Hi our previous session

in the plane of two variables function of the

integrally on the trajectory from the three We have seen how to move dimension.

Further wherein the three types We have identified the problem.

Theorem Stoks second type, the third type Green-

Was the implementation of Gauss's theorem.

We are now, of course, the first We start with the type of problems.

The first problem is a problem as follows, an x, y, z given function

that the integral over a curve We want the integral over ds,

As the function e to the minus z x squared plus y squared multiplied by the

DSi will go from A to B on the integral, How are we going, given the road.

Road along a helix.

He defined how the helix?

A circle with a radius Are you going to step on the b

See x squared plus y squared yi When you receive a frame involved,

Would you say that going on a circle x-y plane in the plane of the projection

but also in terms of the z direction Are you constantly rising down

you can see the top of this DNA in a biology

As one of the double helix,

proteins, such as helical or We saw so much easier than the current issues,

a minaret rising above when riding on a circle longitudinal

can rise, Where If you are asked where to go?

from the zero point zero.

T is zero point zero zero mean point is zero because the intermediate compound in sinus t

t is equal to zero means zero zero When we returned back to zero again in y's

See two pi times the rise b t really say that we're going on two pins,

providing two pins as t When a cosine two pins

sinus is a d are two pi because it is zero is zero.

So we compute this integral t

of zero to two pi We'll find much change.

Do this supremely simple,

where we saw each term Instead of going yereştir.

At first we calculate the DS one.

point x in the plane of square frames years has DSi points.

Dots derivatives with respect to t.

Here, too, there is a point in the z or of the position vector of the vector x to t

do this by taking the derivative We will account for the length of the value,

norms will account.

See derivative of x with respect to t comes from the negative sine cosine,

t is derived from the cosine-sine coming that is staying here for a just a little.

Means taking the length thereof, first component squared plus

The second component of the frame components plus the square of the third component.

See the first two components of the square of a t squared plus a squared cosine squared sine-squared

t is a frame only for because it comes sine cosine squared t

b is a positive frame for the frame t To facilitate the writing of a frame plus

See if b squared c squared square root of the c dt ds will be times that we receive.

We will take them into place, from A to B is equal to T

zero meant the opposite, b t is equal to two pie was coming the opposite.

Where z instead of Z We will have placed him bt,

there is a square cosine of x squared t have a square plus y square,

a squared sine-squared t, easy to get here, of course, he calculates

which is important because this kind of mixed not to fiddle with the accounts to understand the concept

sine squared plus cosine squared t here, as a t

comes a square a square hard Because you're getting out,

c dt ds have found here, too, he also c We take it out to get that fixed,

bt just stayed inside to minus from zero to two pi dt stayed.

Of course, this easy integration; a square c here is the integral of e to the minus CT,

bt minus minus b to divide.

Less integral because it is the order of We're changing plus translation.

Means that t is equal to zero before We'll find t is equal to zero would be in

because it is above zero.

E pide two pi over two times b As you can see very easily found

too much of a bivariate vector No difficulty of this integral in space.

The second integral given the following structure, slightly different structure.

In a previous DSi directly had here in the integral, t

Once the product becomes u.

The first vector component E z times over the year,

The second component to the minus x minus x, the xy third component.

This is where are we going?

Here you need to define the curve.

We will go from A to B, but what ways We need to know that we're going.

This is called double parabola with this function,

components of the position of the x component t y component of the vector

t squared divided by two, z component of the cube at t divided by three.

Here, double parabola, The reason is called double parabola

a parabola in the xy plane There xz projection as well as

There is a parabola percent even There are also a parabola.

Tr y to see that if you say that x y will be x squared divided by two,

a parabola.

When you come here a divided our When the three x cube cubic prune

will be a parabola, geometry, such but just that as the account

where x is also not so important to know, y, z will yereştir what we see,

There dx means that x t would be here by dx dt.

T square by one-half years that the dye, dt dx becomes.

t be dt times, z a t divided by three cubes by that t would be dt dz square too.

DX will put them to take place, dy, dz t.

Easy to them via e z z,

that one-third of our cube by t here you via e y to z, we put

with a higher z to y multiplied by the DX.

Y is a square divided into two t but there are also dt dt

the common factor in all that outside as we hoarded.

Second term e to the minus x, According to the minus x xt that

e to the minus t dy, dy dt is the time we find the t.

So here goes t.

Here again is a common factor dt x times

There are times dz y, x times y t one-half cubic

dz is square and t is t d in five t that's finally an integral,

ha to know the limits We're going to need from A to B.

Look here not given t but the integral over t.

See a zero b initially zero,

a is the y component of the x component zero in the z-component.

Following means that t is zero, a point.

B above us at the point of two thirds given two to eight,

Let's look on this curve?

If This-

that if we do not provide the point b a t not on the curve, but really to t

If you see two data here two t squared divided by two to four or two,

eight divided by three in the z-frame, means that b point of this curve gives the equation,

therefore, say that on the curve the integral from zero to two,

If you want to deal with now that the integral This important aspect of the problem is

but if you say u t cube cube t, d Deliver Please buluyor t is squared.

So here's this integral to do.

Where e is there now minus t times t wherein the partial done by integrals,

If you say that you have the base of this request,

there are times minus any base,

have to be given to a base above t minus the integral means

This conventional fractional integration is done through.

T in five of the integral t easy six divided by six

place them here by mistake have not done this you will find the number.

What is important to establish this integration, After him an account of this integral.

That's a good thing to do it, but this problem is that the basis weight

integral can be edited.

The third integral as follows: A given vector,

Calculation of the integral dx u once asked, but that,

integrally connected to the orbit 'll determine whether,

u is independent from orbit potential are saying.

Here the two variables.

Do you know the fit criteria were negative vx.

The third component is really here was not out of here,

but when the third component rotational mean, you're writing ijk'y,

radians of the components dx, dy, dz are writing.

After that u We are writing to bring components.

The components shown in zxy, zxy.

Now i see here To find components

we are taking the first line, We take the first column.

Y is a derivative of y minus derivative of x by z,

x is derived by z zero So one is staying here.

When we arrived we start with j axis,

plus or minus signs plus As for going.

Gene first row and j We take the column.

See here x by y by derivative minus derivative of z z.

But in the jar so that the negative staying here for a.

Already shows the following criteria, If zero rotational

independent of the orbit, If nonzero dependent orbit.

Already one from 0 enough to be different.

But just to finish third problem here we're calculate component.

Any number already nonzero clear that although the zero vector.

So it's our basic criteria.

Necessity and sufficiency Question

the condition is not provided.

Therefore, this integral depends on the orbit.

Hence there is no potential of the.

Therefore, this problem We do not calculate too.

Is connected to its orbit not given to orbit.

Therefore the problem, integral is undefined, it can not calculate.