Hi there. Open or closed curves in the plane n t and u on the u We saw integration. In particular, closed-loop, In practice very often. Already in the complex-valued function on this closed loop, On closed curves, cycles We have important information on. As now, a little semi-speculative I want to start with something. We have two such integral. u t u n d s and d p. If these integrals values ??independent from scratch, This orbit is independent of integrals will be zero. It stands for u t, To calculate D x, v d y. u n d y, and d stands for n minus x because it has components, In this way they are going to zero. I also know:When these be independent of the orbit of this integral? If u y v x is equal to the first type, If so there is x minus y equals zero. In the second type, see u v's thereof, minus its derivative with respect to x. This is the minus minus minus one there, plus it also to y derivative is equal to zero. We have seen them. Cauchy-Riemann them anyway conditions as complex valued function emerged. We can think of this speculative as:I wonder if we take them While we integrate over an area, Do something useful out? V x minus Y obviously to be zero, which translate this c If we say that the area of, so this will be zero by definition. D will receive a zero on the integral. Similarly, wherein the zero We'd take over an area integrated. Now, what we have found? That the first integral so that on cycle The integration of this cycle turn on the field, wherein using the components is an integral zero. Similarly here. Now that's one I so we s two is equal to that of a base I so this is what we say n'l from Integral is equal to the integral of this. Now this is a coincidence? Or under it more Is there something deeper? Do you have a more general knowledge? Them or are they just Is it because resets? This zero, zero more of here Do not say anything that comes to mind questions. Of course, in science before already After beginning with the appropriate estimation by proving progress has many time. My yalaÅ this inductive, We say that the inductive approach. Describing the deductive is told through but, to see this sort of thing important from an educational point. Now we have two examples Let's try it on. The characteristic of this example the following:This is the first example with x and y components minus We will see in the example the cycle, t projecting cycle different from zero. I wonder if that same value for the right side of Is it going to be different from zero? We want to do this. A second test we want to do more. It is here that an x components with x and y in the plane vector fields. They are coming out of here, for example as a cycle tornado Or a wind cycle Solar storms, such as the Gorse them online How we express? An angular rotation speed If the direction of the k omega in the direction perpendicular to the plane, Combining the starting point Get the vector by vector multiplication See As we come to a conclusion: Means multiplication of vectors i j k We are writing the first line. In the second row vector, k vector subject to x and y components are zero, z directions and x is a vector component in the plane, Because the x and y components. The third component is zero. I see it when we open component i of the row We take temporary. Columns are taking. Zero. If the product on the second diagonal y. But we put it by changing the sign, minus y. The second component, component j, and the column j We take the line temporarily. We start with a minus sign. See the product is reset to zero before. Zero. Minus the product of the merger of x. x. But there is a minus because of this jar. Because the plus signs minus plus he is going. Plus of minus two as we find it. This means that the vector field indicates a rotation types. If we draw here that at each point The angular velocity vectors from speed As you can see judging as vectors In all of these have an angular velocity omega. Apartment for her in this sector area There are parts of equal area. But is increasing with distance The heights of these vectors. In a second area, let us take the x and y, When we stopped at any point in this extending away from a center we see that the vectors. If that physical sample and, an electrical charge to such here about putting the electric field Or water goes in this direction supply and water if I put your fÄ±ÅkÄ±rtsa there is spread in the same way in every direction. In an idealized situation, of course. Now two of them, we can even say that the opposite of each other Two field vector field. Now let's get them. We're taking this first vector field. Here u t will take time. I also had this one. On this account, will this cycle. What is on this cycle? 're Getting an apartment. On the center of this circle, coordinates of the center at the start of the radius a circle with a're getting. This physical connection hani If you want to make this angle, omega-times proportionally over time. Under such a bi There is also a realistic model. Wherein the vector field We have seen that be the case. If we write it in circular coordinates, y is obviously Once sinus t, r times the cosine of x t. r taken out of this circle, other than the circumferential direction tangent vector This is not something negative sine and cosine t. So with such a return, with cycles are concerned. This account is easy to do. Here is a cosine of x t, y t sinus, d * certain, specific dye, d x, take them to the point If there is repositioning or u d x plus y, As you can see here first with the first, with the first, ie with negative y of this second sign when it is substituted, and the second component of x Multiplied by d y as you can see here, There are a sine here. There is a sine of x. In both the minus sign, square is a square sine of theta is happening. ÃbÃ¼rkÃ¼ cosine squared in the square. By combining them A square is going theta. Therefore, a supremely I Calculating easy. Taking out a square, is the theta integrally involved two pi squared. Now this same process once here we see the following:orbit. Although independent, this closed-loop would be zero on. Not independent from orbit integral. Thus there is x minus y will not be zero. It would also have provided already. Y did u see the negative. It is v's of the x,. We v is xa We're taking the derivative. there is an x. Less, it's the end of the first We're taking the derivative with respect to y. There are also a minus here. Plus two going. Close this circle is twice going out of the circle area. Because the area of ??the integral gives d. We know him. PI is a square area of ??the circle, of course. Two times pi squared, As you can see they came out equal. So at the beginning of the semi-speculative Is it just resets wonder if our forecast We generally do not apply We have seen that be true. Of course, this is not a proof. But this is a routing I wanted to give it to. Science is sometimes ensues, going on an estimate of you can see that to be true. This cycle is perpendicular to the flux we take, I base so this is what we call once u n d s, where the integral of the negative d x and d y components because it has once u d y d x plus minus have times going. Here in our chosen fields From this center, the centrifugal field. We will take this place. Similarly, u x v y of this We will also account. If you are independent from orbit have u x plus y would be zero. Therefore, this would be zero, here would be zero. But the orbit is bound to be seen. Because they can not be zero. Now again in the same manner x of y in a cosine A sinus d * d n times of the year and receive When account here again a square d theta involved. Yet out of a square take two theta zero a square when he calculated pi pie turns two. But here's a different area. X minus y in the previous field had If x and y here. In dealing with this area When the orbit is independent from zero v The size of x plus y was happening. This integration account If we refer X X, partial derivative with respect to x X X a. Plus there is y. partial derivative of y to y, then a. As you can see twice the area of this circle close, Turns out that area of ??the circle. The area of ??the circle of the pin We know that a square. So two times pi squared As it equally. Now you have calculated? We zero integral of the two We know that they are equal. Bun, which are independent from orbit This loop integral is zero. Because you are starting from an a, is independent from orbit When you go to a b only The two extreme values ??will be the difference. If you come to a repeat, from the value in a You're doing a mean value in the. Zero was happening to her. This necessary and sufficient on condition that there was x minus y. This integral zero I have to wonder at seeing that I always equal to half a sturdy? So when it's not zero equal? That is independent of orbit when it's not equal? Similarly, the integral of u at s I wonder if these two were equal to the call, it turns equal zero, when it's not zero equal? Now we have to make a general proof but on two specific areas these areas are very basic. One of them shows a complete turning a in the form of a tornado wind or that it returned in a water We model this kind of cycles that. The second example, the opposite. No return here. All currents from a center of all, all vector of a vector field away. Here n u point this with integrally on cycle of that divergence, since X plus there is the divergence of the year it happens. We have seen that it is equal to two. Now we can ask the following question: I have found them. Two examples also on We saw equality. We saw an example of equality on. But what a coincidence that they We can ask whether. Now this is a bit of examples You're distracting. You're a bit suggestive. Hopefully this example of a intuition can achieve. Now take a break and then this We will prove theorems about. One of them is not a coincidence is a group consisting of identities, the first of these identities and so every second of Green's theorem one prove that a separate We will see that a general theorem. Bye for now.