1 00:00:00,180 --> 00:00:06,150 OK, so now you know how to calculate three dimensional integrals in the three dimensional space. 2 00:00:07,110 --> 00:00:13,880 However, since we are now in multiple dimensions, we can also calculate different types of integrals. 3 00:00:14,310 --> 00:00:20,080 For example, we can calculate one dimensional integrals along some path. 4 00:00:20,760 --> 00:00:23,450 So this is a quasi one dimensional problem. 5 00:00:23,460 --> 00:00:30,630 We are still in the three dimensional space, but we only consider a certain line and for simplicity 6 00:00:30,660 --> 00:00:32,800 so that you understand what is happening here. 7 00:00:33,150 --> 00:00:38,310 I want to show you an example of a two dimensional function where we consider a path. 8 00:00:39,060 --> 00:00:44,740 And I found this nice animation on Wikipedia that I want to explain to you on the following slide. 9 00:00:46,080 --> 00:00:46,940 So here you go. 10 00:00:46,980 --> 00:00:50,080 This is a contour plots, as we had previously. 11 00:00:50,220 --> 00:00:55,680 So here the value of the function F is colored by these different colors here. 12 00:00:56,250 --> 00:01:05,280 And now we have a path see from point A to point B, so you can see that this path here is along this 13 00:01:05,280 --> 00:01:12,810 red line and that the blue line indicates the value of the function f with respect to the zero line. 14 00:01:13,890 --> 00:01:19,380 And I can see here the path is projected into one dimensions. 15 00:01:19,410 --> 00:01:26,200 And so what you have to do is you have to integrate over this blue function with respect to the red 16 00:01:26,250 --> 00:01:27,030 zero line. 17 00:01:27,870 --> 00:01:29,940 So let's look at it one more time. 18 00:01:30,570 --> 00:01:36,780 So here again, you can see the colors and the path and of course, the color is just an indication 19 00:01:36,780 --> 00:01:38,260 for the value of the function. 20 00:01:38,280 --> 00:01:43,820 So you could plot it in such a three dimensional plot where the value of the function is the third dimension. 21 00:01:44,830 --> 00:01:46,770 Now here the important thing happens. 22 00:01:46,770 --> 00:01:54,260 Now the path will be projected into one dimension and now it's just a one dimensional problem. 23 00:01:54,960 --> 00:02:01,740 And when we integrate the integral here, we integrate over F, but we also have to account for this 24 00:02:01,740 --> 00:02:06,900 additional term that comes from projecting the path into one dimension. 25 00:02:08,100 --> 00:02:10,300 So this is what we get for the equation. 26 00:02:10,590 --> 00:02:13,650 This was the same as in the animation that you have just seen. 27 00:02:13,660 --> 00:02:16,160 So it's pretty, pretty basic. 28 00:02:16,470 --> 00:02:18,240 We integrate over the function. 29 00:02:18,570 --> 00:02:20,370 But here this is new. 30 00:02:20,400 --> 00:02:25,000 We have the absolute value of the derivative of our path. 31 00:02:26,100 --> 00:02:28,440 So this may look a bit, uh, yeah. 32 00:02:28,440 --> 00:02:29,600 But, uh, theoretical. 33 00:02:29,610 --> 00:02:33,120 Let's let's look at the detail so that you will understand better what's happening here. 34 00:02:34,260 --> 00:02:34,760 OK. 35 00:02:34,800 --> 00:02:37,230 So here in blue, we have our function. 36 00:02:37,260 --> 00:02:40,980 This is a function that we had already considered previously and. 37 00:02:42,310 --> 00:02:50,530 Now, you introduce a pass and here I've just chosen this Redpath here path, see, and our path is 38 00:02:50,530 --> 00:02:56,530 parametrized by a parameter T. So it's a two dimensional quantity. 39 00:02:56,650 --> 00:03:00,750 So we have X and Y component and you change T.. 40 00:03:00,760 --> 00:03:05,800 I have chosen to boundaries for this path to be minus pi over two and PI over two. 41 00:03:06,340 --> 00:03:11,920 And the X component is just T and the Y component is a sign of T. 42 00:03:13,270 --> 00:03:18,010 Now, of course, you may say that this path can be parametrized also in a different way, and that 43 00:03:18,010 --> 00:03:19,140 is absolutely fine. 44 00:03:19,750 --> 00:03:26,050 For example, you could choose a different parameter to not that goes from zero to one, and then you 45 00:03:26,050 --> 00:03:29,110 would have to multiply here a constant and here a constant. 46 00:03:29,110 --> 00:03:31,330 And that would be exactly the same path. 47 00:03:31,330 --> 00:03:32,920 And it would also be correct. 48 00:03:34,030 --> 00:03:36,060 But let's go with what I've chosen here. 49 00:03:36,070 --> 00:03:42,040 So this is our path are of Tea Party is this parameter that has these two boundaries that will later 50 00:03:42,040 --> 00:03:49,050 be A and B, and now we calculate the derivative of this path are with respect to tea. 51 00:03:50,050 --> 00:03:55,840 So we get for the X component one and for the Y component, we get cosine of T. 52 00:03:56,930 --> 00:03:58,650 I guess that's pretty, pretty easy. 53 00:03:59,260 --> 00:04:05,550 And now the absolute value of this is of course the square root of X squared plus Y square. 54 00:04:06,160 --> 00:04:12,280 So we get the absolute value is square root of one plus cosine square of tea. 55 00:04:13,690 --> 00:04:15,310 And no, we are almost done. 56 00:04:15,310 --> 00:04:22,900 The only thing we have to be careful about is that here we do not have any more of our I mean, we still 57 00:04:22,900 --> 00:04:24,100 have half of our pots. 58 00:04:24,100 --> 00:04:25,870 We have our of tea. 59 00:04:25,870 --> 00:04:31,590 So we have to introduce our path here in terms of the argument of this function. 60 00:04:32,590 --> 00:04:40,330 So this is because we later will integrate over the parameter T because it is a quasi one dimensional 61 00:04:40,330 --> 00:04:40,820 problem. 62 00:04:41,320 --> 00:04:47,890 This is also why you have here only one integral symbol and not to even though our function has two 63 00:04:47,890 --> 00:04:49,210 arguments, X and Y. 64 00:04:50,470 --> 00:04:53,740 OK, so here our function is one we had previously already. 65 00:04:54,250 --> 00:04:58,360 It's this polynomial where we have this additional term that makes us X and Y. 66 00:04:58,750 --> 00:05:03,840 And now I introduced a path hour of tea. 67 00:05:04,120 --> 00:05:09,480 So that means our X component will be T and our Y component will be a sign of T. 68 00:05:10,360 --> 00:05:15,220 So this means F of R of T is minus three T to the part three. 69 00:05:16,180 --> 00:05:23,350 And then here we have four times sine square left and here we have four X Y, we get tea times sign 70 00:05:23,350 --> 00:05:23,710 of T. 71 00:05:24,790 --> 00:05:32,230 So this is our parametrized function and now we can calculate this line integral here with the boundaries 72 00:05:32,230 --> 00:05:36,390 minus pi a two and plus pi over two of this function. 73 00:05:37,180 --> 00:05:40,230 And now this is something you just have to calculate. 74 00:05:40,630 --> 00:05:43,060 So to be honest, this is very, very difficult. 75 00:05:43,060 --> 00:05:46,300 I think it's not not even possible to integrate analytically. 76 00:05:46,630 --> 00:05:51,130 So here I want to show you a cool trick that I use quite often that you can use as well. 77 00:05:51,400 --> 00:05:56,130 So if you have an integral, you can go to the website that is called Wolfram Alpha. 78 00:05:56,740 --> 00:06:02,170 This is using the language of Mathematica, which is a program that I'm using quite often. 79 00:06:02,620 --> 00:06:06,460 And there you you do not really have to know the correct syntax. 80 00:06:06,460 --> 00:06:09,340 Even you can just type what you want to do. 81 00:06:09,340 --> 00:06:13,840 And most of the time, Wolfram Alpha will be able to do what you want them to do. 82 00:06:14,350 --> 00:06:17,950 So you just write integrate them in these brackets. 83 00:06:17,950 --> 00:06:20,290 I wrote the function and then the boundaries. 84 00:06:20,290 --> 00:06:23,650 I want to integrate from minus pi over to two pi over to. 85 00:06:24,010 --> 00:06:25,180 I didn't even write. 86 00:06:25,900 --> 00:06:30,660 What's the what's the variable here or if I just understood what I want to do. 87 00:06:30,670 --> 00:06:31,140 That's good. 88 00:06:31,720 --> 00:06:33,910 So here's what do we get as the output. 89 00:06:34,130 --> 00:06:38,730 That's exactly what we want to calculate and the result is four hundred and six. 90 00:06:39,430 --> 00:06:41,600 So yeah, it doesn't really matter what's the result. 91 00:06:42,040 --> 00:06:51,640 The important thing is that you understand that it is possible to integrate over a path and that's why 92 00:06:51,650 --> 00:06:52,540 we want to do this. 93 00:06:52,870 --> 00:07:00,370 We have to introduce the path in the argument of the function and we have to account here for this additional 94 00:07:00,370 --> 00:07:04,930 factor, which comes from projecting this path onto a straight line. 95 00:07:05,380 --> 00:07:08,690 So that really matters if our path is straight or not. 96 00:07:09,670 --> 00:07:13,030 So, of course, if you would have a straight path, then this one here would become one. 97 00:07:15,190 --> 00:07:23,400 OK, so this was the line integral for a scalar function F and we can also calculate the line integrals 98 00:07:23,410 --> 00:07:25,570 for vector functions in two dimensions. 99 00:07:26,350 --> 00:07:30,720 And so here the calculation is quite similarly. 100 00:07:31,120 --> 00:07:32,380 But what happens here? 101 00:07:32,530 --> 00:07:38,130 I want to again show you at the example of this nice animation that I found on Wikipedia. 102 00:07:39,190 --> 00:07:44,050 So here we have a path and our function is now a vector field. 103 00:07:44,260 --> 00:07:46,750 It is indicated by these blue arrows. 104 00:07:47,000 --> 00:07:51,700 So this means at each position we do not have one single value, but we have to. 105 00:07:52,330 --> 00:07:55,750 And a way to visualize this is to use such a vector plot. 106 00:07:56,230 --> 00:08:00,190 So now let's look at this animation from the beginning for restart soon. 107 00:08:00,640 --> 00:08:06,160 So we have here our function capital F, which is just two dimensional vector, and we have once again 108 00:08:06,160 --> 00:08:12,670 our path C from A to B, and now we calculate the DOT product of D-R and. 109 00:08:12,820 --> 00:08:19,470 F, which is the product of these two arrows that you can also see here, so they are almost parallel. 110 00:08:19,480 --> 00:08:22,450 So it's quite a large value, as you can see here. 111 00:08:23,260 --> 00:08:27,880 And now if we go along the line, as was just done in this animation, you will see, for example, 112 00:08:27,880 --> 00:08:36,970 at some point our our arrow for D.R will be pointing in the opposite direction of the value of the function. 113 00:08:37,000 --> 00:08:40,030 This is where we get these negative values that you have just seen. 114 00:08:41,080 --> 00:08:42,430 So let's look at it one more time. 115 00:08:45,310 --> 00:08:48,550 This time, let's let's consider this red arrow here. 116 00:08:48,550 --> 00:08:51,070 You can see it always follows along the line. 117 00:08:51,520 --> 00:08:56,650 And for example, at this point here, it will be pointing along the opposite direction to this blue 118 00:08:56,650 --> 00:08:58,690 arrow here just now. 119 00:08:58,690 --> 00:09:00,740 And this is where we get this negative value. 120 00:09:01,720 --> 00:09:06,360 So once again, we have a quasi one dimensional problem over which we have to integrate. 121 00:09:06,370 --> 00:09:08,260 And, of course, we know how this is done. 122 00:09:08,890 --> 00:09:10,930 And so we end up with this formula here. 123 00:09:12,760 --> 00:09:17,800 So when you compare this scalar function of the vector function, it is quite similarly. 124 00:09:17,800 --> 00:09:21,670 But here you have to take the absolute value and you have to dot product. 125 00:09:21,970 --> 00:09:24,330 So in the end you will get a scalar result. 126 00:09:24,340 --> 00:09:26,170 It would just get a number, of course. 127 00:09:27,550 --> 00:09:33,160 So I think, yeah, since we had this example for the scalar function, it's OK to not have an example 128 00:09:33,160 --> 00:09:37,540 for the vector function because in the end it's just an exercise in math. 129 00:09:37,720 --> 00:09:38,280 What do you have? 130 00:09:38,410 --> 00:09:44,620 The only thing that's a bit new from a conceptual side is that you have to introduce a path and then 131 00:09:44,620 --> 00:09:45,910 you have to parameterize. 132 00:09:46,270 --> 00:09:48,430 And this is exactly the same as in this case. 133 00:09:48,760 --> 00:09:54,040 But here you just take the vector of these derivatives and calculate the DOT product. 134 00:09:56,000 --> 00:10:03,290 OK, so this is basically all I want to tell you about these line integrals, the only final comment 135 00:10:03,290 --> 00:10:09,470 that I have is that quite often, especially in physics, we will integrate over a closed path. 136 00:10:09,650 --> 00:10:10,700 Where are the boundaries? 137 00:10:10,700 --> 00:10:12,270 A and B will be the same. 138 00:10:12,980 --> 00:10:19,260 So whenever this is the case, you typically indicate this is such an integral sign with a circle. 139 00:10:19,430 --> 00:10:23,080 So this means that you integrate of a closed path. 140 00:10:23,930 --> 00:10:31,700 But of course, even though the starting and the ending position will be the same, the integral will 141 00:10:31,700 --> 00:10:34,280 most of the time still be different from zero. 142 00:10:34,640 --> 00:10:40,070 And also the value of the integral will strongly depend on the path itself. 143 00:10:40,850 --> 00:10:46,910 So this is quite different from a purely one dimensional problem where the value of the integral is 144 00:10:46,910 --> 00:10:51,040 just determined by the boundaries and b in multiple dimensions. 145 00:10:51,050 --> 00:10:56,960 When you consider such a line integral, which is a quasi one dimensional problem, the result strongly 146 00:10:56,960 --> 00:10:58,490 depends on the path.