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