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And, well let's see.
So, before we actually start

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00:00:25,000 --> 00:00:30,000
reviewing for the test,
I still have to tell you a few

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00:00:30,000 --> 00:00:34,000
small things because I promised
to say a few words about what's

10
00:00:34,000 --> 00:00:37,000
the difference,
or precisely,

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00:00:37,000 --> 00:00:41,000
what's the difference between
curl being zero and a field

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00:00:41,000 --> 00:00:45,000
being a gradient field,
and why we have this assumption

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00:00:45,000 --> 00:00:49,000
that our vector field had to be
defined everywhere for a field

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00:00:49,000 --> 00:00:53,000
with curl zero to actually be
conservative for our test for

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00:00:53,000 --> 00:01:04,000
gradient fields to be valid?
So -- More about validity of

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00:01:04,000 --> 00:01:17,000
Green's theorem and things like
that.

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00:01:17,000 --> 00:01:29,000
So, we've seen the statement of
Green's theorem in two forms.

18
00:01:29,000 --> 00:01:33,000
Both of them have to do with
comparing a line integral along

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00:01:33,000 --> 00:01:38,000
a closed curve to a double
integral over the region inside

20
00:01:38,000 --> 00:01:41,000
enclosed by the curve.
So,

21
00:01:41,000 --> 00:01:45,000
one of them says the line
integral for the work done by a

22
00:01:45,000 --> 00:01:50,000
vector field along a closed
curve counterclockwise is equal

23
00:01:50,000 --> 00:01:54,000
to the double integral of a curl
of a field over the enclosed

24
00:01:54,000 --> 00:01:58,000
region.
And, the other one says the

25
00:01:58,000 --> 00:02:05,000
total flux out of the region,
so, the flux through the curve

26
00:02:05,000 --> 00:02:11,000
is equal to the double integral
of divergence of a field in the

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00:02:11,000 --> 00:02:13,000
region.
So, in both cases,

28
00:02:13,000 --> 00:02:18,000
we need the vector field to be
defined not only,

29
00:02:18,000 --> 00:02:21,000
I mean, the left hand side
makes sense if a vector field is

30
00:02:21,000 --> 00:02:24,000
just defined on the curve
because it's just a line

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00:02:24,000 --> 00:02:27,000
integral on C.
We don't care what happens

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00:02:27,000 --> 00:02:29,000
inside.
But, for the right-hand side to

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00:02:29,000 --> 00:02:31,000
make sense,
and therefore for the equality

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00:02:31,000 --> 00:02:34,000
to make sense,
we need the vector field to be

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00:02:34,000 --> 00:02:38,000
defined everywhere inside the
region.

36
00:02:38,000 --> 00:02:41,000
So, I said, if there is a point
somewhere in here where my

37
00:02:41,000 --> 00:02:45,000
vector field is not defined,
then it doesn't work.

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00:02:45,000 --> 00:02:53,000
And actually,
we've seen that example.

39
00:02:53,000 --> 00:03:08,000
So, this only works if F and
its derivatives are defined

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00:03:08,000 --> 00:03:17,000
everywhere in the region,
R.

41
00:03:17,000 --> 00:03:20,000
Otherwise, we are in trouble.
OK,

42
00:03:20,000 --> 00:03:29,000
so we've seen for example that
if I gave you the vector field

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00:03:29,000 --> 00:03:35,000
minus yi xj over x squared plus
y squared,

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00:03:35,000 --> 00:03:40,000
so that's the same vector field
that was on that problem set a

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00:03:40,000 --> 00:03:45,000
couple of weeks ago.
Then, well, f is not defined at

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00:03:45,000 --> 00:03:52,000
the origin, but it's defined
everywhere else.

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00:03:52,000 --> 00:04:03,000
And, wherever it's defined,
it's curl is zero.

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00:04:03,000 --> 00:04:14,000
I should say everywhere it's --
And so, if we have a closed

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00:04:14,000 --> 00:04:24,000
curve in the plane,
well, there's two situations.

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00:04:24,000 --> 00:04:28,000
One is if it does not enclose
the origin.

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00:04:28,000 --> 00:04:30,000
Then,
yes,

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00:04:30,000 --> 00:04:36,000
we can apply Green's theorem
and it will tell us that it's

53
00:04:36,000 --> 00:04:42,000
equal to the double integral in
here of curl F dA,

54
00:04:42,000 --> 00:04:46,000
which will be zero because this
is zero.

55
00:04:46,000 --> 00:04:51,000
However,
if I have a curve that encloses

56
00:04:51,000 --> 00:04:54,000
the origin,
let's say like this,

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00:04:54,000 --> 00:05:03,000
for example,
then,

58
00:05:03,000 --> 00:05:07,000
well, I cannot use the same
method because the vector field

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00:05:07,000 --> 00:05:11,000
and its curl are not defined at
the origin.

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00:05:11,000 --> 00:05:14,000
And, in fact,
you know that ignoring the

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00:05:14,000 --> 00:05:17,000
problem and saying,
well, the curl is still zero

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00:05:17,000 --> 00:05:19,000
everywhere,
will give you the wrong answer

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00:05:19,000 --> 00:05:23,000
because we've seen an example.
We've seen that along the unit

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00:05:23,000 --> 00:05:27,000
circle the total work is 2 pi
not zero.

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00:05:27,000 --> 00:05:35,000
So, we can't use Green.
However, we can't use it

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00:05:35,000 --> 00:05:38,000
directly.
So, there is an extended

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00:05:38,000 --> 00:05:44,000
version of Green's theorem that
tells you the following thing.

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00:05:44,000 --> 00:05:49,000
Well,
it tells me that even though I

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00:05:49,000 --> 00:05:54,000
can't do things for just this
region enclosed by C prime,

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00:05:54,000 --> 00:05:58,000
I can still do things for the
region in between two different

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00:05:58,000 --> 00:06:06,000
curves.
OK, so let me show you what I

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00:06:06,000 --> 00:06:11,000
have in mind.
So, let's say that I have my

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00:06:11,000 --> 00:06:15,000
curve C'.
Where's my yellow chalk?

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00:06:15,000 --> 00:06:22,000
Oh, here.
So, I have this curve C'.

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00:06:22,000 --> 00:06:30,000
I can't apply Green's theorem
inside it, but let's get out the

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00:06:30,000 --> 00:06:34,000
smaller thing.
So, that one I'm going to make

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00:06:34,000 --> 00:06:41,000
going clockwise.
You will see why.

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00:06:41,000 --> 00:06:48,000
Then, I could say,
well, let me change my mind.

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00:06:48,000 --> 00:06:49,000
This picture is not very well
prepared.

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00:06:49,000 --> 00:06:53,000
That's because my writer is on
strike.

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00:06:53,000 --> 00:07:02,000
OK, so let's say we have C' and
C'' both going counterclockwise.

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00:07:02,000 --> 00:07:05,000
Then,
I claim that Green's theorem

83
00:07:05,000 --> 00:07:09,000
still applies,
and tells me that the line

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00:07:09,000 --> 00:07:14,000
integral along C prime minus the
line integral along C double

85
00:07:14,000 --> 00:07:19,000
prime is equal to the double
integral over the region in

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00:07:19,000 --> 00:07:23,000
between.
So here, now,

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00:07:23,000 --> 00:07:35,000
it's this region with the hole
of the curve.

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00:07:35,000 --> 00:07:41,000
And, well, in our case,
that will turn out to be zero

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00:07:41,000 --> 00:07:45,000
because curl is zero.
OK, so this doesn't tell us

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00:07:45,000 --> 00:07:47,000
what each of these two line
integrals is.

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00:07:47,000 --> 00:07:49,000
But actually,
it tells us that they are equal

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00:07:49,000 --> 00:07:51,000
to each other.
And so, by computing one,

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00:07:51,000 --> 00:07:53,000
you can see actually that for
this vector field,

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00:07:53,000 --> 00:07:57,000
if you take any curve that goes
counterclockwise around the

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00:07:57,000 --> 00:08:00,000
origin,
you would get two pi no matter

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00:08:00,000 --> 00:08:04,000
what the curve is.
So how do you get to this?

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00:08:04,000 --> 00:08:07,000
Why is this not like
conceptually a new theorem?

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00:08:07,000 --> 00:08:13,000
Well, just think of the
following thing.

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00:08:13,000 --> 00:08:17,000
I'm not going to do it on top
of that because it's going to be

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00:08:17,000 --> 00:08:23,000
messy if I draw too many things.
But, so here I have my C''.

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00:08:23,000 --> 00:08:29,000
Here, I have C'.
Let me actually make a slit

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00:08:29,000 --> 00:08:33,000
that will connect them to each
other like this.

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00:08:33,000 --> 00:08:37,000
So now if I take,
see,

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00:08:37,000 --> 00:08:43,000
I can form a single closed
curve that will enclose all of

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00:08:43,000 --> 00:08:49,000
this region with kind of an
infinitely thin slit here

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00:08:49,000 --> 00:08:52,000
counterclockwise.
And so, if I go

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00:08:52,000 --> 00:08:55,000
counterclockwise around this
region, basically I go

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00:08:55,000 --> 00:08:58,000
counterclockwise along the outer
curve.

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00:08:58,000 --> 00:09:01,000
Then I go along the slit.
Then I go clockwise along the

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00:09:01,000 --> 00:09:04,000
inside curve,
then back along the slit.

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00:09:04,000 --> 00:09:07,000
And then I'm done.
So,

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00:09:07,000 --> 00:09:11,000
if I take the line integral
along this big curve consisting

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00:09:11,000 --> 00:09:15,000
of all these pieces,
now I can apply Green's theorem

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00:09:15,000 --> 00:09:19,000
to that because it is the usual
counterclockwise curve that goes

115
00:09:19,000 --> 00:09:22,000
around a region where my field
is well-defined.

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00:09:22,000 --> 00:09:27,000
See, I've eliminated the origin
from the picture.

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00:09:27,000 --> 00:09:37,000
And, so the total line integral
for this thing is equal to the

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00:09:37,000 --> 00:09:46,000
integral along C prime,
I guess the outer one.

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00:09:46,000 --> 00:09:50,000
Then, I also need to have what
I do along the inner side.

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00:09:50,000 --> 00:09:52,000
And, the inner side is going to
be C double prime,

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00:09:52,000 --> 00:09:57,000
but going backwards because now
I'm going clockwise on C prime

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00:09:57,000 --> 00:10:01,000
so that I'm going
counterclockwise around the

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00:10:01,000 --> 00:10:04,000
shaded region.
Well, of course there will be

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00:10:04,000 --> 00:10:06,000
contributions from the line
integral along this wide

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00:10:06,000 --> 00:10:08,000
segment.
But, I do it twice,

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00:10:08,000 --> 00:10:17,000
once each way.
So, they cancel out.

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00:10:17,000 --> 00:10:21,000
So, the white segments cancel
out.

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00:10:21,000 --> 00:10:23,000
You probably shouldn't,
in your notes,

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00:10:23,000 --> 00:10:25,000
write down white segments
because probably they are not

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00:10:25,000 --> 00:10:29,000
white on your paper.
But, hopefully you get the

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00:10:29,000 --> 00:10:33,000
meaning of what I'm trying to
say.

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00:10:33,000 --> 00:10:36,000
OK, so basically that tells
you, you can still play tricks

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00:10:36,000 --> 00:10:39,000
with Green's theorem when the
region has holes in it.

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00:10:39,000 --> 00:10:44,000
You just had to be careful and
somehow subtract some other

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00:10:44,000 --> 00:10:48,000
curve so that together things
will work out.

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00:10:48,000 --> 00:10:51,000
There is a similar thing with
the divergence theorem,

137
00:10:51,000 --> 00:10:55,000
of course, with flux and double
integral of div f,

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00:10:55,000 --> 00:10:58,000
you can apply exactly the same
argument.

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00:10:58,000 --> 00:11:02,000
OK, so basically you can apply
Green's theorem for a region

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00:11:02,000 --> 00:11:04,000
that has several boundary
curves.

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00:11:04,000 --> 00:11:07,000
You just have to be careful
that the outer boundary must go

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00:11:07,000 --> 00:11:13,000
counterclockwise.
The inner boundary either goes

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00:11:13,000 --> 00:11:19,000
clockwise, or you put a minus
sign.

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00:11:19,000 --> 00:11:26,000
OK,
and the last cultural note,

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00:11:26,000 --> 00:11:34,000
so, the definition,
we say that a region in the

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00:11:34,000 --> 00:11:36,000
plane,
sorry, I should say a connected

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00:11:36,000 --> 00:11:45,000
region in the plane,
so that means -- So,

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00:11:45,000 --> 00:11:47,000
connected means it consists of
a single piece.

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00:11:47,000 --> 00:11:50,000
OK, so, connected,
there is a single piece.

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00:11:50,000 --> 00:11:53,000
These two guys together are not
connected.

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00:11:53,000 --> 00:11:58,000
But, if I join them,
then this is a connected

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00:11:58,000 --> 00:12:08,000
region.
We say it's simply connected --

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00:12:08,000 --> 00:12:17,000
-- if any closed curve in it,
OK,

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00:12:17,000 --> 00:12:18,000
so I need to gave a name to my
region,

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00:12:18,000 --> 00:12:22,000
let's say R,
any closed curve in R,

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00:12:22,000 --> 00:12:29,000
bounds,
no,

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00:12:29,000 --> 00:12:37,000
sorry.
If the interior of any closed

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00:12:37,000 --> 00:12:49,000
curve in R -- -- is also
contained in R.

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00:12:49,000 --> 00:12:51,000
So, concretely,
what does that mean?

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00:12:51,000 --> 00:12:57,000
That means the region,
R, does not have any holes

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00:12:57,000 --> 00:13:02,000
inside it.
Maybe I should draw two

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00:13:02,000 --> 00:13:08,000
pictures to explain what I mean.
So,

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00:13:08,000 --> 00:13:17,000
this guy here is simply
connected while -- -- this guy

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00:13:17,000 --> 00:13:29,000
here is not simply connected
because if I take this curve,

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00:13:29,000 --> 00:13:34,000
that's a curve inside my region.
But, the piece that it bounds

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00:13:34,000 --> 00:13:38,000
is not actually entirely
contained in my origin.

167
00:13:38,000 --> 00:13:41,000
And, so why is that relevant?
Well,

168
00:13:41,000 --> 00:13:45,000
if you know that your vector
field is defined everywhere in a

169
00:13:45,000 --> 00:13:47,000
simply connected region,
then you don't have to worry

170
00:13:47,000 --> 00:13:50,000
about this question of,
can I apply Green's theorem to

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00:13:50,000 --> 00:13:52,000
the inside?
You know it's automatically OK

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00:13:52,000 --> 00:13:54,000
because if you have a closed
curve,

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00:13:54,000 --> 00:13:59,000
then the vector field is,
I mean, if a vector field is

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00:13:59,000 --> 00:14:03,000
defined on the curve it will
also be defined inside.

175
00:14:03,000 --> 00:14:11,000
OK,
so if the domain of definition

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00:14:11,000 --> 00:14:25,000
-- -- of a vector field is
defined and differentiable -- --

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00:14:25,000 --> 00:14:38,000
is simply connected -- -- then
we can always apply -- --

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00:14:38,000 --> 00:14:47,000
Green's theorem -- -- and,
of course,

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00:14:47,000 --> 00:14:49,000
provided that we do it on a
curve where the vector field is

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00:14:49,000 --> 00:14:50,000
defined.
I mean, your line integral

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00:14:50,000 --> 00:14:53,000
doesn't make sense so there's
nothing to compute.

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00:14:53,000 --> 00:14:56,000
But, if you have,
so, again, the argument would

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00:14:56,000 --> 00:14:59,000
be, well, if a vector field is
defined on the curve,

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00:14:59,000 --> 00:15:01,000
it's also defined inside.
So,

185
00:15:01,000 --> 00:15:04,000
see,
the problem with that vector

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00:15:04,000 --> 00:15:07,000
field here is precisely that its
domain of definition is not

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00:15:07,000 --> 00:15:09,000
simply connected because there
is a hole,

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00:15:09,000 --> 00:15:17,000
namely the origin.
OK, so for this guy,

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00:15:17,000 --> 00:15:28,000
domain of definition,
which is plane minus the origin

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00:15:28,000 --> 00:15:39,000
with the origin removed is not
simply connected.

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00:15:39,000 --> 00:15:42,000
And so that's why you have this
line integral that makes perfect

192
00:15:42,000 --> 00:15:45,000
sense, but you can't apply
Green's theorem to it.

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00:15:45,000 --> 00:15:47,000
So now, what does that mean a
particular?

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00:15:47,000 --> 00:15:51,000
Well, we've seen this criterion
that if a curl of the vector

195
00:15:51,000 --> 00:15:55,000
field is zero and it's defined
in the entire plane,

196
00:15:55,000 --> 00:15:58,000
then the vector field is
conservative,

197
00:15:58,000 --> 00:16:01,000
and it's a gradient field.
And, the argument to prove that

198
00:16:01,000 --> 00:16:03,000
is basically to use Green's
theorem.

199
00:16:03,000 --> 00:16:07,000
So, in fact,
the actual optimal statement

200
00:16:07,000 --> 00:16:11,000
you can make is if a vector
field is defined in a simply

201
00:16:11,000 --> 00:16:13,000
connected region,
and its curl is zero,

202
00:16:13,000 --> 00:16:26,000
then it's a gradient field.
So, let me just write that down.

203
00:16:26,000 --> 00:16:29,000
So, the correct statement,
I mean, the previous one we've

204
00:16:29,000 --> 00:16:35,000
seen is also correct.
But this one is somehow better

205
00:16:35,000 --> 00:16:45,000
and closer to what exactly is
needed if curl F is zero and the

206
00:16:45,000 --> 00:16:55,000
domain of definition where F is
defined is simply connected --

207
00:16:55,000 --> 00:17:04,000
-- then F is conservative.
And that means also it's a

208
00:17:04,000 --> 00:17:11,000
gradient field.
It's the same thing.

209
00:17:11,000 --> 00:17:23,000
OK, any questions on this?
No?

210
00:17:23,000 --> 00:17:27,000
OK, some good news.
What I've just said here won't

211
00:17:27,000 --> 00:17:31,000
come up on the test on Thursday.
OK.

212
00:17:31,000 --> 00:17:35,000
(APPLAUSE) Still,
it's stuff that you should be

213
00:17:35,000 --> 00:17:39,000
aware of generally speaking
because it will be useful,

214
00:17:39,000 --> 00:17:42,000
say, on the next week's problem
set.

215
00:17:42,000 --> 00:17:46,000
And,
maybe on the final it would be,

216
00:17:46,000 --> 00:17:48,000
there won't be any really,
really complicated things

217
00:17:48,000 --> 00:17:53,000
probably,
but you might need to be at

218
00:17:53,000 --> 00:18:01,000
least vaguely aware of this
issue of things being simply

219
00:18:01,000 --> 00:18:04,000
connected.
And by the way,

220
00:18:04,000 --> 00:18:08,000
I mean, this is also somehow
the starting point of topology,

221
00:18:08,000 --> 00:18:12,000
which is the branch of math
that studies the shapes of

222
00:18:12,000 --> 00:18:13,000
regions.
So,

223
00:18:13,000 --> 00:18:15,000
in particular,
you can try to distinguish

224
00:18:15,000 --> 00:18:18,000
domains in the plains by looking
at whether they're simply

225
00:18:18,000 --> 00:18:21,000
connected or not,
and what kinds of features they

226
00:18:21,000 --> 00:18:25,000
have in terms of how you can
joint point what kinds of curves

227
00:18:25,000 --> 00:18:28,000
exist in them.
And, since that's the branch of

228
00:18:28,000 --> 00:18:32,000
math in which I work,
I thought I should tell you a

229
00:18:32,000 --> 00:18:41,000
bit about it.
OK, so now back to reviewing

230
00:18:41,000 --> 00:18:47,000
for the exam.
So, I'm going to basically list

231
00:18:47,000 --> 00:18:49,000
topics.
And, if time permits,

232
00:18:49,000 --> 00:18:53,000
I will say a few things about
problems from practice exam 3B.

233
00:18:53,000 --> 00:18:56,000
I'm hoping that you have it or
your neighbor has it,

234
00:18:56,000 --> 00:18:59,000
or you can somehow get it.
Anyway, given time,

235
00:18:59,000 --> 00:19:04,000
I'm not sure how much I will
say about the problems in and of

236
00:19:04,000 --> 00:19:08,000
themselves.
OK, so the main thing to know

237
00:19:08,000 --> 00:19:13,000
about this exam is how to set up
and evaluate double integrals

238
00:19:13,000 --> 00:19:17,000
and line integrals.
OK, if you know how to do these

239
00:19:17,000 --> 00:19:20,000
two things, then you are in much
better shape than if you don't.

240
00:19:26,000 --> 00:19:43,000
And -- So, the first thing
we've seen, just to write it

241
00:19:43,000 --> 00:19:55,000
down, there's two main objects.
And, it's kind of important to

242
00:19:55,000 --> 00:19:57,000
not confuse them with each
other.

243
00:19:57,000 --> 00:20:02,000
OK, there's double integrals of
our regions of some quantity,

244
00:20:02,000 --> 00:20:06,000
dA,
and the other one is the line

245
00:20:06,000 --> 00:20:11,000
integral along a curve of a
vector field,

246
00:20:11,000 --> 00:20:17,000
F.dr or F.Mds depending on
whether it's work or flux that

247
00:20:17,000 --> 00:20:21,000
we are trying to do.
And, so we should know how to

248
00:20:21,000 --> 00:20:24,000
set up these things and how to
evaluate them.

249
00:20:24,000 --> 00:20:27,000
And, roughly speaking,
in this one you start by

250
00:20:27,000 --> 00:20:32,000
drawing a picture of the region,
then deciding which way you

251
00:20:32,000 --> 00:20:34,000
will integrate it.
It could be dx dy,

252
00:20:34,000 --> 00:20:37,000
dy dx,
r dr d theta,

253
00:20:37,000 --> 00:20:41,000
and then you will set up the
bound carefully by slicing it

254
00:20:41,000 --> 00:20:45,000
and studying how the bounds for
the inner variable depend on the

255
00:20:45,000 --> 00:20:51,000
outer variable.
So, the first topic will be

256
00:20:51,000 --> 00:20:57,000
setting up double integrals.
And so, remember,

257
00:20:57,000 --> 00:21:03,000
OK, so maybe I should make this
more explicit.

258
00:21:03,000 --> 00:21:12,000
We want to draw a picture of R
and take slices in the chosen

259
00:21:12,000 --> 00:21:18,000
way so that we get an iterated
integral.

260
00:21:18,000 --> 00:21:25,000
OK, so let's do just a quick
example.

261
00:21:25,000 --> 00:21:38,000
So, if I look at problem one on
the exam 3B,

262
00:21:38,000 --> 00:21:43,000
it says to look at the line
integral from zero to one,

263
00:21:43,000 --> 00:21:46,000
line integral from x to 2x of
possibly something,

264
00:21:46,000 --> 00:21:50,000
but dy dx.
And it says,

265
00:21:50,000 --> 00:21:58,000
let's look at how we would set
this up the other way around by

266
00:21:58,000 --> 00:22:03,000
exchanging x and y.
So, we should get to something

267
00:22:03,000 --> 00:22:06,000
that will be the same integral
dx dy.

268
00:22:06,000 --> 00:22:09,000
I mean, if you have a function
of x and y, then it will be the

269
00:22:09,000 --> 00:22:11,000
same function.
But, of course,

270
00:22:11,000 --> 00:22:14,000
the bounds change.
So, how do we exchange the

271
00:22:14,000 --> 00:22:17,000
order of integration?
Well, the only way to do it

272
00:22:17,000 --> 00:22:20,000
consistently is to draw a
picture.

273
00:22:20,000 --> 00:22:23,000
So, let's see,
what does this mean?

274
00:22:23,000 --> 00:22:28,000
Here, it means we integrate
from y equals x to y equals 2x,

275
00:22:28,000 --> 00:22:32,000
x between zero and one.
So, we should draw a picture.

276
00:22:32,000 --> 00:22:35,000
The lower bound for y is y
equals x.

277
00:22:35,000 --> 00:22:41,000
So, let's draw y equals x.
That seems to be here.

278
00:22:41,000 --> 00:22:47,000
And, we'll go up to y equals
2x, which is a line also but

279
00:22:47,000 --> 00:22:52,000
with bigger slope.
And then, all right,

280
00:22:52,000 --> 00:22:58,000
so for each value of x,
my origin will go from x to 2x.

281
00:22:58,000 --> 00:23:03,000
Well, and I do this for all
values of x that go to x equals

282
00:23:03,000 --> 00:23:06,000
one.
So, I stop at x equals one,

283
00:23:06,000 --> 00:23:10,000
which is here.
And then, my region is

284
00:23:10,000 --> 00:23:15,000
something like this.
OK, so this point here,

285
00:23:15,000 --> 00:23:21,000
in case you are wondering,
well, when x equals one,

286
00:23:21,000 --> 00:23:27,000
y is one.
And that point here is one, two.

287
00:23:27,000 --> 00:23:29,000
OK, any questions about that so
far?

288
00:23:29,000 --> 00:23:33,000
OK, so somehow that's the first
kill, when you see an integral,

289
00:23:33,000 --> 00:23:36,000
how to figure out what it
means, how to draw the region.

290
00:23:36,000 --> 00:23:39,000
And then there's a converse
scale which is given the region,

291
00:23:39,000 --> 00:23:42,000
how to set up the integral for
it.

292
00:23:42,000 --> 00:23:46,000
So, if we want to set up
instead dx dy,

293
00:23:46,000 --> 00:23:50,000
then it means we are going to
actually look at the converse

294
00:23:50,000 --> 00:23:54,000
question which is,
for a given value of y,

295
00:23:54,000 --> 00:23:57,000
what is the range of values of
x?

296
00:23:57,000 --> 00:24:01,000
OK, so if we fix y,
well, where do we enter the

297
00:24:01,000 --> 00:24:04,000
region, and where do we leave
it?

298
00:24:04,000 --> 00:24:08,000
So, we seem to enter on this
side, and we seem to leave on

299
00:24:08,000 --> 00:24:10,000
that side.
At least that seems to be true

300
00:24:10,000 --> 00:24:12,000
for the first few values of y
that I choose.

301
00:24:12,000 --> 00:24:16,000
But, hey, if I take a larger
value of y, then I will enter on

302
00:24:16,000 --> 00:24:19,000
the side, and I will leave on
this vertical side,

303
00:24:19,000 --> 00:24:22,000
not on that one.
So, I seem to have two

304
00:24:22,000 --> 00:24:28,000
different things going on.
OK, the place where enter my

305
00:24:28,000 --> 00:24:38,000
region is always y equals 2x,
which is the same as x equals y

306
00:24:38,000 --> 00:24:45,000
over two.
So, x seems to always start at

307
00:24:45,000 --> 00:24:51,000
y over two.
But, where I leave to be either

308
00:24:51,000 --> 00:24:55,000
x equals y, or here,
x equals y.

309
00:24:55,000 --> 00:24:57,000
And, that depends on the value
of y.

310
00:24:57,000 --> 00:24:59,000
So, in fact,
I have to break this into two

311
00:24:59,000 --> 00:25:03,000
different integrals.
I have to treat separately the

312
00:25:03,000 --> 00:25:07,000
case where y is between zero and
one, and between one and two.

313
00:25:07,000 --> 00:25:15,000
So, what I do in that case is I
just make two integrals.

314
00:25:15,000 --> 00:25:18,000
So, I say, both of them start
at y over two.

315
00:25:18,000 --> 00:25:22,000
But, in the first case,
we'll stop at x equals y.

316
00:25:22,000 --> 00:25:30,000
In the second case,
we'll stop at x equals one.

317
00:25:30,000 --> 00:25:31,000
OK, and now,
what are the values of y for

318
00:25:31,000 --> 00:25:34,000
each case?
Well, the first case is when y

319
00:25:34,000 --> 00:25:38,000
is between zero and one.
The second case is when y is

320
00:25:38,000 --> 00:25:40,000
between one and two,
which I guess this picture now

321
00:25:40,000 --> 00:25:44,000
is completely unreadable,
but hopefully you've been

322
00:25:44,000 --> 00:25:48,000
following what's going on,
or else you can see it in the

323
00:25:48,000 --> 00:25:53,000
solutions to the problem.
And, so that's our final answer.

324
00:25:53,000 --> 00:26:01,000
OK, any questions about how to
set up double integrals in xy

325
00:26:01,000 --> 00:26:04,000
coordinates?
No?

326
00:26:04,000 --> 00:26:07,000
OK, who feels comfortable with
this kind of problem?

327
00:26:07,000 --> 00:26:11,000
OK, good.
I'm happy to see the vast

328
00:26:11,000 --> 00:26:16,000
majority.
So, the bad news is we have to

329
00:26:16,000 --> 00:26:23,000
be able to do it not only in xy
coordinates, but also in polar

330
00:26:23,000 --> 00:26:27,000
coordinates.
So, when you go to polar

331
00:26:27,000 --> 00:26:32,000
coordinates, basically all you
have to remember on the side of

332
00:26:32,000 --> 00:26:36,000
integrand is that x becomes r
cosine theta.

333
00:26:36,000 --> 00:26:45,000
Y becomes r sine theta.
And, dx dy becomes r dr d theta.

334
00:26:45,000 --> 00:26:49,000
In terms of how you slice for
your region, well,

335
00:26:49,000 --> 00:26:52,000
you will be integrating first
over r.

336
00:26:52,000 --> 00:26:57,000
So, that means what you're
doing is you're fixing the value

337
00:26:57,000 --> 00:26:59,000
of theta.
And, for that value of theta,

338
00:26:59,000 --> 00:27:03,000
you ask yourself,
for what range of values of r

339
00:27:03,000 --> 00:27:06,000
am I going to be inside my
origin?

340
00:27:06,000 --> 00:27:09,000
So, if my origin looks like
this, then for this value of

341
00:27:09,000 --> 00:27:13,000
theta, r would go from zero to
whatever this distance is.

342
00:27:13,000 --> 00:27:16,000
And of course I have to find
how this distance depends on

343
00:27:16,000 --> 00:27:18,000
theta.
And then, I will find the

344
00:27:18,000 --> 00:27:20,000
extreme values of theta.
Now, of course,

345
00:27:20,000 --> 00:27:22,000
is the origin is really looking
like this, then you're not going

346
00:27:22,000 --> 00:27:25,000
to do it in polar coordinates.
But, if it's like a circle or a

347
00:27:25,000 --> 00:27:27,000
half circle, or things like
that,

348
00:27:27,000 --> 00:27:31,000
then even if a problem doesn't
tell you to do it in polar

349
00:27:31,000 --> 00:27:34,000
coordinates you might want to
seriously consider it.

350
00:27:34,000 --> 00:27:38,000
OK, so I'm not going to do it
but problem two in the practice

351
00:27:38,000 --> 00:27:43,000
exam is a good example of doing
something in polar coordinates.

352
00:27:43,000 --> 00:27:50,000
OK,
so in terms of things that we

353
00:27:50,000 --> 00:27:56,000
do with double integrals,
there's a few formulas that I'd

354
00:27:56,000 --> 00:28:00,000
like you to remember about
applications that we've seen of

355
00:28:00,000 --> 00:28:04,000
double integrals.
So, quantities that we can

356
00:28:04,000 --> 00:28:10,000
compute with double integrals
include things like the area of

357
00:28:10,000 --> 00:28:13,000
region,
its mass if it has a density,

358
00:28:13,000 --> 00:28:16,000
the average value of some
function,

359
00:28:16,000 --> 00:28:19,000
for example,
the average value of the x and

360
00:28:19,000 --> 00:28:22,000
y coordinates,
which we called the center of

361
00:28:22,000 --> 00:28:31,000
mass or moments of inertia.
So, these are just formulas to

362
00:28:31,000 --> 00:28:35,000
remember.
So, for example,

363
00:28:35,000 --> 00:28:40,000
the area of region is the
double integral of just dA,

364
00:28:40,000 --> 00:28:44,000
or if it helps you,
one dA if you want.

365
00:28:44,000 --> 00:28:47,000
You are integrating the
function 1.

366
00:28:47,000 --> 00:28:49,000
You have to remember formulas
for mass,

367
00:28:49,000 --> 00:28:54,000
for the average value of a
function is the F bar,

368
00:28:54,000 --> 00:29:05,000
in particular x bar y bar,
which is the center of mass,

369
00:29:05,000 --> 00:29:14,000
and the moment of inertia.
OK, so the polar moment of

370
00:29:14,000 --> 00:29:18,000
inertia, which is moment of
inertia about the origin.

371
00:29:18,000 --> 00:29:22,000
OK, so that's double integral
of x squared plus y squared,

372
00:29:22,000 --> 00:29:27,000
density dA,
but also moments of inertia

373
00:29:27,000 --> 00:29:33,000
about the x and y axis,
which are given by just taking

374
00:29:33,000 --> 00:29:36,000
one of these guys.
Don't worry about moments of

375
00:29:36,000 --> 00:29:39,000
inertia about an arbitrary line.
I will ask you for a moment of

376
00:29:39,000 --> 00:29:42,000
inertia for some weird line or
something like that.

377
00:29:42,000 --> 00:29:47,000
OK, but these you should know.
Now, what if you somehow,

378
00:29:47,000 --> 00:29:49,000
on the spur of the moment,
you forget, what's the formula

379
00:29:49,000 --> 00:29:51,000
for moment of inertia?
Well, I mean,

380
00:29:51,000 --> 00:29:54,000
I prefer if you know,
but if you have a complete

381
00:29:54,000 --> 00:29:56,000
blank in your memory,
there will still be partial

382
00:29:56,000 --> 00:29:59,000
credit were setting up the
bounds and everything else.

383
00:29:59,000 --> 00:30:01,000
So,
the general rule for the exam

384
00:30:01,000 --> 00:30:04,000
will be if you're stuck in a
calculation or you're missing a

385
00:30:04,000 --> 00:30:08,000
little piece of the puzzle,
try to do as much as you can.

386
00:30:08,000 --> 00:30:10,000
In particular,
try to at least set up the

387
00:30:10,000 --> 00:30:15,000
bounds of the integral.
There will be partial credit

388
00:30:15,000 --> 00:30:21,000
for that always.
So, while we're at it about

389
00:30:21,000 --> 00:30:26,000
grand rules, how about
evaluation?

390
00:30:26,000 --> 00:30:31,000
How about evaluating integrals?
So, once you've set it up,

391
00:30:31,000 --> 00:30:33,000
you have to sometimes compute
it.

392
00:30:33,000 --> 00:30:36,000
First of all,
check just in case the problem

393
00:30:36,000 --> 00:30:40,000
says set up but do not evaluate.
Then, don't waste your time

394
00:30:40,000 --> 00:30:45,000
evaluating it.
If a problem says to compute

395
00:30:45,000 --> 00:30:50,000
it, then you have to compute it.
So, what kinds of integration

396
00:30:50,000 --> 00:30:54,000
techniques do you need to know?
So, you need to know,

397
00:30:54,000 --> 00:30:57,000
you must know,
well, how to integrate the

398
00:30:57,000 --> 00:31:01,000
usual functions like one over x
or x to the n,

399
00:31:01,000 --> 00:31:05,000
or exponential,
sine, cosine,

400
00:31:05,000 --> 00:31:08,000
things like that,
OK, so the usual integrals.

401
00:31:08,000 --> 00:31:16,000
You must know what I will call
easy trigonometry.

402
00:31:16,000 --> 00:31:17,000
OK, I don't want to give you a
complete list.

403
00:31:17,000 --> 00:31:20,000
And the more you ask me about
which ones are on the list,

404
00:31:20,000 --> 00:31:22,000
the more I will add to the
list.

405
00:31:22,000 --> 00:31:26,000
But, those that you know that
you should know,

406
00:31:26,000 --> 00:31:28,000
you should know.
Those that you think you

407
00:31:28,000 --> 00:31:31,000
shouldn't know,
you don't have to know because

408
00:31:31,000 --> 00:31:36,000
I will say what I will say soon.
You should know also

409
00:31:36,000 --> 00:31:41,000
substitution,
how to set U equals something,

410
00:31:41,000 --> 00:31:45,000
and then see,
oh, this becomes u times du,

411
00:31:45,000 --> 00:31:50,000
and so substitution method.
What do I mean by easy

412
00:31:50,000 --> 00:31:52,000
trigonometrics?
Well, certainly you should know

413
00:31:52,000 --> 00:31:54,000
how to ingrate sine.
You should know how to

414
00:31:54,000 --> 00:31:57,000
integrate cosine.
You should be aware that sine

415
00:31:57,000 --> 00:32:01,000
squared plus cosine squared
simplifies to one.

416
00:32:01,000 --> 00:32:03,000
And, you should be aware of
general things like that.

417
00:32:03,000 --> 00:32:06,000
I would like you to know,
maybe, the double angles,

418
00:32:06,000 --> 00:32:09,000
sine 2x and cosine 2x.
Know what these are,

419
00:32:09,000 --> 00:32:12,000
and the kinds of the easy
things you can do with that,

420
00:32:12,000 --> 00:32:16,000
also things that involve
substitution setting like U

421
00:32:16,000 --> 00:32:19,000
equals sine T or U equals cosine
T.

422
00:32:19,000 --> 00:32:21,000
I mean, let me,
instead, give an example of

423
00:32:21,000 --> 00:32:25,000
hard trig that you don't need to
know, and then I will answer.

424
00:32:25,000 --> 00:32:34,000
OK, so, not needed on Thursday;
it doesn't mean that I don't

425
00:32:34,000 --> 00:32:37,000
want you to know them.
I would love you to know every

426
00:32:37,000 --> 00:32:41,000
single integral formula.
But, that shouldn't be your top

427
00:32:41,000 --> 00:32:44,000
priority.
So, you don't need to know

428
00:32:44,000 --> 00:32:47,000
things like hard trigonometric
ones.

429
00:32:47,000 --> 00:32:52,000
So, let me give you an example.
OK, so if I ask you to do this

430
00:32:52,000 --> 00:32:55,000
one, then actually I will give
you maybe, you know,

431
00:32:55,000 --> 00:32:59,000
I will reprint the formula from
the notes or something like

432
00:32:59,000 --> 00:33:02,000
that.
OK, so that one you don't need

433
00:33:02,000 --> 00:33:04,000
to know.
I would love if you happen to

434
00:33:04,000 --> 00:33:07,000
know it, but if you need it,
it will be given to you.

435
00:33:07,000 --> 00:33:13,000
So, these kinds of things that
you cannot compute by any easy

436
00:33:13,000 --> 00:33:16,000
method.
And, integration by parts,

437
00:33:16,000 --> 00:33:21,000
I believe that I successfully
test-solved all the problems

438
00:33:21,000 --> 00:33:26,000
without doing any single
integration by parts.

439
00:33:26,000 --> 00:33:29,000
Again, in general,
it's something that I would

440
00:33:29,000 --> 00:33:33,000
like you to know,
but it shouldn't be a top

441
00:33:33,000 --> 00:33:40,000
priority for this week.
OK, sorry, you had a question,

442
00:33:40,000 --> 00:33:42,000
or?
Inverse trigonometric

443
00:33:42,000 --> 00:33:45,000
functions: let's say the most
easy ones.

444
00:33:45,000 --> 00:33:50,000
I would like you to know the
easiest inverse trig functions,

445
00:33:50,000 --> 00:33:56,000
but not much.
OK, OK, so be aware that these

446
00:33:56,000 --> 00:34:04,000
functions exist,
but it's not a top priority.

447
00:34:04,000 --> 00:34:06,000
I should say,
the more I tell you I don't

448
00:34:06,000 --> 00:34:08,000
need you to know,
the more your physics and other

449
00:34:08,000 --> 00:34:11,000
teachers might complain that,
oh, these guys don't know how

450
00:34:11,000 --> 00:34:12,000
to integrate.
So, try not to forget

451
00:34:12,000 --> 00:34:19,000
everything.
But, yes?

452
00:34:19,000 --> 00:34:22,000
No, no, here I just mean for
evaluating just a single

453
00:34:22,000 --> 00:34:24,000
variable integral.
I will get to change variables

454
00:34:24,000 --> 00:34:27,000
and Jacobian soon,
but I'm thinking of this as a

455
00:34:27,000 --> 00:34:29,000
different topic.
What I mean by this one is if

456
00:34:29,000 --> 00:34:32,000
I'm asking you to integrate,
I don't know,

457
00:34:32,000 --> 00:34:37,000
what's a good example?
Zero to one t dt over square

458
00:34:37,000 --> 00:34:42,000
root of one plus t squared,
then you should think of maybe

459
00:34:42,000 --> 00:34:44,000
substituting u equals one plus t
squared,

460
00:34:44,000 --> 00:34:55,000
and then it becomes easier.
OK, so this kind of trig,

461
00:34:55,000 --> 00:35:00,000
that's what I have in mind here
specifically.

462
00:35:00,000 --> 00:35:02,000
And again,
if you're stuck,

463
00:35:02,000 --> 00:35:05,000
in particular,
if you hit this dreaded guy,

464
00:35:05,000 --> 00:35:09,000
and you don't actually have a
formula giving you what it is,

465
00:35:09,000 --> 00:35:12,000
it means one of two things.
One is something's wrong with

466
00:35:12,000 --> 00:35:13,000
your solution.
The other option is something

467
00:35:13,000 --> 00:35:16,000
is wrong with my problem.
So, either way,

468
00:35:16,000 --> 00:35:22,000
check quickly what you've done
it if you can't find a mistake,

469
00:35:22,000 --> 00:35:27,000
then just move ahead to the
next problem.

470
00:35:27,000 --> 00:35:30,000
Which one, this one?
Yeah,

471
00:35:30,000 --> 00:35:32,000
I mean if you can do it,
if you know how to do it,

472
00:35:32,000 --> 00:35:33,000
which everything is fair:
I mean,

473
00:35:33,000 --> 00:35:36,000
generally speaking,
give enough of it so that you

474
00:35:36,000 --> 00:35:38,000
found the solution by yourself,
not like,

475
00:35:38,000 --> 00:35:43,000
you know, it didn't somehow
come to you by magic.

476
00:35:43,000 --> 00:35:47,000
But, yeah, if you know how to
integrate this without doing the

477
00:35:47,000 --> 00:35:49,000
substitution,
that's absolutely fine by me.

478
00:35:49,000 --> 00:35:53,000
Just show enough work.
The general rule is show enough

479
00:35:53,000 --> 00:35:58,000
work that we see that you knew
what you are doing.

480
00:35:58,000 --> 00:36:02,000
OK, now another thing we've
seen with double integrals is

481
00:36:02,000 --> 00:36:05,000
how to do more complicated
changes of variables.

482
00:36:18,000 --> 00:36:23,000
So, when you want to replace x
and y by some variables,

483
00:36:23,000 --> 00:36:28,000
u and v, given by some formulas
in terms of x and y.

484
00:36:28,000 --> 00:36:33,000
So, you need to remember
basically how to do them.

485
00:36:33,000 --> 00:36:36,000
So, you need to remember that
the method consists of three

486
00:36:36,000 --> 00:36:43,000
steps.
So, one is you have to find the

487
00:36:43,000 --> 00:36:46,000
Jacobian.
And, you can choose to do

488
00:36:46,000 --> 00:36:50,000
either this Jacobian or the
inverse one depending on what's

489
00:36:50,000 --> 00:36:53,000
easiest given what you're given.
You don't have to worry about

490
00:36:53,000 --> 00:36:55,000
solving for things the other way
around.

491
00:36:55,000 --> 00:36:58,000
Just compute one of these
Jacobians.

492
00:36:58,000 --> 00:37:06,000
And then, the rule is that du
dv is absolute value of the

493
00:37:06,000 --> 00:37:12,000
Jacobian dx dy.
So, that takes care of dx dy,

494
00:37:12,000 --> 00:37:18,000
how to convert that into du dv.
The second thing to know is

495
00:37:18,000 --> 00:37:20,000
that,
well,

496
00:37:20,000 --> 00:37:25,000
you need to of course
substitute any x and y's in the

497
00:37:25,000 --> 00:37:32,000
integrand to convert them to u's
and v's so that you have a valid

498
00:37:32,000 --> 00:37:36,000
integrand involving only u and
v.

499
00:37:36,000 --> 00:37:51,000
And then, the last part is
setting up the bounds.

500
00:37:51,000 --> 00:37:54,000
And you see that,
probably you seen on P-sets and

501
00:37:54,000 --> 00:37:58,000
an example we did in the lecture
that this can be complicated.

502
00:37:58,000 --> 00:38:00,000
But now, in real life,
you do this actually to

503
00:38:00,000 --> 00:38:02,000
simplify the integrals.
So,

504
00:38:02,000 --> 00:38:04,000
probably the one that will be
there on Thursday,

505
00:38:04,000 --> 00:38:07,000
if there's a problem about that
on Thursday,

506
00:38:07,000 --> 00:38:10,000
it will be a situation where
the bounds that you get after

507
00:38:10,000 --> 00:38:13,000
changing variables are
reasonably easy.

508
00:38:13,000 --> 00:38:15,000
OK, I'm not saying that it will
be completely obvious

509
00:38:15,000 --> 00:38:17,000
necessarily, but it will be a
fairly easy situation.

510
00:38:17,000 --> 00:38:22,000
So, the general method is you
look at your region,

511
00:38:22,000 --> 00:38:25,000
R, and it might have various
sides.

512
00:38:25,000 --> 00:38:29,000
Well, on each side you ask
yourself, what do I know about x

513
00:38:29,000 --> 00:38:33,000
and y, and how to convert that
in terms of u and v?

514
00:38:33,000 --> 00:38:37,000
And maybe you'll find that the
equation might be just u equals

515
00:38:37,000 --> 00:38:39,000
zero for example,
or u equals v,

516
00:38:39,000 --> 00:38:42,000
or something like that.
And then, it's up to you to

517
00:38:42,000 --> 00:38:46,000
decide what you want to do.
But, maybe the easiest usually

518
00:38:46,000 --> 00:38:49,000
is to draw a new picture in
terms of u and v coordinates of

519
00:38:49,000 --> 00:38:53,000
what your region will look like
in the new coordinates.

520
00:38:53,000 --> 00:38:55,000
It might be that it will
actually much easier.

521
00:38:55,000 --> 00:39:00,000
It should be easier looking
than what you started with.

522
00:39:00,000 --> 00:39:05,000
OK, so that's the general idea.
There is one change of variable

523
00:39:05,000 --> 00:39:09,000
problem on each of the two
practice exams to give you a

524
00:39:09,000 --> 00:39:13,000
feeling for what's realistic.
The problem that's on practice

525
00:39:13,000 --> 00:39:18,000
exam 3B actually is on the hard
side of things because the

526
00:39:18,000 --> 00:39:21,000
question is kind of hidden in a
way.

527
00:39:21,000 --> 00:39:25,000
So, if you look at problem six,
you might find that it's not

528
00:39:25,000 --> 00:39:28,000
telling you very clearly what
you have to do.

529
00:39:28,000 --> 00:39:34,000
That's because it was meant to
be the hardest problem on that

530
00:39:34,000 --> 00:39:37,000
test.
But, once you've reduced it to

531
00:39:37,000 --> 00:39:41,000
an actual change of variables
problem, I expect you to be able

532
00:39:41,000 --> 00:39:44,000
to know how to do it.
And, on practice exam 3A,

533
00:39:44,000 --> 00:39:48,000
there's also,
I think it's problem five on

534
00:39:48,000 --> 00:39:52,000
the other practice exam.
And, that one is actually

535
00:39:52,000 --> 00:39:55,000
pretty standard and
straightforward.

536
00:39:55,000 --> 00:40:00,000
OK, time to move on, sorry.
So, we've also seen about line

537
00:40:00,000 --> 00:40:00,000
integrals.

538
00:40:21,000 --> 00:40:30,000
OK,
so line integrals,

539
00:40:30,000 --> 00:40:33,000
so the main thing to know about
them,

540
00:40:33,000 --> 00:40:37,000
so the line integral for work,
which is line integral of F.dr,

541
00:40:37,000 --> 00:40:40,000
so let's say that your vector
field has components,

542
00:40:40,000 --> 00:40:49,000
M and N.
So, the line integral for work

543
00:40:49,000 --> 00:40:57,000
becomes in coordinates integral
of Mdx plus Ndy while we've also

544
00:40:57,000 --> 00:41:05,000
seen line integral for flux.
So, line integral of F.n ds

545
00:41:05,000 --> 00:41:13,000
becomes the integral along C
just to make sure that I give it

546
00:41:13,000 --> 00:41:18,000
to you correctly.
So, remember that just,

547
00:41:18,000 --> 00:41:22,000
I don't want to make the
mistake in front of you.

548
00:41:22,000 --> 00:41:30,000
So, T ds is dx, dy.
And, the normal vector,

549
00:41:30,000 --> 00:41:36,000
so, T ds goes along the curve.
Nds goes clockwise

550
00:41:36,000 --> 00:41:41,000
perpendicular to the curve.
So, it's going to be,

551
00:41:41,000 --> 00:41:48,000
well, it's going to be dy and
negative dx.

552
00:41:48,000 --> 00:42:00,000
So, you will be integrating
negative Ndx plus Mdy.

553
00:42:00,000 --> 00:42:04,000
OK, see, if you are blanking
and don't remember the signs,

554
00:42:04,000 --> 00:42:07,000
then you can just draw this
picture and make sure that you

555
00:42:07,000 --> 00:42:10,000
get it right.
So, you should know a little

556
00:42:10,000 --> 00:42:14,000
bit about geometric
interpretation and how to see

557
00:42:14,000 --> 00:42:17,000
easily that it's going to be
zero in some cases.

558
00:42:17,000 --> 00:42:21,000
But, mostly you should know how
to compute, set up and compute

559
00:42:21,000 --> 00:42:23,000
these things.
So, what do we do when we are

560
00:42:23,000 --> 00:42:24,000
here?
Well, it's year,

561
00:42:24,000 --> 00:42:27,000
we have both x and y together,
but we want to,

562
00:42:27,000 --> 00:42:30,000
because it's the line integral,
there should be only one

563
00:42:30,000 --> 00:42:34,000
variable.
So, the important thing to know

564
00:42:34,000 --> 00:42:39,000
is we want to reduce everything
to a single parameter.

565
00:42:39,000 --> 00:42:55,000
OK, so the evaluation method is
always by reducing to a single

566
00:42:55,000 --> 00:43:01,000
parameter.
So, for example,

567
00:43:01,000 --> 00:43:06,000
maybe x and y are both
functions of some variable,

568
00:43:06,000 --> 00:43:10,000
t,
and then express everything in

569
00:43:10,000 --> 00:43:18,000
terms of some integral of,
some quantity involving t dt.

570
00:43:18,000 --> 00:43:21,000
It could be that you will just
express everything in terms of x

571
00:43:21,000 --> 00:43:24,000
or in terms of y,
or in terms of some angle or

572
00:43:24,000 --> 00:43:26,000
something.
It's up to you to choose how to

573
00:43:26,000 --> 00:43:29,000
parameterize things.
And then, when you're there,

574
00:43:29,000 --> 00:43:33,000
it's a usual one variable
integral with a single variable

575
00:43:33,000 --> 00:43:36,000
in there.
OK, so that's the general

576
00:43:36,000 --> 00:43:40,000
method of calculation,
but we've seen a shortcut for

577
00:43:40,000 --> 00:43:45,000
work when we can show that the
field is the gradient of

578
00:43:45,000 --> 00:43:48,000
potential.
So,

579
00:43:48,000 --> 00:43:55,000
one thing to know is if the
curl of F,

580
00:43:55,000 --> 00:44:01,000
which is an x minus My happens
to be zero,

581
00:44:01,000 --> 00:44:03,000
well,
and now I can say,

582
00:44:03,000 --> 00:44:06,000
and the domain is simply
connected,

583
00:44:06,000 --> 00:44:11,000
or if the field is defined
everywhere,

584
00:44:11,000 --> 00:44:19,000
then F is actually a gradient
field.

585
00:44:19,000 --> 00:44:22,000
So, that means,
just to make it more concrete,

586
00:44:22,000 --> 00:44:26,000
that means we can find a
function little f called the

587
00:44:26,000 --> 00:44:30,000
potential such that its
derivative respect to x is M,

588
00:44:30,000 --> 00:44:32,000
and its derivative with respect
to Y is N.

589
00:44:32,000 --> 00:44:37,000
We can solve these two
conditions for the same

590
00:44:37,000 --> 00:44:42,000
function, f, simultaneously.
And, how do we find this

591
00:44:42,000 --> 00:44:46,000
function, little f?
OK, so that's the same as

592
00:44:46,000 --> 00:44:50,000
saying that the field,
big F, is the gradient of

593
00:44:50,000 --> 00:44:52,000
little f.
And, how do we find this

594
00:44:52,000 --> 00:44:54,000
function, little f?
Well, we've seen two methods.

595
00:44:54,000 --> 00:44:58,000
One of them involves computing
a line integral from the origin

596
00:44:58,000 --> 00:45:02,000
to a point in the plane by going
first along the x axis,

597
00:45:02,000 --> 00:45:05,000
then vertically.
The other method was to first

598
00:45:05,000 --> 00:45:09,000
figure out what this one tells
us by integrating it with

599
00:45:09,000 --> 00:45:12,000
respect to x.
And then, we differentiate our

600
00:45:12,000 --> 00:45:17,000
answer with respect to y,
and we compare with that to get

601
00:45:17,000 --> 00:45:20,000
the complete answer.
OK, so I is that relevant?

602
00:45:20,000 --> 00:45:22,000
Well,
first of all it's relevant in

603
00:45:22,000 --> 00:45:25,000
physics,
but it's also relevant just to

604
00:45:25,000 --> 00:45:29,000
calculation of line integrals
because we see the fundamental

605
00:45:29,000 --> 00:45:34,000
theorem of calculus for line
integrals which says if we are

606
00:45:34,000 --> 00:45:39,000
integrating a gradient field and
we know what the potential is.

607
00:45:39,000 --> 00:45:43,000
Then, we just have to,
well, the line integral is just

608
00:45:43,000 --> 00:45:46,000
the change in value of a
potential.

609
00:45:46,000 --> 00:45:49,000
OK, so we take the value of a
potential at the starting point,

610
00:45:49,000 --> 00:45:52,000
sorry, we take value potential
at the endpoint minus the value

611
00:45:52,000 --> 00:45:58,000
at the starting point.
And, that will give us the line

612
00:45:58,000 --> 00:46:00,000
integral, OK?
So, important:

613
00:46:00,000 --> 00:46:05,000
this is only for work.
There's no statement like that

614
00:46:05,000 --> 00:46:09,000
for flux, OK,
so don't tried to fly this in a

615
00:46:09,000 --> 00:46:11,000
problem about flux.
I mean, usually,

616
00:46:11,000 --> 00:46:13,000
if you look at the practice
exams,

617
00:46:13,000 --> 00:46:17,000
you will see it's pretty clear
that there's one problem in

618
00:46:17,000 --> 00:46:20,000
which you are supposed to do
things this way.

619
00:46:20,000 --> 00:46:25,000
It's kind of a dead giveaway,
but it's probably not too bad.

620
00:46:25,000 --> 00:46:29,000
OK, and the other thing we've
seen, so I mentioned it at the

621
00:46:29,000 --> 00:46:32,000
beginning but let me mention it
again.

622
00:46:32,000 --> 00:46:36,000
To compute things,
Green's theorem,

623
00:46:36,000 --> 00:46:42,000
let's just compute,
well, let us forget,

624
00:46:42,000 --> 00:46:45,000
sorry, find the value of a line
integral along the closed curve

625
00:46:45,000 --> 00:46:47,000
by reducing it to double
integral.

626
00:46:47,000 --> 00:46:55,000
So,
the one for work says -- --

627
00:46:55,000 --> 00:46:59,000
this,
and you should remember that in

628
00:46:59,000 --> 00:47:01,000
there,
so C is a closed curve that

629
00:47:01,000 --> 00:47:05,000
goes counterclockwise,
and R is the region inside.

630
00:47:05,000 --> 00:47:08,000
So, the way you would,
if you had to compute both

631
00:47:08,000 --> 00:47:10,000
sides separately,
you would do them in extremely

632
00:47:10,000 --> 00:47:12,000
different ways,
right?

633
00:47:12,000 --> 00:47:15,000
This one is a line integral.
So, you use the method to

634
00:47:15,000 --> 00:47:18,000
explain here,
namely, you express x and y in

635
00:47:18,000 --> 00:47:22,000
terms of a single variable.
See that you're doing a circle.

636
00:47:22,000 --> 00:47:24,000
I want to see a theta.
I don't want to see an R.

637
00:47:24,000 --> 00:47:27,000
R is not a variable.
You are on the circle.

638
00:47:27,000 --> 00:47:30,000
This one is a double integral.
So, if you are doing it,

639
00:47:30,000 --> 00:47:32,000
say, on a disk,
you would have both R and theta

640
00:47:32,000 --> 00:47:34,000
if you're using polar
coordinates.

641
00:47:34,000 --> 00:47:37,000
You would have both x and y.
Here, you have two variables of

642
00:47:37,000 --> 00:47:40,000
integration.
Here, you should have only one

643
00:47:40,000 --> 00:47:42,000
after you parameterize the
curve.

644
00:47:42,000 --> 00:47:46,000
And, the fact that it stays
curl F, I mean,

645
00:47:46,000 --> 00:47:51,000
curl F is just Nx-My is just
like any function of x and y.

646
00:47:51,000 --> 00:47:54,000
OK, the fact that we called it
curl F doesn't change how you

647
00:47:54,000 --> 00:47:56,000
compute it.
You have first to compute the

648
00:47:56,000 --> 00:47:58,000
curl of F.
Say you find,

649
00:47:58,000 --> 00:48:00,000
I don't know,
xy minus x squared,

650
00:48:00,000 --> 00:48:04,000
well, it becomes just the usual
double integral of the usual

651
00:48:04,000 --> 00:48:09,000
function xy minus x squared.
There's nothing special to it

652
00:48:09,000 --> 00:48:15,000
because it's a curl.
And, the other one is the

653
00:48:15,000 --> 00:48:21,000
counterpart for flux.
So, it says this,

654
00:48:21,000 --> 00:48:25,000
and remember this is mx plus
ny.

655
00:48:25,000 --> 00:48:27,000
I mean, what's important about
these statements is not only

656
00:48:27,000 --> 00:48:30,000
remembering, you know,
if you just know this formula

657
00:48:30,000 --> 00:48:32,000
by heart,
you are still in trouble

658
00:48:32,000 --> 00:48:35,000
because you need to know what
actually the symbols in here

659
00:48:35,000 --> 00:48:37,000
mean.
So, you should remember,

660
00:48:37,000 --> 00:48:40,000
what is this line integral,
and what's the divergence of a

661
00:48:40,000 --> 00:48:47,000
field?
So, just something to remember.

662
00:48:47,000 --> 00:48:51,000
And, so I guess I'll let you
figure out practice problems

663
00:48:51,000 --> 00:48:54,000
because it's time,
but I think that's basically

664
00:48:54,000 --> 00:48:59,000
the list of all we've seen.
And, well, that should be it.