AI Assistant
Transcript
00:28:430Paolo Guiotto: Okay, good morning.
00:34:110Paolo Guiotto: It doesn't work.
00:37:300Paolo Guiotto: I'll check the inbox.
00:49:610Paolo Guiotto: It works.
00:50:990Paolo Guiotto: Okay, good morning.
00:55:450Paolo Guiotto: I hope you, you received, you received. You have seen my message of yesterday about an exercise, that contained, and I don't know if you tried.
01:09:940Paolo Guiotto: I will… I will do this.
01:13:780Paolo Guiotto: This 572, the number 2.
01:18:20Paolo Guiotto: It's not particularly complicated, but it contains a little difficulty in the parameterization of the integral.
01:25:850Paolo Guiotto: So, we have integral for X positive, Y positive, and they say that there is a missing condition, Z positive.
01:34:270Paolo Guiotto: And then we have also X plus Y plus Z less or equal than 1.
01:39:90Paolo Guiotto: of XYZ, DX, DYDZ.
01:43:440Paolo Guiotto: It's particularly simple for the function.
01:47:650Paolo Guiotto: But, there is a little complication with the domain.
01:53:30Paolo Guiotto: I don't know if you have tried, and if you…
01:57:10Paolo Guiotto: realize this little difficulty. So, here we have the function F, XYZ…
02:05:230Paolo Guiotto: is X times Y times Z, so it is clearly a continuous function on a tree.
02:12:20Paolo Guiotto: The domain, the integration domain, is a set of points, X, Y, Z, such that X positive, Y positive.
02:22:540Paolo Guiotto: Is that the positive?
02:24:900Paolo Guiotto: X plus Y plus Z less will equal than 1.
02:29:290Paolo Guiotto: There's all this combinations.
02:31:430Paolo Guiotto: So, clearly, it is defined by large inequality, so D is closed.
02:38:620Paolo Guiotto: About, bounded, we can easily see that it is also bounded, because
02:44:930Paolo Guiotto: Is bounded, because, first we see that, because…
02:53:320Paolo Guiotto: Clearly, we have these conditions that say that XYZ are lower bounded, so the three coordinates are positive.
03:02:420Paolo Guiotto: This does not mean that they are bounded, okay? Because there is not bound from above, they can be huge.
03:09:860Paolo Guiotto: But, from this condition, we have X plus Y plus
03:15:220Paolo Guiotto: Z, which is less or equal than 1, combined with this.
03:19:890Paolo Guiotto: We necessarily have that the three coordinates must be less or equal than 1.
03:26:910Paolo Guiotto: Because, for example, if X is greater than 1,
03:32:400Paolo Guiotto: we would have that X plus Y plus Z, These two guys are positive.
03:38:880Paolo Guiotto: So this would be greater than X, that would be greater than 1, so…
03:44:300Paolo Guiotto: It cannot be less or equal than 1. It's impossible, you see?
03:51:430Paolo Guiotto: So you cannot have that this quantity, which is greater or equal than X plus Y plus Z, is
03:57:100Paolo Guiotto: greater, so one strictly greater than 1, that's not possible.
04:02:840Paolo Guiotto: So, in particular.
04:05:20Paolo Guiotto: This says that the three coordinates are positive and are less than 1, so they are all bounded, the D is bounded.
04:12:120Paolo Guiotto: So, now we know that F is continuous on a closed and bounded domain, F is integral on the domain, so there is the integral of F only.
04:23:10Paolo Guiotto: Now, to compute, we apply
04:28:500Paolo Guiotto: reduction formula. Yeah, since we already know that the function is integral, we don't need to check that the function is integral. Actually, we check that by using that condition.
04:40:810Paolo Guiotto: Now,
04:44:250Paolo Guiotto: As you can see, the function is indifferent. The three variables is the same thing. So the first integration could be in X or in Y or Z is the same, so we may decide, let's start with X.
04:58:160Paolo Guiotto: So, X, Y, Z, DX, and then we finish in the other two, the Y and Z.
05:05:700Paolo Guiotto: Now, the point is, what should we write here?
05:09:710Paolo Guiotto: That's the difficult, say, part, which is the beginning of more complicated exercises, by the way.
05:18:250Paolo Guiotto: So, here, let's look at the conditions we have. They are… the problem is also that we have a lot of conditions, so it's apparently a mess.
05:31:940Paolo Guiotto: So, but let's focus on X.
05:34:350Paolo Guiotto: What are the conditions for X? This one is the condition for X.
05:38:720Paolo Guiotto: Y greater or equal than 0 is not the condition for X, there is no X.
05:43:210Paolo Guiotto: And Z greater or equal than 0 is not involving any X, and there is this one also.
05:48:620Paolo Guiotto: So, apparently, we have two conditions.
05:51:800Paolo Guiotto: So, these two conditions says that X must be positive.
05:56:510Paolo Guiotto: And this one, I can see as X less or equal than 1 minus Y plus Z.
06:03:560Paolo Guiotto: So if I glue these two, I get a condition, a unique condition, X positive and less than 1 minus Y plus Z, right?
06:15:190Paolo Guiotto: So, I think that up to this point, everyone can reach this conclusion.
06:22:960Paolo Guiotto: But now, that's the next step that comes the error. Because so, I would say that, at this point, we should write. For X, the integral is from 0 to 1 minus y plus z, right?
06:37:330Paolo Guiotto: And for YZ, there are two conditions here.
06:41:700Paolo Guiotto: So, I should say Y positive and Z positive.
06:47:250Paolo Guiotto: Now, is that correct?
06:53:290Paolo Guiotto: What?
06:57:260Paolo Guiotto: But I used it, it's, yeah, that condition.
07:04:390Paolo Guiotto: Is it correct or not?
07:06:420Paolo Guiotto: And if… If not, what is wrong?
07:15:30Paolo Guiotto: It's not written on my face. It's much better if you look at the whiteboard.
07:26:750Paolo Guiotto: There is another condition that you don't see there, because it's not yet read…
07:31:840Paolo Guiotto: How is possible there is another condition here, right? There is another condition.
07:47:360Paolo Guiotto: Why?
07:50:910Paolo Guiotto: Wipe, that's it?
08:01:570Paolo Guiotto: I don't… I don't understand what you want to do. I… I wouldn't use the X.
08:05:200Paolo Guiotto: X plus Y plus then less or equal than 1 means that it is X between 0, the first condition, and less than that.
08:15:540Paolo Guiotto: If there is something here that should be, checked.
08:21:590Paolo Guiotto: You don't… Which function?
08:28:170Paolo Guiotto: this one.
08:29:880Paolo Guiotto: No.
08:32:299Paolo Guiotto: No, no, it's something that has to do with the domain.
08:35:289Paolo Guiotto: Now, this is not complete. Why?
08:38:250Paolo Guiotto: Because if you look at this condition up here.
08:42:720Paolo Guiotto: Makes sense to have some X. You need that the quantity here, this number.
08:49:690Paolo Guiotto: this one be greater than zero, otherwise you don't have any X.
08:55:520Paolo Guiotto: You see?
08:56:850Paolo Guiotto: If you put the Y… if you leave Y and Z positive, take Y equal 1 million and Z equal 1 million. They are positive.
09:05:160Paolo Guiotto: But X plus Z is 2 million, and 1 minus 2 millionth is negative, so there is no X between 0 and 1 minus YZ. So it means that that condition, Y is greater or equal than 0, Z greater or equal than 0 is not sufficient.
09:21:80Paolo Guiotto: And in fact, to be true, this is when or provided.
09:30:700Paolo Guiotto: Y1 minus Y plus Z.
09:35:580Paolo Guiotto: Well, let's write in red, because…
09:38:830Paolo Guiotto: 1 minus Y plus Z is positive.
09:43:610Paolo Guiotto: And this means that Y plus Z is less than 1.
09:49:630Paolo Guiotto: You see? If this is not true, it means that this number is negative, and there is no X between 0, above zero, and below a negative number.
10:01:940Paolo Guiotto: So, it's not correct to say that I am integrating on this interpaca, there is no interface.
10:08:540Paolo Guiotto: Okay? So you are integrating on the interval between 0 and 1 minus 5 times Z when this is positive.
10:16:640Paolo Guiotto: Otherwise, there is no X.
10:19:300Paolo Guiotto: So, literally, this means that this is the YZ section. The YZZ section is empty if this number is negative, so there is no integral there.
10:32:250Paolo Guiotto: And therefore, this means that we have to add this condition here
10:41:350Paolo Guiotto: Okay, so, let me rewrite, clean, everything. So this is integral for Y positive.
10:48:940Paolo Guiotto: Z positive, they were already there, but implicitly, this means also that Y plus Z must be less than 1.
10:57:560Paolo Guiotto: And then here we have integral from 0 to 1 minus y plus z of XYZ. This is the integral in the X, then we have DY.
11:07:930Paolo Guiotto: Decent.
11:09:930Paolo Guiotto: Okay, now we can do the calculation. We carry outside this. We get integral for Y positive, Z positive, etc.
11:19:800Paolo Guiotto: Y plus Z less or equal 1. Then what remains is, so YZ, it remains the integral of X, no?
11:30:860Paolo Guiotto: Integral from 0 to 1 minus y plus z of X dx. This is the evaluation X squared over 2 between 0 and 1 minus y plus
11:44:110Paolo Guiotto: Zap them?
11:45:660Paolo Guiotto: And then we have to integrate in Y and Z. So after this evaluation, we get… so we can take the factor 1 half
11:54:600Paolo Guiotto: In front of the integral, and we have Y greater than 0, Z greater than equal 0,
12:02:380Paolo Guiotto: Y plus Z less or equal than 1 of YZ that we have here, and then the evaluation is this square, 1 minus Y plus Z
12:15:730Paolo Guiotto: square minus the value at x equals 0, 0. So that's the new integral to be computed.
12:24:820Paolo Guiotto: Okay, now we have this double integral. We apply once more the reduction formula.
12:29:810Paolo Guiotto: So let's see if you have learned the lesson.
12:33:100Paolo Guiotto: So now I do a double integration. Let's say it's in different… we start with Y, so we will copy the function down here, and Z. And now tell me what are the range for these two integrations. What is the range for Y?
12:52:470Paolo Guiotto: 01 minus Z. What is the range for Z?
12:58:00Paolo Guiotto: 0 to 1, because you see.
13:01:340Paolo Guiotto: We have these conditions, Y greater or equal than zero, Z greater or equal than zero, and Y plus Z less or equal 1, right? We decided to do first integration in Y. So, look at why, we have this condition and this.
13:17:570Paolo Guiotto: Or why?
13:18:920Paolo Guiotto: This second can be seen as Y less or equal than 1 minus Z. So together with the first, they give Y between 0 and 1 minus Z. In other words, you see that these two conditions must be verified together.
13:35:990Paolo Guiotto: So Y must be such that it is positive and less than this. If this guy is negative, there is no Y which is positive and less than this, you see?
13:47:300Paolo Guiotto: So it's wrong just to write Z positive, because this condition needs that 1 minus Z be positive, otherwise there is no Y, and so this yields the Z less or equal 1, and that's the condition we have here.
14:06:170Paolo Guiotto: Okay?
14:07:680Paolo Guiotto: Now, since it is a relatively… boring calculation.
14:17:00Paolo Guiotto: I already did the calculations, so…
14:24:720Paolo Guiotto: It's a polynomial. So let me just copy the, I don't know where I wrote it, if that's the problem.
14:32:320Paolo Guiotto: It feels unsafe.
14:33:780Paolo Guiotto: So, wait a second, I… Find your values.
14:41:30Paolo Guiotto: Should be, yeah.
14:43:290Paolo Guiotto: Yeah.
14:45:180Paolo Guiotto: I brought the solution yesterday on the… on the notes.
14:49:900Paolo Guiotto: So, when you do this calculation here, you will get 1 half
14:58:440Paolo Guiotto: So, it remains the integral 0 to 1. I'm doing the calculation inside here.
15:04:920Paolo Guiotto: So… this is, it's different because I did with different letters, okay.
15:14:930Paolo Guiotto: So it should be… Should be, should be, should be what?
15:22:690Paolo Guiotto: This is a Y, so it should be Z times 1 minus Z to power… Hanging.
15:32:420Paolo Guiotto: well, actually, I've not done this calculation, I asked ChatGP to do the calculations. So that's… it's not my fault. So this is what it comes out, this is the value of the integral from 0 to 1 minus Z of
15:51:50Paolo Guiotto: So, of course, you carry the Z outside, so it's… the quantity is Y.
15:57:40Paolo Guiotto: times 1 minus Y minus Z, the 2 power.
16:01:570Paolo Guiotto: 2? Yeah.
16:03:850Paolo Guiotto: in the Y. So if…
16:06:680Paolo Guiotto: If you do these calculations with charge, you get this kind of result. And when you do this one, this is one variable integral, you get… you get something like…
16:19:350Paolo Guiotto: 1 over 720, something like this. Doesn't matter whether. What they wanted to show you was this point, okay? That's the most…
16:30:470Paolo Guiotto: Okay, so… This is another course.
16:37:780Paolo Guiotto: And I'll turn back, okay,
16:42:530Paolo Guiotto: Okay, so let's say that we have basically completed the exploration of the reduction formula.
16:53:890Paolo Guiotto: But before we leave the reduction formula, we have… of course, we will still continue to use this.
17:00:400Paolo Guiotto: The next point will be to introduce the change of variables formula, but before we leave that, I want to introduce a couple of definitions.
17:11:300Paolo Guiotto: So, the definition one is, the area we call,
17:17:460Paolo Guiotto: For a domain D of the Cartesian plane R2, we call area, of D,
17:25:530Paolo Guiotto: As we say, that this is the integral of 1, on the domain D, provided.
17:36:750Paolo Guiotto: Deep.
17:38:40Paolo Guiotto: Integral.
17:40:310Paolo Guiotto: exists.
17:44:750Paolo Guiotto: for, D.
17:47:780Paolo Guiotto: in R3… So, in a two-dimensional space, we define the volume…
17:59:100Paolo Guiotto: of D as still the integral of 1 in DX, DYDZ.
18:04:820Paolo Guiotto: You remind that I explained at the beginning why this should be the values of areas and volume.
18:11:30Paolo Guiotto: No, because the idea is, for the case of the first formula, you have a subset in the Cartesian plane, that's the domain. You consider a function which is constantly equal to 1 on that domain, so the graph of this function is flat.
18:30:760Paolo Guiotto: copy of D at quote equal 1.
18:35:70Paolo Guiotto: And the integral of 1 on the domain D
18:40:940Paolo Guiotto: Should be the volume of the trapezoid
18:45:860Paolo Guiotto: This is the original idea of the integral of the function 1 on the domain D. So the volume of this cylinder with base D and H1. So we expect that it is equal to the area
19:01:250Paolo Guiotto: Yeah, yeah.
19:02:600Paolo Guiotto: of the base times the height, which is 1. That's the… that's why we defined the area with the T integral, okay?
19:12:240Paolo Guiotto: But now, apparently, there is a little coherence problem, because,
19:17:500Paolo Guiotto: For a certain plane figure, we have another method to compute the area that you learned last year, so let's do this remark.
19:28:140Paolo Guiotto: If the domain D contained in RF2, is, the… trapezoid, the…
19:39:600Paolo Guiotto: Of a function of one real variable.
19:43:740Paolo Guiotto: So, it is this, so imagine you have an interval AB,
19:49:670Paolo Guiotto: a function which is positive, defined on an AB,
19:58:330Paolo Guiotto: with positive values.
20:00:550Paolo Guiotto: The trapezoid is this region, delimited above by the graph of the function, and below by the interval AB, okay? So this is in the Cartesian plane, and last year.
20:15:430Paolo Guiotto: In, first… Yeah.
20:20:320Paolo Guiotto: Calculus.
20:23:120Paolo Guiotto: U.
20:25:440Paolo Guiotto: defined.
20:28:760Paolo Guiotto: Area of this, region, let's call it D, the region, this one.
20:38:350Paolo Guiotto: as the integral from A to B, F of X dx, right?
20:43:580Paolo Guiotto: So this was the first general method to compute areas.
20:49:60Paolo Guiotto: Now, in principle, we may see that D is the set of points… I remind you that the trapezoid is defined in this way. It's a set of points of R2. We did this in the first class, but let's repeat.
21:04:500Paolo Guiotto: set of points XY, where DX is here in the interval AB.
21:11:530Paolo Guiotto: And the Y is here, from 0, which is on the axis, to the maximum quota that we can have for abscessa X, which is the value F of X. So for the Y, we have this, from 0 to F of X.
21:30:970Paolo Guiotto: Okay?
21:33:300Paolo Guiotto: But now, with this new definition, in principle, we have a different formula to compute the area. So, we may wonder, do we get the same value, or do we get another value?
21:45:520Paolo Guiotto: So… With the new formula.
21:56:460Paolo Guiotto: area of D…
21:59:20Paolo Guiotto: D is a subset of the Cartesian plane, so it's exactly the situation I have here. Now, for a subset of R2, I define area as the double integral, so it's not an integral in one variable, it's an integral in two variables of one on the domain D. So it is this one.
22:19:470Paolo Guiotto: Now, the question is, is this the same of this?
22:24:890Paolo Guiotto: Because if they are different, it would be a little bit disturbing, huh? We have two different definitions for area.
22:31:950Paolo Guiotto: Now we show that it's the same.
22:34:320Paolo Guiotto: And in fact, this is a nice consequence of the reduction formula, because look how this domain is defined.
22:42:320Paolo Guiotto: And imagine we use the reduction formula here. So we decide to integrate this integral, this function in two variables, through two integrations in one variable. Now, I say, looking at the description of the domain, it could be better to
23:00:930Paolo Guiotto: integrate first in X or first in Y. The function is just constant equal to 1, so it's indifferent which is the first integration, okay? It's a constant, so it's the same integrating first in X or first in Y. But for the domain, it's not the same.
23:18:300Paolo Guiotto: Because if you look at the domain, it is better to integrate first in X or first in Y.
23:25:90Paolo Guiotto: Yeah, because you see, you have a condition Y in function of X. In fact, if you start integrating in Y, so 1 is the function.
23:34:00Paolo Guiotto: What is the range of Y?
23:37:630Paolo Guiotto: 0 to f of x. When x is between A and B, so the other integration is from A to B. And what is the integral of 1
23:48:90Paolo Guiotto: between 0 and FX.
23:52:420Paolo Guiotto: It is exactly now the length of that interval, so final value minus initial value, f of x. So this is the integral from A to B, F of X dx, which is exactly the same as above.
24:06:70Paolo Guiotto: So it means that this new quantity coincides with the old quantity, but actually it extends.
24:13:410Paolo Guiotto: Okay? So… D… That's called New Area.
24:19:780Paolo Guiotto: new definition.
24:21:870Paolo Guiotto: off… Here we are.
24:26:910Paolo Guiotto: Coincides.
24:33:410Paolo Guiotto: We… the… older.
24:39:360Paolo Guiotto: 1.
24:41:520Paolo Guiotto: But actually, it works in more general situations.
24:46:200Paolo Guiotto: And… scenes… It works.
24:56:650Paolo Guiotto: Ian… more.
25:00:120Paolo Guiotto: General.
25:02:70Paolo Guiotto: Situations.
25:05:940Paolo Guiotto: For example, imagine that you want to compute the area of this plane figure.
25:12:770Paolo Guiotto: How can you do with the ordinary integral?
25:16:90Paolo Guiotto: That's not the region included between the interval AB and the graph of the positive function.
25:22:210Paolo Guiotto: So, how do you compute with the old definition? You cannot do that, but you can do with the new definition. So, with this, you can say that the area of this D will be equal to the integral only of 1, DXDY.
25:40:830Paolo Guiotto: Okay? So, it extends the old definition.
25:47:370Paolo Guiotto: Mmm…
25:49:80Paolo Guiotto: Well, another… so this was the first remark. Another second important remark, which is a nice interpretation of the
25:56:630Paolo Guiotto: of the reduction formula for the case of volumes. It works also for the case of areas, but it's less appealing, let's say. If the domain D is contained in R3, so now we have a solid.
26:13:850Paolo Guiotto: So we have to imagine that here we are in space, X, Y, Z. We have a solid, so like a potato.
26:24:670Paolo Guiotto: Something like this.
26:30:890Paolo Guiotto: And we want to compute the volume of this thing, no?
26:34:960Paolo Guiotto: So, the volume of D is the integral of 1,
26:42:310Paolo Guiotto: In this case, it is a triple integral in the three variables DFTYDZ.
26:47:880Paolo Guiotto: Imagine we apply, to do this calculation, the reduction formula, okay?
26:54:320Paolo Guiotto: Now, yeah, let… let's say that,
27:01:290Paolo Guiotto: with this perspective, I will do
27:04:920Paolo Guiotto: I will show you the reduction formula by doing once the integration in Y first, and then XZ. So, you will see why I choose this. So, let's say that we start integrating Y, and then we finish with XZ.
27:19:700Paolo Guiotto: or the alternative is the opposite. We start integrating in X and Z, and then we finish in Y.
27:27:890Paolo Guiotto: But let's see what are the integrations domain, because for the function, there is nothing to say, it's 1. Now, this is the set DXZ. It is the XZ section, so this is a set of Ys.
27:41:970Paolo Guiotto: And then we have, the set of X and Z for which this thing is non-empty.
27:48:270Paolo Guiotto: while this is a DY, and this is DY non-empty.
27:55:110Paolo Guiotto: Now, I prefer to start with this second formula, so volume of D is this one.
28:05:720Paolo Guiotto: Yeah, because if you look at this quantity in the box.
28:11:380Paolo Guiotto: This is a literal of 1.
28:14:80Paolo Guiotto: in two variables, X and Z. So, according to what we have given as definition, this is an area.
28:21:40Paolo Guiotto: This is exactly the area.
28:24:10Paolo Guiotto: of the set you have here, so EY.
28:28:520Paolo Guiotto: So basically, this formula says that you can compute the volume by doing this. You integrate over one variable, the area.
28:38:820Paolo Guiotto: of the DUI section.
28:41:620Paolo Guiotto: along the Ys for which this Y section is non-empty.
28:46:60Paolo Guiotto: This is called the slicing formula.
28:54:920Paolo Guiotto: Because, let's see in a figure what this is saying, so…
29:02:180Paolo Guiotto: This is the x-axis, the y-axis, and the z-axis. So, I won't do a too much complicated object here, so…
29:11:730Paolo Guiotto: But, of course, the, the application is general.
29:16:550Paolo Guiotto: Imagine you have your soul in here.
29:20:190Paolo Guiotto: Now, you take a Y, because this is an integral in Y, so fix a Y. What is the set DY? Remind that this is a set of the other coordinates, so there are two coordinates, X and Z.
29:36:560Paolo Guiotto: Such that the point XYZ belongs to D. How can you visualize this? So when you fix a Y,
29:46:740Paolo Guiotto: Now, you remind me that in two dimensions, if we are in the… if we have only two variables, so let's happen with this special case, so let's say that we have two variables, and let's say that for convenience, I would put Y on the x-axis.
30:03:590Paolo Guiotto: And that's on the vertical axis, okay?
30:06:700Paolo Guiotto: When you compute the dissection DY, remind that here Y is the abscessa, you fix a Y here, you take the line passing through that point parallel to the other axis, so in this case, the x-axis.
30:24:620Paolo Guiotto: If you have an intersection with the domain, this section is made of the ordinates of those green points, so it is accept here on the
30:36:570Paolo Guiotto: vertical axis. So this is… this blue set is the Y section. So, you slice, you get something, you project on the Y on the x-axis, and you get this section, okay?
30:51:590Paolo Guiotto: Now, let's imagine to do the same in this figure. I fix a Y. Now, to slice this potato with something, I have to imagine that this is now a two-dimensional object, and in fact, the set DY is a subset of the Cartesian plane.
31:10:290Paolo Guiotto: The precisely, the capacian plane that you see in the… that is, let's say,
31:17:250Paolo Guiotto: determined by these two axes, the X and the Z axis. So imagine that you have a big knife that cuts
31:26:40Paolo Guiotto: the potato at a point with the ordinate Y. So what you get is a slice, the corresponding green set is a slice, which is a plain figure here.
31:39:520Paolo Guiotto: Okay? So, like, salami is licensed, not a potato chip. And then you project all this in the plane XZ, you get a figure here.
31:50:390Paolo Guiotto: So this is, just the shadow of that slice on the plane XY. That blue set is the DY section.
32:00:610Paolo Guiotto: Now, this is a plane set, you compute its area, and what you do, you do the integral of all these areas. Now, integral is more or less a sum.
32:10:880Paolo Guiotto: So it is like if you are saying that if you slice your potato in small slices here, and for each slice you compute the area, by summing all the areas, you get the volume.
32:25:50Paolo Guiotto: You see? So this is saying that to get the volume, you can do the sum of all the areas of all the slices, and then you get the volume. This is a very old idea, and that's precisely how it works. Now, to show you an example, a non-trivial example.
32:45:480Paolo Guiotto: Example… I'm pretty sure that everyone knows the volume of a sphere, right?
32:54:770Paolo Guiotto: Volume of a sphere… of radius.
33:04:450Paolo Guiotto: Do you know the formula?
33:09:330Paolo Guiotto: 4 children, bye.
33:13:70Paolo Guiotto: October.
33:14:340Paolo Guiotto: Do you know where it comes from? I've never seen the proof of this.
33:20:520Paolo Guiotto: No, it's impossible, because to do this, you need triple integrals, okay? So you have not studied before. And so, probably, you accepted that someone maybe in the primary school told you that the volume of the spirit is this thick, but what is… where it comes from, maybe you never…
33:37:800Paolo Guiotto: ask them where it comes from, but it comes from this. So, this fear is this thing.
33:43:840Paolo Guiotto: So let's imagine that it is centered in the origin.
33:48:950Paolo Guiotto: So, the domain we want to compute the volume is the set of points XYZ in R3,
33:57:10Paolo Guiotto: Such that X squared plus Y squared plus Z squared is less or equal than R squared. That's the way to describe this square, no?
34:09:179Paolo Guiotto: So we have this object here.
34:15:620Paolo Guiotto: And the radius is R. We pull, okay, with the interior. Now, according to the formula, we have, so this is the integral on D of 1, dx, DY, DZ.
34:30:380Paolo Guiotto: Now, for the function, it's entirely indifferent, which is the first integration we do.
34:36:110Paolo Guiotto: For the, for the domain, the same, because the domain is completely symmetric in the three variables, so there is no difference.
34:47:699Paolo Guiotto: So let's say that we decide to start with XY and to finish with Z. So we have integral of 1 in the XDY,
34:56:380Paolo Guiotto: And then in descent.
34:58:350Paolo Guiotto: Now, what is the condition we have to write here for XY and for Z?
35:03:510Paolo Guiotto: Now, we have a unique inequality that defines the domain, so if I look for the condition for X and Y, I read X squared plus Y squared less or equal than R squared minus Z squared. So that's the condition for XY.
35:23:40Paolo Guiotto: But now I'm sure that you learned the lesson. There is another condition here, which is, in some sense.
35:28:630Paolo Guiotto: Hidden, because this quantity must be
35:34:640Paolo Guiotto: must be positive, otherwise there is no XY. So you can say, for XY, there is this condition, X squared plus Y squared less or equal R squared minus Z squared. And for Z, I have that R squared minus Z squared must be positive, otherwise no Y.
35:52:220Paolo Guiotto: So this means Z squared less or equal than R squared, or modulus of Z less or equal than R. We assume that R is the radius, so it is a positive number, okay?
36:06:260Paolo Guiotto: So, this means that, if you want, let's write modulus Z less or equal than R.
36:12:830Paolo Guiotto: But now let's write this thing under this form that we said here. Integral of 1 on a domain is its area.
36:22:630Paolo Guiotto: So this is… this quantity here is the area of this figure of the plane set X squared plus Y squared less or equal than R squared minus Z squared.
36:37:760Paolo Guiotto: But what is this set?
36:40:450Paolo Guiotto: This one is… in the Cartesian plane for X and Y.
36:47:990Paolo Guiotto: No, not a sphere. XYZ is freezed here.
36:53:270Paolo Guiotto: is a disk.
36:56:330Paolo Guiotto: Sanford.
36:59:910Paolo Guiotto: at theology, you see?
37:03:120Paolo Guiotto: 0, 0. And radius… what is the radius?
37:11:240Paolo Guiotto: Goodbye.
37:12:490Paolo Guiotto: Whatever is, this number is the root of that number, root of R squared minus Z squared. Okay.
37:21:720Paolo Guiotto: And what is the area here? You can use what you know about the area of, in this case, pi times the square of the radius, which is that number, R squared minus Z squared.
37:36:50Paolo Guiotto: So, we don't need to do any kind of integration here, because this is basically done by elementary geometry, so I continue down here.
37:46:420Paolo Guiotto: we get the integral for modulus of Z less or equal than R of what? That integral is equal to pi r squared minus z squared, so pi
37:57:970Paolo Guiotto: R squared minus Z squared.
38:01:660Paolo Guiotto: And this is an integral in DZ.
38:04:370Paolo Guiotto: Now, it's easy because this is integral from minus R to R pi r squared minus Z squared EZ.
38:14:490Paolo Guiotto: We may observe, to simplify a bit, the calculation, well, we can do directly here. The function is symmetric, is even. When you change Z into minus Z, nothing changes, and the interval is symmetric. This means that it is twice the integral from half
38:32:330Paolo Guiotto: interval, pi r squared minus Z squared.
38:37:380Paolo Guiotto: or if you don't know this, you just compute directly. So, 2 pi, so 2 pi times the integral from 0 to r of r square, r is a constant, so this becomes R cubed.
38:51:520Paolo Guiotto: minus the integral from 0 to r of z squared, this is Z cubed over 3 to be evaluated between 0 and R.
39:02:520Paolo Guiotto: And when we do the evaluation, we get 2 pi r cubed minus… when Z is R, we get r cubed over 3. When Z is 0, we get 0.
39:14:780Paolo Guiotto: So finally, we have this, and this is exactly 2 pi.
39:19:910Paolo Guiotto: 1 minus 1 3rd is 2 thirds.
39:24:20Paolo Guiotto: are Cuba. So, the famous port, the pie, Cuba.
39:30:370Paolo Guiotto: And that's the formula for the volume of a sphere, okay?
39:36:480Paolo Guiotto: Is that in question?
39:39:520Paolo Guiotto: No.
39:42:40Paolo Guiotto: Okay, now, to increase our, possibilities in computing integrals, we need to introduce the second important technique, which is the change of variables.
40:01:540Paolo Guiotto: Now, here we could say change of variable if we think that the variable is an array, a vector, or if you want change of variables, if we think that there are several variables. It's indifference, it's the same thing.
40:15:730Paolo Guiotto: Now, to understand what is going on here, we refresh what happens with integrals in one single variable, okay?
40:24:120Paolo Guiotto: So, in a dimension, 1… We have this.
40:32:420Paolo Guiotto: Suppose that you want to compute your integral from A to B of
40:37:390Paolo Guiotto: The function f x in the X. And for whatever is the reason, you think that it might be better if we change variables. We introduce some new variable Y, which is a certain function phi of the variable X.
40:53:610Paolo Guiotto: Okay? Because maybe we think that this will simplify the calculations, and so on. This is usually the point, no? We introduce this to simplify.
41:03:980Paolo Guiotto: Now, when we say Y equal phi of X, and we talk about the change of variable, we just do not mean that we set a new variable function of the starting variable X. But this map phi must be… must have some particular properties.
41:21:620Paolo Guiotto: So this fee, first of all, Fee must be… Invertible.
41:35:600Paolo Guiotto: Because the idea is that this says to an X, there is Y, but also that vice versa should be true. To a Y, there is only one X. So it should be possible to invert this, saying that X is the inverse function.
41:51:540Paolo Guiotto: of Y. This is very important because it is exactly this function that enters in the formula I'm going to write.
41:58:910Paolo Guiotto: So this fee must be invertible, and moreover, it must verify some, let's say, technical conditions, like regularity conditions. So, the function fee and also the inverse function must be
42:15:810Paolo Guiotto: differentiable.
42:26:20Paolo Guiotto: So
42:28:510Paolo Guiotto: before I write how this formula is changed here, or maybe we can just write… yeah, let's write the formula. So, then what happens? This integral is transformed into a new integral in the Bible y.
42:43:150Paolo Guiotto: Which is not just a change of letters. I'm not changing letter X with letter Y. I'm changing variable.
42:49:420Paolo Guiotto: So it means that, for example, instead of having f of x, I will have F at the X that corresponds to Y. So this is the X that corresponds to Y. So I will read here F at the phi minus 1 of Y,
43:04:750Paolo Guiotto: which apparently seems to be much more complicated than f of x, but the idea is that this function will be, in fact, easier than F , because I introduced that block Y equals f of x to simplify that thing.
43:19:500Paolo Guiotto: I will show in a moment an example. So, let's do an example in parallel. So, suppose that I want to compute integrals 0 to 1E of root of X.
43:28:750Paolo Guiotto: which is the easiest example I can think of, to do something which is non-trivial at the same time with very low
43:38:80Paolo Guiotto: this technical difficulties. So the idea would be, natural idea, take Y equals root of X, in such a way that you simplify your exponential, because in the new variable y, that e to root of X becomes E2Y, which looks to be easier than e to root of X. So I have e to y.
43:57:490Paolo Guiotto: here. So, as you can see, this is the function
44:00:900Paolo Guiotto: if you want, technically, what I've done, this was F of X, so I inverted this. How do I invert? I take the square, so this Y square is equal to X, this is the inverse.
44:14:700Paolo Guiotto: So, what I have to do here, so this is the phi minus 1 in notations, and this was the phi of X, this was the direct map.
44:25:660Paolo Guiotto: This is the inverse map. So, when you do F of P minus 1 of Y, that you never do when you do the practical calculations, you know that, root of X is Y, but let's be convinced that you get really Y.
44:39:850Paolo Guiotto: Because you do E to root of X is phi minus 1 of Y, which is Y squared.
44:46:920Paolo Guiotto: So technically speaking, you have this, which is apparently more complex than e to the root of X. But the point is that root of y squared is the absolute value of y, right?
45:00:220Paolo Guiotto: And since our Y is root of X, which is positive, this is E toy, and that's why we have A toy here.
45:09:00Paolo Guiotto: Okay.
45:11:220Paolo Guiotto: So we have a new integration in Y, but then there is something here. Unfortunately, the real world is not so easy, it's not just changing the function. There is something to put here after the F in the new Bible. And this something is the derivative of the phi minus 1.
45:32:310Paolo Guiotto: So what you need to put here is the derivative of phi minus 1, which is not a problem here. We have phi minus 1, derivative of phi minus 1 at Y is 2Y. So the, let's say, correction factor is this 2Y.
45:47:750Paolo Guiotto: Moreover, there is something that changed also in the range of the integration variable, because that left axis between A and B
45:57:30Paolo Guiotto: What for Y? But since Y is phi of X, the idea is that when X is between A and B, Y will be between phi of A and phi of B. So you have phi A here and phi B here.
46:12:310Paolo Guiotto: So that's finally the formula. So in this example, when X is 0, you get 0. When x is 1, you get 1. So this is incidental, if I…
46:25:780Paolo Guiotto: just to convince you that it's just a casuality, let's say that we integrated from 0 to 4 for X, then for Y, I will integrate from 0 to 2, okay? So let's use this example in such a way that
46:40:650Paolo Guiotto: You are not pumping back.
46:43:550Paolo Guiotto: Now, the… the…
46:46:370Paolo Guiotto: The point is that if you look at this second integral, it is easier, because you see, we do an integration by parts, and we go back to some elementary calculations.
46:57:600Paolo Guiotto: While on the other side, I don't see what is the function whose derivative is e to root of X. So, that's why we change variable, okay? So, the change of variable formula is the following.
47:10:90Paolo Guiotto: So, integral from A to B, F of X, DX,
47:15:430Paolo Guiotto: When we introduce y equals phi of X,
47:19:520Paolo Guiotto: That is X equals phi minus 1 of Y.
47:24:600Paolo Guiotto: is the integral between Phi A and Phoebe.
47:28:670Paolo Guiotto: of F, written in the new variable.
47:32:320Paolo Guiotto: So this does not mean change X into Y, but change X into what corresponds to X in the Y variable, and this is given by that map C. Here you have the modulus of P-1 prime.
47:47:750Paolo Guiotto: Why? No amount of serving means no modules.
47:51:400Paolo Guiotto: and then DY.
47:55:990Paolo Guiotto: Okay.
47:57:790Paolo Guiotto: We can, take a short break now.
48:01:150Paolo Guiotto: And then we see how this formula becomes…
48:06:170Paolo Guiotto: We need to work first a little bit on this formula, how it becomes for a multiple integral.
48:15:730Paolo Guiotto: Oops.
48:16:510Paolo Guiotto: Opposed?
48:21:440Paolo Guiotto: Okay, Michael.
48:27:960Paolo Guiotto: Take your seats, please.
48:33:720Paolo Guiotto: Okay.
48:35:910Paolo Guiotto: Now, before we can, we can imagine how this formula becomes, for multiple integrals.
48:50:200Paolo Guiotto: We have to do a little restyling of this formula, because when we write a multiple integral, we are used to say it is an integral on a domain.
49:03:590Paolo Guiotto: Okay, we use this notation because there is not anymore the range A to B, okay? This is when we apply the reduction formula, but in general, whenever we have a multiple integral, it is on a domain.
49:17:350Paolo Guiotto: So, let's write this formula with integration on a domain. So, how it could be this one? This could be the integral on the domain which is made by the interval AB of the function F.
49:32:550Paolo Guiotto: about the right-hand side, there is a little, issue, because it is not correct to say that it is the interval PA Phoeb. Let's see why.
49:46:660Paolo Guiotto: Because, basically, this map fee can be of two types.
49:51:800Paolo Guiotto: We can have a fee which is increasing.
49:56:360Paolo Guiotto: So imagine that this is the map… this is the X axis, the Y axis. Let's say that X is from A to B, so this is where we have the X.
50:08:110Paolo Guiotto: And so, imagine that you have your function phi, which is increasing, so you start from this point, this is the value of phi of A,
50:16:920Paolo Guiotto: And the function is made like that.
50:20:340Paolo Guiotto: So this will be the value P of B.
50:24:530Paolo Guiotto: And when X belongs to… between A and B,
50:30:270Paolo Guiotto: P , which is the Y,
50:34:20Paolo Guiotto: will be between phi of A and phi of B.
50:38:400Paolo Guiotto: So…
50:39:770Paolo Guiotto: the range from phi of A to phi of V is exactly the interval with the endpoints phi A and phi B.
50:50:880Paolo Guiotto: But, if the function ph is decreasing, something different happens, because again, we have the two axes, X and Y, the interval AB,
51:03:940Paolo Guiotto: But now imagine that the function is decreasing. So you start from a value, and then you decrease down to another value. So phi of A is here.
51:15:110Paolo Guiotto: And the phi of B is down here.
51:18:150Paolo Guiotto: So you cannot say that you go from phi of A to phi of B, but the contrary.
51:23:30Paolo Guiotto: The interval is PhiB2 phi8.
51:28:330Paolo Guiotto: So we have to distinguish these two situations. So… about,
51:35:200Paolo Guiotto: The right-hand side of the change of variable formula.
51:40:70Paolo Guiotto: We have that. If the function phi is increasing, the integral from phi of A to phi of B
51:49:710Paolo Guiotto: of the function F, phi minus 1, y, Derivative of phi minus 1, Why? Why?
52:00:620Paolo Guiotto: Now, this is really the integral on the interval from phi A to phi B, of FP.
52:13:230Paolo Guiotto: Minus 1YP minus 1 prime.
52:17:890Paolo Guiotto: Why do you want?
52:20:340Paolo Guiotto: And this is clear.
52:22:410Paolo Guiotto: But when fee is decreasing, I do not rewrite this, so let's say star, yeah.
52:32:850Paolo Guiotto: start here. Let's see what is the same integral in this case. Now.
52:39:00Paolo Guiotto: In this second case, when phi is decreasing, phi of A is bigger than phi of B, as you see in the figure.
52:45:760Paolo Guiotto: So we are in the case when the integration is from a number below which is greater than the number above.
52:54:150Paolo Guiotto: So it's like if the two endpoints are inverted in order. We can put in normal order, flipping the two endpoints, but we have to change the size. So this will be minus the integral from phi of B to phi of A of the same thing, I do not copy.
53:12:80Paolo Guiotto: And now this is an interval on an interval. So this is now minus integral on the interval that goes from C of B
53:23:300Paolo Guiotto: Don Fieuve.
53:26:70Paolo Guiotto: of the same quantity, F phi minus 1Y.
53:31:340Paolo Guiotto: phi minus 1 prime YVY.
53:36:610Paolo Guiotto: So now we have this problem that we have two formulas in some sense, no? We have this is the case when P is increasing.
53:47:200Paolo Guiotto: And this is the case for the right-hand side. For the left-hand side, we have a unique formula, but for the right-hand side, we may have two different representations, depending on fee increasing or fee decreasing.
54:02:750Paolo Guiotto: Now, what I want to do is to unify these two in a unique formula.
54:07:130Paolo Guiotto: Well, the first remark is that, Let's unify this.
54:17:130Paolo Guiotto: these… formulas.
54:22:690Paolo Guiotto: So, I noticed that if I am, in the first case, phi increasing, that interval, the interval from phi of A to phi of B,
54:34:320Paolo Guiotto: Let's go back to the figure.
54:36:980Paolo Guiotto: We are in the left field.
54:38:850Paolo Guiotto: No?
54:39:880Paolo Guiotto: It's about PR.
54:43:130Paolo Guiotto: is the image of the interval AD
54:48:90Paolo Guiotto: So I can say it's P of the interval in P.
54:57:630Paolo Guiotto: So, you know what doesn't mean P of a sector is an agreement to say it's a sector made of all C of X when X is between A and B.
55:07:750Paolo Guiotto: And the same happens also for phi decreasing, because for phi decreasing, the interval is the opposite, phi B to phi A,
55:16:800Paolo Guiotto: But again, if you go back to the figure, This is now being blood.
55:22:350Paolo Guiotto: This is, again, the image of AV through the mapped field.
55:27:70Paolo Guiotto: So it's the same thing.
55:29:850Paolo Guiotto: So this one is also phi of AB.
55:34:890Paolo Guiotto: Okay, so this at least says that in these two formulas.
55:40:50Paolo Guiotto: The interval down there can be written in the same way, okay?
55:44:590Paolo Guiotto: So, this means that, the integral on, PA, to Phoebe…
55:56:20Paolo Guiotto: of… I do not copy all the store inside, can be seen as the integral on the set, which is the interval image through the map V of the interval AB of the same thing.
56:08:860Paolo Guiotto: This is for Formula, let's say, 1.
56:14:870Paolo Guiotto: And this is Formula 2.
56:18:680Paolo Guiotto: So this is Formula 1.
56:21:810Paolo Guiotto: while, for Formula 2, I take also the minus with me, so minus the integral, etc.
56:29:380Paolo Guiotto: minus the integral from this case, I would have the…
56:33:90Paolo Guiotto: the end, the interval written in this way, phi B, phi A,
56:37:430Paolo Guiotto: of et cetera, is minus the integral on the set, which is the image of the integral AB of
56:45:200Paolo Guiotto: the same quantity. So now, if you look at these two formulas, there is a unique difference, because what is written inside that line is the same thing. It's F of phi minus 1, y, then there is phi minus 1 prime Y, DY. For both, it's the same here, okay?
57:04:460Paolo Guiotto: So the unique difference is now that you have a plus here and a minus here.
57:10:410Paolo Guiotto: But… If you look carefully.
57:13:560Paolo Guiotto: So let's focus now on… I will rightly write the term we have to, focus.
57:21:450Paolo Guiotto: on, so I will cancel this, I will rewrite in a second.
57:26:460Paolo Guiotto: So let me just write FE minus 1Y,
57:30:740Paolo Guiotto: And the two factors here, phi minus 1 prime, Y,
57:36:310Paolo Guiotto: And the same here, C minus 1 prime y.
57:43:830Paolo Guiotto: Now, let's keep for a second the first one, but let's look at the second line.
57:50:60Paolo Guiotto: This is the case when phi is decreasing. So, first case is phi increasing, second case, phi decreasing.
57:59:710Paolo Guiotto: If B is decreasing, we notice also that
58:04:100Paolo Guiotto: If you rotate your head of 90 degrees, you see the figure of phi minus 1.
58:09:770Paolo Guiotto: So you have to look at the Y-axis as the original. Okay, so go ahead and get it that way.
58:17:430Paolo Guiotto: And what do you see? This is the increasing phase, so the y-axis goes that way.
58:22:800Paolo Guiotto: So this is the FCC. What you see here, the function, this is the plot of P minus 1.
58:29:380Paolo Guiotto: And this is… We think about the quizzing
58:34:320Paolo Guiotto: It is increasing. While this one… now, imagine that you go that way, you see that the function goes down, that's decreasing. So, in other words, both P and P-1 have the same level.
58:46:960Paolo Guiotto: If one is increasing, the other is increasing. If one is decreasing, the other is decreasing.
58:52:560Paolo Guiotto: So, in particular.
58:56:770Paolo Guiotto: Since, let's write here this, also phi minus 1 is increasing, and phi minus 1 here is decreasing.
59:05:660Paolo Guiotto: Well, if I give this minus to the derivative.
59:10:90Paolo Guiotto: Now, the derivative here, since the function is decreasing, the derivative will be Negative.
59:18:30Paolo Guiotto: So, with the minus in front, minus… that value is positive, and what value is this one?
59:26:40Paolo Guiotto: minus… phi minus 1 prime of Y.
59:31:950Paolo Guiotto: Knowing if you have that, alpha is negative, minus alpha is… Yeah, thank you. Is also…
59:48:480Paolo Guiotto: The absolute value of alpha, right?
59:51:760Paolo Guiotto: So this is the absolute value of phi minus 1 prime y.
59:57:700Paolo Guiotto: Can be written in that way.
59:59:610Paolo Guiotto: And the other one can be written the same way, because in this case, this is now positive, so it coincides with its absolute value.
00:08:310Paolo Guiotto: So now I have a unique shape for the two. So I can say that the right-hand side, finally, can be written as an integral on the image of the interval A,
00:20:950Paolo Guiotto: Whatever it is, this interval.
00:23:390Paolo Guiotto: of F evaluated the phi minus 1Y, modulus of the derivative of phi minus 1, at point Y, DY.
00:34:990Paolo Guiotto: So, after this work, we can say that
00:39:410Paolo Guiotto: Our change of variable formula, which was written there, can be written in a shape which is more similar to what it will be for multiple integrals, where we do not have integral from A to B, but integral on domain. So I wrote this, I did this to have these integrals.
00:57:370Paolo Guiotto: as integrals on a domain, where I see clearly what is the domain.
01:01:720Paolo Guiotto: So… So, D.
01:06:190Paolo Guiotto: Change.
01:08:650Paolo Guiotto: of variable, formula.
01:14:300Paolo Guiotto: becomes… this.
01:19:50Paolo Guiotto: integral on AB.
01:21:560Paolo Guiotto: of F of X.
01:24:860Paolo Guiotto: VX, huh?
01:26:610Paolo Guiotto: when I introduce this change of variable, Y equal, D of X.
01:33:750Paolo Guiotto: That means X equals phi minus 1Y. I said above that phi and phi minus 1
01:42:940Paolo Guiotto: So, let's say fee invertible, invertible.
01:52:390Paolo Guiotto: And the 2 phii phi minus 1 differentiable.
01:58:230Paolo Guiotto: Then this becomes the integral on the image of the interval AB through this map P of F, P minus 1Y,
02:08:630Paolo Guiotto: absolute value of V minus 1 prime, Y,
02:13:290Paolo Guiotto: in the Y. That's the unified way to write this book.
02:17:160Paolo Guiotto: And it is exactly in this format that I have the analogous for the multidimensional integral.
02:25:340Paolo Guiotto: So, we have this theorem, so this formula.
02:34:560Paolo Guiotto: Expanse.
02:38:930Paolo Guiotto: to multiple… You eat the grouse?
02:46:440Paolo Guiotto: in this.
02:49:610Paolo Guiotto: way.
02:50:990Paolo Guiotto: So it's pretty much the same, but of course, there will be something that should be adapted, because, for example… well, let's say… let's see, writing the formula, what is it? Theorem, so…
03:05:980Paolo Guiotto: let's imagine that we have an integral on a domain D. It is better if, instead of writing f of x, y, things like this, we keep a vector notation, so F of X dx.
03:20:300Paolo Guiotto: So in this way, you see… How much, is,
03:25:20Paolo Guiotto: looks like this one, okay? So imagine that you are… you have to integrate a function f on a domain D, and for some reason, you think that it might be better to introduce a new variable that will be a vector now.
03:39:430Paolo Guiotto: So it will be a vector, function of the original variable X.
03:45:690Paolo Guiotto: Now, look at this map.
03:49:60Paolo Guiotto: This map fee now should be a map of what type? It's a map that takes a vector X and gives a vector Y, so it's a… it's a vector-valued map of vector variable, and more precisely.
04:06:940Paolo Guiotto: Here, there is a technical point, but whether we accept for a second that this map goes from domain D, contained into, say, RN, to
04:20:600Paolo Guiotto: a domain, its image, phi of D, which will be still a subset of Rn. So, we transform vectors into vectors.
04:33:270Paolo Guiotto: In such a way that this map be invertible, in such a way that… In such.
04:41:790Paolo Guiotto: the way… That, exactly as we have here, famous being vegetable, B.
04:51:200Paolo Guiotto: Ease.
04:52:290Paolo Guiotto: Invertible, let's the arrow. Invertible.
04:58:900Paolo Guiotto: So there exists the inverse.
05:01:620Paolo Guiotto: So this relation can be inverted.
05:05:950Paolo Guiotto: And the inverse will be called, phi minus 1.
05:14:440Paolo Guiotto: And we need also that the two, phi and the inverse, are differentiable, with the phi
05:22:710Paolo Guiotto: And the inverse of phi differentiable.
05:34:130Paolo Guiotto: Now, differentiable, be careful, these are maps from our N to our N.
05:39:250Paolo Guiotto: Okay? So, what is the derivative here?
05:44:490Paolo Guiotto: What kind of object is the derivative?
05:49:420Paolo Guiotto: It's a matrix, okay? And you know what kind of metrics? The size of these metrics?
05:57:330Paolo Guiotto: is an N by n matrix, okay? So, derivative, just for phi, but also for phi minus 1. So for phi, the phi prime of X,
06:08:970Paolo Guiotto: will be a matrix, which is the Jacobian matrix, Jacobian matrix.
06:16:740Paolo Guiotto: Now, to write this metrics, I needed to introduce components, so if phi, if phi has
06:26:550Paolo Guiotto: end components, say, V1, Phi 2, Piena?
06:32:380Paolo Guiotto: This matrix is made of all the partial derivatives, or if you want, the gradients of phi1, the gradient of phi2, etc, the gradient of phi n. Each of the gradients is the array of the partial derivatives. So this is an N by N matrix.
06:54:200Paolo Guiotto: And the same is for the inverse, okay? Also, the inverse is an object of the same nature, because it is a function from Rn to Rn, so the derivative of this
07:07:250Paolo Guiotto: As a function, let's call y the variable for phi minus 1, will be a Jacobian matrix, again.
07:18:330Paolo Guiotto: Which is made of gradients, If the map phi-1 has components,
07:27:560Paolo Guiotto: I don't want to write P1 minus 1, P2 minus 1, because it seems like if those are invertible, they cannot be invertible, because the components, the single component, each single component is a function
07:42:870Paolo Guiotto: n variables, numerical variables. You cannot have a projection between Rn and R1. So the single components are not between those functions. It's the function itself which is invaluable, so let's call them C1,
07:58:330Paolo Guiotto: CN, the components.
08:00:980Paolo Guiotto: So here we will have the gradients of these components, C1, CN, and this one is the, is also an N by N matrix. They are both N by N matrix. This is important because in a moment.
08:14:310Paolo Guiotto: You have to remind that in the formula, there is that quantity, phi minus 1 prime, so we have to understand what becomes that.
08:21:670Paolo Guiotto: However, when I do the change of variable, the new integral will be an integral.
08:28:330Paolo Guiotto: on the domain, as you can see here, which is the image through the map fee of the original domain. So what I have here is the image through the map fee of domain.
08:40:390Paolo Guiotto: About F, I have the same thing, so you have to rewrite F in the new variable, so you evaluate F at phi minus 1 over Y,
08:52:439Paolo Guiotto: And now, the point is, what replaces that modulus of phi minus 1 prime?
08:59:729Paolo Guiotto: Now, we need a number here, because that quantity is a number.
09:03:850Paolo Guiotto: And phi minus 1 prime is a matrix.
09:07:680Paolo Guiotto: So what is the magic number that I have to put there?
09:11:450Paolo Guiotto: The magic number is the modulus of the determinant of the Jacobian matrix, P-1 prime y.
09:22:410Paolo Guiotto: And this is in the Y.
09:25:430Paolo Guiotto: So that's the change of variable formula, okay? So, the idea is the same of the one variable. You start from the integral, you decide that it's convenient to introduce a new variable. The variable for the integral here is an array.
09:43:100Paolo Guiotto: If this is a double integral, you have XY. If it is a triple integral, you have XYZ, okay? So you introduce a new array.
09:52:529Paolo Guiotto: That will be made of two variables if you are integrating in two variables. It will be made of three variables if you are integrating in three variables.
10:00:780Paolo Guiotto: It must have the same number of components
10:04:200Paolo Guiotto: Because let's say you transform a triple integral into another triple integral, that maybe is easier. You want to transform a triple integral into a double integral. That's not…
10:14:650Paolo Guiotto: not feasible, because it means that you have a detection between space, R3, and plane R2. That's not possible.
10:24:720Paolo Guiotto: We didn't judge you.
10:25:980Paolo Guiotto: Right?
10:26:910Paolo Guiotto: So, these, the dimension of the two factors must be the same.
10:34:860Paolo Guiotto: There, you have a new integral, which is an integral in the variable Y, so here you write your function in the new variable, and this is the operation for CPR.
10:45:530Paolo Guiotto: This is the integration domain, which is the image of the original integration domain through this path.
10:51:850Paolo Guiotto: And finally, you have this correction factor, which in the scholar case was the derivative, the modulus of the derivative of this phi minus 1.
11:01:590Paolo Guiotto: Now, the derivative of V minus 1 is now a matrix, so you cannot take the modules of a matrix, okay? But the quantity, the right quantity, turns out to be the determinant of this Jacobian matrix, absolute value of this.
11:18:80Paolo Guiotto: That seems extremely complicated.
11:21:540Paolo Guiotto: And in fact, it is a certain sense, but… It is,
11:28:650Paolo Guiotto: at the end, when you learn how to use it, it becomes easy. It's not this one, the problem of computing integrals. The most relevant problem of computing multiple integrals is this one I've shown this morning, which is the proper parameterization of the domain. That's the most important.
11:47:900Paolo Guiotto: In the sense that, most of you
11:52:180Paolo Guiotto: If you… most of you will do errors, see, in computing integrals, we'll do errors at this stage, when they are too parameterized, not to apply the change of variable, or things like this.
12:05:780Paolo Guiotto: Okay, so now, I do not approve this formula. It's extremely technical, so it's much beyond our scopes.
12:17:730Paolo Guiotto: We just learned how to use them.
12:21:70Paolo Guiotto: There is, let's say, one big difference between the use of this formula in the case of one variable integral and multiple variables integral.
12:34:340Paolo Guiotto: In some sense, in one variable integrals,
12:39:680Paolo Guiotto: You can guess what is the reasonable change of variable, because you see how to simplify the function.
12:46:870Paolo Guiotto: For multiple variable integrals, this is much harder.
12:50:910Paolo Guiotto: Because, it's not a question of introducing a new variable function of the other. Here, the variable is a vector, so you have to imagine that you are going to replace… suppose you are doing a triple integral.
13:03:270Paolo Guiotto: you have to replace 3 variables with three new variables, so you have to imagine how to do. So one variable is maybe easy, but two together, or three together, is much more complicated, okay?
13:15:300Paolo Guiotto: So, in general, here.
13:19:200Paolo Guiotto: The important application of this is with special change of variables, which are not important in applications. So, let's start doing, instead of taking… showing you the application in general, that would be at the beginning a little bit too difficult.
13:39:480Paolo Guiotto: I will take a special case of this formula.
13:44:00Paolo Guiotto: with the spatial change of variables, which are very interesting and useful in applications. So we start, let's say, differently from respect to what you have seen in one variable. So let's start talking about spatial
14:05:740Paolo Guiotto: Change.
14:09:160Paolo Guiotto: off valuable.
14:13:440Paolo Guiotto: So, particular, standard, in fact.
14:17:680Paolo Guiotto: The first one is, the first case is the integration
14:28:270Paolo Guiotto: in polar… coordinates.
14:35:950Paolo Guiotto: Then, a second one will be in three variables, in spherical coordinates or in cylindrical coordinates, we will see next time.
14:44:730Paolo Guiotto: Now, the idea is that, suppose we have,
14:52:710Paolo Guiotto: I have to compute an integral in two variables, so F, X, Y, The X, DY.
15:01:780Paolo Guiotto: Now, XY are the Cartesian coordinates, so the situation is the following. We have, in the Cartesian plane, a certain domain, D,
15:13:390Paolo Guiotto: And we describe the points by giving the coordinates of these points.
15:20:780Paolo Guiotto: But we could describe points in another way, giving two other coordinates.
15:27:290Paolo Guiotto: So, one is the distance to the origin, that we call raw, and the second one is the angle made with the positive direction of the x-axis, and we call this variable theta.
15:38:460Paolo Guiotto: Now, from trigonometry, this means that X is equal to raw cosine theta, and Y is equal to raw sine theta.
15:49:700Paolo Guiotto: So these two numbers, rho and theta, are, in fact, two new coordinates, okay?
15:56:330Paolo Guiotto: And, well, think about that if we are on our planet, for example.
16:03:210Paolo Guiotto: The Cartesian coordinates are never used, so you do not say the position of New York in the globe is X equals this, Y equal that, and Z equal that.
16:13:710Paolo Guiotto: It says latitude and longitude, so two coordinates, which are two angles, in fact, so pretty much similar to polar coordinates, so are not so unusual things, this one.
16:25:370Paolo Guiotto: So now we look at this as a change of variable. Let's see how.
16:29:720Paolo Guiotto: So, first of all, we have to specify that XY
16:35:70Paolo Guiotto: is a point of R2, so it belongs to D, which is a domain of R2.
16:41:680Paolo Guiotto: For the two coordinates rot theta, we cannot say that they are in R2.
16:47:350Paolo Guiotto: Because rah is a distance, so raw belongs to positive numbers.
16:53:860Paolo Guiotto: N theta is an angle.
16:55:990Paolo Guiotto: Of course, we can use infinitely many angles, because 0 is the same of 2 pi, which is the same of 4 pi, which is the same of 6 pi, which is the same of minus 2 pi, but this would mean infinitely many coordinates for the same point.
17:10:940Paolo Guiotto: We have one to do a soft objection, one-to-one correspondence between XY and 1 theta. So that's why we choose theta conventionally between 0 and 2 pi, excluding 2 pi, because 2 pi is the 360 degrees angle, so it means the same point.
17:31:70Paolo Guiotto: So, it means that, if you consider, in Poland.
17:38:50Paolo Guiotto: So the Cartesian plane, where the two coordinates are, the polar coordinates, rho and theta, Whatever is the domain.
17:46:730Paolo Guiotto: Each point of the domain will have a counterpart in this plane, rhet theta. Well, be careful, because rho is only positive, so we discard all this part here, and theta is between 0 and 2 pi, so we also discard… well, actually, let's say that…
18:06:180Paolo Guiotto: This is…
18:08:190Paolo Guiotto: If this is the line at, quote, 2 pi, so we do not consider these points here, and also these points here.
18:16:210Paolo Guiotto: So, let's say that everything in Cartesian plane is actually represented by this special region in the rotita plane. So, the rotita plane, that red strip, is the Cartesian plane.
18:34:200Paolo Guiotto: in polar coordinates, okay? You see this?
18:38:630Paolo Guiotto: Now, if you look at this, this is a map that gives XY as function of raw theta. So this, if I look as a map, it is saying that XY is… I do not put a letter, I will put in a minute, function of the pair raw theta.
18:57:780Paolo Guiotto: So now imagine that, since we are doing the integration in X and Y, the pair, the vector X, Y, is our starting variable. So in the notation of this is the vector X.
19:12:880Paolo Guiotto: So the vector Y is the new variable.
19:17:60Paolo Guiotto: Which is the pair Rotita in this game.
19:20:830Paolo Guiotto: Now, using those correspondence, this says Y function of X and X function of Y, the second line.
19:28:970Paolo Guiotto: Now, just imagine that this X is the pair XY, and Y is the pair raw theta.
19:36:90Paolo Guiotto: So that correspondence that you see here.
19:40:990Paolo Guiotto: which is the typical way to describe polar coordinates, is giving X and Y as function of rho and theta.
19:49:350Paolo Guiotto: So, if you look at these two, Meet these, these swamp.
19:56:30Paolo Guiotto: Because, you see, you have the initial coordinates
20:00:230Paolo Guiotto: function of the new coordinates. So, what I have there is not this, but this. So, I have that this function
20:11:230Paolo Guiotto: is the phi minus 1, in… D.
20:17:750Paolo Guiotto: notations… off.
20:22:980Paolo Guiotto: Change.
20:24:610Paolo Guiotto: of variable formula.
20:29:150Paolo Guiotto: If you want, we could do the opposite. We actually don't need strictly the opposite, because if you look at the formula, what we need mostly is the phi minus 1, not phi. Phi is only there in the domain, okay?
20:44:750Paolo Guiotto: So, and moreover, the vice versa is not easy, because if you want the phi, the phi is the function that, starting from XY gives you rho and theta.
20:58:260Paolo Guiotto: Okay, this is the fee.
21:00:100Paolo Guiotto: Well, we can say what is raw. Raw is easy, because it is the root of X squared plus Y squared.
21:08:860Paolo Guiotto: About theta is a little bit more complicated to write a function of X theta. So, however, let's say that there is this function P, if we will need that, we will write, but we will never need, as you will see.
21:23:540Paolo Guiotto: Okay, so now, since we have a phi minus 1, which is the ingredients we need mostly to write this formula here, because it is in the argument of F, it is in that determinant, let's see, in practice, with this spatial change of coordinates, this formula becomes
21:42:800Paolo Guiotto: In this particular case. So, the integral on D of F in Cartesian coordinates
21:50:210Paolo Guiotto: the XDY will become an integral. Well, let's write for a moment phi of D,
21:57:860Paolo Guiotto: So, here I have to put F evaluated at phi minus 1 of raw theta. Remind that the new variable is the pair raw theta, so the new integration will be in raw and theta.
22:10:220Paolo Guiotto: So what is F of phi minus 1 of rho theta? But phi minus 1 rho theta is raw cosine theta.
22:16:830Paolo Guiotto: raw sine theta. So what you have to do, you have just to plug. In place of X, you put raw cosine theta. In place of Y, you put raw sine theta.
22:29:450Paolo Guiotto: And that's done.
22:31:120Paolo Guiotto: Then we have the modulus of the determinant of the phi minus 1 prime, since we have a very long thing.
22:37:380Paolo Guiotto: As function of the variables rho and theta.
22:40:910Paolo Guiotto: Now, this quantity is always the same. Whenever you use this change of variable, that quantity will be always the same, so you don't have to repeat each time the calculation of this. So we will do once forever. So let's do this calculation. Let's start doing the Jacobian matrix.
22:59:410Paolo Guiotto: So, I remind you that the phi minus 1 of rh theta is this math, rock cosine theta.
23:06:920Paolo Guiotto: Raw Scienta.
23:09:440Paolo Guiotto: So if you want, in the notations we wrote above.
23:14:230Paolo Guiotto: So, you see, phi minus 1 has component C1, P2, CN.
23:19:800Paolo Guiotto: These are deep sea. This is C1, and this is C2.
23:25:330Paolo Guiotto: So my Jacobian matrix is the matrix made of gradient of the first component, which is raw cos theta.
23:34:70Paolo Guiotto: A gradient of second component, raw sine theta.
23:40:580Paolo Guiotto: Gradient respect, too?
23:44:230Paolo Guiotto: XY or raw theta?
23:50:690Paolo Guiotto: raw theta are the variables for this. There is no XY here. So, this means that the two derivatives are the derivative with respect to rho and the derivative with respect to theta. So, when I do the derivative with respect to rho of raw cos theta, I get cos theta.
24:06:620Paolo Guiotto: Derivative with respect to theta is minus raw sine theta.
24:11:710Paolo Guiotto: Second line, derivative with respect to raw… of raw sine theta is sine theta.
24:17:620Paolo Guiotto: The relative respect to theta is raw cos theta.
24:21:990Paolo Guiotto: Okay, that's the Jacobian matrix. Now we have to do the modules of the determinant.
24:26:860Paolo Guiotto: Modulus of determinant of phi minus 1 prime rheta is…
24:33:270Paolo Guiotto: the modules of the determinant of this matrix. What is the determinant? It's cos theta times rho cos theta, so rho cos square theta minus… then I have, you see, minus rho sine theta sine theta, so minus rho sine squared theta.
24:53:500Paolo Guiotto: But if you put together, you can factorize rho times cos squared theta plus sine squared theta.
25:03:150Paolo Guiotto: which is constantly equal to 1, and so we get absolute value of raw. But do not forget that raw here is positive.
25:12:370Paolo Guiotto: So, it's just raw. And so, we got this formula. So, the modulus of the determinant of the phi minus 1 prime raw theta, if phi is the change of variable in polar coordinate, this is raw.
25:31:650Paolo Guiotto: So, at the end, what you have to write back in that integral is raw.
25:39:30Paolo Guiotto: Okay? And this is now the change of variable formula.
25:46:120Paolo Guiotto: Let's… let's do a very quick example in order that we fix this idea.
25:53:30Paolo Guiotto: to show you that it's very easy. So, example.
26:02:730Paolo Guiotto: I want to compute that this is the 4545.
26:08:530Paolo Guiotto: I want to compute the integral on the full plane, R2,
26:13:30Paolo Guiotto: of E minus the root of X squared plus Y squared in the XDY.
26:21:940Paolo Guiotto: Well, as you can see, I am in a case where the function here is good enough to be continuous, but the domain is unbounded.
26:35:350Paolo Guiotto: So I should check integrability.
26:37:880Paolo Guiotto: Okay?
26:39:20Paolo Guiotto: Actually, you notice that since the function f is positive.
26:44:230Paolo Guiotto: If I do the iterated integral for the absolute value, I will have the integral for F, okay? But instead of doing that, we apply the change of variable in polar coordinates.
26:57:720Paolo Guiotto: Now, let's write this integral in the new pair rho and theta.
27:04:260Paolo Guiotto: So what happens to the function? Let's start with the… what is written inside, then we see the domain.
27:11:720Paolo Guiotto: What happens to the quantity E minus the root of X squared plus Y squared?
27:17:920Paolo Guiotto: E to the minus rho, because whatever… sorry, the change of variable is, of course, X equal rho cos theta.
27:24:360Paolo Guiotto: Y equals raw sine theta. So, it is clear that with this change of variables, the quantity root of X squared plus Y squared is the norm of the vector, which is exactly raw. So we get e to minus raw.
27:39:20Paolo Guiotto: That's all for the function. I now need to put the… this factor, this modulus of the determinant of the Jacobian matrix of the inverse, all this to say raw. And that's all for the integration inside. Now the domain.
27:56:740Paolo Guiotto: Well, it's easy because here, my domain in XY is the full Cartesian plane R2, right?
28:06:630Paolo Guiotto: So it's everything.
28:08:760Paolo Guiotto: And, what is the… so I have… I don't need, as you can see, I don't need to write what is exactly this map fee, because if I have all points.
28:21:110Paolo Guiotto: It means that I must have all coordinates here.
28:24:970Paolo Guiotto: So the domain, which is the image in polar coordinates of far2, is the set made of all possible rho and all possible thetas, because they describe all points in the Cartesian plane. So for rho, all positive values, and for theta, all thetas between 0 and 2 pi.
28:44:920Paolo Guiotto: Okay, is that clear?
28:47:190Paolo Guiotto: So what are the polar coordinates of all points of Cartesian plane? Are all polar coordinates.
28:54:210Paolo Guiotto: Okay? Because whatever raw positive means every distance, whatever is theta between 0 and 2 pi means every angle.
29:02:690Paolo Guiotto: So, this means that I have raw positive.
29:05:930Paolo Guiotto: And theta between 0 and 2 pi.
29:10:20Paolo Guiotto: Okay, now I have this new integral. I can here apply the reduction formula.
29:17:700Paolo Guiotto: So decide to integrate this into first integration in TITA, and then involved.
29:23:940Paolo Guiotto: The function is, E minus raw times raw.
29:29:130Paolo Guiotto: Now, about the range. Now, I know that this is the new domain for raw and Tita.
29:36:640Paolo Guiotto: As you can see, it's nice because there is no condition between the coordinates, so I can say that theta is between 0 and 2 pi.
29:45:200Paolo Guiotto: When raw is positive, between 0 and plus infinity.
29:49:760Paolo Guiotto: Now, this is a constant for this integral, so it comes out, and I have integral 0 to plus infinity of rho E minus raw.
29:59:60Paolo Guiotto: The integral is the integral from 0 to 2 pi of 1, so it gets 2 pi, 0, and now you can easily finish this value.
30:09:150Paolo Guiotto: At the end, well, you can do an integration by parts, looking this as derivative with respect to rho of minus e to minus rho.
30:20:90Paolo Guiotto: So, we have that.
30:22:790Paolo Guiotto: This is equal to… stop it… 2 pi…
30:27:50Paolo Guiotto: Then we have the evaluation of, raw
30:31:490Paolo Guiotto: E minus raw, minus, between raw equals 0 and raw equals plus infinity.
30:38:420Paolo Guiotto: minus integral 0 to plus infinity, now derivative moves on this factor, we get 1 times minus E minus 4.
30:46:830Paolo Guiotto: You're wrong.
30:49:710Paolo Guiotto: about the evaluation at plus infinity, the exponential kills the power. At zero, it is 0, so this is just a zero. Minus, minus, plus, so we have 2 pi.
31:01:260Paolo Guiotto: Integrals 0 to plus infinity of E minus 1.
31:05:570Paolo Guiotto: VROC.
31:06:910Paolo Guiotto: And this is,
31:09:490Paolo Guiotto: This is, again, the derivative of minus E minus r, so we get 2 pi, the evaluation of minus E minus rho, between rho equal 0 and rho equal plus infinity. At plus infinity, we get 0, at 0 we get 1 with the minus, so at the end we get 2 pi.
31:29:140Paolo Guiotto: Okay?
31:30:850Paolo Guiotto: So, do please, these, exercises at the end.
31:37:860Paolo Guiotto: So, do.
31:39:310Paolo Guiotto: The exercise of 5.7.3.
31:45:210Paolo Guiotto: Skip the 574.
31:48:210Paolo Guiotto: And do the 575… These are all with polar coordinates. Number 1… Number 2…
32:04:620Paolo Guiotto: And that's it for the moment, okay? So do these three exercises. Please do by Friday, we will see the solution at the beginning of the class, at least two out of 3 of this.
32:17:50Paolo Guiotto: Okay, have a nice day.