds.

280
00:12:48,220 --> 00:12:50,925
So we need to know the absolute
largest and smallest values

281
00:12:50,925 --> 00:12:51,967
that x can take.

282
00:12:51,967 --> 00:12:53,050
Well, what does that mean?

283
00:12:53,050 --> 00:12:55,791
We need to know the absolute
leftmost and absolute rightmost

284
00:12:55,791 --> 00:12:56,290
points.

285
00:12:56,290 --> 00:12:58,930
So the absolute leftmost
point is the origin.

286
00:12:58,930 --> 00:13:01,950
The absolute rightmost
is this vertical line

287
00:13:01,950 --> 00:13:03,540
x equals square root of 2.

288
00:13:03,540 --> 00:13:07,885
So over here, the
value of x is 0.

289
00:13:07,885 --> 00:13:11,570
And at the rightmost boundary
of this triangle, the value of x

290
00:13:11,570 --> 00:13:13,540
is the square root of 2.

291
00:13:13,540 --> 00:13:14,040
OK.

292
00:13:14,040 --> 00:13:16,090
So that's going to give
us the double integral

293
00:13:16,090 --> 00:13:20,130
just over this triangular
part of the region.

294
00:13:20,130 --> 00:13:22,420
Yeah?

295
00:13:22,420 --> 00:13:26,160
So now, we need to
add to this-- but I'm

296
00:13:26,160 --> 00:13:28,473
going to put it down on
this next line-- we need

297
00:13:28,473 --> 00:13:30,310
to add to this the
part of the integral

298
00:13:30,310 --> 00:13:34,580
over this little segment
of the circle here.

299
00:13:34,580 --> 00:13:38,750
The remainder of the region
that's not in that triangle.

300
00:13:38,750 --> 00:13:44,740
So for that, again, we're going
to write down two integrals,

301
00:13:44,740 --> 00:13:49,360
and it's going to be dy dx.

302
00:13:49,360 --> 00:13:50,052
Whew.

303
00:13:50,052 --> 00:13:51,315
We're nearly done, right?

304
00:13:51,315 --> 00:13:54,560


305
00:13:54,560 --> 00:13:59,140
So y is inside, so we need
to know what the bounds on y

306
00:13:59,140 --> 00:14:01,041
are for a given value of x.

307
00:14:01,041 --> 00:14:02,540
So we need to know
for a given value

308
00:14:02,540 --> 00:14:06,820
of x, what are the bottom
and the topmost points

309
00:14:06,820 --> 00:14:08,490
of this region?

310
00:14:08,490 --> 00:14:10,450
So for a given value
of x, that means

311
00:14:10,450 --> 00:14:14,510
that y is going here between the
x-axis and between this circle.

312
00:14:14,510 --> 00:14:19,290
So the x-axis is y equals 0,
so that's the lower bound.

313
00:14:19,290 --> 00:14:22,650
So for the upper bound, we
need to know this circle.

314
00:14:22,650 --> 00:14:24,090
What is y on this circle?

315
00:14:24,090 --> 00:14:25,465
Well, the equation
of this circle

316
00:14:25,465 --> 00:14:27,080
is x squared plus
y squared equals 4,

317
00:14:27,080 --> 00:14:31,080
so y is equal to the square
root of the quantity 4 minus x

318
00:14:31,080 --> 00:14:31,580
squared.

319
00:14:31,580 --> 00:14:36,440


320
00:14:36,440 --> 00:14:38,640
Where again, here we take
the positive square root,

321
00:14:38,640 --> 00:14:42,360
because this is a part of the
circle where y is positive.

322
00:14:42,360 --> 00:14:43,370
Yeah.

323
00:14:43,370 --> 00:14:46,490
If we were somehow on the
bottom part of the circle,

324
00:14:46,490 --> 00:14:49,222
then we would have to take a
negative square root there,

325
00:14:49,222 --> 00:14:51,180
but because we're on the
top part of the circle

326
00:14:51,180 --> 00:14:54,300
where y is positive, we
take a positive square root.

327
00:14:54,300 --> 00:14:54,980
OK, good.

328
00:14:54,980 --> 00:14:56,860
So those are the
bounds on y, and now we

329
00:14:56,860 --> 00:14:59,161
need to know the
absolute bounds on x.

330
00:14:59,161 --> 00:14:59,660
Yeah?

331
00:14:59,660 --> 00:15:01,780
So those are the bounds
on y in terms of x.

332
00:15:01,780 --> 00:15:03,870
And now because x
is the outer thing

333
00:15:03,870 --> 00:15:05,535
we're integrating
with respect to,

334
00:15:05,535 --> 00:15:07,530
we need the absolute
bounds on x.

335
00:15:07,530 --> 00:15:12,290
And you can see
in this circular--

336
00:15:12,290 --> 00:15:16,480
I don't really know what the
name for a shape like that is--

337
00:15:16,480 --> 00:15:20,020
but whatever that thing is, we
need to know what its leftmost

338
00:15:20,020 --> 00:15:21,750
and rightmost points are.

339
00:15:21,750 --> 00:15:25,070
We need to know the smallest and
largest values that x can take.

340
00:15:25,070 --> 00:15:30,620
And so its leftmost edge is this
line x equals square root of 2.

341
00:15:30,620 --> 00:15:33,660
And its rightmost edge is that
rightmost point on the circle

342
00:15:33,660 --> 00:15:35,490
there-- where the
circle hit the x-axis--

343
00:15:35,490 --> 00:15:37,390
and that's the value
when x equals 2.

344
00:15:37,390 --> 00:15:40,710


345
00:15:40,710 --> 00:15:42,150
OK, so there you go.

346
00:15:42,150 --> 00:15:47,180
There's this last integral
written in the dy dx order,

347
00:15:47,180 --> 00:15:50,000
but we can't write it as a
single iterated integral.

348
00:15:50,000 --> 00:15:52,890
We need to write it as a sum of
two iterated integrals because

349
00:15:52,890 --> 00:15:54,930
of the shape of this region.

350
00:15:54,930 --> 00:15:55,430
All right.

351
00:15:55,430 --> 00:15:58,600


352
00:15:58,600 --> 00:16:01,820
Let me just make one
quick, summary comment.

353
00:16:01,820 --> 00:16:05,570
Which is that if you're
doing this, one thing that

354
00:16:05,570 --> 00:16:09,167
should always be true, is
that these integrals, when

355
00:16:09,167 --> 00:16:10,750
you evaluate them--
so here, I haven't

356
00:16:10,750 --> 00:16:12,520
been writing an integrand.

357
00:16:12,520 --> 00:16:14,770
I guess the integrand has
always been 1, or something.

358
00:16:14,770 --> 00:16:19,000
But for any integrand,
the nature of this process

359
00:16:19,000 --> 00:16:26,050
is that it shouldn't matter
which order you integrate.

360
00:16:26,050 --> 00:16:28,810
You should get the same answer
if you integrate dx dy or dy

361
00:16:28,810 --> 00:16:29,680
dx.

362
00:16:29,680 --> 00:16:32,930
So one very low-level
check that you

363
00:16:32,930 --> 00:16:35,660
can make-- that you haven't done
anything horribly, egregiously

364
00:16:35,660 --> 00:16:38,060
wrong when changing the
bounds of integration--

365
00:16:38,060 --> 00:16:43,090
is that you can check that
actually these things evaluate

366
00:16:43,090 --> 00:16:44,010
the same.

367
00:16:44,010 --> 00:16:44,930
Yeah?

368
00:16:44,930 --> 00:16:46,910
Where you can
choose any function

369
00:16:46,910 --> 00:16:48,780
that you happen to
want to put in there--

370
00:16:48,780 --> 00:16:50,884
function of x and y-- and
evaluate this integral,

371
00:16:50,884 --> 00:16:53,550
and choose any function that you
happen to want to put in there,

372
00:16:53,550 --> 00:16:54,840
and evaluate those integrals.

373
00:16:54,840 --> 00:16:57,440
And see that you actually get
the same thing on both sides.

374
00:16:57,440 --> 00:17:00,360
Now one simple example
is that you could just

375
00:17:00,360 --> 00:17:02,180
evaluate the
integral as written,

376
00:17:02,180 --> 00:17:05,100
with a 1 written in there.

377
00:17:05,100 --> 00:17:07,400
And so in both cases,
what you should get

378
00:17:07,400 --> 00:17:09,320
is the area of the
region when you

379
00:17:09,320 --> 00:17:10,839
evaluate an integral like that.

380
00:17:10,839 --> 00:17:13,380
But you can also check with any
other function if you wanted.

381
00:17:13,380 --> 00:17:19,910


382
00:17:19,910 --> 00:17:22,410
It won't show that what
you've done is right,

383
00:17:22,410 --> 00:17:25,980
but it will show if you've
done something wrong.

384
00:17:25,980 --> 00:17:29,517
That method will sometimes
pick it out, right?

385
00:17:29,517 --> 00:17:31,100
Because you'll
actually be integrating

386
00:17:31,100 --> 00:17:32,800
over two different regions,
and there's no reason

387
00:17:32,800 --> 00:17:34,290
you should get the same answer.

388
00:17:34,290 --> 00:17:36,680
So if you were to
compute these integrals

389
00:17:36,680 --> 00:17:38,140
and get different
numbers, then you

390
00:17:38,140 --> 00:17:40,440
would know that something
had gone wrong at some point

391
00:17:40,440 --> 00:17:43,780
for sure, and you'd have to go
and figure out where it was.

392
00:17:43,780 --> 00:17:45,633
I think I'll end there.

393
00:17:45,633 --> 00:17:46,133