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So, basically the last few
weeks, we've been doing

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00:00:27 --> 00:00:33
derivatives.
Now, we're going to integrals.

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00:00:33 --> 00:00:46
So -- OK, so more precisely,
we are going to be talking

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00:00:46 --> 00:00:58
about double integrals.
OK, so just to motivate the

11
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notion,
let me just remind you that

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when you have a function of one
variable -- -- say,

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f of x, 
and you take its integrals

14
00:01:16 --> 00:01:21
from, say,
a to b of f of x dx, 

15
00:01:21 --> 00:01:33
well, that corresponds to the
area below the graph of f over

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00:01:33 --> 00:01:43
the interval from a to b.
OK, so the picture is something

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00:01:43 --> 00:01:47
like you have a;
you have b.

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00:01:47 --> 00:01:55
You have the graph of f,
and then what the integral

19
00:01:55 --> 00:02:02
measures is the area of this
region.

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00:02:02 --> 00:02:04
And, when we say the area of
this region, of course,

21
00:02:04 --> 00:02:06
if f is positive,
that's what happens.

22
00:02:06 --> 00:02:11
If f is negative,
then we count negatively the

23
00:02:11 --> 00:02:15
area below the x axis.
OK, so, now,

24
00:02:15 --> 00:02:19
when you have a function of two
variables, then you can try to

25
00:02:19 --> 00:02:22
do the same thing.
Namely, you can plot its graph.

26
00:02:22 --> 00:02:24
Its graph will be a surface in
space.

27
00:02:24 --> 00:02:28
And then, we can try to look
for the volume below the graph.

28
00:02:28 --> 00:02:35
And that's what we will call
the double integral of the

29
00:02:35 --> 00:02:44
function over a certain region.
OK, so let's say that we have a

30
00:02:44 --> 00:02:51
function of two variables,
x and y.

31
00:02:51 --> 00:03:00
Then, we'll look at the volume
that's below the graph z equals

32
00:03:00 --> 00:03:06
f of xy.
OK, so, let's draw a picture

33
00:03:06 --> 00:03:10
for what this means.
I have a function of x and y.

34
00:03:10 --> 00:03:16
I can draw its graph.
The graph will be the surface

35
00:03:16 --> 00:03:20
with equation z equals f of x
and y.

36
00:03:20 --> 00:03:24
And, well, I have to decide
where I will integrate the

37
00:03:24 --> 00:03:26
function.
So, for that,

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00:03:26 --> 00:03:30
I will choose some region in
the xy plane.

39
00:03:30 --> 00:03:39
And, I will integrate the
function on that region.

40
00:03:39 --> 00:03:46
So, it's over a region,
R, in the xy plane.

41
00:03:46 --> 00:03:51
So, I have this region R and I
look at the piece of the graph

42
00:03:51 --> 00:03:56
that is above this region.
And, we'll try to compute the

43
00:03:56 --> 00:04:02
volume of this solid here.
OK, that's what the double

44
00:04:02 --> 00:04:08
integral will measure.
So, we'll call that the double

45
00:04:08 --> 00:04:13
integral of our region,
R, of f of xy dA and I will

46
00:04:13 --> 00:04:18
have to explain what the
notation means.

47
00:04:18 --> 00:04:22
So, dA here stands for a piece
of area.

48
00:04:22 --> 00:04:26
A stands for area.
And, well, it's a double

49
00:04:26 --> 00:04:28
integral.
So, that's why we have two

50
00:04:28 --> 00:04:30
integral signs.
And, we'll have to indicate

51
00:04:30 --> 00:04:33
somehow the region over which we
are integrating.

52
00:04:33 --> 00:04:37
OK, we'll come up with more
concrete notations when we see

53
00:04:37 --> 00:04:39
how to actually compute these
things.

54
00:04:39 --> 00:04:44
That's the basic definition.
OK, so actually,

55
00:04:44 --> 00:04:45
how do we define it,
that's not really much of a

56
00:04:45 --> 00:04:49
definition yet.
How do we actually define this

57
00:04:49 --> 00:04:52
rigorously?
Well, remember,

58
00:04:52 --> 00:04:56
the integral in one variable,
you probably saw a definition

59
00:04:56 --> 00:04:58
where you take your integral
from a to b,

60
00:04:58 --> 00:05:01
and you cut it into little
pieces.

61
00:05:01 --> 00:05:03
And then, for each little
piece, you take the value of a

62
00:05:03 --> 00:05:06
function, and you multiply by
the width of a piece.

63
00:05:06 --> 00:05:11
That gives you a rectangular
slice, and then you sum all of

64
00:05:11 --> 00:05:14
these rectangular slices
together.

65
00:05:14 --> 00:05:17
So, here we'll do the same
thing.

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00:05:17 --> 00:05:23
So, well, let me put a picture
up and explain what it does.

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00:05:23 --> 00:05:26
So, we're going to cut our
origin into little pieces,

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00:05:26 --> 00:05:30
say, little rectangles or
actually anything we want.

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00:05:30 --> 00:05:35
And then, for each piece, 
with the small area, delta A, 

70
00:05:35 --> 00:05:39
we'll take the area delta a
times the value of a function in

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00:05:39 --> 00:05:43
there that will give us the
volume of a small box that sits

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00:05:43 --> 00:05:46
under the graph.
And then, we'll add all these

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00:05:46 --> 00:05:49
boxes together.
That gives us an estimate of a

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00:05:49 --> 00:05:51
volume.
And then, to get actually the

75
00:05:51 --> 00:05:53
integral,
the integral will be defined as

76
00:05:53 --> 00:05:56
a limit as we subdivide into
smaller and smaller boxes,

77
00:05:56 --> 00:05:59
and we sum more and more
pieces,

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00:05:59 --> 00:06:02
OK?
So, actually,

79
00:06:02 --> 00:06:11
what we do, oh,
I still have a board here.

80
00:06:11 --> 00:06:20
So, the actual definition
involves cutting R into small

81
00:06:20 --> 00:06:29
pieces of area that's called
delta A or maybe delta Ai,

82
00:06:29 --> 00:06:36
the area of the i'th piece.
And then, OK,

83
00:06:36 --> 00:06:42
so maybe in the xy plane,
we have our region,

84
00:06:42 --> 00:06:47
and we'll cut it may be using
some grade.

85
00:06:47 --> 00:06:49
OK, and then we'll have each
small piece.

86
00:06:49 --> 00:07:00
Each small piece will have area
delta Ai and it will be at some

87
00:07:00 --> 00:07:06
point, let's call it xi,
yi ...

88
00:07:06 --> 00:07:13
yi, xi.
And then, we'll consider the

89
00:07:13 --> 00:07:18
sum over all the pieces of f at
that point, xi,

90
00:07:18 --> 00:07:22
yi times the area of a small
piece.

91
00:07:22 --> 00:07:27
So, what that corresponds to in
the three-dimensional picture is

92
00:07:27 --> 00:07:31
just I sum the volumes of all of
these little columns that sit

93
00:07:31 --> 00:07:36
under the graph.
OK, and then,

94
00:07:36 --> 00:07:49
so what I do is actually I take
the limit as the size of the

95
00:07:49 --> 00:07:57
pieces tends to zero.
So, I have more and more

96
00:07:57 --> 00:08:03
smaller and smaller pieces.
And, that gives me the double

97
00:08:03 --> 00:08:07
integral.
OK, so that's not a very good

98
00:08:07 --> 00:08:13
sentence, but whatever.
So, OK, so that's the

99
00:08:13 --> 00:08:16
definition.
Of course, we will have to see

100
00:08:16 --> 00:08:19
how to compute it.
We don't actually compute it.

101
00:08:19 --> 00:08:21
When you compute an integral in
single variable calculus,

102
00:08:21 --> 00:08:23
you don't do that.
You don't cut into little

103
00:08:23 --> 00:08:25
pieces and sum the pieces
together.

104
00:08:25 --> 00:08:28
You've learned how to integrate
functions using various

105
00:08:28 --> 00:08:29
formulas,
and similarly here, 

106
00:08:29 --> 00:08:32
we'll learn how to actually
compute these things without

107
00:08:32 --> 00:08:34
doing that cutting into small
pieces.

108
00:08:34 --> 00:08:38
OK, any questions first about
the concept, or what the

109
00:08:38 --> 00:08:48
definition is?
Yes?

110
00:08:48 --> 00:08:50
Well, so we'll have to learn
which tricks work,

111
00:08:50 --> 00:08:52
and how exactly.
But, so what we'll do actually

112
00:08:52 --> 00:08:56
is we'll reduce the calculation
of a double integral to two

113
00:08:56 --> 00:08:58
calculations of single
integrals.

114
00:08:58 --> 00:09:00
And so, for V,
certainly, all the tricks

115
00:09:00 --> 00:09:03
you've learned in single
variable calculus will come in

116
00:09:03 --> 00:09:05
handy.
OK, so, 

117
00:09:05 --> 00:09:07
yeah that's a strong suggestion
that if you've forgotten

118
00:09:07 --> 00:09:09
everything about single variable
calculus,

119
00:09:09 --> 00:09:17
now would be a good time to
actually brush up on integrals.

120
00:09:17 --> 00:09:21
The usual integrals,
and the usual substitution

121
00:09:21 --> 00:09:26
tricks and easy trig in
particular, these would be very

122
00:09:26 --> 00:09:28
useful.
OK, so, yeah,

123
00:09:28 --> 00:09:31
how do we compute these things?
That's what we would have to

124
00:09:31 --> 00:09:37
come up with.
And, well, 

125
00:09:37 --> 00:09:39
going back to what we did with
derivatives,

126
00:09:39 --> 00:09:42
to understand variations of
functions and derivatives,

127
00:09:42 --> 00:09:45
what we did was really we took
slices parallel to an axis or

128
00:09:45 --> 00:09:46
another one.
So, in fact,

129
00:09:46 --> 00:09:50
here, the key is also the same.
So, what we are going to do is

130
00:09:50 --> 00:09:52
instead of cutting into a lot of
small boxes like that and

131
00:09:52 --> 00:09:57
summing completely at random,
we will actually somehow scan

132
00:09:57 --> 00:10:01
through our region by parallel
planes,

133
00:10:01 --> 00:10:03
OK?
So, let me put up,

134
00:10:03 --> 00:10:08
actually, a slightly different
picture up here.

135
00:10:08 --> 00:10:11
So, what I'm going to do is I'm
going to take planes,

136
00:10:11 --> 00:10:14
say in this picture,
parallel to the yz plane.

137
00:10:14 --> 00:10:18
I'll take a moving plane that
scans from the back to the front

138
00:10:18 --> 00:10:23
or from the front to the back.
So, that means I set the value

139
00:10:23 --> 00:10:27
of x, and I look at the slice,
x equals x0,

140
00:10:27 --> 00:10:31
and then I will do that for all
values of x0.

141
00:10:31 --> 00:10:35
So, now in each slice,
well, I get what looks a lot

142
00:10:35 --> 00:10:40
like a single variable integral.
OK, and that integral will tell

143
00:10:40 --> 00:10:44
me, what is the area in this?
Well, I guess it's supposed to

144
00:10:44 --> 00:10:48
be green, but it all comes as
black, so, let's say the black

145
00:10:48 --> 00:10:51
shaded slice.
And then, when I add all of

146
00:10:51 --> 00:10:55
these areas together,
as the value of x changes,

147
00:10:55 --> 00:11:00
I will get the volume.
OK, let me try to explain that

148
00:11:00 --> 00:11:07
again.
So, to compute this integral,

149
00:11:07 --> 00:11:16
what we do is actually we take
slices.

150
00:11:16 --> 00:11:29
So, let's consider,
let's call s of x the area of a

151
00:11:29 --> 00:11:41
slice, well, by a plane parallel
to the yz plane.

152
00:11:41 --> 00:11:46
OK, so on the picture,
s of x is just the area of this

153
00:11:46 --> 00:11:51
thing in the vertical wall.
Now, if you sum all of these,

154
00:11:51 --> 00:11:55
well, why does that work?
So, if you take the origin

155
00:11:55 --> 00:11:59
between two parallel slices that
are very close to each other,

156
00:11:59 --> 00:12:01
what's the volume in these two
things?

157
00:12:01 --> 00:12:05
Well, it's essentially s of x
times the thickness of this very

158
00:12:05 --> 00:12:09
thin slice,
and the thickness would be

159
00:12:09 --> 00:12:16
delta x0 dx if you take a limit
with more and more slices.

160
00:12:16 --> 00:12:25
OK, so the volume will be the
integral of s of x dx from,

161
00:12:25 --> 00:12:32
well, what should be the range
for x?

162
00:12:32 --> 00:12:35
Well, we would have to start at
the very lowest value of x that

163
00:12:35 --> 00:12:39
ever happens in our origin,
and we'd have to go all the way

164
00:12:39 --> 00:12:43
to the very largest value of x,
from the very far back to the

165
00:12:43 --> 00:12:48
very far front.
So, in this picture,

166
00:12:48 --> 00:12:57
we probably start over here at
the back, and we'd end over here

167
00:12:57 --> 00:13:05
at the front.
So, let me just say from the

168
00:13:05 --> 00:13:13
minimum, x, to the maximum x.
And now, how do we find S of x?

169
00:13:13 --> 00:13:17
Well, S of x will be actually
again an integral.

170
00:13:17 --> 00:13:20
But now, it's an integral of
the variable,

171
00:13:20 --> 00:13:25
y, because when we look at this
slice, what changes from left to

172
00:13:25 --> 00:13:35
right is y.
So, well let me actually write

173
00:13:35 --> 00:13:40
that down.
For a given,

174
00:13:40 --> 00:13:48
x, the area S of x you can
compute as an integral of f of

175
00:13:48 --> 00:13:52
x, y dy.
OK, well, now x is a constant,

176
00:13:52 --> 00:13:56
and y will be the variable of
integration.

177
00:13:56 --> 00:14:00
What's the range for y?
Well, it's from the leftmost

178
00:14:00 --> 00:14:06
point here to the rightmost
point here on the given slice.

179
00:14:06 --> 00:14:10
So, there is a big catch here.
That's a very important thing

180
00:14:10 --> 00:14:12
to remember.
What is the range of

181
00:14:12 --> 00:14:14
integration?
The range of integration for y

182
00:14:14 --> 00:14:18
depends actually on x.
See, if I take the slice that's

183
00:14:18 --> 00:14:22
pictured on that diagram,
then the range for y goes all

184
00:14:22 --> 00:14:26
the way from the very left to
the very right.

185
00:14:26 --> 00:14:30
But, if I take a slice that,
say, near the very front,

186
00:14:30 --> 00:14:32
then in fact,
only a very small segment of it

187
00:14:32 --> 00:14:36
will be in my region.
So, the range of values for y

188
00:14:36 --> 00:14:41
will be much less.
Let me actually draw a 2D

189
00:14:41 --> 00:14:45
picture for that.
So, remember,

190
00:14:45 --> 00:14:51
we fix x, so,
sorry, so we fix a value of x.

191
00:14:51 --> 00:14:55
OK, and for a given value of x,
what we will do is we'll slice

192
00:14:55 --> 00:14:59
our graph by this plane parallel
to the yz plane.

193
00:14:59 --> 00:15:02
So, now we mention the graph is
sitting above that.

194
00:15:02 --> 00:15:06
OK, that's the region R.
We have the region,

195
00:15:06 --> 00:15:12
R, and I have the graph of a
function above this region,

196
00:15:12 --> 00:15:15
R.
And, I'm trying to find the

197
00:15:15 --> 00:15:20
area between this segment and
the graph above it in this

198
00:15:20 --> 00:15:24
vertical plane.
Well, to do that,

199
00:15:24 --> 00:15:33
I have to integrate from y
going from here to here.

200
00:15:33 --> 00:15:37
I want the area of a piece that
sits above this red segment.

201
00:15:37 --> 00:15:40
And, so in particular, 
the endpoints, 

202
00:15:40 --> 00:15:43
the extreme values for y depend
on x because,

203
00:15:43 --> 00:15:47
see,
if I slice here instead, 

204
00:15:47 --> 00:15:54
well, my bounds for y will be
smaller.

205
00:15:54 --> 00:15:59
OK, so now, if I put the two
things together,

206
00:15:59 --> 00:16:15
what I will get -- -- is
actually a formula where I have

207
00:16:15 --> 00:16:30
to integrate -- -- over x -- --
an integral over y.

208
00:16:30 --> 00:16:42
OK, and so this is called an
iterated integral because we

209
00:16:42 --> 00:16:54
iterate twice the process of
taking an integral.

210
00:16:54 --> 00:16:57
OK, so again, 
what's important to realize

211
00:16:57 --> 00:16:59
here,
I mean, I'm going to say that

212
00:16:59 --> 00:17:02
several times over the next few
days but that's because it's the

213
00:17:02 --> 00:17:05
single most important thing to
remember about double integrals,

214
00:17:05 --> 00:17:09
the bounds here are just going
to be numbers,

215
00:17:09 --> 00:17:12
OK, because the question I'm
asking myself here is,

216
00:17:12 --> 00:17:15
what is the first value of x by
which I might want to slice,

217
00:17:15 --> 00:17:21
and what is the last value of x?
Which range of x do I want to

218
00:17:21 --> 00:17:25
look at to take my red slices?
And, the answer is I would go

219
00:17:25 --> 00:17:29
all the way from here,
that's my first slice,

220
00:17:29 --> 00:17:34
to somewhere here.
That's my last slice.

221
00:17:34 --> 00:17:38
For any value in between these,
I will have some red segment,

222
00:17:38 --> 00:17:40
and I will want to integrate
over that that.

223
00:17:40 --> 00:17:45
On the other hand here, 
the bounds will depend on the

224
00:17:45 --> 00:17:47
outer variable,
x,

225
00:17:47 --> 00:17:54
because at a fixed value of x, 
what the values of y will be

226
00:17:54 --> 00:18:01
depends on x in general.
OK, so I think we should do

227
00:18:01 --> 00:18:11
lots of examples to convince
ourselves and see how it works.

228
00:18:11 --> 00:18:15
Yeah, it's called an iterated
integral because first we

229
00:18:15 --> 00:18:18
integrated over y,
and then we integrate again

230
00:18:18 --> 00:18:20
over x, OK?
So, we can do that,

231
00:18:20 --> 00:18:23
well, I mean,
y depends on x or x depends,

232
00:18:23 --> 00:18:27
no, actually x and y vary
independently of each other

233
00:18:27 --> 00:18:31
inside here.
What is more complicated is how

234
00:18:31 --> 00:18:34
the bounds on y depend on x.
But actually,

235
00:18:34 --> 00:18:37
you could also do the other way
around: first integrate over x,

236
00:18:37 --> 00:18:39
and then over y,
and then the bounds for x will

237
00:18:39 --> 00:18:42
depend on y.
We'll see that on an example.

238
00:18:42 --> 00:18:53
Yes?
So, for y, I'm using the range

239
00:18:53 --> 00:18:58
of values for y that corresponds
to the given value of x,

240
00:18:58 --> 00:19:00
OK?
Remember, this is just like a

241
00:19:00 --> 00:19:04
plot in the xy plane.
Above that, we have the graph.

242
00:19:04 --> 00:19:06
Maybe I should draw a picture
here instead.

243
00:19:06 --> 00:19:09
For a given value of x, 
so that's a given slice, 

244
00:19:09 --> 00:19:13
I have a range of values for y, 
that is, 

245
00:19:13 --> 00:19:16
from this picture at the
leftmost point on that slice to

246
00:19:16 --> 00:19:19
the rightmost point on that
slice.

247
00:19:19 --> 00:19:24
So, where start and where I
stop depends on the value of x.

248
00:19:24 --> 00:19:33
Does that make sense?
OK.

249
00:19:33 --> 00:19:45
OK, no more questions?
OK, so let's do our first

250
00:19:45 --> 00:19:52
example.
So, let's say that we want to

251
00:19:52 --> 00:20:00
integrate the function 1-x^2-y^2
over the region defined by x

252
00:20:00 --> 00:20:06
between 0 and 1,
and y between 0 and 1.

253
00:20:06 --> 00:20:09
So, what does that mean
geometrically?

254
00:20:09 --> 00:20:13
Well, z = 1-x^2-y^2,
and it's a variation on,

255
00:20:13 --> 00:20:17
actually I think we plotted
that one, right?

256
00:20:17 --> 00:20:22
That was our first example of a
function of two variables

257
00:20:22 --> 00:20:24
possibly.
And, so, we saw that the graph

258
00:20:24 --> 00:20:27
is this paraboloid pointing
downwards.

259
00:20:27 --> 00:20:31
OK, it's what you get by taking
a parabola and rotating it.

260
00:20:31 --> 00:20:35
And now, what we are asking is, 
what is the volume between the

261
00:20:35 --> 00:20:40
paraboloid and the xy plane over
the square of side one in the xy

262
00:20:40 --> 00:20:43
plane over the square of side
one in the xy plane,

263
00:20:43 --> 00:20:49
x and y between zero and one.
OK, so, what we'll do is we'll,

264
00:20:49 --> 00:20:54
so, see, here I try to
represent the square.

265
00:20:54 --> 00:20:59
And, we'll just sum the areas
of the slices as,

266
00:20:59 --> 00:21:03
say, x varies from zero to one.
And here, of course,

267
00:21:03 --> 00:21:06
setting up the bounds will be
easy because no matter what x I

268
00:21:06 --> 00:21:08
take, y still goes from zero to
one.

269
00:21:08 --> 00:21:13
See, it's easiest to do double
integrals what the region is

270
00:21:13 --> 00:21:18
just a rectangle on the xy plane
because then you don't have to

271
00:21:18 --> 00:21:21
worry too much about what are
the ranges.

272
00:21:21 --> 00:21:29
OK, so let's do it.
Well, that would be the

273
00:21:29 --> 00:21:38
integral from zero to one of the
integral from zero to one of

274
00:21:38 --> 00:21:42
1-x^2-y^2 dy dx.
So, I'm dropping the

275
00:21:42 --> 00:21:44
parentheses.
But, if you still want to see

276
00:21:44 --> 00:21:48
them, I'm going to put that in
very thin so that you see what

277
00:21:48 --> 00:21:50
it means.
But, actually,

278
00:21:50 --> 00:21:53
the convention is we won't put
this parentheses in there

279
00:21:53 --> 00:21:57
anymore.
OK, so what this means is first

280
00:21:57 --> 00:22:02
I will integrate 1-x^2-y^2 over
y, ranging from zero to one with

281
00:22:02 --> 00:22:07
x held fixed.
So, what that represents is the

282
00:22:07 --> 00:22:10
area in this slice.
So, see here,

283
00:22:10 --> 00:22:13
I've drawn, well,
what happens is actually the

284
00:22:13 --> 00:22:16
function takes positive and
negative values.

285
00:22:16 --> 00:22:18
So, in fact,
I will be counting positively

286
00:22:18 --> 00:22:21
this part of the area.
And, I will be counting

287
00:22:21 --> 00:22:25
negatively this part of the
area, I mean,

288
00:22:25 --> 00:22:32
as usual when I do an integral.
OK, so what I will do to

289
00:22:32 --> 00:22:40
evaluate this,
I will first do what's called

290
00:22:40 --> 00:22:48
the inner integral.
So, to do the inner integral,

291
00:22:48 --> 00:22:55
well, it's pretty easy.
How do I integrate this?

292
00:22:55 --> 00:22:58
Well, it becomes,
so, what's the integral of one?

293
00:22:58 --> 00:23:01
It's y.
Just anything to remember is we

294
00:23:01 --> 00:23:03
are integrating this with
respect to y,

295
00:23:03 --> 00:23:08
not to x.
The integral of x^2 is x^2

296
00:23:08 --> 00:23:14
times y.
And, the integral of y^2 is y^3

297
00:23:14 --> 00:23:18
over 3.
OK, and that we plug in the

298
00:23:18 --> 00:23:21
bounds, which are zero and one
in this case.

299
00:23:21 --> 00:23:28
And so, when you plug y equals
one, you will get one minus x^2

300
00:23:28 --> 00:23:34
minus one third minus,
well, for y equals zero you get

301
00:23:34 --> 00:23:41
0,0, 0, so nothing changes.
OK, so you are left with two

302
00:23:41 --> 00:23:45
thirds minus x^2.
OK, and that's a function of x

303
00:23:45 --> 00:23:47
only.
Here, you shouldn't see any y's

304
00:23:47 --> 00:23:49
anymore because y was your
integration variable.

305
00:23:49 --> 00:23:55
But, you still have x.
You still have x because the

306
00:23:55 --> 00:24:01
area of this shaded slice
depends, of course,

307
00:24:01 --> 00:24:05
on the value of x.
And, so now,

308
00:24:05 --> 00:24:11
the second thing to do is to do
the outer integral.

309
00:24:11 --> 00:24:17
So, now we integrate from zero
to one what we got,

310
00:24:17 --> 00:24:22
which is two thirds minus x^2
dx.

311
00:24:22 --> 00:24:26
OK, and we know how to compute
that because that integrates to

312
00:24:26 --> 00:24:31
two thirds x minus one third x^3
between zero and one.

313
00:24:31 --> 00:24:33
And, I'll let you do the
computation.

314
00:24:33 --> 00:24:43
You will find it's one third.
OK, so that's the final answer.

315
00:24:43 --> 00:24:46
So, that's the general pattern.
When we have a double integral

316
00:24:46 --> 00:24:49
to compute, first we want to set
it up carefully.

317
00:24:49 --> 00:24:51
We want to find,
what will be the bounds in x

318
00:24:51 --> 00:24:53
and y?
And here, that was actually

319
00:24:53 --> 00:24:57
pretty easy because our equation
was very simple.

320
00:24:57 --> 00:25:00
Then, we want to compute the
inner integral,

321
00:25:00 --> 00:25:02
and then we compute the outer
integral.

322
00:25:02 --> 00:25:11
And, that's it.
OK, any questions at this point?

323
00:25:11 --> 00:25:16
No?
OK, so, by the way,

324
00:25:16 --> 00:25:20
we started with dA in the
notation, right?

325
00:25:20 --> 00:25:26
Here we had dA.
And, that somehow became a dy

326
00:25:26 --> 00:25:28
dx.
OK, 

327
00:25:28 --> 00:25:35
so, 
dA became dy dx because when we

328
00:25:35 --> 00:25:39
do the iterated integral this
way,

329
00:25:39 --> 00:25:45
what we're actually doing is
that we are slicing our origin

330
00:25:45 --> 00:25:49
into small rectangles.
OK, that was the area of this

331
00:25:49 --> 00:25:52
small rectangle here?
Well, it's the product of its

332
00:25:52 --> 00:25:58
width times its height.
So, that's delta x times delta

333
00:25:58 --> 00:26:03
y.
OK, so, delta a equals delta x

334
00:26:03 --> 00:26:08
delta y becomes...
So actually,

335
00:26:08 --> 00:26:18
it's not just becomes,
it's really equal.

336
00:26:18 --> 00:26:28
So, the small rectangles for.
Now, it became dy dx and not dx

337
00:26:28 --> 00:26:30
dy.
Well, that's a question of,

338
00:26:30 --> 00:26:32
in which order we do the
iterated integral?

339
00:26:32 --> 00:26:36
It's up to us to decide whether
we want to integrate x first,

340
00:26:36 --> 00:26:38
then y, or y first,
then x.

341
00:26:38 --> 00:26:41
But, as we'll see very soon,
that is an important decision

342
00:26:41 --> 00:26:44
when it comes to setting up the
bounds of integration.

343
00:26:44 --> 00:26:47
Here, it doesn't matter,
but in general we have to be

344
00:26:47 --> 00:26:51
very careful about in which
order we will do things.

345
00:26:51 --> 00:26:57
Yes?
Well, in principle it always

346
00:26:57 --> 00:27:00
works both ways.
Sometimes it will be that

347
00:27:00 --> 00:27:04
because the region has a strange
shape, you can actually set it

348
00:27:04 --> 00:27:06
up more easily one way or the
other.

349
00:27:06 --> 00:27:09
Sometimes it will also be that
the function here,

350
00:27:09 --> 00:27:12
you actually know how to
integrate in one way,

351
00:27:12 --> 00:27:15
but not the other.
So, the theory is that it

352
00:27:15 --> 00:27:18
should work both ways.
In practice,

353
00:27:18 --> 00:27:25
one of the two calculations may
be much harder.

354
00:27:25 --> 00:27:33
OK.
Let's do another example.

355
00:27:33 --> 00:27:39
Let's say that what I wanted to
know was not actually what I

356
00:27:39 --> 00:27:42
computed,
namely, the volume below the

357
00:27:42 --> 00:27:45
paraboloid,
but also the negative of some

358
00:27:45 --> 00:27:48
part that's now in the corner
towards me.

359
00:27:48 --> 00:27:51
But let's say really what I
wanted was just the volume

360
00:27:51 --> 00:27:54
between the paraboloid and the
xy plane,

361
00:27:54 --> 00:27:58
so looking only at the part of
it that sits above the xy plane.

362
00:27:58 --> 00:28:01
So, that means,
instead of integrating over the

363
00:28:01 --> 00:28:05
entire square of size one,
I should just integrate over

364
00:28:05 --> 00:28:08
the quarter disk.
I should stop integrating where

365
00:28:08 --> 00:28:18
my paraboloid hits the xy plane.
So, let me draw another picture.

366
00:28:18 --> 00:28:28
So, let's say I wanted to
integrate, actually -- So,

367
00:28:28 --> 00:28:36
let's call this example two.
So, we are going to do the same

368
00:28:36 --> 00:28:39
function but over a different
region.

369
00:28:39 --> 00:28:44
And, the region will just be,
now, this quarter disk here.

370
00:28:44 --> 00:28:55
OK, so maybe I should draw a
picture on the xy plane.

371
00:28:55 --> 00:29:14
That's your region, R.
OK, so in principle,

372
00:29:14 --> 00:29:19
it will be the same integral.
But what changes is the bounds.

373
00:29:19 --> 00:29:23
Why do the bounds change?
Well, the bounds change because

374
00:29:23 --> 00:29:27
now if I set,
if I fixed some value of x, 

375
00:29:27 --> 00:29:31
then I want to integrate this
part of the slice that's above

376
00:29:31 --> 00:29:36
the xy plane and I don't want to
take this part that's actually

377
00:29:36 --> 00:29:39
outside of my disk.
So, I should stop integrating

378
00:29:39 --> 00:29:42
over y when y reaches this value
here.

379
00:29:42 --> 00:29:46
OK, on that picture here,
on this picture,

380
00:29:46 --> 00:29:52
it tells me for a fixed value
of x, the range of values for y

381
00:29:52 --> 00:29:55
should go only from here to
here.

382
00:29:55 --> 00:30:00
So, that's from here to less
than one.

383
00:30:00 --> 00:30:07
OK, so for a given,
x, the range of y is,

384
00:30:07 --> 00:30:16
well, so what's the lowest
value of y that we want to look

385
00:30:16 --> 00:30:20
at?
It's still zero.

386
00:30:20 --> 00:30:24
From y equals zero to,what's
the value of y here?

387
00:30:24 --> 00:30:28
Well, I have to solve in the
equation of a circle,

388
00:30:28 --> 00:30:31
OK?
So, if I'm here,

389
00:30:31 --> 00:30:39
this is x^2 y^2 equals one.
That means y is square root of

390
00:30:39 --> 00:30:44
one minus x^2.
OK, so I will integrate from y

391
00:30:44 --> 00:30:48
equals zero to y equals square
root of one minus x^2.

392
00:30:48 --> 00:30:55
And, now you see how the bound
from y will depend on the value

393
00:30:55 --> 00:30:59
of x.
OK, so while I erase,

394
00:30:59 --> 00:31:05
I will let you think about,
what is the bound for x now?

395
00:31:05 --> 00:31:06
It's a trick question.

396
00:31:06 --> 00:31:35

397
00:31:35 --> 00:31:48
OK, so I claim that what we
will do -- We write this as an

398
00:31:48 --> 00:31:57
iterated integral first dy then
dx.

399
00:31:57 --> 00:32:04
And, we said for a fixed value
of x, the range for y is from

400
00:32:04 --> 00:32:09
zero to square root of one minus
x^2.

401
00:32:09 --> 00:32:13
What about the range for x?
Well, the range for x should

402
00:32:13 --> 00:32:15
just be numbers.
OK, remember,

403
00:32:15 --> 00:32:18
the question I have to ask now
is if I look at all of these

404
00:32:18 --> 00:32:21
yellow slices,
which one is the first one that

405
00:32:21 --> 00:32:24
I will consider?
Which one is the last one that

406
00:32:24 --> 00:32:27
I want to consider?
So, the smallest value of x

407
00:32:27 --> 00:32:30
that I want to consider is zero
again.

408
00:32:30 --> 00:32:33
And then, I will have actually
a pretty big slice.

409
00:32:33 --> 00:32:35
And I will get smaller,
and smaller,

410
00:32:35 --> 00:32:36
and smaller slices.
And, it stops.

411
00:32:36 --> 00:32:39
I have to stop when x equals
one.

412
00:32:39 --> 00:32:42
Afterwards, there's nothing
else to integrate.

413
00:32:42 --> 00:32:49
So, x goes from zero to one.
OK, and now,

414
00:32:49 --> 00:32:52
see how in the inner integral,
the bounds depend on in the

415
00:32:52 --> 00:32:55
inner integral,
the bounds depend on x.

416
00:32:55 --> 00:32:58
In the outer one,
you just get numbers because

417
00:32:58 --> 00:33:02
the questions that you have to
ask to set up this one and set

418
00:33:02 --> 00:33:05
up that one are different.
Here, the question is,

419
00:33:05 --> 00:33:07
if I fix a given,
x, if I look at a given slice,

420
00:33:07 --> 00:33:10
what's the range for y?
Here, the question is,

421
00:33:10 --> 00:33:16
what's the first slice?
What is the last slice?

422
00:33:16 --> 00:33:21
Does that make sense?
Everyone happy with that?

423
00:33:21 --> 00:33:28
OK, very good.
So, now, how do we compute that?

424
00:33:28 --> 00:33:36
Well, we do the inner integral.
So, that's an integral from

425
00:33:36 --> 00:33:42
zero to square root of one minus
x^2 of one minus x^2 minus y^2

426
00:33:42 --> 00:33:48
dy.
And, well, that integrates to

427
00:33:48 --> 00:34:00
y-x^2y-y^3 over three from zero
to square root of one minus x^2.

428
00:34:00 --> 00:34:06
And then, that becomes,
well, the root of one minus x^2

429
00:34:06 --> 00:34:13
minus x^2 root of one minus x^2
minus y minus x^2 to the three

430
00:34:13 --> 00:34:16
halves over three.
And actually,

431
00:34:16 --> 00:34:20
if you look at it for long
enough, see, this says one minus

432
00:34:20 --> 00:34:23
x^2 times square root of one
minus x^2.

433
00:34:23 --> 00:34:26
So, actually,
that's also,

434
00:34:26 --> 00:34:30
so, in fact,
that simplifies to two thirds

435
00:34:30 --> 00:34:35
of one minus x^2 to the three
halves.

436
00:34:35 --> 00:34:40
OK, let me redo that,
maybe, slightly differently.

437
00:34:40 --> 00:34:50
This was one minus x^2 times y.
So -- -- one minus x^2 times y

438
00:34:50 --> 00:35:02
becomes square root of one minus
x^2 minus y^3 over three.

439
00:35:02 --> 00:35:04
And then, when I take y equals
zero, I get zero.

440
00:35:04 --> 00:35:08
So, I don't subtract anything.
OK, so now you see this is one

441
00:35:08 --> 00:35:11
minus x^2 to the three halves
minus a third of it.

442
00:35:11 --> 00:35:21
So, you're left with two thirds.
OK, so, that's the integral.

443
00:35:21 --> 00:35:27
The outer integral is the
integral from zero to one of two

444
00:35:27 --> 00:35:31
thirds of one minus x^2 to the
three halves dx.

445
00:35:31 --> 00:35:37
And, 
well, 

446
00:35:37 --> 00:35:42
I let you see if you remember
single variable integrals by

447
00:35:42 --> 00:35:47
trying to figure out what this
actually comes out to be is it

448
00:35:47 --> 00:35:54
pi over two,
or pi over eight, actually?

449
00:35:54 --> 00:36:12
I think it's pi over eight.
OK, well I guess we have to do

450
00:36:12 --> 00:36:15
it then.
I wrote something on my notes,

451
00:36:15 --> 00:36:16
but it's not very clear,
OK?

452
00:36:16 --> 00:36:18
So, how do we compute this
thing?

453
00:36:18 --> 00:36:20
Well, we have to do trig
substitution.

454
00:36:20 --> 00:36:24
That's the only way I know to
compute an integral like that,

455
00:36:24 --> 00:36:27
OK?
So, we'll set x equal sine

456
00:36:27 --> 00:36:35
theta, and then square root of
one minus x^2 will be cosine

457
00:36:35 --> 00:36:41
theta.
We are using sine squared plus

458
00:36:41 --> 00:36:51
cosine squared equals one.
And, so that will become -- --

459
00:36:51 --> 00:36:59
so, two thirds remains two
thirds.

460
00:36:59 --> 00:37:04
One minus x^2 to the three
halves becomes cosine cubed

461
00:37:04 --> 00:37:08
theta.
dx, well, if x is sine theta,

462
00:37:08 --> 00:37:13
then dx is cosine theta d
theta.

463
00:37:13 --> 00:37:17
So, that's cosine theta d theta.
And, 

464
00:37:17 --> 00:37:19
well, if you do things with
substitution,

465
00:37:19 --> 00:37:22
which is the way I do them,
then you should worry about the

466
00:37:22 --> 00:37:25
bounds for theta which will be
zero to pi over two.

467
00:37:25 --> 00:37:30
Or, you can also just plug in
the bounds at the end.

468
00:37:30 --> 00:37:36
So, now you have the two thirds
times the integral from zero to

469
00:37:36 --> 00:37:41
pi over two of cosine to the
fourth theta d theta.

470
00:37:41 --> 00:37:48
And, how do you integrate that?
Well, you have to use double

471
00:37:48 --> 00:37:55
angle formulas.
OK, so cosine to the fourth,

472
00:37:55 --> 00:38:02
remember, cosine squared theta
is one plus cosine two theta

473
00:38:02 --> 00:38:14
over two.
And, we want the square of that.

474
00:38:14 --> 00:38:23
And, 
so that will give us -- -- of, 

475
00:38:23 --> 00:38:29
well, we'll have, 
it's actually one quarter plus

476
00:38:29 --> 00:38:37
one half cosine to theta plus
one quarter cosine square to

477
00:38:37 --> 00:38:41
theta d theta.
And, how will you handle this

478
00:38:41 --> 00:38:42
guy?
Well, using,

479
00:38:42 --> 00:38:45
again, the double angle
formula.

480
00:38:45 --> 00:38:51
OK, so it's getting slightly
nasty.

481
00:38:51 --> 00:38:54
So, but I don't know any
simpler solution except for one

482
00:38:54 --> 00:38:56
simpler solution,
which is you have a table of

483
00:38:56 --> 00:38:59
integrals of this form inside
the notes.

484
00:38:59 --> 00:39:08
Yes?
No, I don't think so because if

485
00:39:08 --> 00:39:12
you take one half times cosine
half times two,

486
00:39:12 --> 00:39:15
you will still have half,
OK?

487
00:39:15 --> 00:39:18
So, if you do,
again, the double angle

488
00:39:18 --> 00:39:22
formula, I think I'm not going
to bother to do it.

489
00:39:22 --> 00:39:37
I claim you will get,
at the end, pi over eight

490
00:39:37 --> 00:39:46
because I say so.
OK, so exercise,

491
00:39:46 --> 00:39:54
continue calculating and get pi
over eight.

492
00:39:54 --> 00:39:56
OK, now what does the show us?
Well, this shows us,

493
00:39:56 --> 00:40:00
actually, that this is probably
not the right way to do this.

494
00:40:00 --> 00:40:02
OK, the right way to do this
will be to integrate it in polar

495
00:40:02 --> 00:40:05
coordinates.
And, that's what we will learn

496
00:40:05 --> 00:40:09
how to do tomorrow.
So, we will actually see how to

497
00:40:09 --> 00:40:11
do it with much less trig.

498
00:40:11 --> 00:40:36

499
00:40:36 --> 00:40:46
So, that will be easier in
polar coordinates.

500
00:40:46 --> 00:40:51
So, we will see that tomorrow.
OK, so we are almost there.

501
00:40:51 --> 00:40:53
I mean, here you just use a
double angle again and then you

502
00:40:53 --> 00:40:55
can get it.
And, it's pretty

503
00:40:55 --> 00:41:00
straightforward.
OK, so one thing that's kind of

504
00:41:00 --> 00:41:07
interesting to know is we can
exchange the order of

505
00:41:07 --> 00:41:11
integration.
Say we have an integral given

506
00:41:11 --> 00:41:14
to us in the order dy dx,
we can switch it to dx dy.

507
00:41:14 --> 00:41:18
But, we have to be extremely
careful with the bounds.

508
00:41:18 --> 00:41:22
So, you certainly cannot just
swap the bounds of the inner and

509
00:41:22 --> 00:41:26
outer because there you would
end up having this square root

510
00:41:26 --> 00:41:30
of one minus x^2 on the outside,
and you would never get a

511
00:41:30 --> 00:41:33
number out of that.
So, that cannot work.

512
00:41:33 --> 00:41:37
It's more complicated than that.
OK, so, well,

513
00:41:37 --> 00:41:42
here's a first baby example.
Certainly, if I do integral

514
00:41:42 --> 00:41:46
from zero to one,
integral from zero to two dx

515
00:41:46 --> 00:41:51
dy, there, I can certainly
switch the bounds without

516
00:41:51 --> 00:41:54
thinking too much.
What's the reason for that?

517
00:41:54 --> 00:41:59
Well, the reason for that is
this corresponds in both cases

518
00:41:59 --> 00:42:04
to integrating x from zero to
two, and y from zero to one.

519
00:42:04 --> 00:42:08
It's a rectangle.
So, if I slice it this way,

520
00:42:08 --> 00:42:11
you see that y goes from zero
to one for any x between zero

521
00:42:11 --> 00:42:13
and two.
It's this guy.

522
00:42:13 --> 00:42:18
If I slice it that way,
then x goes from zero to two

523
00:42:18 --> 00:42:22
for any value of y between zero
and one.

524
00:42:22 --> 00:42:25
And, it's this one.
So, here it works.

525
00:42:25 --> 00:42:28
But in general,
I have to draw picture of my

526
00:42:28 --> 00:42:32
region, and see how the slices
look like both ways.

527
00:42:32 --> 00:42:41
OK, so let's do a more
interesting one.

528
00:42:41 --> 00:42:48
Let's say that I want to
compute an integral from zero to

529
00:42:48 --> 00:42:55
one of integral from x to square
root of x of e^y over y dy dx.

530
00:42:55 --> 00:42:59
So, why did I choose this guy?
Which is the guy because as far

531
00:42:59 --> 00:43:03
as I can tell,
there's no way to integrate e^y

532
00:43:03 --> 00:43:06
over y.
So, this is an integral that

533
00:43:06 --> 00:43:10
you cannot compute this way.
So, it's a good example for why

534
00:43:10 --> 00:43:13
this can be useful.
So, if you do it this way,

535
00:43:13 --> 00:43:14
you are stuck immediately.
So, instead,

536
00:43:14 --> 00:43:15
we will try to switch the
order.

537
00:43:15 --> 00:43:20
But, to switch the order,
we have to understand,

538
00:43:20 --> 00:43:25
what do these bounds mean?
OK, so let's draw a picture of

539
00:43:25 --> 00:43:29
the region.
Well what I am saying is y

540
00:43:29 --> 00:43:33
equals x to y equals square root
of x.

541
00:43:33 --> 00:43:48
Well, let's draw y equals x,
and y equals square root of x.

542
00:43:48 --> 00:43:52
Well, maybe I should actually
put this here,

543
00:43:52 --> 00:43:56
y equals x to y equals square
root of x.

544
00:43:56 --> 00:44:01
OK, and so I will go,
for each value of x I will go

545
00:44:01 --> 00:44:07
from y equals xo to y equals
square root of x.

546
00:44:07 --> 00:44:11
And then, we'll do that for
values of x that go from x

547
00:44:11 --> 00:44:15
equals zero to x equals one,
which happens to be exactly

548
00:44:15 --> 00:44:22
where these things intersect.
So, my region will consist of

549
00:44:22 --> 00:44:26
all this, OK?
So now, if I want to do it the

550
00:44:26 --> 00:44:29
other way around,
I have to decompose my region.

551
00:44:29 --> 00:44:32
The other way around,
I have to, so my goal,

552
00:44:32 --> 00:44:35
now, is to rewrite this as an
integral.

553
00:44:35 --> 00:44:36
Well, it's still the same
function.

554
00:44:36 --> 00:44:41
It's still e to the y over y.
But now, I want to integrate dx

555
00:44:41 --> 00:44:45
dy.
So, how do I integrate over x?

556
00:44:45 --> 00:44:50
Well, I fix a value of y.
And, for that value of y,

557
00:44:50 --> 00:44:54
what's the range of x?
Well, the range for x is from

558
00:44:54 --> 00:44:57
here to here.
OK, what's the value of x here?

559
00:44:57 --> 00:45:02
Let's start with an easy one.
This is x equals y.

560
00:45:02 --> 00:45:08
What about this one?
It's x equals y^2.

561
00:45:08 --> 00:45:16
OK, so, x goes from y2 to y,
and then what about y?

562
00:45:16 --> 00:45:19
Well, I have to start at the
bottom of my region.

563
00:45:19 --> 00:45:24
That's y equals zero to the
top, which is at y equals one.

564
00:45:24 --> 00:45:31
So, y goes from zero to one.
So, switching the bounds is not

565
00:45:31 --> 00:45:36
completely obvious.
That took a little bit of work.

566
00:45:36 --> 00:45:40
But now that we've done that, 
well, 

567
00:45:40 --> 00:45:44
just to see how it goes, 
it's actually going to be much

568
00:45:44 --> 00:45:48
easier to integrate because the
inner integral,

569
00:45:48 --> 00:45:52
well, what's the integral of
e^y over y with respect to x?

570
00:45:52 --> 00:46:01
It's just that times x,
right, from x equals y^2 to y.

571
00:46:01 --> 00:46:04
So, that will be, 
well, if I plug x equals y, 

572
00:46:04 --> 00:46:09
I will get e to the y minus, 
if I plug x equals y^2, 

573
00:46:09 --> 00:46:16
I will get e to the y over y
times y^2 into the y times y,

574
00:46:16 --> 00:46:20
OK?
So, now, if I do the outer

575
00:46:20 --> 00:46:27
integral, I will have the
integral from zero to one of e

576
00:46:27 --> 00:46:33
to the y minus y^e to the y dy.
And, that one actually is a

577
00:46:33 --> 00:46:37
little bit easier.
So, we know how to integrate

578
00:46:37 --> 00:46:39
e^y.
We don't quite know how to

579
00:46:39 --> 00:46:43
integrate ye^y.
But, let's try.

580
00:46:43 --> 00:46:49
So, let's see,
what's the derivative of ye^y?

581
00:46:49 --> 00:46:55
Well, there's a product rule
that's one times e^y plus y

582
00:46:55 --> 00:46:59
times the derivative of e^y is
ye^y.

583
00:46:59 --> 00:47:03
So, if we do,
OK, let's put a minus sign in

584
00:47:03 --> 00:47:06
front.
Well, that's almost what we

585
00:47:06 --> 00:47:09
want, except we have a minus e^y
instead of a plus e^y.

586
00:47:09 --> 00:47:15
So, we need to add 2e^y.
And, I claim that's the

587
00:47:15 --> 00:47:18
antiderivative.
OK, if you got lost,

588
00:47:18 --> 00:47:21
you can also integrate by
integrating by parts,

589
00:47:21 --> 00:47:25
by taking the derivative of y
and integrating these guys.

590
00:47:25 --> 00:47:29
Or, but, you know, that works.
Just, your first guess would

591
00:47:29 --> 00:47:32
be, maybe, let's try minus y^e
to the y.

592
00:47:32 --> 00:47:35
Take the derivative of that,
compare, see what you need to

593
00:47:35 --> 00:47:39
do to fix.
And so, if you take that

594
00:47:39 --> 00:47:45
between zero and one,
you'll actually get e minus

595
00:47:45 --> 00:47:47
two.
OK, 

596
00:47:47 --> 00:47:50
so, 
tomorrow we are going to see

597
00:47:50 --> 00:47:53
how to do double integrals in
polar coordinates,

598
00:47:53 --> 00:47:55
and also applications of double
integrals,

599
00:47:55 --> 00:47:58
how to use them for interesting
things.

600
00:47:58 --> 00:47:59