1
00:00:00,000 --> 00:00:07,080

2
00:00:07,080 --> 00:00:07,540
JOEL LEWIS: Hi.

3
00:00:07,540 --> 00:00:09,070
Welcome back to recitation.

4
00:00:09,070 --> 00:00:10,750
In lecture, you've been
learning about triple

5
00:00:10,750 --> 00:00:13,730
integration, and I have a nice
average value problem for you

6
00:00:13,730 --> 00:00:15,610
using triple integration here.

7
00:00:15,610 --> 00:00:19,670
So what I'd like you to do is
to consider the tetrahedron

8
00:00:19,670 --> 00:00:22,670
that has vertices at the origin,
and at the points 1,

9
00:00:22,670 --> 00:00:25,750
0, 0, and 0, 1, 0,
and 0, 0, 1.

10
00:00:25,750 --> 00:00:29,400
So that's one point on each of
the positive axes at distance

11
00:00:29,400 --> 00:00:30,360
1 from the origin.

12
00:00:30,360 --> 00:00:34,920
So I've taken the liberty of
drawing it here for you.

13
00:00:34,920 --> 00:00:38,620
So consider that solid
tetrahedron.

14
00:00:38,620 --> 00:00:42,580
And what I'd like you to do is
find the average distance of

15
00:00:42,580 --> 00:00:45,900
the points in that tetrahedron
from the xy plane.

16
00:00:45,900 --> 00:00:46,640
All right.

17
00:00:46,640 --> 00:00:49,820
So I'd like you to compute the
average value of the distance

18
00:00:49,820 --> 00:00:54,100
as the point ranges over the
whole tetrahedron of its

19
00:00:54,100 --> 00:00:56,580
distance from the xy plane.

20
00:00:56,580 --> 00:01:00,590
So why don't you pause the
video, spend some time working

21
00:01:00,590 --> 00:01:02,470
that out, come back, and we
can work it out together.

22
00:01:02,470 --> 00:01:11,390

23
00:01:11,390 --> 00:01:13,250
Hopefully, you had some luck
with this problem.

24
00:01:13,250 --> 00:01:14,500
Let's get started.

25
00:01:14,500 --> 00:01:19,660

26
00:01:19,660 --> 00:01:25,800
So the average value of a
function F over a region R is

27
00:01:25,800 --> 00:01:30,370
going to be 1 over the volume of
the region times the triple

28
00:01:30,370 --> 00:01:36,150
integral over your whole region
R of the function value

29
00:01:36,150 --> 00:01:43,540
f of x, y, z with respect to
volume, in any order you want.

30
00:01:43,540 --> 00:01:46,990
So I guess I'll write dv here,
and then to evaluate this, you

31
00:01:46,990 --> 00:01:48,730
set it up as an iterated
integral.

32
00:01:48,730 --> 00:01:50,740
So in our case, the function
we're trying to find the

33
00:01:50,740 --> 00:01:53,810
average value of is the distance
between a point and

34
00:01:53,810 --> 00:01:55,070
the xy plane.

35
00:01:55,070 --> 00:01:56,770
But that's an easy
function, right?

36
00:01:56,770 --> 00:01:58,310
That's just z.

37
00:01:58,310 --> 00:02:01,950
For any point in space, its
distance from the xy plane is

38
00:02:01,950 --> 00:02:02,770
just its height.

39
00:02:02,770 --> 00:02:03,790
It's z-value.

40
00:02:03,790 --> 00:02:05,620
So the function that we're
seeking to find the average

41
00:02:05,620 --> 00:02:08,250
value of is z, and so most of
the work of this problem then

42
00:02:08,250 --> 00:02:10,560
is going to be in figuring out
what the bounds are and then

43
00:02:10,560 --> 00:02:13,850
doing the actual integral
after that.

44
00:02:13,850 --> 00:02:14,830
So OK.

45
00:02:14,830 --> 00:02:17,890
So let's think about
the bounds.

46
00:02:17,890 --> 00:02:20,430
This tetrahedron is
a nice, reasonably

47
00:02:20,430 --> 00:02:24,810
simple, geometric object.

48
00:02:24,810 --> 00:02:28,570
So in fact, it doesn't matter
too much which order you take

49
00:02:28,570 --> 00:02:29,320
your bounds.

50
00:02:29,320 --> 00:02:33,460
So I think I'm going to do
it in the order dz dy dx.

51
00:02:33,460 --> 00:02:36,320
You know, I'll do z first,
and then y, and then x.

52
00:02:36,320 --> 00:02:37,350
But it doesn't matter.

53
00:02:37,350 --> 00:02:40,500
If you do it a different way,
it'll probably work out very

54
00:02:40,500 --> 00:02:44,890
similarly overall, and you'll
still be able to compare the

55
00:02:44,890 --> 00:02:46,330
overall process.

56
00:02:46,330 --> 00:02:48,610
So let's think about z.

57
00:02:48,610 --> 00:02:49,000
Yeah?

58
00:02:49,000 --> 00:02:55,750
So in this tetrahedron, we want
to integrate with respect

59
00:02:55,750 --> 00:03:00,780
to z first. So we look at this
tetrahedron and we say, OK, at

60
00:03:00,780 --> 00:03:04,490
a point when we choose the x-
and y-values, what's the

61
00:03:04,490 --> 00:03:07,580
lowest surface-- what's the
smallest value z can take--

62
00:03:07,580 --> 00:03:09,550
and what's the upper surface--
what's the

63
00:03:09,550 --> 00:03:11,060
largest value z can take?

64
00:03:11,060 --> 00:03:13,500
So the lowest surface here
is the xy plane.

65
00:03:13,500 --> 00:03:15,490
That's the bottom face
of this tetrahedron.

66
00:03:15,490 --> 00:03:20,030
And for any choice of x and y,
the lowest value z can take is

67
00:03:20,030 --> 00:03:21,710
when it's in the xy plane.

68
00:03:21,710 --> 00:03:22,960
So when it's equal to 0.

69
00:03:22,960 --> 00:03:25,660

70
00:03:25,660 --> 00:03:27,530
So this is going to be equal,
so in our case, so

71
00:03:27,530 --> 00:03:28,230
let's set this up.

72
00:03:28,230 --> 00:03:31,241
So it's going to be an
iterated integral.

73
00:03:31,241 --> 00:03:33,760
The function we're integrating
is z, and I said

74
00:03:33,760 --> 00:03:38,500
we'll do dz dy dx.

75
00:03:38,500 --> 00:03:42,070
And we just said that the lowest
value that z takes--

76
00:03:42,070 --> 00:03:42,970
the dz--

77
00:03:42,970 --> 00:03:43,660
is 0.

78
00:03:43,660 --> 00:03:46,330
So the up-highest value that
z takes is when it

79
00:03:46,330 --> 00:03:48,250
hits this top surface.

80
00:03:48,250 --> 00:03:52,660
This plane that passes through
the point (1, 0, 0), (0, 1,

81
00:03:52,660 --> 00:03:54,760
0), and (0, 0, 1).

82
00:03:54,760 --> 00:03:55,030
All right.

83
00:03:55,030 --> 00:03:57,030
So we need to know the equation
of that plane.

84
00:03:57,030 --> 00:03:58,990
Luckily, that's a pretty easy
plane to write down the

85
00:03:58,990 --> 00:04:00,470
equation for.

86
00:04:00,470 --> 00:04:09,050
So this slanted plane passing
through the three vertices

87
00:04:09,050 --> 00:04:13,680
other than the origin is
the plane x plus y

88
00:04:13,680 --> 00:04:16,465
plus z equals 1.

89
00:04:16,465 --> 00:04:17,160
All right.

90
00:04:17,160 --> 00:04:19,500
So it's a nice, easy
plane to work with.

91
00:04:19,500 --> 00:04:21,180
And so what we want to know
is what's the value

92
00:04:21,180 --> 00:04:22,450
of z on that plane.

93
00:04:22,450 --> 00:04:24,620
So we isolate z on one
side and we bring

94
00:04:24,620 --> 00:04:25,200
everything else over.

95
00:04:25,200 --> 00:04:29,810
So we have that top value of z,
the largest value of z that

96
00:04:29,810 --> 00:04:33,680
z can take when x and y are
fixed, is 1 minus x minus 1.

97
00:04:33,680 --> 00:04:34,930
So that's what goes up here.

98
00:04:34,930 --> 00:04:37,830

99
00:04:37,830 --> 00:04:40,360
So that's the biggest
value z can take.

100
00:04:40,360 --> 00:04:41,460
OK, good.

101
00:04:41,460 --> 00:04:45,410
So now we need to figure out
what the bounds on y are in

102
00:04:45,410 --> 00:04:46,880
terms of x.

103
00:04:46,880 --> 00:04:50,880
So what I like to do in this
case is I like to draw a

104
00:04:50,880 --> 00:04:53,110
projection of your surface.

105
00:04:53,110 --> 00:04:56,090
So then you're in a
two-dimensional world, and

106
00:04:56,090 --> 00:04:58,370
then you can look at that
image more easily.

107
00:04:58,370 --> 00:04:59,950
So what we're going to do is
we're going to look at this

108
00:04:59,950 --> 00:05:03,330
tetrahedron and we're going to
imagine projecting it down

109
00:05:03,330 --> 00:05:04,480
into the xy plane.

110
00:05:04,480 --> 00:05:06,450
So for every point in the
tetrahedron, we're going to

111
00:05:06,450 --> 00:05:08,800
draw a dot below it
in the xy plane.

112
00:05:08,800 --> 00:05:11,070
And then we're going to look
at that set of dots.

113
00:05:11,070 --> 00:05:16,280
So what that's going to give us
is this bottom face of the

114
00:05:16,280 --> 00:05:17,060
tetrahedron.

115
00:05:17,060 --> 00:05:19,520
Every point of the tetrahedron
is above its bottom face.

116
00:05:19,520 --> 00:05:21,460
That's not true for every
tetrahedron, but it's

117
00:05:21,460 --> 00:05:23,160
true for this one.

118
00:05:23,160 --> 00:05:24,890
So the region that we're
interested in

119
00:05:24,890 --> 00:05:25,930
is that bottom face.

120
00:05:25,930 --> 00:05:28,270
So I'm going to draw another
picture of it

121
00:05:28,270 --> 00:05:29,780
over on my left here.

122
00:05:29,780 --> 00:05:39,390
So that region is the region
that has vertices (0, 0), and

123
00:05:39,390 --> 00:05:41,370
(1, 0), and (0, 1).

124
00:05:41,370 --> 00:05:44,050
So it's this triangle.

125
00:05:44,050 --> 00:05:48,540
And this bottom edge of
the triangle is the

126
00:05:48,540 --> 00:05:49,860
line y equals 0.

127
00:05:49,860 --> 00:05:52,770
This left edge is the
line x equals 0.

128
00:05:52,770 --> 00:05:58,490
And this slanted line is the
line x plus y equals 1.

129
00:05:58,490 --> 00:06:03,310
OK, so this is that same bottom
face that we just drew.

130
00:06:03,310 --> 00:06:06,970
But now I've changed my axes to
our usual two-dimensional

131
00:06:06,970 --> 00:06:11,550
direction with x to the
right and y up.

132
00:06:11,550 --> 00:06:13,560
OK.

133
00:06:13,560 --> 00:06:15,000
So we're doing dy next.

134
00:06:15,000 --> 00:06:17,370
So we need to figure out for a
fixed value of x, what are the

135
00:06:17,370 --> 00:06:19,270
bounds on y?

136
00:06:19,270 --> 00:06:22,580
So we see from this picture that
for any fixed value of x,

137
00:06:22,580 --> 00:06:25,130
y goes from 0--

138
00:06:25,130 --> 00:06:26,140
the x-axis--

139
00:06:26,140 --> 00:06:27,760
up to this line.

140
00:06:27,760 --> 00:06:28,330
OK?

141
00:06:28,330 --> 00:06:32,790
So on the x-axis, y takes the
value 0, and on this line, y

142
00:06:32,790 --> 00:06:36,640
takes the value 1 minus x.

143
00:06:36,640 --> 00:06:40,910
Finally, our outermost variable
of integration is x,

144
00:06:40,910 --> 00:06:43,180
and so we need to know what are
the absolute largest and

145
00:06:43,180 --> 00:06:48,400
smallest values that x takes.

146
00:06:48,400 --> 00:06:50,500
So we can do that looking either
at this picture or

147
00:06:50,500 --> 00:06:51,650
looking at our original
picture.

148
00:06:51,650 --> 00:06:54,010
Either way, it's not hard
to see that x just goes

149
00:06:54,010 --> 00:06:55,250
between 0 and 1.

150
00:06:55,250 --> 00:06:56,770
The smallest value that
x takes in this

151
00:06:56,770 --> 00:06:57,680
tetrahedron is 0.

152
00:06:57,680 --> 00:07:01,320
The largest value
it takes is 1.

153
00:07:01,320 --> 00:07:03,430
So that's our integral that
we have to compute.

154
00:07:03,430 --> 00:07:05,350
So that's not that bad at all.

155
00:07:05,350 --> 00:07:07,050
So now you have to go through
and you have to actually

156
00:07:07,050 --> 00:07:08,310
integrate this.

157
00:07:08,310 --> 00:07:08,800
Yeah?

158
00:07:08,800 --> 00:07:11,880
And so I'm going to look at my
notes just to make sure I

159
00:07:11,880 --> 00:07:14,620
don't make any big arithmetic
mistakes.

160
00:07:14,620 --> 00:07:16,930
So let's see.

161
00:07:16,930 --> 00:07:18,920
Now we do these one at a time.

162
00:07:18,920 --> 00:07:23,470
So this innermost integral
is an integral of z dz.

163
00:07:23,470 --> 00:07:24,310
OK, well that's easy.

164
00:07:24,310 --> 00:07:26,020
That's z squared over 2.

165
00:07:26,020 --> 00:07:30,010
And then we're taking z squared
over 2 between 0 and 1

166
00:07:30,010 --> 00:07:32,250
minus x minus y.

167
00:07:32,250 --> 00:07:42,520
So this innermost integral is z
squared over 2 between 0 and

168
00:07:42,520 --> 00:07:45,270
1 minus x minus y.

169
00:07:45,270 --> 00:07:48,210
So that's equal to what the
innermost integral gives us,

170
00:07:48,210 --> 00:07:54,170
which is 1 minus x minus
y squared over 2.

171
00:07:54,170 --> 00:07:56,040
So that's what we get for
the innermost integral.

172
00:07:56,040 --> 00:08:00,470
So our integral that we're
looking at then is equal to

173
00:08:00,470 --> 00:08:04,430
the integral as x goes from 0
to 1, of the integral as y

174
00:08:04,430 --> 00:08:09,190
goes from 0 to 1 minus
x of this integrand.

175
00:08:09,190 --> 00:08:10,250
So this is the inner one.

176
00:08:10,250 --> 00:08:11,500
Let me write that.

177
00:08:11,500 --> 00:08:15,070

178
00:08:15,070 --> 00:08:16,170
That's what I've got here.

179
00:08:16,170 --> 00:08:18,420
Just integrating z with respect
to z gives me z

180
00:08:18,420 --> 00:08:19,420
squared over 2.

181
00:08:19,420 --> 00:08:23,010
And then I evaluated at the
bounds of the integral.

182
00:08:23,010 --> 00:08:25,080
OK, so now I need to
do the middle one.

183
00:08:25,080 --> 00:08:26,570
So let's do that up here.

184
00:08:26,570 --> 00:08:31,980

185
00:08:31,980 --> 00:08:33,540
So I need to compute
the integral.

186
00:08:33,540 --> 00:08:36,625
So now I take the bounds for the
middle one is y, and the

187
00:08:36,625 --> 00:08:44,940
bounds are from 0 to 1 minus
x of the inner integral.

188
00:08:44,940 --> 00:08:46,270
This thing that I
just computed.

189
00:08:46,270 --> 00:08:56,820
So that's of 1 minus x minus
y squared over 2, dy.

190
00:08:56,820 --> 00:08:57,390
OK.

191
00:08:57,390 --> 00:08:59,060
So All right.

192
00:08:59,060 --> 00:08:59,940
So this isn't that bad.

193
00:08:59,940 --> 00:09:04,170
This is a quadratic
polynomial in y.

194
00:09:04,170 --> 00:09:06,440
And so it's not terribly
hard to see.

195
00:09:06,440 --> 00:09:08,990
I'm running a little bit
out of board space.

196
00:09:08,990 --> 00:09:14,090
So I'm not going to give you a
full, detailed explanation.

197
00:09:14,090 --> 00:09:23,900
But it's not hard to see, I
think, that this integral of 1

198
00:09:23,900 --> 00:09:32,360
minus x minus y squared over 2
with respect to y is 1 minus x

199
00:09:32,360 --> 00:09:35,540
minus y cubed over 3, but it's
negative, because the sign

200
00:09:35,540 --> 00:09:36,150
here is negative.

201
00:09:36,150 --> 00:09:37,970
And you could check by
differentiating this and

202
00:09:37,970 --> 00:09:39,030
seeing that you get that.

203
00:09:39,030 --> 00:09:42,290
And so we have to evaluate
that as y goes from

204
00:09:42,290 --> 00:09:46,350
0 to 1 minus x.

205
00:09:46,350 --> 00:09:47,040
So what do we get?

206
00:09:47,040 --> 00:09:52,050
Well, when y is equal to
one minus x, this is 0.

207
00:09:52,050 --> 00:09:54,160
So we get 0 minus--

208
00:09:54,160 --> 00:09:58,840
and when y is equal to 0, this
is minus 1 minus x quantity

209
00:09:58,840 --> 00:10:00,090
cubed over 6--

210
00:10:00,090 --> 00:10:04,840

211
00:10:04,840 --> 00:10:09,760
so it's minus minus, so that's
just 1 minus x cubed over 6.

212
00:10:09,760 --> 00:10:18,530
And so finally the outermost
integral is we take the inner

213
00:10:18,530 --> 00:10:22,280
two integrals and we integrate
them with respect to x as x

214
00:10:22,280 --> 00:10:23,600
goes from 0 to 1.

215
00:10:23,600 --> 00:10:28,540
So it's the integral from
0 to 1 of 1 minus x

216
00:10:28,540 --> 00:10:33,590
cubed over 6, dx.

217
00:10:33,590 --> 00:10:38,640
And that's going to equal--

218
00:10:38,640 --> 00:10:39,890
I've run out of space--

219
00:10:39,890 --> 00:10:42,010

220
00:10:42,010 --> 00:10:44,430
1 over 24.

221
00:10:44,430 --> 00:10:44,760
All right.

222
00:10:44,760 --> 00:10:47,080
Except I've done something wrong
right at the beginning.

223
00:10:47,080 --> 00:10:47,980
I hope you all caught me.

224
00:10:47,980 --> 00:10:48,540
Right?

225
00:10:48,540 --> 00:10:52,100
I had this 1 over v factor
here, and it disappeared.

226
00:10:52,100 --> 00:10:52,340
Right?

227
00:10:52,340 --> 00:10:55,730
I forgot about this 1 over v,
so over here, I should have

228
00:10:55,730 --> 00:10:58,170
written 1 over v right in
front of that integral.

229
00:10:58,170 --> 00:11:03,130

230
00:11:03,130 --> 00:11:06,700
I've correctly computed 1 over
24 as the value of my triple

231
00:11:06,700 --> 00:11:10,750
integral, but the average height
here isn't 1 over 24.

232
00:11:10,750 --> 00:11:14,720
It's 1 over 24v. All right.

233
00:11:14,720 --> 00:11:18,610
So the average height is 1 over
24v, and so we need to go

234
00:11:18,610 --> 00:11:20,190
and we need to look at our
tetrahedron and figure out

235
00:11:20,190 --> 00:11:21,430
what its volume is.

236
00:11:21,430 --> 00:11:23,560
So if we come over here and
see our tetrahedron.

237
00:11:23,560 --> 00:11:25,960
Now this is nice, simple
tetrahedron.

238
00:11:25,960 --> 00:11:31,320
The volume of a tetrahedron is
1/3 the area of the base times

239
00:11:31,320 --> 00:11:34,170
the height, right?

240
00:11:34,170 --> 00:11:36,540
So this is a nice,
easy tetrahedron.

241
00:11:36,540 --> 00:11:37,860
Its height is 1.

242
00:11:37,860 --> 00:11:40,610
Its base is a right triangle
whose legs are both 1.

243
00:11:40,610 --> 00:11:44,990
So the base area is 1/2,
so the volume is 1/6.

244
00:11:44,990 --> 00:11:49,070
So if the volume is 1/6, and we
said the average value is 1

245
00:11:49,070 --> 00:11:52,020
over 24v, so that works
out to 1 over 4.

246
00:11:52,020 --> 00:11:55,170
So let me write that
just in this space.

247
00:11:55,170 --> 00:12:03,500
So the average height
then is 1 over 4.

248
00:12:03,500 --> 00:12:05,860
So that's going to be
our final answer.

249
00:12:05,860 --> 00:12:09,700
OK, so let's just recap
briefly what we did.

250
00:12:09,700 --> 00:12:12,950
We had an average value problem
that we started with.

251
00:12:12,950 --> 00:12:16,250
So we use this general formula
for average value problems.

252
00:12:16,250 --> 00:12:19,030
When you have a function f
that you want to take its

253
00:12:19,030 --> 00:12:22,790
average value of over a region
R, what do you do?

254
00:12:22,790 --> 00:12:26,120
Well, you take 1 over the volume
of the region times the

255
00:12:26,120 --> 00:12:29,360
triple integral of your function
f with respect to

256
00:12:29,360 --> 00:12:31,800
volume over that region.

257
00:12:31,800 --> 00:12:32,110
OK.

258
00:12:32,110 --> 00:12:34,850
So this is the average
value in general.

259
00:12:34,850 --> 00:12:37,845
In our particular case, the
function was the height.

260
00:12:37,845 --> 00:12:39,690
It was z.

261
00:12:39,690 --> 00:12:44,240
And then you have to set it out
choosing the proper bounds

262
00:12:44,240 --> 00:12:45,660
for your integrals.

263
00:12:45,660 --> 00:12:50,370
So in this case, you choose
some order of integration

264
00:12:50,370 --> 00:12:51,250
based on the region.

265
00:12:51,250 --> 00:12:53,270
In this particular case, it's
a nice, simple region.

266
00:12:53,270 --> 00:12:55,015
It doesn't matter too much
what order you choose.

267
00:12:55,015 --> 00:12:57,760

268
00:12:57,760 --> 00:13:00,380
So I chose dz dy dx.

269
00:13:00,380 --> 00:13:01,410
And then what does that mean?

270
00:13:01,410 --> 00:13:04,580
So for the innermost one, you
look at your original solid.

271
00:13:04,580 --> 00:13:07,080
So I'm going to go back and look
at this picture again.

272
00:13:07,080 --> 00:13:10,510
So for your innermost variable
you say-- so if it's z-- you

273
00:13:10,510 --> 00:13:14,400
say, so when I fix x and y,
what's the bottom surface and

274
00:13:14,400 --> 00:13:18,210
what's the top surface when
I solve that for z in

275
00:13:18,210 --> 00:13:19,210
terms of x and y?

276
00:13:19,210 --> 00:13:22,690
So here, that was the plane z
equals 0, and the plane z

277
00:13:22,690 --> 00:13:25,250
equals 1 minus x minus y.

278
00:13:25,250 --> 00:13:27,020
So that explains my
bounds over here.

279
00:13:27,020 --> 00:13:29,940
Why they were 0 and
1 minus x minus y.

280
00:13:29,940 --> 00:13:31,880
Then, when you go to your
next variable-- in this

281
00:13:31,880 --> 00:13:32,980
case, it was y--

282
00:13:32,980 --> 00:13:33,740
what do you do?

283
00:13:33,740 --> 00:13:37,310
Well, first you project to
eliminate this first variable.

284
00:13:37,310 --> 00:13:40,190
So you project your
region down.

285
00:13:40,190 --> 00:13:42,010
Down in this case,
because it's z.

286
00:13:42,010 --> 00:13:44,350
So you project in
the z-direction.

287
00:13:44,350 --> 00:13:47,600
And you draw this shadow
of your region.

288
00:13:47,600 --> 00:13:48,850
So this is what I drew here.

289
00:13:48,850 --> 00:13:51,610
This is the shadow of my region
in the xy plane after I

290
00:13:51,610 --> 00:13:52,490
projected it.

291
00:13:52,490 --> 00:13:53,630
And then you do the
same thing.

292
00:13:53,630 --> 00:13:57,070
So now this just like what you
did when you had to find

293
00:13:57,070 --> 00:14:00,600
bounds for double integrals,
when you wrote them as

294
00:14:00,600 --> 00:14:04,550
iterated integrals a
few lectures ago.

295
00:14:04,550 --> 00:14:07,120

296
00:14:07,120 --> 00:14:08,380
And so then you do
the same thing.

297
00:14:08,380 --> 00:14:10,490
So then in this case, I
was next integrating

298
00:14:10,490 --> 00:14:11,290
with respect to y.

299
00:14:11,290 --> 00:14:14,180
So I needed to find the bounds
on y with respect to x.

300
00:14:14,180 --> 00:14:17,410
So I needed to look in this
picture at the bottom edge and

301
00:14:17,410 --> 00:14:20,050
the top edge of this region.

302
00:14:20,050 --> 00:14:22,220
And then your outermost
variable, you look at its

303
00:14:22,220 --> 00:14:23,440
absolute bounds.

304
00:14:23,440 --> 00:14:26,460
So the largest and smallest
value it takes on the region.

305
00:14:26,460 --> 00:14:29,030
OK, then you have an iterated
integral and you evaluate it

306
00:14:29,030 --> 00:14:32,330
by successive integrations.

307
00:14:32,330 --> 00:14:33,090
OK.

308
00:14:33,090 --> 00:14:34,110
So that was what we did.

309
00:14:34,110 --> 00:14:36,840
We just did the three integrals,
starting from the

310
00:14:36,840 --> 00:14:38,300
inside and working our way out.

311
00:14:38,300 --> 00:14:39,550