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OK, so last time we've seen the
curl of the vector field with
00:00:32.000 --> 00:00:39.000
components M and N.
We defined that to be N sub x
00:00:39.000 --> 00:00:43.000
minus M sub y.
And, we said this measures how
00:00:43.000 --> 00:00:47.000
far that vector field is from
being conservative.
00:00:47.000 --> 00:00:50.000
If the curl is zero,
and if the field is defined
00:00:50.000 --> 00:00:53.000
everywhere, then it's going to
be conservative.
00:00:53.000 --> 00:00:55.000
And so, when I take the line
integral along a closed curve,
00:00:55.000 --> 00:00:59.000
I don't have to compute it.
I notes going to be zero.
00:00:59.000 --> 00:01:04.000
But now, let's say that I have
a general vector field.
00:01:04.000 --> 00:01:07.000
So, the curl will not be zero.
And, I still want to compute
00:01:07.000 --> 00:01:10.000
the line integral along a closed
curve.
00:01:10.000 --> 00:01:14.000
Well, I could compute it
directly or there's another way.
00:01:14.000 --> 00:01:17.000
And that's what we are going to
see today.
00:01:17.000 --> 00:01:26.000
So, say that I have a closed
curve, C, and I want to find the
00:01:26.000 --> 00:01:30.000
work.
So, there's two options.
00:01:30.000 --> 00:01:40.000
One is direct calculation,
and the other one is Green's
00:01:40.000 --> 00:01:47.000
theorem.
So, Green's theorem is another
00:01:47.000 --> 00:01:56.000
way to avoid calculating line
integrals if we don't want to.
00:01:56.000 --> 00:02:04.000
OK, so what does it say?
It says if C is a closed curve
00:02:04.000 --> 00:02:16.000
enclosing a region R in the
plane, and I have to insist C
00:02:16.000 --> 00:02:28.000
should go counterclockwise.
And, if I have a vector field
00:02:28.000 --> 00:02:37.000
that's defined and
differentiable everywhere not
00:02:37.000 --> 00:02:42.000
only on the curve,
C, which is what I need to
00:02:42.000 --> 00:02:48.000
define the line integral,
but also on the region inside.
00:02:48.000 --> 00:02:57.000
Then -- -- the line integral
for the work done along C is
00:02:57.000 --> 00:03:07.000
actually equal to a double
integral over the region inside
00:03:07.000 --> 00:03:15.000
of curl F dA.
OK, so that's the conclusion.
00:03:15.000 --> 00:03:19.000
And, if you want me to write it
in coordinates,
00:03:19.000 --> 00:03:24.000
maybe I should do that.
So, the line integral in terms
00:03:24.000 --> 00:03:28.000
of the components,
that's the integral of M dx
00:03:28.000 --> 00:03:36.000
plus N dy.
And, the curl is (Nx-My)dA.
00:03:36.000 --> 00:03:42.000
OK, so that's the other way to
state it.
00:03:42.000 --> 00:03:47.000
So, that's a really strange
statement if you think about it
00:03:47.000 --> 00:03:51.000
because the left-hand side is a
line integral.
00:03:51.000 --> 00:03:57.000
OK, so the way we compute it is
we take this expression Mdx Ndy
00:03:57.000 --> 00:04:01.000
and we parameterize the curve.
We express x and y in terms of
00:04:01.000 --> 00:04:04.000
some variable,
t, maybe, or whatever you want
00:04:04.000 --> 00:04:07.000
to call it.
And then, you'll do a one
00:04:07.000 --> 00:04:11.000
variable integral over t.
This right-hand side here,
00:04:11.000 --> 00:04:13.000
it's a double integral,
dA.
00:04:13.000 --> 00:04:16.000
So, we do it the way that we
learn how to couple of weeks
00:04:16.000 --> 00:04:18.000
ago.
You take your region,
00:04:18.000 --> 00:04:22.000
you slice it in the x direction
or in the y direction,
00:04:22.000 --> 00:04:26.000
and you integrate dx dy after
setting up the bounds carefully,
00:04:26.000 --> 00:04:29.000
or maybe in polar coordinates r
dr d theta.
00:04:29.000 --> 00:04:32.000
But, see, the way you compute
these things is completely
00:04:32.000 --> 00:04:36.000
different.
This one on the left-hand side
00:04:36.000 --> 00:04:42.000
lives only on the curve,
while the right-hand side lives
00:04:42.000 --> 00:04:46.000
everywhere in this region
inside.
00:04:46.000 --> 00:04:49.000
So, here, x and y are related,
they live on the curve.
00:04:49.000 --> 00:04:53.000
Here, x and y are independent.
There just are some bounds
00:04:53.000 --> 00:04:54.000
between them.
And, of course,
00:04:54.000 --> 00:04:56.000
what you're integrating is
different.
00:04:56.000 --> 00:05:01.000
It's a line integral for work.
Here, it's a double integral of
00:05:01.000 --> 00:05:08.000
some function of x and y.
So, it's a very perplexing
00:05:08.000 --> 00:05:14.000
statement at first.
But, it's a very powerful tool.
00:05:14.000 --> 00:05:18.000
So, we're going to try to see
how it works concretely,
00:05:18.000 --> 00:05:20.000
what it says,
what are the consequences,
00:05:20.000 --> 00:05:23.000
how we could convince ourselves
that, yes,
00:05:23.000 --> 00:05:26.000
this works, and so on.
That's going to be the topic
00:05:26.000 --> 00:05:32.000
for today.
Any questions about the
00:05:32.000 --> 00:05:38.000
statement first?
No?
00:05:38.000 --> 00:05:43.000
OK, yeah, one remark, sorry.
So, here, it stays
00:05:43.000 --> 00:05:46.000
counterclockwise.
What if I have a curve that
00:05:46.000 --> 00:05:49.000
goes clockwise?
Well, you could just take the
00:05:49.000 --> 00:05:52.000
negative, and integrate
counterclockwise.
00:05:52.000 --> 00:05:57.000
Why does the theorem choose
counterclockwise over clockwise?
00:05:57.000 --> 00:06:00.000
How doesn't know that it's
counterclockwise rather than
00:06:00.000 --> 00:06:02.000
clockwise?
Well, the answer is basically
00:06:02.000 --> 00:06:05.000
in our convention for curl.
See, we've said curl is Nx
00:06:05.000 --> 00:06:08.000
minus My, and not the other way
around.
00:06:08.000 --> 00:06:10.000
And, that's a convention as
well.
00:06:10.000 --> 00:06:13.000
So, somehow,
the two conventions match with
00:06:13.000 --> 00:06:15.000
each other.
That's the best answer I can
00:06:15.000 --> 00:06:18.000
give you.
So, if you met somebody from a
00:06:18.000 --> 00:06:20.000
different planet,
they might have Green's theorem
00:06:20.000 --> 00:06:23.000
with the opposite conventions,
with curves going clockwise,
00:06:23.000 --> 00:06:27.000
and the curl defined the other
way around.
00:06:27.000 --> 00:06:35.000
Probably if you met an alien,
I'm not sure if you would be
00:06:35.000 --> 00:06:43.000
discussing Green's theorem
first, but just in case.
00:06:43.000 --> 00:06:53.000
OK, so that being said,
there is a warning here which
00:06:53.000 --> 00:07:00.000
is that this is only for closed
curves.
00:07:00.000 --> 00:07:03.000
OK, so if I give you a curve
that's not closed,
00:07:03.000 --> 00:07:05.000
and I tell you,
well, compute the line
00:07:05.000 --> 00:07:08.000
integral, then you have to do it
by hand.
00:07:08.000 --> 00:07:10.000
You have to parameterize the
curve.
00:07:10.000 --> 00:07:12.000
Or, if you really don't like
that line integral,
00:07:12.000 --> 00:07:16.000
you could close the path by
adding some other line integral
00:07:16.000 --> 00:07:19.000
to it,
and then compute using Green's
00:07:19.000 --> 00:07:23.000
theorem.
But, you can't use Green's
00:07:23.000 --> 00:07:30.000
theorem directly if the curve is
not closed.
00:07:30.000 --> 00:07:42.000
OK, so let's do a quick example.
So, let's say that I give you
00:07:42.000 --> 00:07:52.000
C, the circle of radius one,
centered at the point (2,0).
00:07:52.000 --> 00:08:00.000
So, it's out here.
That's my curve, C.
00:08:00.000 --> 00:08:09.000
And, let's say that I do it
counterclockwise so that it will
00:08:09.000 --> 00:08:16.000
match with the statement of the
theorem.
00:08:16.000 --> 00:08:24.000
And, let's say that I want you
to compute the line integral
00:08:24.000 --> 00:08:34.000
along C of ye^(-x) dx plus (one
half of x squared minus e^(-x))
00:08:34.000 --> 00:08:37.000
dy.
And, that's a kind of sadistic
00:08:37.000 --> 00:08:41.000
example, but maybe I'll ask you
to do that.
00:08:41.000 --> 00:08:44.000
So, how would you do it
directly?
00:08:44.000 --> 00:08:48.000
Well, to do it directly you
would have to parameterize this
00:08:48.000 --> 00:08:53.000
curve.
So that would probably involve
00:08:53.000 --> 00:09:03.000
setting x equals two plus cosine
theta y equals sine theta.
00:09:03.000 --> 00:09:06.000
But, I'm using as parameter of
the angle around the circle,
00:09:06.000 --> 00:09:09.000
it's like the unit circle,
the usual ones that shifted by
00:09:09.000 --> 00:09:15.000
two in the x direction.
And then, I would set dx equals
00:09:15.000 --> 00:09:21.000
minus sine theta d theta.
I would set dy equals cosine
00:09:21.000 --> 00:09:24.000
theta d theta.
And, I will substitute,
00:09:24.000 --> 00:09:26.000
and I will integrate from zero
to 2pi.
00:09:26.000 --> 00:09:29.000
And, I would probably run into
a bit of trouble because I would
00:09:29.000 --> 00:09:32.000
have these e to the minus x,
which would give me something
00:09:32.000 --> 00:09:35.000
that I really don't want to
integrate.
00:09:35.000 --> 00:09:44.000
So, instead of doing that,
which looks pretty much doomed,
00:09:44.000 --> 00:09:51.000
instead, I'm going to use
Green's theorem.
00:09:51.000 --> 00:09:58.000
So, using Green's theorem,
the way we'll do it is I will,
00:09:58.000 --> 00:10:03.000
instead, compute a double
integral.
00:10:03.000 --> 00:10:16.000
So, I will -- -- compute the
double integral over the region
00:10:16.000 --> 00:10:27.000
inside of curl F dA.
So, I should say probably what
00:10:27.000 --> 00:10:30.000
F was.
So, let's call this M.
00:10:30.000 --> 00:10:37.000
Let's call this N.
And, then I will actually just
00:10:37.000 --> 00:10:45.000
choose the form coordinates,
(Nx minus My) dA.
00:10:45.000 --> 00:10:53.000
And, what is R here?
Well, R is the disk in here.
00:10:53.000 --> 00:10:56.000
OK, so, of course,
it might not be that pleasant
00:10:56.000 --> 00:10:58.000
because we'll also have to set
up this double integral.
00:10:58.000 --> 00:11:02.000
And, for that,
we'll have to figure out a way
00:11:02.000 --> 00:11:05.000
to slice this region nicely.
We could do it dx dy.
00:11:05.000 --> 00:11:08.000
We could do it dy dx.
Or, maybe we will want to
00:11:08.000 --> 00:11:11.000
actually make a change of
variables to first shift this to
00:11:11.000 --> 00:11:14.000
the origin,
you know, change x to x minus
00:11:14.000 --> 00:11:17.000
two and then switch to polar
coordinates.
00:11:17.000 --> 00:11:22.000
Well, let's see what happens
later.
00:11:22.000 --> 00:11:33.000
OK, so what is, so this is R.
So, what is N sub x?
00:11:33.000 --> 00:11:42.000
Well, N sub x is x plus e to
the minus x minus,
00:11:42.000 --> 00:11:48.000
what is M sub y,
e to the minus x,
00:11:48.000 --> 00:11:52.000
OK?
This is Nx.
00:11:52.000 --> 00:11:58.000
This is My dA.
Well, it seems to simplify a
00:11:58.000 --> 00:12:02.000
bit.
I will just get double integral
00:12:02.000 --> 00:12:06.000
over R of x dA,
which looks certainly a lot
00:12:06.000 --> 00:12:10.000
more pleasant.
Of course, I made up the
00:12:10.000 --> 00:12:14.000
example in that way so that it
simplifies when you use Green's
00:12:14.000 --> 00:12:16.000
theorem.
But, you know,
00:12:16.000 --> 00:12:20.000
it gives you an example where
you can turn are really hard
00:12:20.000 --> 00:12:23.000
line integral into an easier
double integral.
00:12:23.000 --> 00:12:28.000
Now, how do we compute that
double integral?
00:12:28.000 --> 00:12:32.000
Well, so one way would be to
set it up.
00:12:32.000 --> 00:12:41.000
Or, let's actually be a bit
smarter and observe that this is
00:12:41.000 --> 00:12:50.000
actually the area of the region
R, times the x coordinate of its
00:12:50.000 --> 00:12:55.000
center of mass.
If I look at the definition of
00:12:55.000 --> 00:12:59.000
the center of mass,
it's the average value of x.
00:12:59.000 --> 00:13:03.000
So, it's one over the area
times the double integral of x
00:13:03.000 --> 00:13:07.000
dA, well, possibly with the
density, but here I'm thinking
00:13:07.000 --> 00:13:11.000
uniform density one.
And, now, I think I know just
00:13:11.000 --> 00:13:15.000
by looking at the picture where
the center of mass of this
00:13:15.000 --> 00:13:16.000
circle will be,
right?
00:13:16.000 --> 00:13:19.000
I mean, it would be right in
the middle.
00:13:19.000 --> 00:13:24.000
So, that is two,
if you want,
00:13:24.000 --> 00:13:29.000
by symmetry.
And, the area of the guy is
00:13:29.000 --> 00:13:33.000
just pi because it's a disk of
radius one.
00:13:33.000 --> 00:13:37.000
So, I will just get 2pi.
I mean, of course,
00:13:37.000 --> 00:13:40.000
if you didn't see that,
then you can also compute that
00:13:40.000 --> 00:13:43.000
double integral directly.
It's a nice exercise.
00:13:43.000 --> 00:13:47.000
But see, here,
using geometry helps you to
00:13:47.000 --> 00:13:50.000
actually streamline the
calculation.
00:13:50.000 --> 00:14:01.000
OK, any questions?
Yes?
00:14:01.000 --> 00:14:04.000
OK, yes, let me just repeat the
last part.
00:14:04.000 --> 00:14:10.000
So, I said we had to compute
the double integral of x dA over
00:14:10.000 --> 00:14:14.000
this region here,
which is a disk of radius one,
00:14:14.000 --> 00:14:18.000
centered at,
this point is (2,0).
00:14:18.000 --> 00:14:22.000
So, instead of setting up the
integral with bounds and
00:14:22.000 --> 00:14:26.000
integrating dx dy or dy dx or in
polar coordinates,
00:14:26.000 --> 00:14:30.000
I'm just going to say, well,
let's remember the definition
00:14:30.000 --> 00:14:32.000
of a center of mass.
It's the average value of a
00:14:32.000 --> 00:14:37.000
function, x in the region.
So, it's one over the area of
00:14:37.000 --> 00:14:42.000
origin times the double integral
of x dA.
00:14:42.000 --> 00:14:46.000
If you look,
again, at the definition of x
00:14:46.000 --> 00:14:51.000
bar, it's one over area of
double integral x dA.
00:14:51.000 --> 00:14:54.000
Well, maybe if there's a
density, then it's one over mass
00:14:54.000 --> 00:14:57.000
times double integral of x
density dA.
00:14:57.000 --> 00:15:02.000
But, if density is one,
then it just becomes this.
00:15:02.000 --> 00:15:06.000
So, switching the area,
moving the area to the other
00:15:06.000 --> 00:15:08.000
side,
I'll get double integral of x
00:15:08.000 --> 00:15:12.000
dA is the area of origin times
the x coordinate of the center
00:15:12.000 --> 00:15:14.000
of mass.
The area of origin is pi
00:15:14.000 --> 00:15:18.000
because it's a unit disk.
And, the center of mass is the
00:15:18.000 --> 00:15:23.000
center of a disk.
So, its x bar is two,
00:15:23.000 --> 00:15:27.000
and I get 2 pi.
OK, that I didn't actually have
00:15:27.000 --> 00:15:30.000
to do this in my example today,
but of course that would be
00:15:30.000 --> 00:15:36.000
good review.
It will remind you of center of
00:15:36.000 --> 00:15:47.000
mass and all that.
OK, any other questions?
00:15:47.000 --> 00:15:50.000
No?
OK, so let's see,
00:15:50.000 --> 00:15:54.000
now that we've seen how to use
it practice, how to avoid
00:15:54.000 --> 00:15:58.000
calculating the line integral if
we don't want to.
00:15:58.000 --> 00:16:04.000
Let's try to convince ourselves
that this theorem makes sense.
00:16:04.000 --> 00:16:09.000
OK, so, well,
let's start with an easy case
00:16:09.000 --> 00:16:15.000
where we should be able to know
the answer to both sides.
00:16:15.000 --> 00:16:23.000
So let's look at the special
case.
00:16:23.000 --> 00:16:32.000
Let's look at the case where
curl F is zero.
00:16:32.000 --> 00:16:45.000
Then, well, we'd like to
conclude that F is conservative.
00:16:45.000 --> 00:16:53.000
That's what we said.
Well let's see what happens.
00:16:53.000 --> 00:16:59.000
So, Green's theorem says that
if I have a closed curve,
00:16:59.000 --> 00:17:06.000
then the line integral of F is
equal to the double integral of
00:17:06.000 --> 00:17:12.000
curl on the region inside.
And, if the curl is zero,
00:17:12.000 --> 00:17:15.000
then I will be integrating
zero.
00:17:15.000 --> 00:17:22.000
I will get zero.
OK, so this is actually how you
00:17:22.000 --> 00:17:26.000
prove that if your vector field
has curve zero,
00:17:26.000 --> 00:17:27.000
then it's conservative.
00:17:54.000 --> 00:17:57.000
OK, so in particular,
if you have a vector field
00:17:57.000 --> 00:18:01.000
that's defined everywhere the
plane, then you take any closed
00:18:01.000 --> 00:18:04.000
curve.
Well, you will get that the
00:18:04.000 --> 00:18:06.000
line integral will be zero.
Straightly speaking,
00:18:06.000 --> 00:18:10.000
that will only work here if the
curve goes counterclockwise.
00:18:10.000 --> 00:18:13.000
But otherwise,
just look at the various loops
00:18:13.000 --> 00:18:16.000
that it makes,
and orient each of them
00:18:16.000 --> 00:18:19.000
counterclockwise and sum things
together.
00:18:19.000 --> 00:18:22.000
So let me state that again.
00:18:45.000 --> 00:18:51.000
So,
OK,
00:18:51.000 --> 00:19:02.000
so a consequence of Green's
theorem is that if F is defined
00:19:02.000 --> 00:19:12.000
everywhere in the plane -- --
and the curl of F is zero
00:19:12.000 --> 00:19:24.000
everywhere,
then F is conservative.
00:19:24.000 --> 00:19:29.000
And so, this actually is the
input we needed to justify our
00:19:29.000 --> 00:19:33.000
criterion.
The test that we saw last time
00:19:33.000 --> 00:19:35.000
saying,
well, to check if something is
00:19:35.000 --> 00:19:37.000
a gradient field if it's
conservative,
00:19:37.000 --> 00:19:40.000
we just have to compute the
curl and check whether it's
00:19:40.000 --> 00:19:45.000
zero.
OK, so how do we prove that now
00:19:45.000 --> 00:19:49.000
carefully?
Well, you just take a closed
00:19:49.000 --> 00:19:52.000
curve in the plane.
You switch the orientation if
00:19:52.000 --> 00:19:55.000
needed so it becomes
counterclockwise.
00:19:55.000 --> 00:19:59.000
And then you look at the region
inside.
00:19:59.000 --> 00:20:05.000
And then you know that the line
integral inside will be equal to
00:20:05.000 --> 00:20:12.000
the double integral of curl,
which is the double integral of
00:20:12.000 --> 00:20:16.000
zero.
Therefore, that's zero.
00:20:16.000 --> 00:20:19.000
But see, OK,
so now let's say that we try to
00:20:19.000 --> 00:20:23.000
do that for the vector field
that was on your problems that
00:20:23.000 --> 00:20:27.000
was not defined at the origin.
So if you've done the problem
00:20:27.000 --> 00:20:30.000
sets and found the same answers
that I did, then you will have
00:20:30.000 --> 00:20:33.000
found that this vector field had
curve zero everywhere.
00:20:33.000 --> 00:20:36.000
But still it wasn't
conservative because if you went
00:20:36.000 --> 00:20:39.000
around the unit circle,
then you got a line integral
00:20:39.000 --> 00:20:42.000
that was 2pi.
Or, if you compared the two
00:20:42.000 --> 00:20:45.000
halves, you got different
answers for two parts that go
00:20:45.000 --> 00:20:47.000
from the same point to the same
point.
00:20:47.000 --> 00:20:51.000
So, it fails this property but
that's because it's not defined
00:20:51.000 --> 00:20:54.000
everywhere.
So, what goes wrong with this
00:20:54.000 --> 00:20:57.000
argument?
Well, if I take the vector
00:20:57.000 --> 00:21:03.000
field that was in the problem
set, and if I do things,
00:21:03.000 --> 00:21:07.000
say that I look at the unit
circle.
00:21:07.000 --> 00:21:10.000
That's a closed curve.
So, I would like to use Green's
00:21:10.000 --> 00:21:13.000
theorem.
Green's theorem would tell me
00:21:13.000 --> 00:21:17.000
the line integral along this
loop is equal to the double
00:21:17.000 --> 00:21:21.000
integral of curl over this
region here, the unit disk.
00:21:21.000 --> 00:21:25.000
And, of course the curl is
zero, well, except at the
00:21:25.000 --> 00:21:27.000
origin.
At the origin,
00:21:27.000 --> 00:21:29.000
the vector field is not
defined.
00:21:29.000 --> 00:21:32.000
You cannot take the
derivatives, and the curl is not
00:21:32.000 --> 00:21:34.000
defined.
And somehow that messes things
00:21:34.000 --> 00:21:38.000
up.
You cannot apply Green's
00:21:38.000 --> 00:21:49.000
theorem to the vector field.
So, you cannot apply Green's
00:21:49.000 --> 00:22:02.000
theorem to the vector field on
problem set eight problem two
00:22:02.000 --> 00:22:12.000
when C encloses the origin.
And so, that's why this guy,
00:22:12.000 --> 00:22:16.000
even though it has curl zero,
is not conservative.
00:22:16.000 --> 00:22:20.000
There's no contradiction.
And somehow,
00:22:20.000 --> 00:22:23.000
you have to imagine that,
well, the curl here is really
00:22:23.000 --> 00:22:26.000
not defined.
But somehow it becomes infinite
00:22:26.000 --> 00:22:30.000
so that when you do the double
integral, you actually get 2 pi
00:22:30.000 --> 00:22:37.000
instead of zero.
I mean, that doesn't make any
00:22:37.000 --> 00:22:46.000
sense, of course,
but that's one way to think
00:22:46.000 --> 00:22:51.000
about it.
OK, any questions?
00:22:51.000 --> 00:23:02.000
Yes?
Well, though actually it's not
00:23:02.000 --> 00:23:06.000
defined because the curl is zero
everywhere else.
00:23:06.000 --> 00:23:08.000
So, if a curl was well defined
at the origin,
00:23:08.000 --> 00:23:11.000
you would try to,
then, take the double integral.
00:23:11.000 --> 00:23:12.000
no matter what value you put
for a function,
00:23:12.000 --> 00:23:15.000
if you have a function that's
zero everywhere except at the
00:23:15.000 --> 00:23:17.000
origin,
and some other value at the
00:23:17.000 --> 00:23:20.000
origin,
the integral is still zero.
00:23:20.000 --> 00:23:24.000
So, it's worse than that.
It's not only that you can't
00:23:24.000 --> 00:23:29.000
compute it, it's that is not
defined.
00:23:29.000 --> 00:23:36.000
OK, anyway, that's like a
slightly pathological example.
00:23:36.000 --> 00:23:44.000
Yes?
Well, we wouldn't be able to
00:23:44.000 --> 00:23:46.000
because the curl is not defined
at the origin.
00:23:46.000 --> 00:23:49.000
So, you can actually integrate
it.
00:23:49.000 --> 00:23:52.000
OK, so that's the problem.
I mean, if you try to
00:23:52.000 --> 00:23:55.000
integrate, we've said everywhere
where it's defined,
00:23:55.000 --> 00:23:57.000
the curl is zero.
So, what you would be
00:23:57.000 --> 00:24:01.000
integrating would be zero.
But, that doesn't work because
00:24:01.000 --> 00:24:09.000
at the origin it's not defined.
Yes?
00:24:09.000 --> 00:24:11.000
Ah, so if you take a curve that
makes a figure 8,
00:24:11.000 --> 00:24:14.000
then indeed my proof over there
is false.
00:24:14.000 --> 00:24:19.000
So, I kind of tricked you.
It's not actually correct.
00:24:19.000 --> 00:24:24.000
So, if the curve does a figure
8, then what you do is you would
00:24:24.000 --> 00:24:27.000
actually cut it into its two
halves.
00:24:27.000 --> 00:24:30.000
And for each of them,
you will apply Green's theorem.
00:24:30.000 --> 00:24:32.000
And then, you'd still get,
if a curl is zero then this
00:24:32.000 --> 00:24:35.000
line integral is zero.
That one is also zero.
00:24:35.000 --> 00:24:38.000
So this one is zero.
OK, small details that you
00:24:38.000 --> 00:24:41.000
don't really need to worry too
much about,
00:24:41.000 --> 00:24:47.000
but indeed if you want to be
careful with details then my
00:24:47.000 --> 00:24:54.000
proof is not quite complete.
But the computation is still
00:24:54.000 --> 00:24:58.000
true.
Let's move on.
00:24:58.000 --> 00:25:06.000
So, I want to tell you how to
prove Green's theorem because
00:25:06.000 --> 00:25:15.000
it's such a strange formula that
where can it come from possibly?
00:25:15.000 --> 00:25:21.000
I mean,
so let me remind you first of
00:25:21.000 --> 00:25:26.000
all the statement we want to
prove is that the line integral
00:25:26.000 --> 00:25:31.000
along a closed curve of Mdx plus
Ndy is equal to the double
00:25:31.000 --> 00:25:36.000
integral over the region inside
of (Nx minus My)dA.
00:25:36.000 --> 00:25:40.000
And, let's simplify our lives a
bit by proving easier
00:25:40.000 --> 00:25:43.000
statements.
So actually,
00:25:43.000 --> 00:25:53.000
the first observation will
actually prove something easier,
00:25:53.000 --> 00:25:58.000
namely, that the line integral,
let's see,
00:25:58.000 --> 00:26:03.000
of Mdx along a closed curve is
equal to the double integral
00:26:03.000 --> 00:26:08.000
over the region inside of minus
M sub y dA.
00:26:08.000 --> 00:26:13.000
OK, so that's the special case
where N is zero,
00:26:13.000 --> 00:26:19.000
where you have only an x
component for your vector field.
00:26:19.000 --> 00:26:23.000
Now, why is that good enough?
Well, the claim is if I can
00:26:23.000 --> 00:26:28.000
prove this, I claim you will be
able to do the same thing to
00:26:28.000 --> 00:26:33.000
prove the other case where there
is only the y component.
00:26:33.000 --> 00:26:38.000
And then, if the other
together, you will get the
00:26:38.000 --> 00:26:40.000
general case.
So, let me explain.
00:27:00.000 --> 00:27:06.000
OK, so a similar argument which
I will not do,
00:27:06.000 --> 00:27:11.000
to save time,
will show, so actually it's
00:27:11.000 --> 00:27:15.000
just the same thing but
switching the roles of x and y,
00:27:15.000 --> 00:27:20.000
that if I integrate along a
closed curve N dy,
00:27:20.000 --> 00:27:29.000
then I'll get the double
integral of N sub x dA.
00:27:29.000 --> 00:27:36.000
And so, now if I have proved
these two formulas separately,
00:27:36.000 --> 00:27:44.000
then if you sum them together
will get the correct statement.
00:27:44.000 --> 00:27:52.000
Let me write it.
We get Green's theorem.
00:27:52.000 --> 00:27:55.000
OK, so we've simplified our
task a little bit.
00:27:55.000 --> 00:28:00.000
We'll just be trying to prove
the case where there's only an x
00:28:00.000 --> 00:28:04.000
component.
So, let's do it.
00:28:04.000 --> 00:28:07.000
Well, we have another problem
which is the region that we are
00:28:07.000 --> 00:28:10.000
looking at, the curve that we're
looking at might be very
00:28:10.000 --> 00:28:12.000
complicated.
If I give you,
00:28:12.000 --> 00:28:17.000
let's say I give you,
I don't know,
00:28:17.000 --> 00:28:22.000
a curve that does something
like this.
00:28:22.000 --> 00:28:26.000
Well, it will be kind of tricky
to set up a double integral over
00:28:26.000 --> 00:28:29.000
the region inside.
So maybe we first want to look
00:28:29.000 --> 00:28:33.000
at curves that are simpler,
that will actually allow us to
00:28:33.000 --> 00:28:36.000
set up the double integral
easily.
00:28:36.000 --> 00:28:42.000
So, the second observation,
so that was the first
00:28:42.000 --> 00:28:51.000
observation.
The second observation is that
00:28:51.000 --> 00:29:02.000
we can decompose R into simpler
regions.
00:29:02.000 --> 00:29:10.000
So what do I mean by that?
Well, let's say that I have a
00:29:10.000 --> 00:29:13.000
region and I'm going to cut it
into two.
00:29:13.000 --> 00:29:18.000
So, I'll have R1 and R2.
And then, of course,
00:29:18.000 --> 00:29:22.000
I need to have the curves that
go around them.
00:29:22.000 --> 00:29:29.000
So, I had my initial curve,
C, was going around everybody.
00:29:29.000 --> 00:29:41.000
They have curves C1 that goes
around R1, and C2 goes around
00:29:41.000 --> 00:29:46.000
R2.
OK, so,
00:29:46.000 --> 00:29:55.000
what I would like to say is if
we can prove that the statement
00:29:55.000 --> 00:30:07.000
is true, so let's see,
for C1 and also for C2 -- --
00:30:07.000 --> 00:30:23.000
then I claim we can prove the
statement for C.
00:30:23.000 --> 00:30:26.000
How do we do that?
Well, we just add these two
00:30:26.000 --> 00:30:28.000
equalities together.
OK, why does that work?
00:30:28.000 --> 00:30:31.000
There's something fishy going
on because C1 and C2 have this
00:30:31.000 --> 00:30:35.000
piece here in the middle.
That's not there in C.
00:30:35.000 --> 00:30:39.000
So, if you add the line
integral along C1 and C2,
00:30:39.000 --> 00:30:44.000
you get these unwanted pieces.
But, the good news is actually
00:30:44.000 --> 00:30:47.000
you go twice through that edge
in the middle.
00:30:47.000 --> 00:30:51.000
See, it appears once in C1
going up, and once in C2 going
00:30:51.000 --> 00:30:52.000
down.
So, in fact,
00:30:52.000 --> 00:30:55.000
when you will do the work,
when you will sum the work,
00:30:55.000 --> 00:30:57.000
you will add these two guys
together.
00:30:57.000 --> 00:31:06.000
They will cancel.
OK, so the line integral along
00:31:06.000 --> 00:31:14.000
C will be, then,
it will be the sum of the line
00:31:14.000 --> 00:31:21.000
integrals on C1 and C2.
And, that will equal,
00:31:21.000 --> 00:31:29.000
therefore, the double integral
over R1 plus the double integral
00:31:29.000 --> 00:31:36.000
over R2, which is the double
integral over R of negative My.
00:31:36.000 --> 00:31:47.000
OK and the reason for this
equality here is because we go
00:31:47.000 --> 00:31:56.000
twice through the inner part.
What do I want to say?
00:31:56.000 --> 00:32:15.000
Along the boundary between R1
and R2 -- -- with opposite
00:32:15.000 --> 00:32:25.000
orientations.
So, the extra things cancel out.
00:32:25.000 --> 00:32:29.000
OK, so that means I just need
to look at smaller pieces if
00:32:29.000 --> 00:32:34.000
that makes my life easier.
So, now, will make my life easy?
00:32:34.000 --> 00:32:41.000
Well, let's say that I have a
curve like that.
00:32:41.000 --> 00:32:45.000
Well, I guess I should really
draw a pumpkin or something like
00:32:45.000 --> 00:32:48.000
that because it would be more
seasonal.
00:32:48.000 --> 00:32:53.000
But, well, I don't really know
how to draw a pumpkin.
00:32:53.000 --> 00:32:57.000
OK, so what I will do is I will
cut this into smaller regions
00:32:57.000 --> 00:33:01.000
for which I have a well-defined
lower and upper boundary so that
00:33:01.000 --> 00:33:05.000
I will be able to set up a
double integral,
00:33:05.000 --> 00:33:10.000
dy dx, easily.
So, a region like this I will
00:33:10.000 --> 00:33:17.000
actually cut it here and here
into five smaller pieces so that
00:33:17.000 --> 00:33:23.000
each small piece will let me set
up the double integral,
00:33:23.000 --> 00:33:31.000
dy dx.
OK, so we'll cut R in to what I
00:33:31.000 --> 00:33:41.000
will call vertically simple --
-- regions.
00:33:41.000 --> 00:33:43.000
So, what's a vertically simple
region?
00:33:43.000 --> 00:33:48.000
That's a region that's given by
looking at x between a and b for
00:33:48.000 --> 00:33:53.000
some values of a and b.
And, for each value of x,
00:33:53.000 --> 00:34:00.000
y is between some function of x
and some other function of x.
00:34:00.000 --> 00:34:03.000
OK, so for example,
this guy is vertically simple.
00:34:03.000 --> 00:34:07.000
See, x runs from this value of
x to that value of x.
00:34:07.000 --> 00:34:13.000
And, for each x,
y goes between this value to
00:34:13.000 --> 00:34:16.000
that value.
And, same with each of these.
00:34:39.000 --> 00:34:49.000
OK, so now we are down to the
main step that we have to do,
00:34:49.000 --> 00:35:05.000
which is to prove this identity
if C is, sorry,
00:35:05.000 --> 00:35:23.000
if -- -- if R is vertically
simple -- -- and C is the
00:35:23.000 --> 00:35:36.000
boundary of R going
counterclockwise.
00:35:36.000 --> 00:35:40.000
OK, so let's look at how we
would do it.
00:35:40.000 --> 00:35:46.000
So, we said vertically simple
region looks like x goes between
00:35:46.000 --> 00:35:52.000
a and b, and y goes between two
values that are given by
00:35:52.000 --> 00:35:57.000
functions of x.
OK, so this is y equals f2 of x.
00:35:57.000 --> 00:36:02.000
This is y equals f1 of x.
This is a.
00:36:02.000 --> 00:36:09.000
This is b.
Our region is this thing in
00:36:09.000 --> 00:36:13.000
here.
So, let's compute both sides.
00:36:13.000 --> 00:36:15.000
And, when I say compute,
of course we will not get
00:36:15.000 --> 00:36:17.000
numbers because we don't know
what M is.
00:36:17.000 --> 00:36:19.000
We don't know what f1 and f2
are.
00:36:19.000 --> 00:36:24.000
But, I claim we should be able
to simplify things a bit.
00:36:24.000 --> 00:36:28.000
So, let's start with the line
integral.
00:36:28.000 --> 00:36:35.000
How do I compute the line
integral along the curve that
00:36:35.000 --> 00:36:40.000
goes all around here?
Well, it looks like there will
00:36:40.000 --> 00:36:45.000
be four pieces.
OK, so we actually have four
00:36:45.000 --> 00:36:50.000
things to compute,
C1, C2, C3, and C4.
00:36:50.000 --> 00:37:01.000
OK?
Well, let's start with C1.
00:37:01.000 --> 00:37:06.000
So, if we integrate on C1 Mdx,
how do we do that?
00:37:06.000 --> 00:37:10.000
Well, we know that on C1,
y is given by a function of x.
00:37:10.000 --> 00:37:15.000
So, we can just get rid of y
and express everything in terms
00:37:15.000 --> 00:37:21.000
of x.
OK, so, we know y is f1 of x,
00:37:21.000 --> 00:37:27.000
and x goes from a to b.
So, that will be the integral
00:37:27.000 --> 00:37:30.000
from a to b of,
well, I have to take the
00:37:30.000 --> 00:37:33.000
function, M.
And so, M depends normally on x
00:37:33.000 --> 00:37:38.000
and y.
Maybe I should put x and y here.
00:37:38.000 --> 00:37:46.000
And then, I will plug y equals
f1 of x dx.
00:37:46.000 --> 00:37:49.000
And, then I have a single
variable integral.
00:37:49.000 --> 00:37:51.000
And that's what I have to
compute.
00:37:51.000 --> 00:37:54.000
Of course, I cannot compute it
here because I don't know what
00:37:54.000 --> 00:37:59.000
this is.
So, it has to stay this way.
00:37:59.000 --> 00:38:06.000
OK, next one.
The integral along C2,
00:38:06.000 --> 00:38:13.000
well, let's think for a second.
On C2, x equals b.
00:38:13.000 --> 00:38:16.000
It's constant.
So, dx is zero,
00:38:16.000 --> 00:38:20.000
and you would integrate,
actually, above a variable,
00:38:20.000 --> 00:38:23.000
y.
But, well, we don't have a y
00:38:23.000 --> 00:38:26.000
component.
See, this is the reason why we
00:38:26.000 --> 00:38:30.000
made the first observation.
We got rid of the other term
00:38:30.000 --> 00:38:33.000
because it's simplifies our life
here.
00:38:33.000 --> 00:38:38.000
So, we just get zero.
OK, just looking quickly ahead,
00:38:38.000 --> 00:38:40.000
there's another one that would
be zero as well,
00:38:40.000 --> 00:38:42.000
right?
Which one?
00:38:42.000 --> 00:38:52.000
Yeah, C4.
This one gives me zero.
00:38:52.000 --> 00:38:55.000
What about C3?
Well, C3 will look a lot like
00:38:55.000 --> 00:38:57.000
C1.
So, we're going to use the same
00:38:57.000 --> 00:38:59.000
kind of thing that we did with
C.
00:39:22.000 --> 00:39:27.000
OK, so along C3,
well, let's see,
00:39:27.000 --> 00:39:34.000
so on C3, y is a function of x,
again.
00:39:34.000 --> 00:39:40.000
And so we are using as our
variable x, but now x goes down
00:39:40.000 --> 00:39:45.000
from b to a.
So, it will be the integral
00:39:45.000 --> 00:39:51.000
from b to a of M of (x and f2 of
x) dx.
00:39:51.000 --> 00:39:57.000
Or, if you prefer,
that's negative integral from a
00:39:57.000 --> 00:40:04.000
to b of M of (x and f2 of x) dx.
OK, so now if I sum all these
00:40:04.000 --> 00:40:10.000
pieces together,
I get that the line integral
00:40:10.000 --> 00:40:20.000
along the closed curve is the
integral from a to b of M(x1f1
00:40:20.000 --> 00:40:30.000
of x) dx minus the integral from
a to b of M(x1f2 of x) dx.
00:40:30.000 --> 00:40:39.000
So, that's the left hand side.
Next, I should try to look at
00:40:39.000 --> 00:40:47.000
my double integral and see if I
can make it equal to that.
00:40:47.000 --> 00:40:58.000
So, let's look at the other
guy, double integral over R of
00:40:58.000 --> 00:41:02.000
negative MydA.
Well, first,
00:41:02.000 --> 00:41:05.000
I'll take the minus sign out.
It will make my life a little
00:41:05.000 --> 00:41:09.000
bit easier.
And second, so I said I will
00:41:09.000 --> 00:41:14.000
try to set this up in the way
that's the most efficient.
00:41:14.000 --> 00:41:20.000
And, my choice of this kind of
region means that it's easier to
00:41:20.000 --> 00:41:22.000
set up dy dx,
right?
00:41:22.000 --> 00:41:30.000
So, if I set it up dy dx,
then I know for a given value
00:41:30.000 --> 00:41:36.000
of x, y goes from f1 of x to f2
of x.
00:41:36.000 --> 00:41:49.000
And, x goes from a to b, right?
Is that OK with everyone?
00:41:49.000 --> 00:41:53.000
OK, so now if I compute the
inner integral,
00:41:53.000 --> 00:41:58.000
well, what do I get if I get
partial M partial y with respect
00:41:58.000 --> 00:42:02.000
to y?
I'll get M back, OK?
00:42:02.000 --> 00:42:19.000
So -- So, I will get M at the
point x f2 of x minus M at the
00:42:19.000 --> 00:42:27.000
point x f1 of x.
And so, this becomes the
00:42:27.000 --> 00:42:35.000
integral from a to b.
I guess that was a minus sign,
00:42:35.000 --> 00:42:45.000
of M of (x1f2 of x) minus M of
(x1f1 of x) dx.
00:42:45.000 --> 00:42:50.000
And so, that's the same as up
there.
00:42:50.000 --> 00:42:54.000
And so, that's the end of the
proof because we've checked that
00:42:54.000 --> 00:42:58.000
for this special case,
when we have only an x
00:42:58.000 --> 00:43:01.000
component and a vertically
simple region,
00:43:01.000 --> 00:43:04.000
things work.
Then, we can remove the
00:43:04.000 --> 00:43:07.000
assumption that things are
vertically simple using this
00:43:07.000 --> 00:43:10.000
second observation.
We can just glue the various
00:43:10.000 --> 00:43:13.000
pieces together,
and prove it for any region.
00:43:13.000 --> 00:43:17.000
Then, we do same thing with the
y component.
00:43:17.000 --> 00:43:22.000
That's the first observation.
When we add things together,
00:43:22.000 --> 00:43:29.000
we get Green's theorem in its
full generality.
00:43:29.000 --> 00:43:39.000
OK, so let me finish with a
cool example.
00:43:39.000 --> 00:43:47.000
So, there's one place in real
life where Green's theorem used
00:43:47.000 --> 00:43:51.000
to be extremely useful.
I say used to because computers
00:43:51.000 --> 00:43:53.000
have actually made that
obsolete.
00:43:53.000 --> 00:44:02.000
But, so let me show you a
picture of this device.
00:44:02.000 --> 00:44:12.000
This is called a planimeter.
And what it does is it measures
00:44:12.000 --> 00:44:17.000
areas.
So, it used to be that when you
00:44:17.000 --> 00:44:23.000
were an experimental scientist,
you would run your chemical or
00:44:23.000 --> 00:44:27.000
biological experiment or
whatever.
00:44:27.000 --> 00:44:29.000
And, you would have all of
these recording devices.
00:44:29.000 --> 00:44:32.000
And, the data would go,
well, not onto a floppy disk or
00:44:32.000 --> 00:44:35.000
hard disk or whatever because
you didn't have those at the
00:44:35.000 --> 00:44:37.000
time.
You didn't have a computer in
00:44:37.000 --> 00:44:39.000
your lab.
They would go onto a piece of
00:44:39.000 --> 00:44:42.000
graph paper.
So, you would have your graph
00:44:42.000 --> 00:44:46.000
paper, and you would have some
curve on it.
00:44:46.000 --> 00:44:48.000
And, very often,
you wanted to know,
00:44:48.000 --> 00:44:51.000
what's the total amount of
product that you have
00:44:51.000 --> 00:44:54.000
synthesized, or whatever the
question might be.
00:44:54.000 --> 00:44:58.000
It might relate with the area
under your curve.
00:44:58.000 --> 00:45:01.000
So, you'd say, oh, it's easy.
Let's just integrate,
00:45:01.000 --> 00:45:02.000
except you don't have a
function.
00:45:02.000 --> 00:45:05.000
You can put that into
calculator.
00:45:05.000 --> 00:45:07.000
The next thing you could do is,
well, let's count the little
00:45:07.000 --> 00:45:09.000
squares.
But, if you've seen a piece of
00:45:09.000 --> 00:45:12.000
graph paper, that's kind of
time-consuming.
00:45:12.000 --> 00:45:14.000
So, people invented these
things called planimeters.
00:45:14.000 --> 00:45:19.000
It's something where there is a
really heavy thing based at one
00:45:19.000 --> 00:45:23.000
corner, and there's a lot of
dials and gauges and everything.
00:45:23.000 --> 00:45:25.000
And, there's one arm that you
move.
00:45:25.000 --> 00:45:30.000
And so, what you do is you take
the moving arm and you just
00:45:30.000 --> 00:45:35.000
slide it all around your curve.
And, you look at one of the
00:45:35.000 --> 00:45:37.000
dials.
And, suddenly what comes,
00:45:37.000 --> 00:45:41.000
as you go around,
it gives you complete garbage.
00:45:41.000 --> 00:45:45.000
But when you come back here,
that dial suddenly gives you
00:45:45.000 --> 00:45:48.000
the value of the area of this
region.
00:45:48.000 --> 00:45:51.000
So, how does it work?
This gadget never knows about
00:45:51.000 --> 00:45:55.000
the region inside because you
don't take it all over here.
00:45:55.000 --> 00:45:57.000
You only take it along the
curve.
00:45:57.000 --> 00:46:00.000
So, what it does actually is it
computes a line integral.
00:46:00.000 --> 00:46:04.000
OK, so it has this system of
wheels and everything that
00:46:04.000 --> 00:46:08.000
compute for you the line
integral along C of,
00:46:08.000 --> 00:46:11.000
well, it depends on the model.
But some of them compute the
00:46:11.000 --> 00:46:14.000
line integral of x dy.
Some of them compute different
00:46:14.000 --> 00:46:17.000
line integrals.
But, they compute some line
00:46:17.000 --> 00:46:21.000
integral, OK?
And, now, if you apply Green's
00:46:21.000 --> 00:46:26.000
theorem, you see that when you
have a counterclockwise curve,
00:46:26.000 --> 00:46:31.000
this will be just the area of
the region inside.
00:46:31.000 --> 00:46:34.000
And so, that's how it works.
I mean, of course,
00:46:34.000 --> 00:46:36.000
now you use a computer and it
does the sums.
00:46:36.000 --> 00:46:39.000
Yes?
That costs several thousand
00:46:39.000 --> 00:46:43.000
dollars, possibly more.
So, that's why I didn't bring
00:46:43.000 --> 00:46:44.000
one.