Our previous session three We have talked about such a total. For a given curve A density function i.e. on the sum of the integral curve springs. From this total, there were integral. A second vector of the curve tangentially onto the sum. This gave us an integral. Or curves on the perpendicular vector sum. This gave an integral. In the previous session the second and third We brought in the integral structure. And why did we do? Because I make a transaction with d s not easy to direct. Because the curves arc length not in terms of data. We discussed earlier you t in terms of parameters you is given as an explicit function. t parameters in terms of d s is what we have calculated that. Insert here it can be done. Here in terms of t or y can be placed, can be made. But why did it? Because the left side of the tractor and n'l explains the concept of integration is fine. t of the projection, n on the projection, it totals. But not conducive to account. Whereas the right-hand If the expressions to calculate the much easier and we will see going a convenient structure. We made it work for him. A final point to attract attention, see the second and If we look at the third integral d in both x's. There are d y. u and v, the vector components. x component y component. But here the different have multiplied in order. Here u dun have here multiplied by x, is multiplied by x. Even here again in a second term u and v When changing the sign of a changed place. This physically thereof very different concepts. We saw one along the curve of the total, One by one the other curve side gives the total transition. However, this mathematically v instead of the second integral you'll get if I put this term. If you just put there instead of missing this you'll get a second term. Therefore, the mathematical operations to be carried out in this the same thing in the second or third integral. For this reason, our calculations more than we We will do this out of the first kind. What is the integral of the second kind, if necessary It is clear, does not change the technical account. Now a first job I want to start with examples. Now we say a point a and b point we choose. small components of a point a x component y component is zero. So a point on the x-axis. the x component of the zero point b. Means that a point on the y-axis and in a long distance. And all this in three ways always the same selected point b. from a point on is being in a. This is obviously a b a a point point in infinite ways We can connect trajectory. Olabisi such an orbit was crooked. We now choose three examples We will bee question:I wonder if this integral How we go from A to B always Does interests may be different or the same? To understand this, these examples are doing. Both accounts technique to accustom our hands, to consolidate. The first orbit on a circle. The second orbit of a square on the edges. The third orbit from A to B. on a straight line going up. Parametric equation of a circle We can do with representation or an We can do with function display. Parametric representation here because we prefer parametric representation of the circular When using coordinates where any of a point from the center Because a radius a, wherein an angle theta, wherein When we get the point of it The projection on the x-axis times the cosine of theta. That the projection on the y axis the opposed open height, A once-sine theta. This is easy to get theta spelling spelling as we show the parameter t. So cosine theta x y de a times a times sine of theta curve, circle equation. d s is so infinitesimal arc length icon point x squared plus y We found that the square point. Who is possessed by it to the d t. According to that a fixed x t give negative points sinus. y cos t gives the point. Sine squared for the future A squared plus cosine squared Because under the square root remains just a square. Taking the square root of a square would be a. There are also a subject d t. Let's put them to take it back. Our integral has a x plus x to y. instead of a sine instead of a cosine of x t y t, instead of DSi are writing a d t. As you can see at a time See here got an A, Here comes one a square. There are an A here, a cube. Similarly, in the term of a cube If we take out the common factor, cosine integral in t's staying here. Here's the sine t, t remains kosÃ¼nÃ¼s. This angle theta is zero at point a. b to the point in time when the angle theta pi divided by two that's going ninety degrees. If these limits also write t is equal to zero pi divided into two parts, came integral structure can be calculated. See this fine expression on the left to integrate what is what. A digital function h we will do the integral over. But I do account not possible at this stage. We have placed these variables a time when the we're going on the circle bears the traces of these statements. And d s t in terms of d is opened. Integral easy. The integral sine cosine. Because the derivative of the sine If you remove ALSA cosine. Control. Because the derivative of the sine cosine here is a derivative of the sine cosine t d t. Di pattern here consists of sinus. So the integration of these two terms t be sine-squared divided by two. So u squared divided by two u would have to say t sinus. That t is equal to zero and p We are divided into two accounts. t is equal to half a Boi when it comes to a sinus. When it comes to zero zero because the sinus sinus was measuring the height of the opposite angle. Here we see the common factor, such as a cube. From a sinus. In one-half have here. Three divided by two is happening. Open the account with the function Can We open function, but the the square root of x when y is expressed in terms of will be released in a squared minus x squared. So karekÃ¶kl ??do not like the term. These respects, a difficult thing when you get the derivative but they need to deal fairly, hand he holds. His notation for the parametric We chose to work. Because the parametric representation natural coordinates for the circle. The second problem from A to B as We're going over the edge of the square. So before we go from A to C. We are also going from c to b. Now let's see it. Gene integral're writing. from A to C. We're going from C to B plus. If you look over x is equal to a constant c. All these points x is equal to a fixed because the x-axis perpendicular to the right. x is equal to a. Here is the arc length d y. Because this is true for d y is perpendicular to that right. Where x will take place this summer. A square is going on. Where x instead of going to a summer. y stands. d s a d y. Our first term of this. c As there are a delicacy. Once the first part is easy. From above we c a b c We're going to b. Wherein y is equal to a. There is not a problem, but see by definition, a length d s. Where x is the length of surplus value but When we look at this in terms of a b c only on x is changing. from C to B when the value of x is shrinking. So it is a valuable thing x minus. To fix it, plus to be able to For we put a minus sign. That I'm fine points. When we took this place We have a minus sign. Wherein y is equal to a. Not to say that x is a change. a x. X in the second term still remains. y Dee. X also comes from a will. If you look at the border from c to b here we went in for x is equal to c. b is zero. Its also a limit to the We are writing to zero. This is a subtlety that we could go from c to b. we do from b to c, d s plus valuable and were going to be a scratch. Already on him as an integral We'll see changing. The first term of the integral squared y. a squared year coming here. From the second term at a constant. y is integral, then y squared divided by two. y squared divided by two. It's here. Reset y is equal to We are a cross-account. There are two times a x squared. The integral of x squared, x squared divided by two. Thus, the dilemma is more simple, remains only a x squared. And we know where the integral We can use the feature. Integral limits When you change the minus sign appears. Here is a minus Coming to a plus. When in this account until a zero here also consists of a cube. Here divided by two to three cubes, here just a cube. Consequently the value of the integral A cube is divided by two to five turns. Now here we are now doing the following observation. The edge of the square from A to B We found five divided by two when you get a cube. Gone over the edge of the circle divided by two to three cubes were found. So it turned out differently. So we went by the way has changed. In the third example, from A to B. We're going to direct. It's true the equation x is a plus y equals. Let's just said to provide. When x is equal to y is equal to zero. Is here. When x is equal to zero is equal to the y. If x minus y is equal to arrange it. Our formula of a square base was plus years. But we are faced with the same situation again. Here is a length h When valuable as a plus, the value of x where we go from A to B. a negative value. To fix it We're putting a minus. Of course, y is equal to a negative x, y be the base minus one. Here is also a plus one square root of two incomes, d x. Let's put them. a x, x times y, x, we know. x minus y from here. We put him. d s an d x minus square root of two. Less're getting here. Here is also the square root of two. Now this is a simple integral an integration course. X has one here. Have here an a x. There are two X minus x squared. The integral has a zero up. Modifying the boundaries of this integral and that would be a minus sign We get rid of the negative. Here the square root there, As you can see we make this account We have reached a different value here again. Now if we consider the results of two previous results were different from each other. This is the third different results from them. So the value of the integral, the The integral, depending on the path from A to B. Had we taken a different path, I do not know such a crooked bi If we go out of the way from A to B, would find yet another value. So how the value of the integral we go depends, depends on the path, Depending on the output trajectory in this example. Now two examples, I'll do one more example. Here you'll still get the same orbit. Yet on the circle, going on and right on the square. Single function this time We have received a different function. for x squared minus y squared component. there is also minus two times x and y components. This is multiplied to y'yl. Integral method is the same. We are writing again x y on the circle. d x x x the derivative with respect to t. derivative with respect to t y y again. Both are multiplied by d t. We put them into place. See where x is a square a square Coming from the square in the same way year. a to d of x is coming. Here x is also an a, y a a, a a dye. A cube is a common happening again. Gene en pi divided by zero Up to two going. Here we are writing these terms. Appeared slightly longer, we're compiling. Here you can see the minus four time t the output sine cosine squared t. One output sine t and the two were collected. Four had been. If we account for this integration, here t is the cosine derivative of the sine t with a minus sign coming here because it is four divided by three. Here comes a a a split, in sinus pi divided by two. We take these differences When a cube divided by three turns negative. This detail, what is certain accounts. Now again, going over the edge of the square If we account for the same integral, ie x squared minus y squared d x, d y in both x and y, see on a car again x is equal to x is equal to zero, it is hard to say. d x is zero when the first term is decreased. x is equal to a. So would a squared minus y squared, but d * We put here to be zero, zero, This term falls. Again in the second term of x is equal to minus two There are also a y d y times. while going from C to B, This time of year is here again equal. When we put x squared minus y going on a square, x squared minus a square. Here are just changing x, d x. Where d s that are not We do not need In thinking about the previous minus signs. In the second term is the term y'l there. But where y is fixed. c b a y is equal to the edge. Therefore d y is zero. Here are putting this zero term. This term is the first one that we stayed in collecting, y d y minus two left. As you can see happening minus y squared, y y squared divided by the integral of the two. Here are the two that is going to here plus the square of x integral x cube divided by three. Less had a square, the integral of x squared is also a square. We're also bringing a zero. See terms of a cube outdoor type common terms is already happening all the time. Here comes one, a cube that is coming. Head, but has a minus sign at the beginning. Here x is equal to zeros, going will be missing this time is coming. Minus one divided by three. Here is a minus plus going for it. As you can see the cancel out. Gene found minus a cube divided by three. Refer again to the previous example We found a cube minus divided by three. Now here's the final orbit Let's go from A to B on the right. The equation x minus y. Therefore, d x d y would be negative. We take it into place. That was our integral. Where y instead of x are writing a minus. There where x squared minus y squared. Instead comes a minus x squared. X squared minus two, We are writing a negative x instead of y. d y has a minus sign, the minus d x. Here is a minus plus happening, plus d x. When we arrange them here we have x squared. There ax minus two times. But because of the minus sign in the front plus. I also have x squared. X squared minus happening. Look here when we have two a x. X also had a couple of four was here. Minus a squared minus a squared from here. Here there was a minus x squared. Two x squared minus one comes here. These are the x squared minus Compiling herein. This initial x squares took each other. Here you have an x squared. Here are a minus x squared. They had taken each other. This term stays. It also integrally, X of work in a manner known integral of x squared is divided by two. The four following by two sadeleÅerek it becomes. A frame is fixed. X it integrally. the integral of x squared is also x divided by three cubes is happening. Gene reset at zero between falls came minus coming. If you collect these numbers again A cube is divided by three is negative. Now, the interpretation of these results, With this function, which We do this in three orbital integral According gave the same result. That the previous function the same, Also in the examples herein Although we do with orbit produced different values??. Now you are faced with a situation like this. Now, in the first instance dependent on the output trajectory. In the second example the output independent of the orbit. The following question may come to mind immediately. Independent from orbit integral Is that a coincidence? One can ask the following question. Is this important? No insignificant as not worth the effort. The first answer is very important, we'll see. Therefore, worth checking out. Is this a coincidence? So things like this would not be a coincidence. Underneath is something deeper lies. As this question a little more changing We'll find it easier to answer if asked. The conditions under which the integration of a b the way to go, is independent from orbit? Now here we get to these examples and We are satisfied with the present observations. Let's try to understand them, which means some cases, functions integrally connected to orbit, some of them not connected. Which does not depend of If conditions are attached provides that there is an important thing to understand. Our next session of this will look to the subject bye.