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Hi there.

Now our very important issues we proceed to the proof of the theorem.

Recently we have seen thus on a closed curve

Spot t is clear that the integral of s

Once there, type u d x d y plus there once.

One area where this cycle off equal to the integral comes on.

The importance of this a little more We will highlight later.

Suffice it now to prove.

A simple way of proving want to go to.

Of course a general cycle It is such an irregular region.

We rectangle is on the proof will do.

I will say that in general such.

As we take a rectangular region.

from A to B, from B to C, C to D, d in a closed loop to the Dee.

On the x and b its boundaries on y of between c and d.

And the point t We will account for the integral.

Now let's draw this shape again here.

The first half of Green's theorem we

we observed as speculative intended to prove.

have u d y d x plus the more we did before the first

x minus y in the sample have to the area of ??integration.

I say again.

If x is equal to y that you have here As seen would be zero.

This orbit is already consistent b is independent of the integral

potential was happening, then this The term is an exact differential was.

Zero was giving.

We prove this theorem in with.

Now that we start from a b

cycle goes on the wheel Let's examine one by these terms.

Now, b is above request, v d d x plus y.

Then we're going to b c, We're going to d c.

We wrote it separately.

Nothing we do here.

Now they only each individual What's on this one confirmed

We will see that.

As you can see on a b y fixed.

d y to y is constant is zero.

Therefore, this second term is reduced.

The first term will remain.

u x, y, where y d y, but no random Not one year, not a variable y.

y also equal to c.

We're doing that too.

So it's quite an easy thing.

Wherein a and b are in integral the value of x at the points,

going a little smaller than b.

As you can see single storey was reduced to a simple integral.

We have already seen this kind of integration.

In the second part we are going from b to c.

this time going from b to c, x fixed.

Therefore x fixed which is zero for d *.

Here we come.

Where d x is zero this term is falling.

That just leaves the second term and that only the integral of y.

y from c to d also going up.

It's easy.

There is an extra.

where x is a random thing.

x is equal to b.

We are writing to bring it on.

As you can see each integral a single variable, includes.

Now we're looking to d c.

y is equal to a constant c d again.

So on a br br equivalent quinine.

D y where y is fixed at zero.

So the second term this fall.

First term stays.

Y in the first term not variable anymore.

Y represents a d.

Is equal to the value given.

Aa lastly than d, while the where x is still not fixed.

equal to Xa.

Therefore, d x is zero.

Hence, our integral in our first fall term, the second term stays.

We are putting in place of x in the second term.

Therefore, we closed curve that on a closed rectangular

The total on this cycle The sum of these four integral.

I need to pay attention to the little things.

We're going from A to B.

from A to B, from a greater than b.

But in a place you're going to a from b.

Why?

Because of this cycle, we According to an observer in

plus the return cycle trigonometric direction.

Hence, we are going from A to B, from c to d while here we go.

B of A came here for him.

However, from A to B here.

So it by a minus sign will be merged with the other integrals.

B because both boundaries.

In this we make a b, it will come with a minus sign.

A similar situation exists in the second pair.

First, we're going from b to c.

I.e., from C to B value going our borders.

However, while on d Coming from d to c.

See here different limits.

But the sequence from c to d better go.

We know it's also an integral feature When you change the boundaries thus a

revenue minus sign.

In this way it will combine.

Now we do these four integral to combine twos.

One is of x, up to a value of x b.

Where u x c had plus,

have u x is minus.

To get out of this negative attribute, You will see why soon.

From c to d wherein Y when the first term

We have changed the sign in the second term.

I remind again.

From here refer to, There are two integral a'yl of b b.

But one of Ã¶bÃ¼rkÃ¼ from A to B. B of A where this

d'lysine minus sign the need to take.

Indeed d'liu minus we put out negative mark

Ã¶bÃ¼rkÃ¼ therefore also positive.

In this case there are two similar terms.

Already been great here, a greater than b great, greater than c.

Now we like them We can evaluate.

Inasmuch as these integrals under the terms of the two difference

y x y is the integral of C to D is the value.

x do not do anything on anything.

However, before the year is We take the x're getting.

This is u to y integral means of derivatives.

We know that.

A derivative of the integral If we will get him.

The upper limit value to minus would be worth following.

Wherein the first term means d, is replaced by y.

Here are a well d *.

It is next to a future year.

Let's look at the second term it has the same structure.

See wherein B and A's.

V x and y that make the value of x in b Get the value of the first term we produce.

Less in the following.

When a sheep instead of x We are producing a second term.

Yet we know that the derivative the same function of the integral of

As produced this analysis, second fundamental theorem of differential calculus.

The first one will recognize the if you receive an integral,

derivative if you receive it again it gives the same function.

But here we do not use it, We use the latter.

Gives the integral of the derivative function.

Gives the integral of the derivative function.

Now let's put them to take place.

In the first, see x an integral over there.

The second term in the square brackets If the second term he has been an integral.

d u d y y integral over.

So here, two-storey we're getting into an integral.

Our current second term of this integral.

The minus sign here is obvious.

d u d y divide.

Here there was already an d x.

The other came a year.

We're looking at the second term.

This is the second in brackets There was an integral y on d.

It's in the brackets as an integral over x.

He d v d * is happening.

But this comes plus valuable.

For this reason, it is particularly negative outside We bought it so that it fits comfortable place.

As you can see from the right terms theorem we want to find expression.

I repeat, if this integral if it was independent from orbit,

d v d d u d y x minus would be zero.

This would be zero.

Orbit is an independent,

In the integral on closed curves, In the integral cycle, it is zero.

I.e. zero or zero We have provided the back.

But we have observed something much more important.

Are independent from orbit, sorry orbit If it is, this integration will be zero.

This time is not zero two integrals are equal.

I.e., the integral over a cycle,

Close this cycle onto the field We make an integral equal.

There is a second way to prove it.

He will not go into that detail.

But for those interested in a good exercise in terms of differential calculus.

This region p k, say p q r s.

This p q r s in Let's take small sections.

Each time, when the integral calculation We were already doing this.

Let any one of them.

a b c d say its name.

Let's account of this cycle.

A typical one of them.

Then let's get them side by side.

As juxtaposes by An interesting thing happens.

Refer next to a b c d A similar gene

on the rectangle that Let us online account.

This cycle is always the According to observers inside

to navigate in the trigonometric sense, Because returned from B to C

rectangle on the left when the We're going upward.

But if the second rectangle going down to the bottom.

So, in this intervening the same value to each other.

Destroying each other.

Consequently, all these rectangles we did on account

At the end of all of the intervening cancel out.

Only those walls remain.

This expression of individual regions of the small

thinking that the Taylor series When opened, i.e. from an x k,

plus delta x k x k, j y j from one year plus delta

distances up to y j, expansions,

from thinking that they were small The first term of the expansion work

for we can see them as they collected gives the same effect as the rest.

It to the attention of interested I want to leave.

This is slightly longer, but In differential calculus

To insert thinking important.

We're getting the frontmost terms,

of the delta x and delta y thinking it was small.

When you collect them, This comes from the Taylor series,

a second term, The first takes each other.

Here you have one.

Bi, do u have a minus.

Here you have a van.

Necessarily have a negative V'la.

These first-order terms cancel out.

But back to the Taylor series expansion The term therefore more second deltal

come on an area of ??an infinite occurs spontaneously small area.

When they collect this area of ??integration is coming.

Here there's the x by x v is negative partial derivative of u

According to y on an area of partial derivatives The sum of these gives the integral.

This Ã§ettek on walls integrals of the intervening

Because no external sounds and theorem can be proved in this way.

In that the overall may prove to be used.

Restricted to the rectangle We have not impair the general.

Now we've done on u t d s.

We did this on the environment Touring.

n components minus d

Because the x and d y in this way,

I'm sorry I said it wrong, Because minus d y and d * D

with a plus sign in y'yl multiplied, have multiplied with X minus d.

We've seen it many times.

It is also possible to prove theorems.

It is also possible to separate proof.

But this two Green's theorem from each other as shown.

Because of the integral on the V. If u translation,

If you turn on u to v this would be something to be gained.

So are you putting in place there, rather than You can also put a minus there, it turns into.

We also do not need a proof.

Because one plane able to obtain from the other.

The variable component to the transformation.

Now a combination of these two theorems In order to examine the differences we see.

From the projection on the tangent The integral, the first,

the inner product of u'yl t you know, gives a projection of the inner product.

And it stands, the

t is the expansion of s d.times.d for giving the vector y,

d x and d y components of the vectors, The first component is the product of x,

y with the second component is multiplied and have achieved this theorem.

Look to the left side,

When we look at the following, where a In terms have a minus sign on the left.

There and in terms of the right and displaced is also the case.

See u come here instead of there.

So it will have X instead of x.

Minus any place, there instead of here comes the negative.

So instead of it there is a minus Put a minus as well have it here.

v x is happening.

As you can see both of them There are similarities as well as differences.

Mathematically, it looks similar an area where it finally integral,

this is an area integral.

A conversion integral, This integral bi cycle.

But behind their sizes are very different.

In one of these for the environment You're going integrally account.

One example of the environment by the components to the other side.

Here is a given vector.

This is the vector t The projection of the time t.

This gives the first integral of this.

The second of these vectors on the projection.

What's on the limit which means that, How is it will give the application.

Now two simple examples I'll cover that bet.

But this theory thoroughly a place in mind

order to review I would advise you to spend.

Our next session some examples will do.

Bye for now.