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

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derivatives.
Now, we're going to integrals.

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So -- OK, so more precisely,
we are going to be talking

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about double integrals.
OK, so just to motivate the

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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

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00:01:16,000 --> 00:01:21,000
from, say,
a to b of f of x dx,

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00:01:21,000 --> 00:01:33,000
well, that corresponds to the
area below the graph of f over

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00:01:33,000 --> 00:01:43,000
the interval from a to b.
OK, so the picture is something

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00:01:43,000 --> 00:01:47,000
like you have a;
you have b.

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00:01:47,000 --> 00:01:55,000
You have the graph of f,
and then what the integral

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00:01:55,000 --> 00:02:02,000
measures is the area of this
region.

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00:02:02,000 --> 00:02:04,000
And, when we say the area of
this region, of course,

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00:02:04,000 --> 00:02:06,000
if f is positive,
that's what happens.

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00:02:06,000 --> 00:02:11,000
If f is negative,
then we count negatively the

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00:02:11,000 --> 00:02:15,000
area below the x axis.
OK, so, now,

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00:02:15,000 --> 00:02:19,000
when you have a function of two
variables, then you can try to

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00:02:19,000 --> 00:02:22,000
do the same thing.
Namely, you can plot its graph.

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00:02:22,000 --> 00:02:24,000
Its graph will be a surface in
space.

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00:02:24,000 --> 00:02:28,000
And then, we can try to look
for the volume below the graph.

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00:02:28,000 --> 00:02:35,000
And that's what we will call
the double integral of the

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00:02:35,000 --> 00:02:44,000
function over a certain region.
OK, so let's say that we have a

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00:02:44,000 --> 00:02:51,000
function of two variables,
x and y.

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00:02:51,000 --> 00:03:00,000
Then, we'll look at the volume
that's below the graph z equals

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00:03:00,000 --> 00:03:06,000
f of xy.
OK, so, let's draw a picture

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00:03:06,000 --> 00:03:10,000
for what this means.
I have a function of x and y.

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00:03:10,000 --> 00:03:16,000
I can draw its graph.
The graph will be the surface

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00:03:16,000 --> 00:03:20,000
with equation z equals f of x
and y.

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00:03:20,000 --> 00:03:24,000
And, well, I have to decide
where I will integrate the

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00:03:24,000 --> 00:03:26,000
function.
So, for that,

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00:03:26,000 --> 00:03:30,000
I will choose some region in
the xy plane.

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00:03:30,000 --> 00:03:39,000
And, I will integrate the
function on that region.

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00:03:39,000 --> 00:03:46,000
So, it's over a region,
R, in the xy plane.

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00:03:46,000 --> 00:03:51,000
So, I have this region R and I
look at the piece of the graph

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00:03:51,000 --> 00:03:56,000
that is above this region.
And, we'll try to compute the

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00:03:56,000 --> 00:04:02,000
volume of this solid here.
OK, that's what the double

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00:04:02,000 --> 00:04:08,000
integral will measure.
So, we'll call that the double

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00:04:08,000 --> 00:04:13,000
integral of our region,
R, of f of xy dA and I will

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00:04:13,000 --> 00:04:18,000
have to explain what the
notation means.

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00:04:18,000 --> 00:04:22,000
So, dA here stands for a piece
of area.

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00:04:22,000 --> 00:04:26,000
A stands for area.
And, well, it's a double

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00:04:26,000 --> 00:04:28,000
integral.
So, that's why we have two

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00:04:28,000 --> 00:04:30,000
integral signs.
And, we'll have to indicate

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00:04:30,000 --> 00:04:33,000
somehow the region over which we
are integrating.

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00:04:33,000 --> 00:04:37,000
OK, we'll come up with more
concrete notations when we see

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00:04:37,000 --> 00:04:39,000
how to actually compute these
things.

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00:04:39,000 --> 00:04:44,000
That's the basic definition.
OK, so actually,

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00:04:44,000 --> 00:04:45,000
how do we define it,
that's not really much of a

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definition yet.
How do we actually define this

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00:04:49,000 --> 00:04:52,000
rigorously?
Well, remember,

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00:04:52,000 --> 00:04:56,000
the integral in one variable,
you probably saw a definition

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00:04:56,000 --> 00:04:58,000
where you take your integral
from a to b,

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00:04:58,000 --> 00:05:01,000
and you cut it into little
pieces.

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00:05:01,000 --> 00:05:03,000
And then, for each little
piece, you take the value of a

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00:05:03,000 --> 00:05:06,000
function, and you multiply by
the width of a piece.

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00:05:06,000 --> 00:05:11,000
That gives you a rectangular
slice, and then you sum all of

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00:05:11,000 --> 00:05:14,000
these rectangular slices
together.

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00:05:14,000 --> 00:05:17,000
So, here we'll do the same
thing.

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00:05:17,000 --> 00:05:23,000
So, well, let me put a picture
up and explain what it does.

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00:05:23,000 --> 00:05:26,000
So, we're going to cut our
origin into little pieces,

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00:05:26,000 --> 00:05:30,000
say, little rectangles or
actually anything we want.

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00:05:30,000 --> 00:05:35,000
And then, for each piece,
with the small area, delta A,

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00:05:35,000 --> 00:05:39,000
we'll take the area delta a
times the value of a function in

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00:05:39,000 --> 00:05:43,000
there that will give us the
volume of a small box that sits

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00:05:43,000 --> 00:05:46,000
under the graph.
And then, we'll add all these

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boxes together.
That gives us an estimate of a

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00:05:49,000 --> 00:05:51,000
volume.
And then, to get actually the

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integral,
the integral will be defined as

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00:05:53,000 --> 00:05:56,000
a limit as we subdivide into
smaller and smaller boxes,

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00:05:56,000 --> 00:05:59,000
and we sum more and more
pieces,

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00:05:59,000 --> 00:06:02,000
OK?
So, actually,

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00:06:02,000 --> 00:06:11,000
what we do, oh,
I still have a board here.

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00:06:11,000 --> 00:06:20,000
So, the actual definition
involves cutting R into small

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00:06:20,000 --> 00:06:29,000
pieces of area that's called
delta A or maybe delta Ai,

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00:06:29,000 --> 00:06:36,000
the area of the i'th piece.
And then, OK,

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00:06:36,000 --> 00:06:42,000
so maybe in the xy plane,
we have our region,

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00:06:42,000 --> 00:06:47,000
and we'll cut it may be using
some grade.

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00:06:47,000 --> 00:06:49,000
OK, and then we'll have each
small piece.

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00:06:49,000 --> 00:07:00,000
Each small piece will have area
delta Ai and it will be at some

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00:07:00,000 --> 00:07:06,000
point, let's call it xi,
yi ...

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00:07:06,000 --> 00:07:13,000
yi, xi.
And then, we'll consider the

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00:07:13,000 --> 00:07:18,000
sum over all the pieces of f at
that point, xi,

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00:07:18,000 --> 00:07:22,000
yi times the area of a small
piece.

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00:07:22,000 --> 00:07:27,000
So, what that corresponds to in
the three-dimensional picture is

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00:07:27,000 --> 00:07:31,000
just I sum the volumes of all of
these little columns that sit

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under the graph.
OK, and then,

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00:07:36,000 --> 00:07:49,000
so what I do is actually I take
the limit as the size of the

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pieces tends to zero.
So, I have more and more

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smaller and smaller pieces.
And, that gives me the double

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00:08:03,000 --> 00:08:07,000
integral.
OK, so that's not a very good

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00:08:07,000 --> 00:08:13,000
sentence, but whatever.
So, OK, so that's the

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00:08:13,000 --> 00:08:16,000
definition.
Of course, we will have to see

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how to compute it.
We don't actually compute it.

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00:08:19,000 --> 00:08:21,000
When you compute an integral in
single variable calculus,

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00:08:21,000 --> 00:08:23,000
you don't do that.
You don't cut into little

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pieces and sum the pieces
together.

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00:08:25,000 --> 00:08:28,000
You've learned how to integrate
functions using various

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formulas,
and similarly here,

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we'll learn how to actually
compute these things without

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doing that cutting into small
pieces.

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OK, any questions first about
the concept, or what the

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definition is?
Yes?

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00:08:48,000 --> 00:08:50,000
Well, so we'll have to learn
which tricks work,

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00:08:50,000 --> 00:08:52,000
and how exactly.
But, so what we'll do actually

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00:08:52,000 --> 00:08:56,000
is we'll reduce the calculation
of a double integral to two

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00:08:56,000 --> 00:08:58,000
calculations of single
integrals.

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00:08:58,000 --> 00:09:00,000
And so, for V,
certainly, all the tricks

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00:09:00,000 --> 00:09:03,000
you've learned in single
variable calculus will come in

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00:09:03,000 --> 00:09:05,000
handy.
OK, so,

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00:09:05,000 --> 00:09:07,000
yeah that's a strong suggestion
that if you've forgotten

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00:09:07,000 --> 00:09:09,000
everything about single variable
calculus,

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00:09:09,000 --> 00:09:17,000
now would be a good time to
actually brush up on integrals.

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00:09:17,000 --> 00:09:21,000
The usual integrals,
and the usual substitution

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00:09:21,000 --> 00:09:26,000
tricks and easy trig in
particular, these would be very

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useful.
OK, so, yeah,

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how do we compute these things?
That's what we would have to

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00:09:31,000 --> 00:09:37,000
come up with.
And, well,

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00:09:37,000 --> 00:09:39,000
going back to what we did with
derivatives,

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00:09:39,000 --> 00:09:42,000
to understand variations of
functions and derivatives,

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00:09:42,000 --> 00:09:45,000
what we did was really we took
slices parallel to an axis or

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00:09:45,000 --> 00:09:46,000
another one.
So, in fact,

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00:09:46,000 --> 00:09:50,000
here, the key is also the same.
So, what we are going to do is

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instead of cutting into a lot of
small boxes like that and

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00:09:52,000 --> 00:09:57,000
summing completely at random,
we will actually somehow scan

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00:09:57,000 --> 00:10:01,000
through our region by parallel
planes,

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00:10:01,000 --> 00:10:03,000
OK?
So, let me put up,

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00:10:03,000 --> 00:10:08,000
actually, a slightly different
picture up here.

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00:10:08,000 --> 00:10:11,000
So, what I'm going to do is I'm
going to take planes,

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00:10:11,000 --> 00:10:14,000
say in this picture,
parallel to the yz plane.

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00:10:14,000 --> 00:10:18,000
I'll take a moving plane that
scans from the back to the front

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00:10:18,000 --> 00:10:23,000
or from the front to the back.
So, that means I set the value

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00:10:23,000 --> 00:10:27,000
of x, and I look at the slice,
x equals x0,

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00:10:27,000 --> 00:10:31,000
and then I will do that for all
values of x0.

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00:10:31,000 --> 00:10:35,000
So, now in each slice,
well, I get what looks a lot

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like a single variable integral.
OK, and that integral will tell

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me, what is the area in this?
Well, I guess it's supposed to

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00:10:44,000 --> 00:10:48,000
be green, but it all comes as
black, so, let's say the black

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00:10:48,000 --> 00:10:51,000
shaded slice.
And then, when I add all of

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00:10:51,000 --> 00:10:55,000
these areas together,
as the value of x changes,

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00:10:55,000 --> 00:11:00,000
I will get the volume.
OK, let me try to explain that

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00:11:00,000 --> 00:11:07,000
again.
So, to compute this integral,

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00:11:07,000 --> 00:11:16,000
what we do is actually we take
slices.

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00:11:16,000 --> 00:11:29,000
So, let's consider,
let's call s of x the area of a

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00:11:29,000 --> 00:11:41,000
slice, well, by a plane parallel
to the yz plane.

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00:11:41,000 --> 00:11:46,000
OK, so on the picture,
s of x is just the area of this

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thing in the vertical wall.
Now, if you sum all of these,

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00:11:51,000 --> 00:11:55,000
well, why does that work?
So, if you take the origin

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between two parallel slices that
are very close to each other,

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what's the volume in these two
things?

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00:12:01,000 --> 00:12:05,000
Well, it's essentially s of x
times the thickness of this very

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00:12:05,000 --> 00:12:09,000
thin slice,
and the thickness would be

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00:12:09,000 --> 00:12:16,000
delta x0 dx if you take a limit
with more and more slices.

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00:12:16,000 --> 00:12:25,000
OK, so the volume will be the
integral of s of x dx from,

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00:12:25,000 --> 00:12:32,000
well, what should be the range
for x?

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00:12:32,000 --> 00:12:35,000
Well, we would have to start at
the very lowest value of x that

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00:12:35,000 --> 00:12:39,000
ever happens in our origin,
and we'd have to go all the way

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00:12:39,000 --> 00:12:43,000
to the very largest value of x,
from the very far back to the

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00:12:43,000 --> 00:12:48,000
very far front.
So, in this picture,

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00:12:48,000 --> 00:12:57,000
we probably start over here at
the back, and we'd end over here

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00:12:57,000 --> 00:13:05,000
at the front.
So, let me just say from the

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00:13:05,000 --> 00:13:13,000
minimum, x, to the maximum x.
And now, how do we find S of x?

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00:13:13,000 --> 00:13:17,000
Well, S of x will be actually
again an integral.

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00:13:17,000 --> 00:13:20,000
But now, it's an integral of
the variable,

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00:13:20,000 --> 00:13:25,000
y, because when we look at this
slice, what changes from left to

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00:13:25,000 --> 00:13:35,000
right is y.
So, well let me actually write

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00:13:35,000 --> 00:13:40,000
that down.
For a given,

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00:13:40,000 --> 00:13:48,000
x, the area S of x you can
compute as an integral of f of

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00:13:48,000 --> 00:13:52,000
x, y dy.
OK, well, now x is a constant,

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00:13:52,000 --> 00:13:56,000
and y will be the variable of
integration.

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00:13:56,000 --> 00:14:00,000
What's the range for y?
Well, it's from the leftmost

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00:14:00,000 --> 00:14:06,000
point here to the rightmost
point here on the given slice.

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00:14:06,000 --> 00:14:10,000
So, there is a big catch here.
That's a very important thing

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00:14:10,000 --> 00:14:12,000
to remember.
What is the range of

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00:14:12,000 --> 00:14:14,000
integration?
The range of integration for y

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00:14:14,000 --> 00:14:18,000
depends actually on x.
See, if I take the slice that's

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00:14:18,000 --> 00:14:22,000
pictured on that diagram,
then the range for y goes all

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the way from the very left to
the very right.

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00:14:26,000 --> 00:14:30,000
But, if I take a slice that,
say, near the very front,

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00:14:30,000 --> 00:14:32,000
then in fact,
only a very small segment of it

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will be in my region.
So, the range of values for y

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00:14:36,000 --> 00:14:41,000
will be much less.
Let me actually draw a 2D

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00:14:41,000 --> 00:14:45,000
picture for that.
So, remember,

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00:14:45,000 --> 00:14:51,000
we fix x, so,
sorry, so we fix a value of x.

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00:14:51,000 --> 00:14:55,000
OK, and for a given value of x,
what we will do is we'll slice

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00:14:55,000 --> 00:14:59,000
our graph by this plane parallel
to the yz plane.

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00:14:59,000 --> 00:15:02,000
So, now we mention the graph is
sitting above that.

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00:15:02,000 --> 00:15:06,000
OK, that's the region R.
We have the region,

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00:15:06,000 --> 00:15:12,000
R, and I have the graph of a
function above this region,

196
00:15:12,000 --> 00:15:15,000
R.
And, I'm trying to find the

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00:15:15,000 --> 00:15:20,000
area between this segment and
the graph above it in this

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00:15:20,000 --> 00:15:24,000
vertical plane.
Well, to do that,

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00:15:24,000 --> 00:15:33,000
I have to integrate from y
going from here to here.

200
00:15:33,000 --> 00:15:37,000
I want the area of a piece that
sits above this red segment.

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00:15:37,000 --> 00:15:40,000
And, so in particular,
the endpoints,

202
00:15:40,000 --> 00:15:43,000
the extreme values for y depend
on x because,

203
00:15:43,000 --> 00:15:47,000
see,
if I slice here instead,

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00:15:47,000 --> 00:15:54,000
well, my bounds for y will be
smaller.

205
00:15:54,000 --> 00:15:59,000
OK, so now, if I put the two
things together,

206
00:15:59,000 --> 00:16:15,000
what I will get -- -- is
actually a formula where I have

207
00:16:15,000 --> 00:16:30,000
to integrate -- -- over x -- --
an integral over y.

208
00:16:30,000 --> 00:16:42,000
OK, and so this is called an
iterated integral because we

209
00:16:42,000 --> 00:16:54,000
iterate twice the process of
taking an integral.

210
00:16:54,000 --> 00:16:57,000
OK, so again,
what's important to realize

211
00:16:57,000 --> 00:16:59,000
here,
I mean, I'm going to say that

212
00:16:59,000 --> 00:17:02,000
several times over the next few
days but that's because it's the

213
00:17:02,000 --> 00:17:05,000
single most important thing to
remember about double integrals,

214
00:17:05,000 --> 00:17:09,000
the bounds here are just going
to be numbers,

215
00:17:09,000 --> 00:17:12,000
OK, because the question I'm
asking myself here is,

216
00:17:12,000 --> 00:17:15,000
what is the first value of x by
which I might want to slice,

217
00:17:15,000 --> 00:17:21,000
and what is the last value of x?
Which range of x do I want to

218
00:17:21,000 --> 00:17:25,000
look at to take my red slices?
And, the answer is I would go

219
00:17:25,000 --> 00:17:29,000
all the way from here,
that's my first slice,

220
00:17:29,000 --> 00:17:34,000
to somewhere here.
That's my last slice.

221
00:17:34,000 --> 00:17:38,000
For any value in between these,
I will have some red segment,

222
00:17:38,000 --> 00:17:40,000
and I will want to integrate
over that that.

223
00:17:40,000 --> 00:17:45,000
On the other hand here,
the bounds will depend on the

224
00:17:45,000 --> 00:17:47,000
outer variable,
x,

225
00:17:47,000 --> 00:17:54,000
because at a fixed value of x,
what the values of y will be

226
00:17:54,000 --> 00:18:01,000
depends on x in general.
OK, so I think we should do

227
00:18:01,000 --> 00:18:11,000
lots of examples to convince
ourselves and see how it works.

228
00:18:11,000 --> 00:18:15,000
Yeah, it's called an iterated
integral because first we

229
00:18:15,000 --> 00:18:18,000
integrated over y,
and then we integrate again

230
00:18:18,000 --> 00:18:20,000
over x, OK?
So, we can do that,

231
00:18:20,000 --> 00:18:23,000
well, I mean,
y depends on x or x depends,

232
00:18:23,000 --> 00:18:27,000
no, actually x and y vary
independently of each other

233
00:18:27,000 --> 00:18:31,000
inside here.
What is more complicated is how

234
00:18:31,000 --> 00:18:34,000
the bounds on y depend on x.
But actually,

235
00:18:34,000 --> 00:18:37,000
you could also do the other way
around: first integrate over x,

236
00:18:37,000 --> 00:18:39,000
and then over y,
and then the bounds for x will

237
00:18:39,000 --> 00:18:42,000
depend on y.
We'll see that on an example.

238
00:18:42,000 --> 00:18:53,000
Yes?
So, for y, I'm using the range

239
00:18:53,000 --> 00:18:58,000
of values for y that corresponds
to the given value of x,

240
00:18:58,000 --> 00:19:00,000
OK?
Remember, this is just like a

241
00:19:00,000 --> 00:19:04,000
plot in the xy plane.
Above that, we have the graph.

242
00:19:04,000 --> 00:19:06,000
Maybe I should draw a picture
here instead.

243
00:19:06,000 --> 00:19:09,000
For a given value of x,
so that's a given slice,

244
00:19:09,000 --> 00:19:13,000
I have a range of values for y,
that is,

245
00:19:13,000 --> 00:19:16,000
from this picture at the
leftmost point on that slice to

246
00:19:16,000 --> 00:19:19,000
the rightmost point on that
slice.

247
00:19:19,000 --> 00:19:24,000
So, where start and where I
stop depends on the value of x.

248
00:19:24,000 --> 00:19:33,000
Does that make sense?
OK.

249
00:19:33,000 --> 00:19:45,000
OK, no more questions?
OK, so let's do our first

250
00:19:45,000 --> 00:19:52,000
example.
So, let's say that we want to

251
00:19:52,000 --> 00:20:00,000
integrate the function 1-x^2-y^2
over the region defined by x

252
00:20:00,000 --> 00:20:06,000
between 0 and 1,
and y between 0 and 1.

253
00:20:06,000 --> 00:20:09,000
So, what does that mean
geometrically?

254
00:20:09,000 --> 00:20:13,000
Well, z = 1-x^2-y^2,
and it's a variation on,

255
00:20:13,000 --> 00:20:17,000
actually I think we plotted
that one, right?

256
00:20:17,000 --> 00:20:22,000
That was our first example of a
function of two variables

257
00:20:22,000 --> 00:20:24,000
possibly.
And, so, we saw that the graph

258
00:20:24,000 --> 00:20:27,000
is this paraboloid pointing
downwards.

259
00:20:27,000 --> 00:20:31,000
OK, it's what you get by taking
a parabola and rotating it.

260
00:20:31,000 --> 00:20:35,000
And now, what we are asking is,
what is the volume between the

261
00:20:35,000 --> 00:20:40,000
paraboloid and the xy plane over
the square of side one in the xy

262
00:20:40,000 --> 00:20:43,000
plane over the square of side
one in the xy plane,

263
00:20:43,000 --> 00:20:49,000
x and y between zero and one.
OK, so, what we'll do is we'll,

264
00:20:49,000 --> 00:20:54,000
so, see, here I try to
represent the square.

265
00:20:54,000 --> 00:20:59,000
And, we'll just sum the areas
of the slices as,

266
00:20:59,000 --> 00:21:03,000
say, x varies from zero to one.
And here, of course,

267
00:21:03,000 --> 00:21:06,000
setting up the bounds will be
easy because no matter what x I

268
00:21:06,000 --> 00:21:08,000
take, y still goes from zero to
one.

269
00:21:08,000 --> 00:21:13,000
See, it's easiest to do double
integrals what the region is

270
00:21:13,000 --> 00:21:18,000
just a rectangle on the xy plane
because then you don't have to

271
00:21:18,000 --> 00:21:21,000
worry too much about what are
the ranges.

272
00:21:21,000 --> 00:21:29,000
OK, so let's do it.
Well, that would be the

273
00:21:29,000 --> 00:21:38,000
integral from zero to one of the
integral from zero to one of

274
00:21:38,000 --> 00:21:42,000
1-x^2-y^2 dy dx.
So, I'm dropping the

275
00:21:42,000 --> 00:21:44,000
parentheses.
But, if you still want to see

276
00:21:44,000 --> 00:21:48,000
them, I'm going to put that in
very thin so that you see what

277
00:21:48,000 --> 00:21:50,000
it means.
But, actually,

278
00:21:50,000 --> 00:21:53,000
the convention is we won't put
this parentheses in there

279
00:21:53,000 --> 00:21:57,000
anymore.
OK, so what this means is first

280
00:21:57,000 --> 00:22:02,000
I will integrate 1-x^2-y^2 over
y, ranging from zero to one with

281
00:22:02,000 --> 00:22:07,000
x held fixed.
So, what that represents is the

282
00:22:07,000 --> 00:22:10,000
area in this slice.
So, see here,

283
00:22:10,000 --> 00:22:13,000
I've drawn, well,
what happens is actually the

284
00:22:13,000 --> 00:22:16,000
function takes positive and
negative values.

285
00:22:16,000 --> 00:22:18,000
So, in fact,
I will be counting positively

286
00:22:18,000 --> 00:22:21,000
this part of the area.
And, I will be counting

287
00:22:21,000 --> 00:22:25,000
negatively this part of the
area, I mean,

288
00:22:25,000 --> 00:22:32,000
as usual when I do an integral.
OK, so what I will do to

289
00:22:32,000 --> 00:22:40,000
evaluate this,
I will first do what's called

290
00:22:40,000 --> 00:22:48,000
the inner integral.
So, to do the inner integral,

291
00:22:48,000 --> 00:22:55,000
well, it's pretty easy.
How do I integrate this?

292
00:22:55,000 --> 00:22:58,000
Well, it becomes,
so, what's the integral of one?

293
00:22:58,000 --> 00:23:01,000
It's y.
Just anything to remember is we

294
00:23:01,000 --> 00:23:03,000
are integrating this with
respect to y,

295
00:23:03,000 --> 00:23:08,000
not to x.
The integral of x^2 is x^2

296
00:23:08,000 --> 00:23:14,000
times y.
And, the integral of y^2 is y^3

297
00:23:14,000 --> 00:23:18,000
over 3.
OK, and that we plug in the

298
00:23:18,000 --> 00:23:21,000
bounds, which are zero and one
in this case.

299
00:23:21,000 --> 00:23:28,000
And so, when you plug y equals
one, you will get one minus x^2

300
00:23:28,000 --> 00:23:34,000
minus one third minus,
well, for y equals zero you get

301
00:23:34,000 --> 00:23:41,000
0,0, 0, so nothing changes.
OK, so you are left with two

302
00:23:41,000 --> 00:23:45,000
thirds minus x^2.
OK, and that's a function of x

303
00:23:45,000 --> 00:23:47,000
only.
Here, you shouldn't see any y's

304
00:23:47,000 --> 00:23:49,000
anymore because y was your
integration variable.

305
00:23:49,000 --> 00:23:55,000
But, you still have x.
You still have x because the

306
00:23:55,000 --> 00:24:01,000
area of this shaded slice
depends, of course,

307
00:24:01,000 --> 00:24:05,000
on the value of x.
And, so now,

308
00:24:05,000 --> 00:24:11,000
the second thing to do is to do
the outer integral.

309
00:24:11,000 --> 00:24:17,000
So, now we integrate from zero
to one what we got,

310
00:24:17,000 --> 00:24:22,000
which is two thirds minus x^2
dx.

311
00:24:22,000 --> 00:24:26,000
OK, and we know how to compute
that because that integrates to

312
00:24:26,000 --> 00:24:31,000
two thirds x minus one third x^3
between zero and one.

313
00:24:31,000 --> 00:24:33,000
And, I'll let you do the
computation.

314
00:24:33,000 --> 00:24:43,000
You will find it's one third.
OK, so that's the final answer.

315
00:24:43,000 --> 00:24:46,000
So, that's the general pattern.
When we have a double integral

316
00:24:46,000 --> 00:24:49,000
to compute, first we want to set
it up carefully.

317
00:24:49,000 --> 00:24:51,000
We want to find,
what will be the bounds in x

318
00:24:51,000 --> 00:24:53,000
and y?
And here, that was actually

319
00:24:53,000 --> 00:24:57,000
pretty easy because our equation
was very simple.

320
00:24:57,000 --> 00:25:00,000
Then, we want to compute the
inner integral,

321
00:25:00,000 --> 00:25:02,000
and then we compute the outer
integral.

322
00:25:02,000 --> 00:25:11,000
And, that's it.
OK, any questions at this point?

323
00:25:11,000 --> 00:25:16,000
No?
OK, so, by the way,

324
00:25:16,000 --> 00:25:20,000
we started with dA in the
notation, right?

325
00:25:20,000 --> 00:25:26,000
Here we had dA.
And, that somehow became a dy

326
00:25:26,000 --> 00:25:28,000
dx.
OK,

327
00:25:28,000 --> 00:25:35,000
so,
dA became dy dx because when we

328
00:25:35,000 --> 00:25:39,000
do the iterated integral this
way,

329
00:25:39,000 --> 00:25:45,000
what we're actually doing is
that we are slicing our origin

330
00:25:45,000 --> 00:25:49,000
into small rectangles.
OK, that was the area of this

331
00:25:49,000 --> 00:25:52,000
small rectangle here?
Well, it's the product of its

332
00:25:52,000 --> 00:25:58,000
width times its height.
So, that's delta x times delta

333
00:25:58,000 --> 00:26:03,000
y.
OK, so, delta a equals delta x

334
00:26:03,000 --> 00:26:08,000
delta y becomes...
So actually,

335
00:26:08,000 --> 00:26:18,000
it's not just becomes,
it's really equal.

336
00:26:18,000 --> 00:26:28,000
So, the small rectangles for.
Now, it became dy dx and not dx

337
00:26:28,000 --> 00:26:30,000
dy.
Well, that's a question of,

338
00:26:30,000 --> 00:26:32,000
in which order we do the
iterated integral?

339
00:26:32,000 --> 00:26:36,000
It's up to us to decide whether
we want to integrate x first,

340
00:26:36,000 --> 00:26:38,000
then y, or y first,
then x.

341
00:26:38,000 --> 00:26:41,000
But, as we'll see very soon,
that is an important decision

342
00:26:41,000 --> 00:26:44,000
when it comes to setting up the
bounds of integration.

343
00:26:44,000 --> 00:26:47,000
Here, it doesn't matter,
but in general we have to be

344
00:26:47,000 --> 00:26:51,000
very careful about in which
order we will do things.

345
00:26:51,000 --> 00:26:57,000
Yes?
Well, in principle it always

346
00:26:57,000 --> 00:27:00,000
works both ways.
Sometimes it will be that

347
00:27:00,000 --> 00:27:04,000
because the region has a strange
shape, you can actually set it

348
00:27:04,000 --> 00:27:06,000
up more easily one way or the
other.

349
00:27:06,000 --> 00:27:09,000
Sometimes it will also be that
the function here,

350
00:27:09,000 --> 00:27:12,000
you actually know how to
integrate in one way,

351
00:27:12,000 --> 00:27:15,000
but not the other.
So, the theory is that it

352
00:27:15,000 --> 00:27:18,000
should work both ways.
In practice,

353
00:27:18,000 --> 00:27:25,000
one of the two calculations may
be much harder.

354
00:27:25,000 --> 00:27:33,000
OK.
Let's do another example.

355
00:27:33,000 --> 00:27:39,000
Let's say that what I wanted to
know was not actually what I

356
00:27:39,000 --> 00:27:42,000
computed,
namely, the volume below the

357
00:27:42,000 --> 00:27:45,000
paraboloid,
but also the negative of some

358
00:27:45,000 --> 00:27:48,000
part that's now in the corner
towards me.

359
00:27:48,000 --> 00:27:51,000
But let's say really what I
wanted was just the volume

360
00:27:51,000 --> 00:27:54,000
between the paraboloid and the
xy plane,

361
00:27:54,000 --> 00:27:58,000
so looking only at the part of
it that sits above the xy plane.

362
00:27:58,000 --> 00:28:01,000
So, that means,
instead of integrating over the

363
00:28:01,000 --> 00:28:05,000
entire square of size one,
I should just integrate over

364
00:28:05,000 --> 00:28:08,000
the quarter disk.
I should stop integrating where

365
00:28:08,000 --> 00:28:18,000
my paraboloid hits the xy plane.
So, let me draw another picture.

366
00:28:18,000 --> 00:28:28,000
So, let's say I wanted to
integrate, actually -- So,

367
00:28:28,000 --> 00:28:36,000
let's call this example two.
So, we are going to do the same

368
00:28:36,000 --> 00:28:39,000
function but over a different
region.

369
00:28:39,000 --> 00:28:44,000
And, the region will just be,
now, this quarter disk here.

370
00:28:44,000 --> 00:28:55,000
OK, so maybe I should draw a
picture on the xy plane.

371
00:28:55,000 --> 00:29:14,000
That's your region, R.
OK, so in principle,

372
00:29:14,000 --> 00:29:19,000
it will be the same integral.
But what changes is the bounds.

373
00:29:19,000 --> 00:29:23,000
Why do the bounds change?
Well, the bounds change because

374
00:29:23,000 --> 00:29:27,000
now if I set,
if I fixed some value of x,

375
00:29:27,000 --> 00:29:31,000
then I want to integrate this
part of the slice that's above

376
00:29:31,000 --> 00:29:36,000
the xy plane and I don't want to
take this part that's actually

377
00:29:36,000 --> 00:29:39,000
outside of my disk.
So, I should stop integrating

378
00:29:39,000 --> 00:29:42,000
over y when y reaches this value
here.

379
00:29:42,000 --> 00:29:46,000
OK, on that picture here,
on this picture,

380
00:29:46,000 --> 00:29:52,000
it tells me for a fixed value
of x, the range of values for y

381
00:29:52,000 --> 00:29:55,000
should go only from here to
here.

382
00:29:55,000 --> 00:30:00,000
So, that's from here to less
than one.

383
00:30:00,000 --> 00:30:07,000
OK, so for a given,
x, the range of y is,

384
00:30:07,000 --> 00:30:16,000
well, so what's the lowest
value of y that we want to look

385
00:30:16,000 --> 00:30:20,000
at?
It's still zero.

386
00:30:20,000 --> 00:30:24,000
From y equals zero to,what's
the value of y here?

387
00:30:24,000 --> 00:30:28,000
Well, I have to solve in the
equation of a circle,

388
00:30:28,000 --> 00:30:31,000
OK?
So, if I'm here,

389
00:30:31,000 --> 00:30:39,000
this is x^2 y^2 equals one.
That means y is square root of

390
00:30:39,000 --> 00:30:44,000
one minus x^2.
OK, so I will integrate from y

391
00:30:44,000 --> 00:30:48,000
equals zero to y equals square
root of one minus x^2.

392
00:30:48,000 --> 00:30:55,000
And, now you see how the bound
from y will depend on the value

393
00:30:55,000 --> 00:30:59,000
of x.
OK, so while I erase,

394
00:30:59,000 --> 00:31:05,000
I will let you think about,
what is the bound for x now?

395
00:31:05,000 --> 00:31:06,000
It's a trick question.

396
00:31:35,000 --> 00:31:48,000
OK, so I claim that what we
will do -- We write this as an

397
00:31:48,000 --> 00:31:57,000
iterated integral first dy then
dx.

398
00:31:57,000 --> 00:32:04,000
And, we said for a fixed value
of x, the range for y is from

399
00:32:04,000 --> 00:32:09,000
zero to square root of one minus
x^2.

400
00:32:09,000 --> 00:32:13,000
What about the range for x?
Well, the range for x should

401
00:32:13,000 --> 00:32:15,000
just be numbers.
OK, remember,

402
00:32:15,000 --> 00:32:18,000
the question I have to ask now
is if I look at all of these

403
00:32:18,000 --> 00:32:21,000
yellow slices,
which one is the first one that

404
00:32:21,000 --> 00:32:24,000
I will consider?
Which one is the last one that

405
00:32:24,000 --> 00:32:27,000
I want to consider?
So, the smallest value of x

406
00:32:27,000 --> 00:32:30,000
that I want to consider is zero
again.

407
00:32:30,000 --> 00:32:33,000
And then, I will have actually
a pretty big slice.

408
00:32:33,000 --> 00:32:35,000
And I will get smaller,
and smaller,

409
00:32:35,000 --> 00:32:36,000
and smaller slices.
And, it stops.

410
00:32:36,000 --> 00:32:39,000
I have to stop when x equals
one.

411
00:32:39,000 --> 00:32:42,000
Afterwards, there's nothing
else to integrate.

412
00:32:42,000 --> 00:32:49,000
So, x goes from zero to one.
OK, and now,

413
00:32:49,000 --> 00:32:52,000
see how in the inner integral,
the bounds depend on in the

414
00:32:52,000 --> 00:32:55,000
inner integral,
the bounds depend on x.

415
00:32:55,000 --> 00:32:58,000
In the outer one,
you just get numbers because

416
00:32:58,000 --> 00:33:02,000
the questions that you have to
ask to set up this one and set

417
00:33:02,000 --> 00:33:05,000
up that one are different.
Here, the question is,

418
00:33:05,000 --> 00:33:07,000
if I fix a given,
x, if I look at a given slice,

419
00:33:07,000 --> 00:33:10,000
what's the range for y?
Here, the question is,

420
00:33:10,000 --> 00:33:16,000
what's the first slice?
What is the last slice?

421
00:33:16,000 --> 00:33:21,000
Does that make sense?
Everyone happy with that?

422
00:33:21,000 --> 00:33:28,000
OK, very good.
So, now, how do we compute that?

423
00:33:28,000 --> 00:33:36,000
Well, we do the inner integral.
So, that's an integral from

424
00:33:36,000 --> 00:33:42,000
zero to square root of one minus
x^2 of one minus x^2 minus y^2

425
00:33:42,000 --> 00:33:48,000
dy.
And, well, that integrates to

426
00:33:48,000 --> 00:34:00,000
y-x^2y-y^3 over three from zero
to square root of one minus x^2.

427
00:34:00,000 --> 00:34:06,000
And then, that becomes,
well, the root of one minus x^2

428
00:34:06,000 --> 00:34:13,000
minus x^2 root of one minus x^2
minus y minus x^2 to the three

429
00:34:13,000 --> 00:34:16,000
halves over three.
And actually,

430
00:34:16,000 --> 00:34:20,000
if you look at it for long
enough, see, this says one minus

431
00:34:20,000 --> 00:34:23,000
x^2 times square root of one
minus x^2.

432
00:34:23,000 --> 00:34:26,000
So, actually,
that's also,

433
00:34:26,000 --> 00:34:30,000
so, in fact,
that simplifies to two thirds

434
00:34:30,000 --> 00:34:35,000
of one minus x^2 to the three
halves.

435
00:34:35,000 --> 00:34:40,000
OK, let me redo that,
maybe, slightly differently.

436
00:34:40,000 --> 00:34:50,000
This was one minus x^2 times y.
So -- -- one minus x^2 times y

437
00:34:50,000 --> 00:35:02,000
becomes square root of one minus
x^2 minus y^3 over three.

438
00:35:02,000 --> 00:35:04,000
And then, when I take y equals
zero, I get zero.

439
00:35:04,000 --> 00:35:08,000
So, I don't subtract anything.
OK, so now you see this is one

440
00:35:08,000 --> 00:35:11,000
minus x^2 to the three halves
minus a third of it.

441
00:35:11,000 --> 00:35:21,000
So, you're left with two thirds.
OK, so, that's the integral.

442
00:35:21,000 --> 00:35:27,000
The outer integral is the
integral from zero to one of two

443
00:35:27,000 --> 00:35:31,000
thirds of one minus x^2 to the
three halves dx.

444
00:35:31,000 --> 00:35:37,000
And,
well,

445
00:35:37,000 --> 00:35:42,000
I let you see if you remember
single variable integrals by

446
00:35:42,000 --> 00:35:47,000
trying to figure out what this
actually comes out to be is it

447
00:35:47,000 --> 00:35:54,000
pi over two,
or pi over eight, actually?

448
00:35:54,000 --> 00:36:12,000
I think it's pi over eight.
OK, well I guess we have to do

449
00:36:12,000 --> 00:36:15,000
it then.
I wrote something on my notes,

450
00:36:15,000 --> 00:36:16,000
but it's not very clear,
OK?

451
00:36:16,000 --> 00:36:18,000
So, how do we compute this
thing?

452
00:36:18,000 --> 00:36:20,000
Well, we have to do trig
substitution.

453
00:36:20,000 --> 00:36:24,000
That's the only way I know to
compute an integral like that,

454
00:36:24,000 --> 00:36:27,000
OK?
So, we'll set x equal sine

455
00:36:27,000 --> 00:36:35,000
theta, and then square root of
one minus x^2 will be cosine

456
00:36:35,000 --> 00:36:41,000
theta.
We are using sine squared plus

457
00:36:41,000 --> 00:36:51,000
cosine squared equals one.
And, so that will become -- --

458
00:36:51,000 --> 00:36:59,000
so, two thirds remains two
thirds.

459
00:36:59,000 --> 00:37:04,000
One minus x^2 to the three
halves becomes cosine cubed

460
00:37:04,000 --> 00:37:08,000
theta.
dx, well, if x is sine theta,

461
00:37:08,000 --> 00:37:13,000
then dx is cosine theta d
theta.

462
00:37:13,000 --> 00:37:17,000
So, that's cosine theta d theta.
And,

463
00:37:17,000 --> 00:37:19,000
well, if you do things with
substitution,

464
00:37:19,000 --> 00:37:22,000
which is the way I do them,
then you should worry about the

465
00:37:22,000 --> 00:37:25,000
bounds for theta which will be
zero to pi over two.

466
00:37:25,000 --> 00:37:30,000
Or, you can also just plug in
the bounds at the end.

467
00:37:30,000 --> 00:37:36,000
So, now you have the two thirds
times the integral from zero to

468
00:37:36,000 --> 00:37:41,000
pi over two of cosine to the
fourth theta d theta.

469
00:37:41,000 --> 00:37:48,000
And, how do you integrate that?
Well, you have to use double

470
00:37:48,000 --> 00:37:55,000
angle formulas.
OK, so cosine to the fourth,

471
00:37:55,000 --> 00:38:02,000
remember, cosine squared theta
is one plus cosine two theta

472
00:38:02,000 --> 00:38:14,000
over two.
And, we want the square of that.

473
00:38:14,000 --> 00:38:23,000
And,
so that will give us -- -- of,

474
00:38:23,000 --> 00:38:29,000
well, we'll have,
it's actually one quarter plus

475
00:38:29,000 --> 00:38:37,000
one half cosine to theta plus
one quarter cosine square to

476
00:38:37,000 --> 00:38:41,000
theta d theta.
And, how will you handle this

477
00:38:41,000 --> 00:38:42,000
guy?
Well, using,

478
00:38:42,000 --> 00:38:45,000
again, the double angle
formula.

479
00:38:45,000 --> 00:38:51,000
OK, so it's getting slightly
nasty.

480
00:38:51,000 --> 00:38:54,000
So, but I don't know any
simpler solution except for one

481
00:38:54,000 --> 00:38:56,000
simpler solution,
which is you have a table of

482
00:38:56,000 --> 00:38:59,000
integrals of this form inside
the notes.

483
00:38:59,000 --> 00:39:08,000
Yes?
No, I don't think so because if

484
00:39:08,000 --> 00:39:12,000
you take one half times cosine
half times two,

485
00:39:12,000 --> 00:39:15,000
you will still have half,
OK?

486
00:39:15,000 --> 00:39:18,000
So, if you do,
again, the double angle

487
00:39:18,000 --> 00:39:22,000
formula, I think I'm not going
to bother to do it.

488
00:39:22,000 --> 00:39:37,000
I claim you will get,
at the end, pi over eight

489
00:39:37,000 --> 00:39:46,000
because I say so.
OK, so exercise,

490
00:39:46,000 --> 00:39:54,000
continue calculating and get pi
over eight.

491
00:39:54,000 --> 00:39:56,000
OK, now what does the show us?
Well, this shows us,

492
00:39:56,000 --> 00:40:00,000
actually, that this is probably
not the right way to do this.

493
00:40:00,000 --> 00:40:02,000
OK, the right way to do this
will be to integrate it in polar

494
00:40:02,000 --> 00:40:05,000
coordinates.
And, that's what we will learn

495
00:40:05,000 --> 00:40:09,000
how to do tomorrow.
So, we will actually see how to

496
00:40:09,000 --> 00:40:11,000
do it with much less trig.

497
00:40:36,000 --> 00:40:46,000
So, that will be easier in
polar coordinates.

498
00:40:46,000 --> 00:40:51,000
So, we will see that tomorrow.
OK, so we are almost there.

499
00:40:51,000 --> 00:40:53,000
I mean, here you just use a
double angle again and then you

500
00:40:53,000 --> 00:40:55,000
can get it.
And, it's pretty

501
00:40:55,000 --> 00:41:00,000
straightforward.
OK, so one thing that's kind of

502
00:41:00,000 --> 00:41:07,000
interesting to know is we can
exchange the order of

503
00:41:07,000 --> 00:41:11,000
integration.
Say we have an integral given

504
00:41:11,000 --> 00:41:14,000
to us in the order dy dx,
we can switch it to dx dy.

505
00:41:14,000 --> 00:41:18,000
But, we have to be extremely
careful with the bounds.

506
00:41:18,000 --> 00:41:22,000
So, you certainly cannot just
swap the bounds of the inner and

507
00:41:22,000 --> 00:41:26,000
outer because there you would
end up having this square root

508
00:41:26,000 --> 00:41:30,000
of one minus x^2 on the outside,
and you would never get a

509
00:41:30,000 --> 00:41:33,000
number out of that.
So, that cannot work.

510
00:41:33,000 --> 00:41:37,000
It's more complicated than that.
OK, so, well,

511
00:41:37,000 --> 00:41:42,000
here's a first baby example.
Certainly, if I do integral

512
00:41:42,000 --> 00:41:46,000
from zero to one,
integral from zero to two dx

513
00:41:46,000 --> 00:41:51,000
dy, there, I can certainly
switch the bounds without

514
00:41:51,000 --> 00:41:54,000
thinking too much.
What's the reason for that?

515
00:41:54,000 --> 00:41:59,000
Well, the reason for that is
this corresponds in both cases

516
00:41:59,000 --> 00:42:04,000
to integrating x from zero to
two, and y from zero to one.

517
00:42:04,000 --> 00:42:08,000
It's a rectangle.
So, if I slice it this way,

518
00:42:08,000 --> 00:42:11,000
you see that y goes from zero
to one for any x between zero

519
00:42:11,000 --> 00:42:13,000
and two.
It's this guy.

520
00:42:13,000 --> 00:42:18,000
If I slice it that way,
then x goes from zero to two

521
00:42:18,000 --> 00:42:22,000
for any value of y between zero
and one.

522
00:42:22,000 --> 00:42:25,000
And, it's this one.
So, here it works.

523
00:42:25,000 --> 00:42:28,000
But in general,
I have to draw picture of my

524
00:42:28,000 --> 00:42:32,000
region, and see how the slices
look like both ways.

525
00:42:32,000 --> 00:42:41,000
OK, so let's do a more
interesting one.

526
00:42:41,000 --> 00:42:48,000
Let's say that I want to
compute an integral from zero to

527
00:42:48,000 --> 00:42:55,000
one of integral from x to square
root of x of e^y over y dy dx.

528
00:42:55,000 --> 00:42:59,000
So, why did I choose this guy?
Which is the guy because as far

529
00:42:59,000 --> 00:43:03,000
as I can tell,
there's no way to integrate e^y

530
00:43:03,000 --> 00:43:06,000
over y.
So, this is an integral that

531
00:43:06,000 --> 00:43:10,000
you cannot compute this way.
So, it's a good example for why

532
00:43:10,000 --> 00:43:13,000
this can be useful.
So, if you do it this way,

533
00:43:13,000 --> 00:43:14,000
you are stuck immediately.
So, instead,

534
00:43:14,000 --> 00:43:15,000
we will try to switch the
order.

535
00:43:15,000 --> 00:43:20,000
But, to switch the order,
we have to understand,

536
00:43:20,000 --> 00:43:25,000
what do these bounds mean?
OK, so let's draw a picture of

537
00:43:25,000 --> 00:43:29,000
the region.
Well what I am saying is y

538
00:43:29,000 --> 00:43:33,000
equals x to y equals square root
of x.

539
00:43:33,000 --> 00:43:48,000
Well, let's draw y equals x,
and y equals square root of x.

540
00:43:48,000 --> 00:43:52,000
Well, maybe I should actually
put this here,

541
00:43:52,000 --> 00:43:56,000
y equals x to y equals square
root of x.

542
00:43:56,000 --> 00:44:01,000
OK, and so I will go,
for each value of x I will go

543
00:44:01,000 --> 00:44:07,000
from y equals xo to y equals
square root of x.

544
00:44:07,000 --> 00:44:11,000
And then, we'll do that for
values of x that go from x

545
00:44:11,000 --> 00:44:15,000
equals zero to x equals one,
which happens to be exactly

546
00:44:15,000 --> 00:44:22,000
where these things intersect.
So, my region will consist of

547
00:44:22,000 --> 00:44:26,000
all this, OK?
So now, if I want to do it the

548
00:44:26,000 --> 00:44:29,000
other way around,
I have to decompose my region.

549
00:44:29,000 --> 00:44:32,000
The other way around,
I have to, so my goal,

550
00:44:32,000 --> 00:44:35,000
now, is to rewrite this as an
integral.

551
00:44:35,000 --> 00:44:36,000
Well, it's still the same
function.

552
00:44:36,000 --> 00:44:41,000
It's still e to the y over y.
But now, I want to integrate dx

553
00:44:41,000 --> 00:44:45,000
dy.
So, how do I integrate over x?

554
00:44:45,000 --> 00:44:50,000
Well, I fix a value of y.
And, for that value of y,

555
00:44:50,000 --> 00:44:54,000
what's the range of x?
Well, the range for x is from

556
00:44:54,000 --> 00:44:57,000
here to here.
OK, what's the value of x here?

557
00:44:57,000 --> 00:45:02,000
Let's start with an easy one.
This is x equals y.

558
00:45:02,000 --> 00:45:08,000
What about this one?
It's x equals y^2.

559
00:45:08,000 --> 00:45:16,000
OK, so, x goes from y2 to y,
and then what about y?

560
00:45:16,000 --> 00:45:19,000
Well, I have to start at the
bottom of my region.

561
00:45:19,000 --> 00:45:24,000
That's y equals zero to the
top, which is at y equals one.

562
00:45:24,000 --> 00:45:31,000
So, y goes from zero to one.
So, switching the bounds is not

563
00:45:31,000 --> 00:45:36,000
completely obvious.
That took a little bit of work.

564
00:45:36,000 --> 00:45:40,000
But now that we've done that,
well,

565
00:45:40,000 --> 00:45:44,000
just to see how it goes,
it's actually going to be much

566
00:45:44,000 --> 00:45:48,000
easier to integrate because the
inner integral,

567
00:45:48,000 --> 00:45:52,000
well, what's the integral of
e^y over y with respect to x?

568
00:45:52,000 --> 00:46:01,000
It's just that times x,
right, from x equals y^2 to y.

569
00:46:01,000 --> 00:46:04,000
So, that will be,
well, if I plug x equals y,

570
00:46:04,000 --> 00:46:09,000
I will get e to the y minus,
if I plug x equals y^2,

571
00:46:09,000 --> 00:46:16,000
I will get e to the y over y
times y^2 into the y times y,

572
00:46:16,000 --> 00:46:20,000
OK?
So, now, if I do the outer

573
00:46:20,000 --> 00:46:27,000
integral, I will have the
integral from zero to one of e

574
00:46:27,000 --> 00:46:33,000
to the y minus y^e to the y dy.
And, that one actually is a

575
00:46:33,000 --> 00:46:37,000
little bit easier.
So, we know how to integrate

576
00:46:37,000 --> 00:46:39,000
e^y.
We don't quite know how to

577
00:46:39,000 --> 00:46:43,000
integrate ye^y.
But, let's try.

578
00:46:43,000 --> 00:46:49,000
So, let's see,
what's the derivative of ye^y?

579
00:46:49,000 --> 00:46:55,000
Well, there's a product rule
that's one times e^y plus y

580
00:46:55,000 --> 00:46:59,000
times the derivative of e^y is
ye^y.

581
00:46:59,000 --> 00:47:03,000
So, if we do,
OK, let's put a minus sign in

582
00:47:03,000 --> 00:47:06,000
front.
Well, that's almost what we

583
00:47:06,000 --> 00:47:09,000
want, except we have a minus e^y
instead of a plus e^y.

584
00:47:09,000 --> 00:47:15,000
So, we need to add 2e^y.
And, I claim that's the

585
00:47:15,000 --> 00:47:18,000
antiderivative.
OK, if you got lost,

586
00:47:18,000 --> 00:47:21,000
you can also integrate by
integrating by parts,

587
00:47:21,000 --> 00:47:25,000
by taking the derivative of y
and integrating these guys.

588
00:47:25,000 --> 00:47:29,000
Or, but, you know, that works.
Just, your first guess would

589
00:47:29,000 --> 00:47:32,000
be, maybe, let's try minus y^e
to the y.

590
00:47:32,000 --> 00:47:35,000
Take the derivative of that,
compare, see what you need to

591
00:47:35,000 --> 00:47:39,000
do to fix.
And so, if you take that

592
00:47:39,000 --> 00:47:45,000
between zero and one,
you'll actually get e minus

593
00:47:45,000 --> 00:47:47,000
two.
OK,

594
00:47:47,000 --> 00:47:50,000
so,
tomorrow we are going to see

595
00:47:50,000 --> 00:47:53,000
how to do double integrals in
polar coordinates,

596
00:47:53,000 --> 00:47:55,000
and also applications of double
integrals,

597
00:47:55,000 --> 00:47:58,000
how to use them for interesting
things.