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# Green's theorem proof part 1 | Multivariable Calculus | Khan Academy

Let's say we have a path in the xy plane.
That's my y-axis, that is my x-axis, in my path
will look like this.
Let's say it looks like that; trying to draw a bit of an
arbitrary path, and let's say we go in a counter clockwise
direction like that along our path.
And we could call this path-- so we're going in a counter
clockwise direction --we could call that path c.
And let's say we also have a vector field.
And our vector field is going to be a little unusual;
I'll call it p.
p of xy.
It only has an i component, or all of its vectors are only
multiples of the i-unit vector.
So it's capital P of xy times the unit vector i.
There is no j component, so if you have to visualize this
vector field, all of the vectors, they're all multiples
of the i-unit vector.
Or they could be negative multiples, so they could
also go in that direction.
But they don't go diagonal or they don't go up.
They all go left to right or right to left.
That's what this vector field would look like.
Now what I'm interested in doing is figuring out the line
integral over a closed loop-- the closed loop c, or the
closed path c --of p dot dr, which is just our standard
kind of way of solving for a line integral.
And we've seen what dr is in the past.
dr is equal to dx times i plus dy times the j-unit vector.
And you might say, isn't it dx, dt times dt?
Let me write that: can't dr be written as dx, dt times
dti plus dy, dt times dtj?
And it could, but if you imagine these differentials
could cancel out, and you're just left with the dx and
a dy, and we've seen that multiple times.
And I'm going to leave it in this form because hopefully, if
we're careful, we won't have to deal with the third
parameter, t.
So let's just look at it in this form right here with
just the dx's and the dy's.
So this integral can be rewritten as the line integral,
the curve c-- actually let me do it over down here.
The line integral over the path of the curve c of p dot dr. So
we take the product of each of the coefficients, let's say the
coefficient of the i component, so we get p-- I'll do that in
green, actually do that purple color --so we get p of xy times
dx plus-- well there's no 0 times j times dy; 0 times dy id
just going to be 0 --so this our line integral simplified
to this right here.
This is equal to this original integral up here, so we're
literally just taking the line integral around this path.
Now I said that we play our cards right, we're not going to
have to deal with the third variable, t; that we might be
able just solve this integral only in terms of x.
And so let's see if we can do that.
So let's look at our minimum and maximum x points.
That looks like our minimum x point.
Let's call that a.
Let's call that our maximum x point; let's call that b.
What we could do is we can break up this curve into
two functions of x. y is functions of x.
So this bottom one right here we could call as y1 of x.
This is just a standard curve; you know when we were just
dealing with standard calculus, this is just you can
imagine this is f of x and it's a function of x.
And this is y2 of x.
Just like that.
So you can imagine two paths; one path defined by y1 of x--
let me do that in a different color; magenta --one path
defined by y1 of x as we go from x is equal to a to x is
equal to b, and then another path defined by y2 of x as we
go from x is equal to b to x is equal to a.
That is our curve.
So what we could do is, we could rewrite this integral--
which is the same thing as that integral --as this is equal to
the integral-- we'll first do this first path --of x going
from a to b of p of x.
And I could to say p of x and y, but we know along this
path y is a function of x.
So we say x and y1 of x.
Wherever we see a y we substitute it with
a y1 of x, dx.
So that covers that first path; I'll do it in the same color.
We could imagine this is c1.
This is kind of the first half of our curve-- well it's not
exactly the half --but that takes us right from that
point to that point.
And then we want to complete the circle.
Maybe I'll do that, and I'll do that in yellow.
That's going to be equal to-- sorry we're going to have to
add these two --plus the integral from x is equal to b
to x is equal to a of-- do it in that same color --of p of x.
And now y is going to be y2 of x.
Wherever you see a y, you can substitute with y2
of x along this curve.
y2 of x, dx.
This is already getting interesting and you
might already see where I'm going with this.
So this is the curve c2.
too I think you appreciate if you take the union of c1 and
c2, we've got our whole curve.
So let's see if we can simplify this integral a little bit.
Well one thing we want to do, we might want to make
their end points the same.
So if you swap a and b here, it just turns
the integral negative.
So you make this into a b, that into an a, and then make that
plus sign into a minus sign.
And now we can rewrite this whole thing as being equal to
the integral from a to b of this thing, of p of x and y1 of
x minus this thing, minus p of x and y2 of x, and then
all of that times dx.
I'll write it in a third color.
Now, I'm going to do something a little bit arbitrary, but I
think you'll appreciate why I did this by the end of this
video, and it's just a very simple operation.
What I'm going to do is I'm going to swap these two.
So I'm essentially going to multiply this whole thing by
negative 1, or essentially multiply and divide
by negative 1.
So I can multiply this by negative 1 and then multiply
the outside by negative 1, and I will not have changed the
integral; I'm multiplying by negative 1 twice.
So if I swap these two things, if I multiply the inside times
negative 1, so this is going to be equal to-- do the outside
of the integral, a to b.
If I multiply the inside-- I'll do a dx out here --if I
multiply the inside of the integral by negative 1,
these two guys switch.
So it becomes p of x of y2 of x.
And then you're going to have minus p of x and y1 of x.
My handwriting's getting a little messy.
But I can't just multiply just the inside by minus 1.
I don't want to change the integral, so I multiplied the
inside by minus 1, let me multiply the outside
by minus 1.
And since I multiplied by minus 1 twice, these two
things are equivalent.
Or you could say this is the negative of that.
Either way, I think you appreciate that I haven't
changed the integral at all, numerically.
I multiplied the inside and the outside by minus 1.
And now the next step I'm going to do, it might look a little
bit foreign to you, but I think you'll appreciate it.
It might be obvious to you if you've recently done
some double integrals.
So this thing can be rewritten as minus the integral from a to
b of-- and let me do a new color --of the function p of x,
y evaluated at y2 of x minus-- and let me make it very clear;
this is y is equal to y2 of x --minus this function evaluated
at y is equal to y1 of x.
And of course all of that times dx.
This statement and what we saw right here-- this
statement right here --are completely identical.
And then if we assume that a partial derivative of capital P
with respect to y exists, hopefully you'll realize-- and
I'll focus on this a little bit because I don't want to
confuse you on this step.
Let me write the outside of this integral.
So this is going to be equal to-- and this is kind of a neat
outcome, and we're starting to build up to a very neat
outcome, which will probably have to take the next video to
do --so we do the outside dx.
If we assume that capital P has a partial derivative, this
right here is the exact same thing.
This right here is the exact same thing as the partial
derivative of P with respect to y, dy, the antiderivative of
that from y1 of x to y2 of x.
I want to make you feel comfortable that these two
things are equivalent.
And to realize they're equivalent, you'll probably
just have to start here and then go to that.
We're used to seeing this; we're used to seeing a double
integral like this, and then the very first step we say, OK
to solve this double integral we start on the inside integral
right there, and we say, OK let's take the antiderivative
of this with respect to y.
So if you take the antiderivative of the partial
of p with respect to y, you're going to end up with p.
And since this is a definite integral, the boundaries are
going to be in terms of x, you're going to evaluate that
from y is equal to y2 of x, and you're going to subtract from
that y is equal to y1 of x.
Normally we start with something like this, and we
go to something like this.
This is kind of unusual that we started, we kind of solved, we
started with the solution of the definite integral, and then
we slowly built back to the definite integral.
So hopefully you realize that this is true, that this is
just we're kind of going in a reverse direction
than we normally do.
And if you do realize that, then we've just established
a pretty neat outcome.
Because what is this right here?
Let me go back, let me see if I can fit everything.
I have some function-- and I'm assuming that the partial of
P with respect to y exists --but I have some function
defined over the xy plane.
You know, you could imagine we're dealing in
three dimensions now.
We'll draw a little bit neater.
So that's y, that's x, that's z, so this, you could imagine,
is some surface; it just happens to be the partial
of P with respect to x.
So it's some surface on the xy plane like that.
And what are we doing?
We're taking the double integral under that surface,
around this region.
The region's boundaries in terms of y are defined
by y2 and y1 of x.
So you literally have that curve.
That's y2 on top, y1 on the bottom.
And so we're essentially taking the volume above.
So if you imagine with the base is-- the whole floor of this is
going to be the area inside of this curve, and then the height
is going to be the function partial of P with respect to y.
It's going to be a little hard for me to draw, but this is
essentially some type of a volume, if you want to
visualize it that way.
But the really neat outcome here is if you call this region
r, we've just simplified this line integral.
And this was a special one.
It only had an x-component, the vector field, but we've just
simplified this line integral to being equivalent to-- maybe
I should write this line integral because that's what's
the really neat outcome.
We've just established that this thing right here, which
is the same as our original one, so let me write that.
The closed line integral around the curve c of p of xy, dx,
we've just established that that's the same thing as the
double integral over the region r-- this is the region
r --of the partial of P with respect to y.
And we could write dy, dx, or we could write da, whatever you
want to write, but this is the double integral
over that region.
The neat thing here is using a vector field that only had an
x-component, we were able to connect its line integral to
the double integral over region-- oh, and I forgot
something very important.
We had a negative sign out here.
So this was a minus sign out here.
Or we could even put the minus in here, but I think you
get the general idea.
In the next video, I'm going to do the same exact thing with
the vector field that only has vectors in the y-direction.
And then we'll connect the two and we'll end up
with Green's theorem.