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00:08:560Paolo Guiotto: Okay.
00:11:390Paolo Guiotto: Some of the exercises, but let me first refresh.
00:15:710Paolo Guiotto: This formula, integration formula in polar coordinates.
00:21:640Paolo Guiotto: This is a particular case of change of variable formula.
00:25:840Paolo Guiotto: So… Integration…
00:39:330Paolo Guiotto: is polar.
00:42:520Paolo Guiotto: Oh, neat.
00:46:670Paolo Guiotto: So we have an integral on a domain D of a function of two variables.
00:52:910Paolo Guiotto: Say X, Y.
00:55:770Paolo Guiotto: We can transform this into an integral.
01:00:930Paolo Guiotto: of, well… Let me just remind the general formula, first up.
01:09:370Paolo Guiotto: This would be F evaluated at P minus 1 of
01:13:760Paolo Guiotto: Let's say, new variable, so new.
01:17:240Paolo Guiotto: modules of the determinant of the Jacobian matrix of T minus 1, you mean.
01:27:570Paolo Guiotto: If, our change of variable if the… Change it.
01:39:150Paolo Guiotto: of… valuable.
01:41:940Paolo Guiotto: Ease.
01:44:870Paolo Guiotto: X equals raw cos theta Why? It was a raw scientifica?
01:52:870Paolo Guiotto: well, Y and P are the polar coordinates. In this case, you see that
01:56:980Paolo Guiotto: So, with this notation, the map phi is the map that introduces new coordinates UV as function of the initial coordinates XY.
02:09:389Paolo Guiotto: It's a map that maps XY into UV.
02:13:540Paolo Guiotto: And the phi minus 1 is the, is the map that does the opposite, so it gives XY as function of the new variables UV.
02:23:720Paolo Guiotto: So, in this notation, what you see here is XY function of raw theta. So, the starting variable, function of the new variable. So this is not P, but P-1, alright.
02:39:00Paolo Guiotto: And we have seen that the modulus of the determinant of the Jacobian matrix of this is always equal to raw, so we don't need to compute
02:49:40Paolo Guiotto: every time.
02:50:620Paolo Guiotto: With this change of variable, the formula becomes integral of F, X, Y, is equal to…
03:01:30Paolo Guiotto: So we have a… it's not easy to have a… it is possible, but it is not needed, necessarily, to have the direct definition of this P. We will see in a moment how to interpret this P of D.
03:17:780Paolo Guiotto: Then, the function F becomes F of phosph, phi minus 1 is already there, so it means that in place of X, you have to put raw cos theta. In place of Y, you have to put raw sine theta.
03:31:530Paolo Guiotto: Then the modulus of the determinant of E minus 1 prime is just raw, and it becomes an integration in raw and theta.
03:39:360Paolo Guiotto: Now, what is this, P of D?
03:44:160Paolo Guiotto: Because, as you see, the change of variable in the integral is now easy. Now, it's just a question of replacing X with drawc of theta, why we draw from theta, and multiply by that. So, do not forget there are the Y's
03:59:140Paolo Guiotto: you, you miss something important. Now, what is that phi of D?
04:03:240Paolo Guiotto: Now, fee of D is literally the image through the map fee of the domain D.
04:09:110Paolo Guiotto: So it means that we start for the domain D in the plane XY,
04:15:600Paolo Guiotto: But we have a certain subset, yeah.
04:19:540Paolo Guiotto: Now, the map that transforms Cartesian coordinates in polar coordinates, it's not written there. However, we know that it gives, in image, the set fee of D. So it gives the polar coordinates of points of D.
04:36:810Paolo Guiotto: We know that polar coordinates, the raw and theta, are not any reals, because, for example, raw cannot be negative.
04:45:560Paolo Guiotto: And theta is just between 0 and 2 pi, so this is to have a dejection.
04:51:150Paolo Guiotto: So, actually, we have to discard all this part of the plane rotator, because phi of D will be some set in the… in this strip, and that's literally phi of D
05:08:790Paolo Guiotto: Ease.
05:10:50Paolo Guiotto: Indeed.
05:11:420Paolo Guiotto: Set.
05:12:970Paolo Guiotto: Off.
05:14:540Paolo Guiotto: Paula… coordinates.
05:19:590Paolo Guiotto: Off.
05:20:700Paolo Guiotto: points… of the… So, in other words, is the vaccine in polar coordinates?
05:29:320Paolo Guiotto: So… It is.
05:32:270Paolo Guiotto: Well, you may say, It is.
05:37:770Paolo Guiotto: D… written… in… fall out.
05:46:650Paolo Guiotto: coordinates.
05:49:80Paolo Guiotto: Remind that for Rotita, we have these constraints. Raw is any positive number.
05:57:510Paolo Guiotto: And theta is any number between 0 and 2 power.
06:02:910Paolo Guiotto: Okay?
06:06:40Paolo Guiotto: So, let's start doing a simple example where we do not have particular difficulty to determining what is the description of the domain in polar coordinates. I told you to do some of the exercises. I will take some of the 575.
06:22:160Paolo Guiotto: The number one is compute the integral on this domain, x squared plus Y squared, less or equal 4 then of the root of 4 minus X squared minus Y squared dx dy.
06:43:430Paolo Guiotto: So, for the moment, of course.
06:48:120Paolo Guiotto: When you have a problem, like computing an integral, you don't know exactly how you have to proceed.
06:55:480Paolo Guiotto: So, you have to… you need to develop a little bit of experience in such a way you know what kind of strategy you could employ here. But let's say that for the moment, the goal is to use a polar integration, so we have to start learning out towards this polar.
07:13:710Paolo Guiotto: So, this is for the first time, because,
07:16:840Paolo Guiotto: It's easy, so we can draw figures and things like that. But in general, you don't need to do any figure, and you have to
07:26:250Paolo Guiotto: determine the description of domain B in polar coordinates just by looking at the constraints you have there in Cartesian coordinates. So, for this first example, I will do
07:39:160Paolo Guiotto: first a solution with, by doing figures, and then we will review this by working directly on the constant, okay?
07:48:150Paolo Guiotto: So now, first of all, the function is this one.
07:52:720Paolo Guiotto: root of 4 minus, if you want, X squared plus Y squared.
07:58:820Paolo Guiotto: which is a well-defined function on that domain. You see, X squared plus Y square less or equal than 4 means that the argument of the square root is positive, so it is well defined.
08:15:620Paolo Guiotto: And, continuous.
08:19:990Paolo Guiotto: on domain D, which is the set where X squared plus Y squared is less or equal than 4, which is closed.
08:31:580Paolo Guiotto: And bounded, clearly.
08:34:490Paolo Guiotto: So it is compact, and therefore the function is integral, there exists the integral on X squared plus Y squared… well, let's say integral only of M.
08:48:650Paolo Guiotto: Now, to compute the integral.
08:57:730Paolo Guiotto: we apply… Not the reduction formula, as we read so far, but… Change. Off.
09:08:460Paolo Guiotto: Volleyball.
09:10:20Paolo Guiotto: variables, if you think to XY, in… polar.
09:17:100Paolo Guiotto: coordinates.
09:19:280Paolo Guiotto: Now, what happens here? We have that the integral only of f, this is in the Cartesian coordinates XY,
09:28:30Paolo Guiotto: becomes the integral on the image of D. So, how is D in polar coordinates?
09:36:270Paolo Guiotto: of the function, which is the root of 4 minus… you see that in place of X, you have to put raw cos theta. In place of Y, you have to put raw sine theta. Since there is X squared plus Y squared, you already know that X squared plus Y squared will be
09:54:930Paolo Guiotto: raw square. So this is raw square. So this is how the function is. So this is F of raw cosine theta raw sine theta.
10:06:670Paolo Guiotto: Then, it is not yet finished, because we have to put the raw. This is the modulus of the determinant of P minus 1 prime.
10:16:650Paolo Guiotto: And then we have the new variables raw intended.
10:20:520Paolo Guiotto: Now, what is that phi of D?
10:23:230Paolo Guiotto: This guy here.
10:25:240Paolo Guiotto: So, in this case, we can easily understand, because we can even do a figure for the domain here.
10:31:650Paolo Guiotto: In the Cartesian plane, D is what?
10:35:700Paolo Guiotto: What is that domain?
10:42:260Paolo Guiotto: I want to know what is the domain in the Cartesian plane?
10:47:570Paolo Guiotto: Yeah
10:48:580Paolo Guiotto: It's a disk centered in the origin and radius 2, right? So this is 2. It's the full disk.
10:57:950Paolo Guiotto: Now, if you think, what are the Cartesian coordinators? The polar coordinates of points of this disk. The polar coordinates are… so if you take a point here, are distances to the origin and angles with the positive direction of x-axis.
11:15:480Paolo Guiotto: It should be clear that the image of D is the set of raw and pitas in 0 plus infinity for raw.
11:26:500Paolo Guiotto: And 0 to pi for theta.
11:30:150Paolo Guiotto: Such that, for raw, I have to take, r is the distance to the urging, I have to take any number between 0 and 2.
11:38:630Paolo Guiotto: And for theta, I have to take any number between 0 and 2.5.
11:43:430Paolo Guiotto: Well, less or equal or equal is the same, because the integral does not fill one single part.
11:50:170Paolo Guiotto: So this is the description, so if you want to see
11:54:290Paolo Guiotto: what is this in the raw data plane, huh?
11:59:490Paolo Guiotto: So we said that, in any case, this will be a subset of a strip, so we discard all this part of the plane, huh?
12:08:500Paolo Guiotto: Now, this is the set meta of points rah theta, where rice between 0 and 2, so say that 2 is here, and theta is between 0 and 2 pi, so this is…
12:21:910Paolo Guiotto: Rectangle of PR.
12:24:600Paolo Guiotto: This is P.O.
12:29:580Paolo Guiotto: Okay?
12:30:640Paolo Guiotto: So, this guy here, the disc, Is mapped by the,
12:37:370Paolo Guiotto: Change your variable in polar coordinate into that rectangle.
12:41:950Paolo Guiotto: So this means that the new integral will be…
12:46:770Paolo Guiotto: The integral for rho between 0 and 2,
12:50:340Paolo Guiotto: and theta between 0 and 2 pi.
12:54:120Paolo Guiotto: Of raw root of 4 minus raw square.
12:58:460Paolo Guiotto: D row d theta.
13:01:50Paolo Guiotto: It is now this integral that we are going to compute, okay? So, now, this is a new double integral, doesn't matter if the letters are not XY, but raw theta, I could call Cartagian polar coordinate X and Y.
13:15:590Paolo Guiotto: Of course, it wouldn't be a good choice, but let's keep raw and theta. So we apply the reduction formula. It says we do a double integration. Here, it is indifferent because of the description you have here. So you could say that when rih is between 0 and 2, theta will be between 0 and 2 pi.
13:35:140Paolo Guiotto: The function is raw times root of 4 minus raw square. The first integration is now in theta, and the second is in raw.
13:44:450Paolo Guiotto: Now, in this particular case, the function does not depend on theta, so it's a constant, so it can carry outside of that integral, and it remains the integral from 0 to pi of 1, so it gives 2 pi that I just write outside of
13:59:970Paolo Guiotto: The second integration, integral 0 to 2 of raw times root of 4 minus raw square 0.
14:07:880Paolo Guiotto: Now, it's time to integrate this one. You see that normally it is better to transform this in powers, so this is the power 4 minus rho squared to exponent 1 half.
14:21:120Paolo Guiotto: Now, we recognize that this rho is more or less the derivative of the argument, so probably this comes from doing the derivative with respect to rho of the power 4 minus rho squared with the exponent added by 1, so 1 half plus 1.
14:39:740Paolo Guiotto: In fact, if you do the derivative of this, you get… so this exponent would be 3 halves.
14:46:560Paolo Guiotto: You get 3 hubs.
14:48:850Paolo Guiotto: D, 4 minus raw square.
14:52:750Paolo Guiotto: 2 exponent 3 halves minus 1, so again, 1 half. Then we have the derivative of the argument with respect to rho, which is minus 2 rho.
15:03:150Paolo Guiotto: So, at the end, you can see that we get minus 3 raw times 4 minus raw squared to exponent 1 half. So, if we have that minus 3,
15:14:550Paolo Guiotto: We have a derivative, so we put the minus 3 here.
15:18:460Paolo Guiotto: and we divide by minus 3, we write the other one outside, and this is now the formula. So we have minus 2 pi over 3, and then we have the evaluation. Since this guy is now the derivative of this.
15:32:860Paolo Guiotto: We will do the evaluation of 4 minus raw square to exponent 3 halves between rh equals 0 and raw equals 2.
15:44:740Paolo Guiotto: At rho equals 2, we get 0, no, because rho square is 4, so 4 minus 4 is 0 to 3 are 0, so 0 minus the value at rho equals 0 is a 4 to 3…
15:57:780Paolo Guiotto: Half.
15:59:100Paolo Guiotto: So, at the end, we get the minus, minus, plus, so 2 pi over 3,
16:05:420Paolo Guiotto: 4 to 3 half… well, 4 to 3 halves is the root of 4, so 2 to power 3 8th. So it is 8 times 216 pi over 3, and that's the value of the integral.
16:23:70Paolo Guiotto: Okay?
16:24:800Paolo Guiotto: Now.
16:26:50Paolo Guiotto: Let's go back to the domain, one second.
16:29:680Paolo Guiotto: And let's determine the description in polar coordinate without looking at a specific figure.
16:38:540Paolo Guiotto: In the example.
16:47:510Paolo Guiotto: D is the set of points in Cartoon coordinates, XY, such that X squared plus Y squared is less or equal than 4.
16:57:390Paolo Guiotto: How can I determine the description and polar coordinate of this set?
17:02:760Paolo Guiotto: Now, we may notice that XY
17:08:890Paolo Guiotto: belongs to the if and only if
17:12:359Paolo Guiotto: Well, let's write XY in polar coordinate, means raw cost theta raw scientifica, right?
17:19:940Paolo Guiotto: Now, this point here belongs to the if and a if there is a unique condition there, it means that a raw square is less or equal than 4.
17:31:560Paolo Guiotto: Okay? So it means that the point raw cosine theta, raw sine theta, belongs to D,
17:37:740Paolo Guiotto: If, finally, if raw square is less than 4, that is, since raw for us is positive, raw is less than 2, so it is between 0 and 2.
17:51:970Paolo Guiotto: So this is the condition on raw and theta, because there is also theta. If there is no condition, it means all possible thetas. So here, I can add theta between 0 and 2 pi, because there is
18:07:80Paolo Guiotto: There is no restriction for titas. And what are these conditions? And this is exactly the condition that the pair raw titta must verify to be in such a way that… if you want, this is also phi of raw titta.
18:23:720Paolo Guiotto: So I'm saying that P, so, P of rotita.
18:31:90Paolo Guiotto: belongs to the domain DE, I'm sorry, this was phi-1, sorry.
18:37:380Paolo Guiotto: V-1 rotita belongs to domain D,
18:40:620Paolo Guiotto: If and only if raw is between 0 and 2, and theta is between 0 and 2 pi. And why this is saying
18:51:470Paolo Guiotto: the image. Because saying that phi minus 1 of rotita belongs to D is the same of saying that rotita is… you apply phi to both sides to phi belongs to phi of D.
19:06:60Paolo Guiotto: So, saying that rot eta is in phi of D is the same of saying that P minus 1 rat eta is in D. So, at the end, forgetting this intermediate step, you read that
19:18:270Paolo Guiotto: raw theta is in phi of B, if and on. If raw is between 0 and 2, and theta is between 0 and 2 pi.
19:27:280Paolo Guiotto: So that's the way I determine that set phi of D, okay? So from this, phi of D is the set of raw enthetas.
19:38:450Paolo Guiotto: Such that raw is between 0 and 2,
19:42:510Paolo Guiotto: And theta is between 0 and 2 pi.
19:46:290Paolo Guiotto: So, in other words, how do you determine these conditions? You impose that the point Ra cosine theta, raw sine theta verifies the condition
19:56:450Paolo Guiotto: The conditions, if there are more than one, that must be verified by the Cartesian point XY, because X is raw data, Y is raw and theta. You will get conditions for rho and theta, and this condition will describe the new domain, integration domain, in raw and theta.
20:17:30Paolo Guiotto: Okay.
20:18:320Paolo Guiotto: Let's do a second example.
20:21:300Paolo Guiotto: Betty C, yeah.
20:36:620Paolo Guiotto: This is the value of the integral, has nothing to do with the value of the variables, okay?
20:43:260Paolo Guiotto: This is… can be whatever. Well, not whatever, because, for example, in this case, we are integrating a positive function, so the value must be positive, cannot be negative, but…
20:55:920Paolo Guiotto: Let's do… this is pretty much the same. Let's compute the number 2 integral. Here we have on the domain, which is X squared plus Y squared, less or equal than 1, 1 over 1 plus X squared.
21:14:180Paolo Guiotto: That's wasteful. Yeah.
21:15:910Paolo Guiotto: Yes, DY.
21:20:650Paolo Guiotto: Okay, now let's take this example.
21:24:520Paolo Guiotto: as an opportunity to show that in this case, it wouldn't be reasonable to start doing the reduction formula in X and Y.
21:33:250Paolo Guiotto: And it would be much better to change variable, okay?
21:38:410Paolo Guiotto: So,
21:41:270Paolo Guiotto: let's imagine that we start trying with the reduction formula, and we see that maybe it's possible, I don't know, but the calculation is pretty long. If I use the reduction formula, so I do not pass to the polar coordinate.
22:05:80Paolo Guiotto: So let's see, which seems the same to start integrating in X or in Y. The function depends on X and Y in the same way, the domain is the same, so one is equivalent to the other. So let's start integrating, for example, in X. So we will do this. We have a double integration, 1 over 1 plus
22:24:590Paolo Guiotto: X squared plus Y squared, first in X, then in Y.
22:29:680Paolo Guiotto: And now we have to write the range for X, Y.
22:33:710Paolo Guiotto: Now, let's look at this condition. This condition says X squared plus Y squared is less or equal 1. So, if I isolate X, and I need to do that because the first indication is in X, I read X squared
22:48:460Paolo Guiotto: less or equal than 1 minus Y squared.
22:52:250Paolo Guiotto: So, I would say that X… here, X is a Cartesian coordinate.
22:58:60Paolo Guiotto: If it is not…
23:00:90Paolo Guiotto: specify that here x is any real, so it must be modulus of X is less or equal than the root of 1 minus Y squared.
23:10:570Paolo Guiotto: You understand better here that this is possible.
23:15:350Paolo Guiotto: When?
23:20:340Paolo Guiotto: Yeah, there is this… is a hidden condition. 1 minus 1 square must be positive, because if it is negative, if you look at this condition, for example, this number, negative.
23:34:170Paolo Guiotto: X squared cannot be less than a negative number. This means there are no X.
23:39:170Paolo Guiotto: For the Y for which this is negative, no X, So the section is empty.
23:46:290Paolo Guiotto: This is the Y section.
23:48:470Paolo Guiotto: For the Y for which 1 minus Y squared is positive, this means
23:54:400Paolo Guiotto: Y squared less equal 1, this means Y between minus 1 and 1. Then you take X, Equivalently, between…
24:07:710Paolo Guiotto: minus plus the root, the root of 1 minus Y squared. You see?
24:14:830Paolo Guiotto: So, be careful, because that can be…
24:19:430Paolo Guiotto: implicit conditions, not hidden, it seems something mysterious, but there are implicit conditions to write this. You cannot write this condition without importing this one. Otherwise, if you want, if that quantity is negative, the root of a negative number is not defined, at least in reals.
24:38:140Paolo Guiotto: So, this means that we are doing to get from minus the root of 1 minus Y squared to plus the root of 1 minus Y squared, when Y is between minus 1 and 1, and this is the
24:52:720Paolo Guiotto: the right application of the reduction form.
24:57:220Paolo Guiotto: Now, let's do the integration. Now, starting with this integral from minus root of 1 minus y squared root of 1 plus Y squared… I'm sorry.
25:11:840Paolo Guiotto: 1 minus.
25:19:870Paolo Guiotto: This is Janelle.
25:21:940Paolo Guiotto: Well, maybe I can get little simplification. I noticed that that function is symmetric in X is even. If you change X with minus X, the function does not change. And the interval is symmetric with respect to the odds, so this can be written as two times
25:40:990Paolo Guiotto: the integral from 0 to the root, 1 minus y square, of the same function, 1 over x squared plus Y squared. Now, we have to integrate this one. This, as a function of X, should remind us
26:00:420Paolo Guiotto: DX tangent, no? Because if you look, it's 1 plus X squared. Yeah, there is also this Y squared. So, but we can rearrange this in this way.
26:11:940Paolo Guiotto: So, as you can see, it's going to be very long, and we'll see that by changing variable weight, polar coordinates, it becomes very easy.
26:19:350Paolo Guiotto: So, this means that we are doing the… not the optimal way. So, let me factorize 1 plus Y squared in such a way that I get a 1 plus, I would say, X squared divided 1 plus Y squared.
26:35:160Paolo Guiotto: So DX, carry this outside, so 2 over 1 plus Y squared, integral 0 to the root, 1 minus y squared.
26:46:300Paolo Guiotto: then I have to transform this into 1 plus a square, which is the square of X divided by the root of 1 plus Y squared. Now this is in VX.
26:59:960Paolo Guiotto: Now, to have the tangent, I take a factor
27:04:320Paolo Guiotto: put 1 plus Y squared here, and I put a root here, in such a way that this is now the derivative, this box here, is the derivative with respect to X of the arc tangent
27:17:450Paolo Guiotto: of X divided the root of 1 plus Y squared.
27:22:640Paolo Guiotto: Because if you do the derivative of the arctangent, you get 1 over 1 plus the square of the argument, that's there, times the derivative with respect to X of the argument, and it comes just 1 over the root, because that's a constant for X.
27:37:320Paolo Guiotto: Okay, so this says that this is 2 over a root of 1 plus Y squared.
27:43:350Paolo Guiotto: times. I have to do the evaluation of the arctangent X divided the root of 1 plus Y squared between 0 and the root of 1 minus Y squared.
27:57:540Paolo Guiotto: So… I get 2 over root of 1 plus Y squared.
28:04:150Paolo Guiotto: Then, when I put X equal that root, I get arctangent of…
28:10:890Paolo Guiotto: A root, if we want, 1 minus
28:14:400Paolo Guiotto: Y squared divided 1 plus Y squared, which unfortunately
28:19:540Paolo Guiotto: does not simplify. Minus, when we put X equals 0, we get 0. So this is the function.
28:26:540Paolo Guiotto: Okay?
28:28:130Paolo Guiotto: And this is just the first integration. So, going back.
28:32:300Paolo Guiotto: To this step, we computed this part here.
28:37:70Paolo Guiotto: You can be this guy.
28:40:80Paolo Guiotto: Now we have to do the second integration, so let's put the equal with…
28:45:610Paolo Guiotto: star here. So, continuing down here, we have to do the integral from minus 1 to 1 of this nice thing, 2 over root of 1 plus Y squared, the arctangent
29:01:290Paolo Guiotto: of the root of 1 minus Y squared divided 1 plus Y squared.
29:13:600Paolo Guiotto: Who knows? No? So… I don't see easily. Let's say that it doesn't look Easy.
29:22:620Paolo Guiotto: Okay?
29:24:380Paolo Guiotto: So, alternative… solution, so that's not a solution, in fact. We… use… Paula.
29:36:10Paolo Guiotto: formulating.
29:38:800Paolo Guiotto: So, let's rewrite. Integral X squared plus Y squared less or equal 1, 1 over 1 plus X squared plus Y squared.
29:49:910Paolo Guiotto: Now, at the end, it will become
29:53:720Paolo Guiotto: Easy, but let's say there is a reason why this happens.
30:00:180Paolo Guiotto: And here, there is a double reason. First of all, the function depends on X squared plus Y squared, this quantity here, which is exactly raw square, so it means that instead of carrying around the two variables, you will carry one variable.
30:17:930Paolo Guiotto: So the function gets easier. But also the domain, as you can see, depends on the same quantity
30:25:300Paolo Guiotto: So the description of the domain becomes also easier.
30:29:170Paolo Guiotto: in the raw, theta coordinates. So that's the way this should be computed by using polar coordinates. So if we do that, the function becomes 1 over 1 plus raw square. That's the function. Then you don't have to forget that there is a raw.
30:47:310Paolo Guiotto: that comes from the modules of the determinant of the Jacobian magic, blah blah blah, neurotic theta. Then what about rho and theta? Well, that's the disk, so we already have seen how to do, but in any case, if we write this in polar coordinates, this becomes rho square less or equal 1.
31:06:350Paolo Guiotto: So it means that for raw, it's between 0, 1. Remind this is a very big error if you write raw between minus 1 and 1, because it seems like if you are not thinking about it, what is the meaning of raw. Raw is… raw is a positive quantity.
31:23:70Paolo Guiotto: And since there is no condition for theta, it means all thetas. This means theta between 0 and 2 pi.
31:32:00Paolo Guiotto: And that integral we can do in two steps, because now, applying the reduction formula, we have the iterated integral. We can decide to do, this time, last, the integration in theta, and first integration in a row.
31:45:890Paolo Guiotto: So, for raw, we have 1 over raw divided by raw 1 plus raw square, which is a simple rational function, so there is an algorithm to compute this. In this case, it is immediate.
31:58:240Paolo Guiotto: And then we have the second integration in theta. Actually, we already know that once we are done with this first integral, we get a number that is independent of theta. So we can write out of the other integrals.
32:12:550Paolo Guiotto: And so, in the integral, in detail, it will remain 1. So, this will be 2 pi.
32:19:930Paolo Guiotto: Now, about this, this is a derivative, more or less. It is the derivative of…
32:27:890Paolo Guiotto: what could be the candidate to be the derivative with respect to Roth?
32:35:680Paolo Guiotto: Huh?
32:38:320Paolo Guiotto: No.
32:41:680Paolo Guiotto: It is the log.
32:43:230Paolo Guiotto: If you do derivative with respect to raw of log 1 plus raw square.
32:48:560Paolo Guiotto: You get 1 over 1 plus rho squared, then you have the derivative of 1 plus rho squared with respect to row this is 2 rho.
32:57:490Paolo Guiotto: So we need a 2. We put a 2 and 1 half, so we divide by 2 here, and therefore we have evaluation of log 1 plus raw square between raw equals 0 and raw equals 1.
33:14:380Paolo Guiotto: So we have, at the end, pi times… when raw is 1, we get log of 2. When raw is 0, we get log of 1, which is 0. So at the end, pi log 2.
33:25:460Paolo Guiotto: And that's it.
33:27:680Paolo Guiotto: You see… the difference, okay?
33:32:20Paolo Guiotto: In fact, I want to show you… now, let's take a parenthesis.
33:38:280Paolo Guiotto: To show a remarkable example application of this, of this.
33:45:300Paolo Guiotto: So, this is… I do not… don't know now what is the number in notes, but this is the example, very important example for probability is the so-called Gaussian…
33:58:230Paolo Guiotto: Integral.
34:03:840Paolo Guiotto: The Gaussian integral is the integral from minus infinity to plus infinity of E minus X squared divided 2 sigma squared.
34:14:710Paolo Guiotto: in the X.
34:17:719Paolo Guiotto: This is a very important integral in probability.
34:21:60Paolo Guiotto: And as… very soon, you will do some probability.
34:25:780Paolo Guiotto: you will have to deal with this. Now, it can be computed explicitly. That's not trivial, because…
34:33:449Paolo Guiotto: Well, that sigma is just a constant, okay? And the value of this turns out to be the square root of 2 pi sigma squared.
34:41:800Paolo Guiotto: For every… He's my… Different from zero.
34:49:510Paolo Guiotto: Normally, it is written with sigma squared, because that constant must be positive. So, instead of writing sigma and saying sigma positive.
34:59:120Paolo Guiotto: You write sigma squared, so you know that it's a positive number. Sigma different from zero, because otherwise, it's a nonsense.
35:07:180Paolo Guiotto: Okay. Now, a particular case that is, in fact, what we will compute is the case sigma equal 1. We can always reduce to that case, because we may notice that integral from minus infinity plus infinity of P minus
35:24:590Paolo Guiotto: X squared over 2 sigma squared.
35:28:560Paolo Guiotto: If you do just a simple change of variable, you call U equal X over sigma, so in such a way that you can write this as E2 minus the square of X over sigma.
35:44:670Paolo Guiotto: divided by 2, hmm? If you call U that…
35:49:440Paolo Guiotto: new variable. This becomes the integral from minus infinity plus infinity E minus mu square over 2.
35:58:290Paolo Guiotto: Now, as you can see, DX is sigma del, so you have a sigma here, du, and you can see this as the root of sigma squared.
36:12:570Paolo Guiotto: If sigma is positive.
36:17:350Paolo Guiotto: Okay? Otherwise, you have plus-minus, that's the same. So the point is that we can always reduce to that calculation, so that's the integral we compute. So…
36:29:180Paolo Guiotto: Wheat.
36:30:550Paolo Guiotto: Will.
36:32:310Paolo Guiotto: prove, huh?
36:35:00Paolo Guiotto: that the integral from minus infinity plus infinity of E minus X squared over 2
36:41:320Paolo Guiotto: in the acts is Ruth of Tupal.
36:45:340Paolo Guiotto: Now… I repeat, the calculation of this integral cannot be done with the elemental tools.
36:52:310Paolo Guiotto: Because, this is the function E minus X squared.
36:56:840Paolo Guiotto: The half is just a steady factor.
37:00:850Paolo Guiotto: Cannot be seen as the derivative of something elementary.
37:04:890Paolo Guiotto: It's a quite complicated function, even if it seems to be very easy. So, we cannot say this is the derivative with respect to X of something, so we do evaluation at plus infinity minus evaluation at minus infinity, we get the integral. That's not…
37:22:200Paolo Guiotto: decade. It can be done with one variable techniques, but it would take about two pages with
37:29:500Paolo Guiotto: A very, very, very dense calculations and non-trivial, non-elementary, something that you cannot do by your own, unless you are a genius, because these are calculations that were made by
37:42:10Paolo Guiotto: what guys like, Gauss, and, and, and so on.
37:48:580Paolo Guiotto: But, with multiple integrals, when you don't see where is the connection with multiple integrals, this becomes a very easy calculation. And the trick is the following. Let i be the value of the integral, from minus infinity plus infinity.
38:04:210Paolo Guiotto: of E minus X squared over 2EX.
38:09:220Paolo Guiotto: Now, we do this trick. We write 2 times I, but with two different letters. The first one with letter X, and the second one with letter Y. You can, of course, do that.
38:24:520Paolo Guiotto: And then we multiply these two, identities. So we have I square.
38:30:270Paolo Guiotto: It's equal to the integral from minus infinity
38:34:100Paolo Guiotto: of E minus X squared over 2, EX, times the integral minus infinity, the class infinity.
38:41:920Paolo Guiotto: E minus Y squared over 2DY.
38:46:650Paolo Guiotto: Right?
38:47:900Paolo Guiotto: Now, you do this trick. Since these two integrals are constants with respect to the other, we can always do this carrying inside this integral, as if it is a number, okay? So we put this inside, and we have this formula. Minus infinity plus infinity, well, let's put,
39:07:900Paolo Guiotto: Here, that's just for aesthetical reasons, there is no particular…
39:11:750Paolo Guiotto: Let's put here, after the factor, E minus X squared over 2 times the integral minus infinity plus infinity, E minus y squared over 2BY 10 dx.
39:25:720Paolo Guiotto: You see the point? This, once you have computed this, becomes a number in character and out.
39:32:980Paolo Guiotto: It's just a fact.
39:34:980Paolo Guiotto: Now, as you see here, this looks like an iterated integral.
39:40:500Paolo Guiotto: Well, you know, first integration is Y, and then an integration next. And in fact, we can carry…
39:46:770Paolo Guiotto: this function inside this one. We can do that because the integration inside is in Y, so for the variable X is a constant for Y, so this becomes integral from minus infinity plus infinity, or fintechal minus infinity plus infinity.
40:04:700Paolo Guiotto: of E2, we can also put together the two exponentials, because when you multiply the exponential, you sum the exponents, so you get this finally.
40:14:220Paolo Guiotto: X squared plus Y squared over 2,
40:17:790Paolo Guiotto: First, in Y, and then in X.
40:21:470Paolo Guiotto: And now, this is an iterated integral, so it's the same of an integral on what domain for X and Y? Well, you see that Y varies between minus infinity and plus infinity when x varies between minus infinity and plus infinity, so this means that the point XY is in the interior Cartesian plane R2.
40:42:360Paolo Guiotto: So the same function, X squared plus Y squared divided by 2 minus DX, DY.
40:51:340Paolo Guiotto: So, this… we have not yet computed anything practically, but it's a way to rewrite that I square, where i is the integral we want to compute. So, let's say that this is now i square equal to this quantity.
41:09:350Paolo Guiotto: The point is that this integral can be easily computed by using polar coordinates, so let's pass in polar coordinates here.
41:21:160Paolo Guiotto: So…
41:22:400Paolo Guiotto: What happens to the function? Now, you see, there is the quantity X squared plus Y squared that becomes raw square, so raw square over 2 minus raw square over 2 at the X plane. Then we have the,
41:36:60Paolo Guiotto: The modulus of the determinant, etc, which is raw.
41:39:310Paolo Guiotto: Then we have the new variables, deroted them. What about the domain? Well, here, for XY, we have the full plane, R2.
41:47:730Paolo Guiotto: So all the coordinates, so we will have all the polar coordinates. So it means that rh is between 0 and plus infinity, and theta is between 0 and 2 pi.
42:00:880Paolo Guiotto: So this is the new integral.
42:03:10Paolo Guiotto: Now, let's apply the reduction formula.
42:05:980Paolo Guiotto: Let's do first integration, for example, in theta, that means 0 to 2 pi, when raw is between 0 and slash infinta. The function is raw E minus raw square over 2,
42:18:500Paolo Guiotto: This is DTA DROT.
42:21:310Paolo Guiotto: Now, this is a constant for this integral, so it goes outside, and it remains the integral from 0 to 2 pi of 1, so it is equal to 2 pi.
42:32:760Paolo Guiotto: We write outside times the integral from 0 to plus infinity rho E minus raw square of 2, 0. But now this is easy. It's very different than having only the
42:46:420Paolo Guiotto: It looks like… It's not the same.
42:50:840Paolo Guiotto: And in fact, yeah, we should recognize that there is a derivative.
42:55:930Paolo Guiotto: In fact, if you take the exponential E minus raw square over 2, and you do the derivative with respect to raw, you get the exponential E minus raw square over 2 times the derivative of the exponent, which is
43:14:860Paolo Guiotto: Just minus 1. So apart from the minus, we have a derivative. So we put the minus here and the minus here.
43:22:350Paolo Guiotto: In such a way that now we have minus 2 pi the evaluation of minus raw square, exponential minus raw square over 2, between rh equals 0 and raw equals plus infinity.
43:36:120Paolo Guiotto: At plus infinity, we get 0, because the exponent goes to minus infinity.
43:42:340Paolo Guiotto: At rho equals 0, we get 1.
43:44:600Paolo Guiotto: So, at the end, this is equal to 2 pi. So, we obtained now that I square that was this…
43:53:230Paolo Guiotto: Double integral.
43:55:470Paolo Guiotto: at the end is equal to 2 pi, and this means that I
44:01:810Paolo Guiotto: which is positive, because i is this integral, is definitely a positive quantity, so we do not say it is plus or minus root of 2 pi. It's just plus the square root of 2 pi.
44:15:700Paolo Guiotto: And this means that, going back here, you see, we will prove that this integral is equal to root of 2 pi, because the other one is already root of sigma squared times this one, which is root of 2 pi. You glue together the two, and you get that formula, okay?
44:35:90Paolo Guiotto: Okay.
44:36:470Paolo Guiotto: So, I would say that it's a good point, we are exactly Mean it.
44:42:420Paolo Guiotto: 11.15, so we take, 10 minutes back now.
44:56:290Paolo Guiotto: Okay, before we move to triple integrals and to spatial coordinates for triple integrals, I want to show you one more example to show you something here. We return back to the exercise 575.
45:14:590Paolo Guiotto: And we do the number 3.
45:19:480Paolo Guiotto: which is the integral on a 2 of exponential
45:24:140Paolo Guiotto: minus X squared plus 2Y squared.
45:33:420Paolo Guiotto: Because here, the point is you have to adapt ideas, okay?
45:39:130Paolo Guiotto: So, we may think that we could compute this by using polar coordinates, so we could say that with polar coordinates.
45:50:120Paolo Guiotto: This becomes the integral of E… well, you see that we do not have X squared plus Y squared, but rather X squared plus 2Y squared, so I could say that
46:00:820Paolo Guiotto: Yes, there is X squared plus Y squared, but there is another Y squared, so it is a raw square, then I have plus raw square sine theta squared.
46:13:460Paolo Guiotto: So I do not simplify, actually, the function here.
46:18:490Paolo Guiotto: Then I have the raw.
46:21:640Paolo Guiotto: Dirodi pita.
46:24:810Paolo Guiotto: And then, about the domain, since the Cartesian domain, the full plane are true for raw, I will have raw between 0 and plus infinity, and 40 tiered theta, so from 0 to 2 pi.
46:39:800Paolo Guiotto: Now, here, the point is that, let's see if we can do something, and what can be done.
46:47:260Paolo Guiotto: So now we apply the reduction formula. We have to decide if we integrate first in raw or first in theta. If we look at the domain, it's indifferent. If we look at the function.
46:59:350Paolo Guiotto: Would you integrate in raw first, or in Theta first?
47:04:980Paolo Guiotto: Indeed, E theta is a function like E… e to constant sine squared theta.
47:13:910Paolo Guiotto: Now, what do you integrate here? In raw, we could have more or less, if you look at raw squared, the coefficient is 1 plus sine squared theta.
47:24:450Paolo Guiotto: Then we have down here the raw, so that could be more or less the derivative of the exponent. If I look as a function of raw, the quantity 1 plus sine squared is a number, it's a constant.
47:35:670Paolo Guiotto: So I would say, let's start integrating in raw from 0 to plus infinity, then we see if we can do it in theta.
47:43:720Paolo Guiotto: So for raw, we have E minus raw square times 1 plus sine square theta.
47:52:250Paolo Guiotto: Draw, then DTAP.
47:56:110Paolo Guiotto: Now, as I said, I want to look at this as a derivative with respect to rho, so probably I have to do the derivative of E minus rho squared times 1 plus sine squared theta, because the exponential has derivative itself, so I have E minus raw square
48:15:900Paolo Guiotto: 1 plus sine squared theta. Then I have the derivative with respect to rho of the exponent. This is minus 2 rho times that coefficient 1 plus sine squared theta.
48:29:990Paolo Guiotto: Now, I do not have that minus 2, so I can put minus 2, that's no problem, because I can multiply and divide, so I can put the minus 1 half out here. I do not have also 1 plus sine squared, which is a coefficient for the innermost integral, so I can write here times 1 plus
48:47:950Paolo Guiotto: sine squared theta, I can divide…
48:52:430Paolo Guiotto: And since this is depending only on theta, I can directly put outside, so I have 1 over 1 plus sine squared theta out here. You see this?
49:04:600Paolo Guiotto: It's a variable, yes, but not for raw. Raw is a constant. So at this point, I have,
49:11:130Paolo Guiotto: Minus 1 half integral 0 to 2 pi.
49:15:570Paolo Guiotto: 1 over… 1 plus sine squared theta.
49:20:810Paolo Guiotto: And then we have the evaluation of E minus raw square 1 plus sine
49:26:780Paolo Guiotto: Square theta is between rho when equals 0 or equal plus infinity.
49:34:110Paolo Guiotto: And this will be integrated in QETA.
49:38:90Paolo Guiotto: When we do rah equal plus infinity, you see this goes to plus infinity.
49:44:110Paolo Guiotto: This is definitely positive, so the exponent goes to minus infinity, the evaluation yields 0.
49:51:550Paolo Guiotto: Minus. The value when raw is 0, there's raw equals 0, kills all the exponents, so it comes e to 0, 1.
49:58:530Paolo Guiotto: So at the end, that gives, so I can give this the minus outside… 1 half integral 0 to pi of 1 over 1 plus sine square theta d theta.
50:16:10Paolo Guiotto: Which is not immediate.
50:18:590Paolo Guiotto: At least to me.
50:21:340Paolo Guiotto: Because you may think it's, yeah, tangent of sine, but you need the derivative of sine.
50:26:260Paolo Guiotto: outside to have that. However, it's a standard type, because in fact, it's a rational function of sine theta cos theta, so there are the parametric change of variables.
50:38:810Paolo Guiotto: Since I do not remind those parameters change of variables, you finish.
50:46:220Paolo Guiotto: Because now we do it in a different way, in a much more smart way.
50:52:120Paolo Guiotto: Alternative… solution.
50:55:640Paolo Guiotto: What is the alternative solution? Is… To use adapted polar coordinates.
51:03:140Paolo Guiotto: So, if we have here, this is the integral we just computed for a second, let's cancel this too. If we have this, we do polar coordinates, and this simplifies that function that becomes zero square.
51:18:900Paolo Guiotto: Should these two danger… bother too much?
51:23:920Paolo Guiotto: No, the idea is that I have this…
51:27:490Paolo Guiotto: So let me just copy the integral.
51:30:260Paolo Guiotto: E minus X squared plus 2Y squared.
51:34:830Paolo Guiotto: So, visa… Take the message, not the specific example.
51:40:470Paolo Guiotto: I want to make this a raw square after the change of variable.
51:45:390Paolo Guiotto: However, I do. So, I have to say, something like X is rock cos theta.
51:54:90Paolo Guiotto: And I want that 2Y squared will be raw square sine theta, so the Y, if you want root of 2Y, should be raw sine theta.
52:07:50Paolo Guiotto: You see this?
52:08:730Paolo Guiotto: Because if you do X squared plus the square, see, if you do X squared plus 2Y squared, this is exactly X squared plus the square of root of 2
52:21:150Paolo Guiotto: Why? So at the end, it is raw square, cos square theta plus raw square sine squared theta.
52:29:290Paolo Guiotto: So it is equal to raw square, which is what I want.
52:32:920Paolo Guiotto: So the change of variable is not the standard polar coordinate, but rather, it is this one.
52:39:890Paolo Guiotto: You have a scaling factor on one of the two coordinates.
52:44:370Paolo Guiotto: So, what changed?
52:47:370Paolo Guiotto: Huh?
52:50:340Paolo Guiotto: Sorry, I don't…
52:55:340Paolo Guiotto: Yeah, there will be just a little change at the end, not a dramatic change, because this is, again, the phi minus 1, okay? So, I don't need to redo the calculation of the determinant of the phi minus 1 prime.
53:13:50Paolo Guiotto: Why?
53:15:160Paolo Guiotto: Because you see that with respect to the… that one of the polar coordinates, which is raw cos theta, raw cent theta, one of the two components is multiplied by the factor 1 over root of 2, okay?
53:28:620Paolo Guiotto: Scientita.
53:33:70Paolo Guiotto: So when I will do the gradient, you know that here I have the gradient of raw cos theta, the gradient in raw theta, and the gradient of 1 over root of 2 raw sine theta.
53:47:420Paolo Guiotto: Then I have to take the determinant of all this.
53:50:300Paolo Guiotto: Now, you see that that's part of one of the…
53:53:630Paolo Guiotto: like this conference, so it will multiply all the derivatives. So I have one other company in the first company, and one of the
54:02:210Paolo Guiotto: Keep that in the technical.
54:04:440Paolo Guiotto: Now, you know that determinant is a linear function of lines and columns.
54:12:610Paolo Guiotto: So, if a line is multiplied by the same factor, that's 1 over root of 2,
54:20:30Paolo Guiotto: That factor can be carried outside, and you can say that this is the same of 1 over root of 2, the determinant of the matrix gradient rho cos theta gradient rh sine theta.
54:37:70Paolo Guiotto: which is the quantity we already computed, it is raw. So at the end, you see that that factor 1 over root of 2 is just a scary factor for this raw.
54:48:530Paolo Guiotto: So, it means that with this change of variable, I can say that my integral, so let's put a star here.
54:57:510Paolo Guiotto: becomes just.
55:00:720Paolo Guiotto: An integral in is exponential. I did this to make the quantity X squared plus 2Y squared equal to rho squared. That's right, right.
55:09:330Paolo Guiotto: E minus raw square. Now, the modulus of the determinant of phi minus 1 prime becomes 1 over root of 2 raw, so not such a big change, just a factor. Then I have bureau dita.
55:23:150Paolo Guiotto: Now, be careful, because rho and data are not the standard polar coordinates. However.
55:30:540Paolo Guiotto: Since XY variables in are two necessary, rho must be positive, any positive number and theta between 0 and 2 pi, because you can see, basically, rho and theta are the polar coordinates
55:46:990Paolo Guiotto: Paula coordinates.
55:50:130Paolo Guiotto: of this point, X and root of 2Y,
55:55:510Paolo Guiotto: So it is clear that when XY varies in the full plane, also X root of 2Y will be in the full Carpeasian plane. So for R and theta, I will have always the same condition. Greater than 0, less than plus infinity, and theta between 0 and 2 pi.
56:14:680Paolo Guiotto: So again, 1 over root of 2, and now this is an elementary integral. We apply the reduction formula, we integrate first in raw, let's say 0 to plus infinity, raw E minus raw square, B raw, then we finish with theta from 0 to 2 pi.
56:35:10Paolo Guiotto: To have the derivative, I needed to put the minus 2 here, so I will put the minus…
56:40:410Paolo Guiotto: to down there. So now this is the derivative with respect to rho of t minus rho square, and therefore this integration is the evaluation of E minus rho square between rho equals 0, rho equals plus infinity.
56:57:240Paolo Guiotto: At plus infinity, we get 0, at 0, we get 1, so the output is minus 1, so all this is equal to minus 1.
57:06:340Paolo Guiotto: And therefore, we have to integrate minus 1 from 0 to 2 pi, this gives minus 2 pi. So, minus 1 over 2 root of 2 times minus 2 pi, and so the final result is pi over root of 2.
57:22:590Paolo Guiotto: As you can see, I hope that you will obtain the same results by doing directly this one, okay? Here, you need to remind… I do not remind, unfortunately.
57:34:500Paolo Guiotto: What are the parametric formula? Those that says,
57:41:410Paolo Guiotto: It's something that is beyond my possibilities, paramedic.
57:46:690Paolo Guiotto: Something, like, with the tangent of,
57:50:270Paolo Guiotto: Of theta divided to parametric formulas.
57:55:10Paolo Guiotto: But you can go back in first year, Kai goes, okay, you're fine. What are the standards that change of Bible to cook with this thing?
58:03:80Paolo Guiotto: Okay, now… Let's move to… 3 poly integrals.
58:15:970Paolo Guiotto: So now, let's discuss the problem of computing an integral of a function of three variables, XYZ,
58:24:380Paolo Guiotto: By using some… Special system of coordinates.
58:29:310Paolo Guiotto: there are analogous of polar coordinates in R3, and there are two main analogous, which are the spherical.
58:43:640Paolo Guiotto: coordinates.
58:47:450Paolo Guiotto: And the other one is the cylindrical.
58:54:790Paolo Guiotto: coordinates.
58:57:180Paolo Guiotto: They respond to two different situations.
59:01:850Paolo Guiotto: Now, in spherical coordinates.
59:04:790Paolo Guiotto: These are more similar, apparently, to the polar coordinates, even if, at the end, what will be similar… more similar are the cylindrical, in fact.
59:16:970Paolo Guiotto: So, you have in space XYZ, point XYZ, which can be represented by 3 other numbers. We've already seen this doing limits for functions of three variables.
59:34:280Paolo Guiotto: So we use… one is the distance to the RG, and this is raw.
59:39:970Paolo Guiotto: If the raw is a positive number.
59:44:550Paolo Guiotto: The second could be the angle with one of the axes.
59:49:350Paolo Guiotto: It's indifferent. Normally, we choose the z-axis, but you can choose the X or the y-axis if you prefer. So let's call this angle phi. Now, this angle phi normally varies between 0, you are on the z-axis above, and pi, you are the z-axis below.
00:07:900Paolo Guiotto: So, let's say North Pole, South Pole, because this is the…
00:11:550Paolo Guiotto: is not the, what's the name of this? It's the latitude, the longitude… I never understood what's the name of this, but it should be the latitude, right? Actually, in geography.
00:24:770Paolo Guiotto: Here, we have latitude 0 means the North Pole, latitude 180 means the South Pole, while in geography, latitude 0 is the quarter, and then we have plus 90 is the North Pole, minus 90 is the South Pole, something like this.
00:40:500Paolo Guiotto: I don't know. However, then we have a second angle, which is the angle theta that is made by the projection of this point on the plane XY.
00:50:430Paolo Guiotto: So the angle theta is this one, theta, and this theta is between 0 and 2 pi. This is…
00:59:600Paolo Guiotto: In longitude, okay?
01:03:70Paolo Guiotto: Okay, so now we have formulas that gives XY theta, XYZ in function of raw theta phi, exactly as for polar coordinates. You have XY function of raw theta.
01:17:90Paolo Guiotto: Now, the formulas with each choice are…
01:21:690Paolo Guiotto: For Z, the Z is the projection on the z-axis of this point, so as you can see, it is raw cosine of angle phi.
01:32:260Paolo Guiotto: a rock.
01:33:530Paolo Guiotto: cosine of angle phi, while this one, this projection has length raw sine phi, so we will have raw sine phi here. That has to be multiplied by cosine theta, and here, raw sine phi
01:52:180Paolo Guiotto: Which are, which we multiply for sine theta.
01:56:440Paolo Guiotto: Okay.
01:57:560Paolo Guiotto: Now, again, this map is giving XYZ functions of raw phi and theta. So this is, with the notation of change of variable formula, this is the phi minus 1 that gives the starting variables
02:12:690Paolo Guiotto: function of the new variables. So you will have that the integral on D of the function F in Cartesian coordinates
02:22:500Paolo Guiotto: DX, D, Y, DZ.
02:25:680Paolo Guiotto: Again, the general formula says it becomes the integral on the image. Well, let's put this in horizontal, because it's the same for the two cases.
02:35:410Paolo Guiotto: So, I just write the general shape. So, is the integral on t of B of F,
02:42:430Paolo Guiotto: On phi minus 1, Well, let's use the new letters, U, V, and W.
02:51:960Paolo Guiotto: the new variables, because for this case, they will be raw P and theta, for the other case, will be different, okay? So I want to give a formula, which is in general, 3 new variables, UVW, modulus of the determinant of the phi minus 1 prime UVW,
03:13:270Paolo Guiotto: in DUDBW. So the UVW are the new variables.
03:19:60Paolo Guiotto: how this formula, looks like in this case. So we have the integral on phi of D,
03:27:890Paolo Guiotto: Now, exactly as it happens for polar coordinates, as we said this morning, phi of D is what? It's domain D written in spherical coordinates, okay? So this is the set of raw
03:44:210Paolo Guiotto: phi and theta, such that this point, XYZ, so phi minus 1 of raw theta and phi belongs to D. Sorry, the order is raw phi and theta.
04:01:650Paolo Guiotto: So…
04:02:970Paolo Guiotto: much more easily. This is the set of polar cord… oh, polar… of spherical coordinates of points of diff, okay?
04:12:240Paolo Guiotto: So… Set… off.
04:16:910Paolo Guiotto: Spatty call.
04:20:550Paolo Guiotto: coordinates.
04:23:290Paolo Guiotto: of points.
04:25:750Paolo Guiotto: Off.
04:26:660Paolo Guiotto: B.
04:28:170Paolo Guiotto: Now, the function… well, the phi minus 1 is the map you see here. It means that in place of X, you have to plug all these things, rho sine, cosine theta.
04:40:60Paolo Guiotto: So, raw sine P cos theta. This is DX. The Y will be raw sine P sine theta.
04:51:350Paolo Guiotto: This is the Y, and the Z will be raw cosine P.
04:56:430Paolo Guiotto: And then, now we have to understand what we have to write here. And as you may expect, the quantity we have to put exactly as for the polar coordinates is a standard object that we always have in the same way. So let's do this calculation. It's a bit longer, because now we have a
05:14:860Paolo Guiotto: that phi minus 1 depends on three variables, so the matrix, the Jacobian matrix, is a 3 by 3 matrix, so the determinant is a little bit more complicated, but let's see what is it.
05:26:270Paolo Guiotto: So, determinant of phi minus 1 prime is… I remind you that,
05:34:620Paolo Guiotto: the P minus 1 prime is the Jacobian matrix of this map. So, this is the first component, you see? And we have to do the gradient of this thing. So, gradient of raw,
05:50:280Paolo Guiotto: Wrong.
05:51:940Paolo Guiotto: sign… P cosine theta.
05:56:340Paolo Guiotto: The gradient is a gradient in derivative with respect to raw, derivative with respect to phi, derivative with respect to theta. These are the variables, okay? Then we have the second component, which is raw sine phi.
06:11:30Paolo Guiotto: sine theta.
06:13:200Paolo Guiotto: And finally, the third component, which is raw, cosine P.
06:22:270Paolo Guiotto: So… Be patient.
06:26:750Paolo Guiotto: Okay, derivative with respect to rho is sine phi
06:32:480Paolo Guiotto: We have not to pray, yeah? Since we are in the church praying, you know.
06:41:230Paolo Guiotto: Then, derivative with respect to phi is, you'd add the little sine phi, which is cosine, so raw, cos…
06:48:490Paolo Guiotto: Because… Feedback.
06:51:270Paolo Guiotto: Derivative with respect to theta, we have the derivative of cos theta, which is minus sine, so minus rho sine P sine
07:01:460Paolo Guiotto: And that's the first line.
07:04:20Paolo Guiotto: The second line, derivative with respect to rho is sine P, sine theta.
07:12:810Paolo Guiotto: Derivative with respect to phi is, again, rho cosine phi sine theta.
07:21:500Paolo Guiotto: Derivative with respect to theta is rho sine P and cosine theta.
07:29:730Paolo Guiotto: to the line, derivative with respect to rho is cos P.
07:33:570Paolo Guiotto: The divot with respect to phi is minus raw sine phi.
07:40:20Paolo Guiotto: And, luckily, the third one is zero.
07:44:990Paolo Guiotto: Okay, now we have to do the determinant of this beautiful matrix. Of course, we use the last line, there is a zero.
07:53:660Paolo Guiotto: So, designer, plus, minus for this, and plus for this, okay?
08:02:140Paolo Guiotto: You know, this, right?
08:04:150Paolo Guiotto: So we have cosine phi times… we have to do the determinant of these metrics, let's do directly. I don't write data in yet.
08:14:210Paolo Guiotto: So, raw cos T cos theta times rho sine t, cos theta, so we get a raw square.
08:21:680Paolo Guiotto: Costs, fee, sign, P times cos squared theta.
08:29:319Paolo Guiotto: Then we have minus, but there is another minus, becomes a plus.
08:33:850Paolo Guiotto: We have raw square, again, sine P cos P, and then sine square theta.
08:44:470Paolo Guiotto: Now, if you look carefully, we have a common factor, which is this one.
08:51:890Paolo Guiotto: And this multiplies cos square plus Sine squared, which is 1.
08:58:770Paolo Guiotto: So, this simplifies a bit. This expression gives raw square cos phi sine phi.
09:08:180Paolo Guiotto: times cos phi, which is out here, so cos square.
09:13:330Paolo Guiotto: Okay, then we have the second. Be careful, because there is a minus, and it counts with minus, so it's a plus.
09:20:260Paolo Guiotto: Plus raw sine phi, And now we have to do the determinant of this Submetrics, this plus this, right?
09:30:960Paolo Guiotto: So we have sine t cos theta times rho sine t cos theta, everything is squared, so we have…
09:37:800Paolo Guiotto: Raw sine square feet cost square theta.
09:45:80Paolo Guiotto: Then we have minus with minus plus raw sine phi sine theta, sine p sine theta, so again, everything is squared, so plus raw sine.
09:55:730Paolo Guiotto: Phi square costs… No, sorry, sine theta.
10:03:940Paolo Guiotto: Square.
10:05:970Paolo Guiotto: Also, here we have some simplification, because if we factorize this
10:12:140Paolo Guiotto: common factor will remain with the cos square theta plus sine squared theta, which is 1. So, this means that this is equal raw sine square P.
10:25:510Paolo Guiotto: Which is multiplied by another raw sine phi. So this, at the end, all this yields raw square sine
10:34:650Paolo Guiotto: cube P.
10:36:870Paolo Guiotto: Okay, so, now we can…
10:40:560Paolo Guiotto: We can say, and of course, there is a final factor, but it is zero times, so we write plus zero times something that matters.
10:49:310Paolo Guiotto: So at the end, we have… you see that we can factorize raw square.
10:54:550Paolo Guiotto: We can factorize a sine phi.
10:58:570Paolo Guiotto: And what remains from the first is cos square phi.
11:03:530Paolo Guiotto: And from the second, a sine squared P.
11:06:750Paolo Guiotto: Which, fortunately, is equal to 1. So we get raw square sine P. Oh, so this is the formula.
11:15:910Paolo Guiotto: For the determinant. So, the modulus of the determinant of phi minus 1 prime for the spherical coordinates, so function of raw P and theta, is equal to the absolute value of that quantity. Now, raw square is positive.
11:34:360Paolo Guiotto: And also, sine phi is positive, because remember that phi is between 0 and pi, so sine is positive, so it remains like that.
11:43:320Paolo Guiotto: So the quantity we have to plug into this formula is… Now, raw square sine P.
11:52:940Paolo Guiotto: And this will be 0 d phi, and d theta. And this is the formula.
11:58:930Paolo Guiotto: Okay?
12:02:270Paolo Guiotto: However, what are you to retain?
12:06:250Paolo Guiotto: That is the first part, which is similar, but more complicated than polar coordinate.
12:11:270Paolo Guiotto: And the unique point is to remind that this is the equation you have to
12:17:570Paolo Guiotto: Before the eroticated the fish.
12:20:310Paolo Guiotto: the… this fee of B is, as usual, the image of the domain B, or how you write the domain B in this verified column.
12:28:910Paolo Guiotto: Let's do some example.
12:31:610Paolo Guiotto: Let's redo an example.
12:34:520Paolo Guiotto: The volume of the spear.
12:38:710Paolo Guiotto: X squared plus Y squared plus Z squared less or equal than R squared. Well, R is not… we already computed, I know, but let's redo with this formula. Let's see how.
12:51:60Paolo Guiotto: Now, you remind that the volume of a domain, D, if we are in R3, is the integral on D of the function 1, dx d YEZ.
13:03:470Paolo Guiotto: So, basically, we do the volume, we compute the volume by computing that integral.
13:09:170Paolo Guiotto: which is the integral on the domain X squared plus Y squared plus z squared less or equal than R squared. The function is 1, dx, dy, dz.
13:21:620Paolo Guiotto: Now, we will use this change of variable formula that we have written here.
13:27:460Paolo Guiotto: Okay?
13:28:710Paolo Guiotto: Now, clearly, the function f is very easy. It's constant, so whatever is the change of variable, it will be always equal to 1.
13:38:400Paolo Guiotto: while the domain, it's a lot simplified in spherical coordinates, no? Because this quantity is exactly the raw square, so in the, let's say, the triplet, raw phi, and theta.
13:54:680Paolo Guiotto: of spherical coordinates belongs to phi of D, so are the spherical coordinates for points of D, if and only if you get this condition, rho square less or equal than R squared. That is, since, also here rho is positive, this means rho between 0 and R.
14:15:00Paolo Guiotto: There is no condition about theta and phi, it means that they vary in the natural range. So, for phi, you will have from 0 to pi, and for theta, you will have 0 to 2 pi.
14:29:540Paolo Guiotto: You don't have to forget that there are also these two coordinates.
14:33:860Paolo Guiotto: So, in particular, this will become what? In the new coordinate, it will be still the integral of 1. Here, we don't have to do anything, because the function is already… it's the same, whatever is the
14:47:510Paolo Guiotto: system of variables, I have. Then, I have to put that coefficient ras square sine p here.
14:57:110Paolo Guiotto: So, raw square sign… B.
15:01:560Paolo Guiotto: there are the variables, the raw d phi, and there is the range for these variables, so raw between 0 and r.
15:10:210Paolo Guiotto: Phi between 0 and pi.
15:14:30Paolo Guiotto: And theta between 0 and 2 pi.
15:19:960Paolo Guiotto: And now, of course, we apply reduction formula.
15:25:160Paolo Guiotto: You see that this function does not depend on theta, so let's first eliminate theta, integrating first in theta.
15:32:860Paolo Guiotto: So I do the first integration in theta. Since the function is constant in theta, I just carry outside. I write here raw square sine phi.
15:42:660Paolo Guiotto: And this remains one, and then I will do the integration in raw and feed.
15:48:400Paolo Guiotto: Now, in this way, for theta, you have 0 to 2 pi.
15:53:250Paolo Guiotto: And for the other two, I still have raw between 0 and R, and the other guy is phi between 0 and pi.
16:03:460Paolo Guiotto: this integral is equal to 2 pi, so it's a constant, I carry outside, and get 2 pi.
16:12:140Paolo Guiotto: times now this double integral in row entry. I apply the reduction formula.
16:17:700Paolo Guiotto: Here, also here. For example, I could integrate… it's indifferent here, because the function is multiplied… is a multiplication of a function of rho times a function of P. There is the domain, which is…
16:31:650Paolo Guiotto: independent, so I can… I can choose whatever I want. So let's say that I… we do first the integration in a row. So we integrate from 0 to R, raw square, sine phi, this first row, then we continue and finish in
16:50:110Paolo Guiotto: P from 0 to pi.
16:53:690Paolo Guiotto: Now, this sine phi is independent of rh, so I can carry outside, so I have 2 pi.
17:00:490Paolo Guiotto: Integrals, 0 to pi of sine phi.
17:04:350Paolo Guiotto: As you can see, once I put this sine phi in the other integral, what remains inside is a constant in phi, so I can just write this way. It's not always the case, but in this case, I can do that.
17:19:00Paolo Guiotto: It is correct, because this integral was inside, and it is independent of phi. Now, what is this? This is, sine is the derivative of minus cos phi, so I have to evaluate between 0 and pi this, and this is the derivative of raw cube divided 3.
17:36:550Paolo Guiotto: between 0 and R.
17:39:650Paolo Guiotto: So, cos pi is minus 1, right? So, I get minus, minus 1, plus 1.
17:46:770Paolo Guiotto: Minus cos 0 is 1, so minus, minus 1.
17:52:640Paolo Guiotto: So at the end, this is 2.
17:55:440Paolo Guiotto: This is R cubed over 3 minus the value at 0, so final value is 2 pi times 2 times R cubed over 3. So, 4 pi r cubed over 3.
18:11:640Paolo Guiotto: Of course, we got the same formula, nothing out of whack.
18:15:410Paolo Guiotto: It seems that it works, what happened.
18:18:680Paolo Guiotto: Okay.
18:20:780Paolo Guiotto: Now…
18:32:210Paolo Guiotto: I want to do cylindica because, well, what is the advantage in using spherical coordinates?
18:39:280Paolo Guiotto: You can see a bit in this example.
18:42:200Paolo Guiotto: You see, the domain gets simplified if we introduce the spherical coordinates, because 3 variables becomes 1.
18:49:870Paolo Guiotto: Okay? The function here is 1, so whatever is the change of variable, it remains the same, so no better, no worse than that.
18:58:440Paolo Guiotto: But let's say that you should use spherical coordinates if your domain and the functions, they both depend on this quantity, X squared plus Y squared plus X squared, because this means that
19:12:250Paolo Guiotto: the domain has an invariance by rotations around the origin, like the sphere. And so this simplifies both the integral and the domain… integration domain.
19:25:80Paolo Guiotto: But generally, this won't be the case, so sometimes we need to use a different system of coordinates, and that's the cylindrical system, which is actually easier than this one.
19:39:140Paolo Guiotto: Let's see how it works. Cylindrical coordinates…
19:44:210Paolo Guiotto: In fact, at the end, these are very close to the polar coordinates, even if in the spirit, it seems this is not the case.
19:54:950Paolo Guiotto: So we still have the Cartesian plane, and our point, XYZ,
20:01:860Paolo Guiotto: So, for which we need the three numbers to identify the position of this bond.
20:08:580Paolo Guiotto: Now, actually, there are 3 different systems of cylindrical coordinates, depending on,
20:16:570Paolo Guiotto: which axis you choose. I will do the cylindrical coordinates with respect to the z-axis. What does it mean? That…
20:28:10Paolo Guiotto: To characterize this point, I will use three numbers that are the following. One of the three is the Z itself. So, one of the new coordinates will be Z, so I will say X equal, Y equal, Z equal, Z will be equal to Z, because one of the coordinates is Z.
20:48:30Paolo Guiotto: then you project this point in the plane XY, And you got,
20:55:460Paolo Guiotto: a point in the Cartesian plane. This point is the point XY, simply Z equals 0.
21:02:190Paolo Guiotto: Now, at this point, we'll have a distance to the origin that we call raw.
21:08:200Paolo Guiotto: And an angle with the positive direction of the x-axis that we call theta.
21:14:740Paolo Guiotto: So the new coordinates we are going to use are these three numbers, rho, theta, and Z.
21:21:900Paolo Guiotto: Well, they have different… it's a mixed system, because we use two R typical coordinates, so here rho is greater or equal than zero, theta is between 0 and 2 pi.
21:35:120Paolo Guiotto: These are the natural ranges for these variables, while Z is a Cartagian coordinate, so Z varies from minus infinity to plus infinity.
21:46:250Paolo Guiotto: Okay, it's not positive.
21:48:890Paolo Guiotto: Now… Well, this figure seems that everything is, it's not in the correct… perspective, maybe. So, no.
22:01:900Paolo Guiotto: Not this one.
22:03:520Paolo Guiotto: So we should do something like that.
22:05:590Paolo Guiotto: This is…
22:08:380Paolo Guiotto: Now, if this is the figure, this pointer X, the X and Y are exactly as the… for the polar coordinates, so they are raw cosine theta, rh, sine theta.
22:21:140Paolo Guiotto: So this is the, the correspondence that builds Cartesian coordinates, XYZ, function of cylindrical coordinates, which are these cylindrical
22:35:970Paolo Guiotto: coordinates, huh?
22:37:790Paolo Guiotto: So again, you have the map phi minus 1 here.
22:43:560Paolo Guiotto: Now, you have to be careful, because raw is no more the distance to the origin, but rather, raw is the distance to the z-axis, you see?
22:53:720Paolo Guiotto: Because raw equals 0 does not mean that you are in the origin. Any point on the z-axis will have raw equals 0. And in fact, you can see, because with this position, raw square is exactly X squared plus Y squared. So, raw is the distance
23:11:320Paolo Guiotto: to Z axis.
23:15:660Paolo Guiotto: So this is the meaning of raw.
23:17:830Paolo Guiotto: And the idea is that, because just of this identity, this is a nice choice when you have that the function, or the domain, or both, they depend on the quantity X squared plus Y squared, which is the distance to D.
23:32:620Paolo Guiotto: a z-axis, but not on X squared plus Y squared plus Z squared, which is the distance to the origin.
23:41:750Paolo Guiotto: The quantity X squared plus Y squared plus Z squared gets simplified when we use spherical coordinates.
23:48:590Paolo Guiotto: while…
23:49:970Paolo Guiotto: If we have only X squared plus Y squared, using spherical coordinates won't lead to a simplification. You will still have two variables. You can do the calculation. If you take this… where is it?
24:02:880Paolo Guiotto: This is X… this is why, imagine you do X3 platform.
24:07:990Paolo Guiotto: Raw square, sine squared, raw square.
24:10:990Paolo Guiotto: plus, plus squared, sine squared is the same plus sine squared. The unique advantage is that cos squared plus sine squared here, beta in pi.
24:21:210Paolo Guiotto: So when you do an experiment, science fair.
24:26:920Paolo Guiotto: So you move from two variables to other two variables, which is not necessarily better. While, if we use the spherical… the cylindrical coordinates in quantity, X squared plus Y squared becomes one of the coordinates, raw square, the square of one of the coordinates.
24:45:270Paolo Guiotto: Another nice feature of this is that when you do the determinant, What is this?
24:52:130Paolo Guiotto: the determinant of the Jacobian matrix of phi minus 1.
24:58:250Paolo Guiotto: That's very easy, because if you look at this.
25:01:490Paolo Guiotto: If you… if you stop at the first two components, that's exactly the polar coordinates.
25:07:10Paolo Guiotto: So, you get exactly the gradient, the gradient now is, D raw d theta DZ of raw cos theta, gradient of raw sine theta, and gradient of Z. Now, what happens here?
25:26:860Paolo Guiotto: If you do the gradient of raw cos theta in the three variables raw theta, and Z, the first two components are exactly the first two components of this metric, so we have seen
25:39:540Paolo Guiotto: Yesterday.
25:41:380Paolo Guiotto: No.
25:42:400Paolo Guiotto: on Wednesday.
25:45:960Paolo Guiotto: You see? Gradient of raw cos theta.
25:49:580Paolo Guiotto: So, we have these two.
25:51:460Paolo Guiotto: Components are the same that we find here.
25:55:760Paolo Guiotto: So we have, derivative with respect to rho is cosine theta, derivative with respect to theta is minus raw sine theta. And then the third component, which is derivative with respect to Z, is 0. Same for the second line.
26:08:990Paolo Guiotto: sine theta.
26:10:740Paolo Guiotto: raw cos theta and 0. And for the third component, when we do now the derivative with respect to raw and theta, we get 0. So 0, 0. And when we do the derivative with respect to 0, we get 1.
26:24:610Paolo Guiotto: So, at the end, when you compute the determinant of these metrics, if you expand the determinant by the last line, you get 0 times something. Minus 0 times something else, plus 1,
26:38:840Paolo Guiotto: So it is 1 times the determinant of that matrix here.
26:45:240Paolo Guiotto: Which is exactly the same of the polar coordinates, so at the end, you get rough.
26:50:830Paolo Guiotto: So, this is nice, because the determinant for this transformation
26:56:590Paolo Guiotto: The Jacobian matrix of C minus 1 is just raw.
27:00:980Paolo Guiotto: So this is this formula.
27:05:370Paolo Guiotto: So… the triple integral.
27:09:900Paolo Guiotto: of F of XYZ.
27:13:980Paolo Guiotto: In the X, the Y, DZ, the… In cylindrical coordinates, Becomes once more.
27:24:00Paolo Guiotto: is an integral on the domain which is the image to cylindrical coordinates of domain D. So this means you look at the domain, how is it in cylindrical coordinates? So what are the cylindrical coordinates of points of D?
27:39:480Paolo Guiotto: Then, for the function, you have to replace just X is raw plus theta.
27:47:90Paolo Guiotto: Why is Rosa NT times Z?
27:50:880Paolo Guiotto: And for the modulus of the determinant, you are raw.
27:54:530Paolo Guiotto: the raw?
27:55:800Paolo Guiotto: dichita DZ.
27:59:740Paolo Guiotto: Example, and this is the exercise you have also to do. 5-7-4.
28:07:580Paolo Guiotto: Be careful, because there are two or three of these, which are terrible, in very high difficulty. So, do, first of all, those I suggest you to do, which are numbers.
28:26:620Paolo Guiotto: Number 3?
28:30:720Paolo Guiotto: Number 4…
28:40:180Paolo Guiotto: And let's say number 5.
28:44:240Paolo Guiotto: 1, 2, 3, 4, 5. Number 6 and 7 are a little bit harder. I want to show you, for example, the number 1, which is…
28:54:790Paolo Guiotto: sufficiently fast.
28:56:630Paolo Guiotto: We have to compute, in all these exercises, you have to compute volumes, okay?
29:03:10Paolo Guiotto: Lastly, a couple of minutes.
29:05:830Paolo Guiotto: So, this is the volume of this set, the domain defined by these two conditions. So, Z greater or equal than X squared plus Y squared.
29:17:90Paolo Guiotto: And, is that done.
29:20:620Paolo Guiotto: Less or equal than 18 minus X squared minus Y squared.
29:27:940Paolo Guiotto: Or at least let's start this. Now, here it's a typical case where you should use cylindrical coordinates, because you see that in the domain, also in this condition, it appears the quantity X squared plus Y squared.
29:45:390Paolo Guiotto: Okay?
29:46:370Paolo Guiotto: So this quantity is simplified when we pass through cylindrical coordinates, those that we have seen here, respect to the z-axis.
29:56:00Paolo Guiotto: So what we do is, let's call D this domain. The volume of D is the integral on D of 1 in coordinates XYZ.
30:08:720Paolo Guiotto: And we transformed this, by, cylindrical… coordinates.
30:16:550Paolo Guiotto: 2.
30:17:620Paolo Guiotto: The integral of the function remains 1.
30:21:140Paolo Guiotto: Okay? Because it is one, I don't have to do anything.
30:25:280Paolo Guiotto: I have to put the modulus of the determinant, which is just raw, so raw, D wrong.
30:33:150Paolo Guiotto: D theta DZ, and now we have to see what we have to write here.
30:38:830Paolo Guiotto: But, let's look.
30:40:540Paolo Guiotto: The point XYZ belongs to the if and only if the two… there are two conditions.
30:48:260Paolo Guiotto: that we could put together. In fact, we could say that Z is greater than X squared plus Y squared, less or equal than 18 minus X squared plus Y squared, right?
31:04:770Paolo Guiotto: Now, this means what in terms of raw theta and z?
31:10:580Paolo Guiotto: If you plug the cylindrical coordinates, the quantity X squared plus Y squared becomes raw square, so you get raw square, less or equal than Z, less or equal than 18 minus raw square.
31:25:780Paolo Guiotto: That's how that condition becomes in cylindrical coordinates. You don't see theta.
31:32:540Paolo Guiotto: But there is Theta, because we are integrating also in theta.
31:36:780Paolo Guiotto: So what does it mean? That there is no condition for theta, so theta is in its natural range. 0 to pi.
31:47:790Paolo Guiotto: Okay, now, if we look at this, so we could just copy and say, so we have to integrate raw
31:55:690Paolo Guiotto: Iraq.
31:57:130Paolo Guiotto: theta and z on this domain. So let's write what is simple. First, theta between 0 and 2 pi.
32:05:330Paolo Guiotto: And then, let's look at the Z between raw square and 18 minus raw square.
32:12:910Paolo Guiotto: Okay, now let's apply a reduction formula. Well, we would first eliminate that theta, because the function is independent of theta. It's wrong.
32:23:230Paolo Guiotto: The theta does not affect the other conditions, so we could say reduction formula, let's wash out this theta, and we are,
32:33:370Paolo Guiotto: This second integration is here, then the first integration will be between theta and 2 pi, raw d theta, then we have zero EZ.
32:45:190Paolo Guiotto: This raw comes out. It's a constant for that integration.
32:50:100Paolo Guiotto: So we remain with integral 0 to pi of 1 against 2 pi. So we have 2 pi, and then this, which is a double integral, raw d raw DZ,
33:02:260Paolo Guiotto: In this domain, Z between raw square and 18 minus raw square.
33:09:450Paolo Guiotto: And now, we apply the reduction formula.
33:13:590Paolo Guiotto: So we integrate first in raw or inside here.
33:21:20Paolo Guiotto: It seems Z better, because we have already the range in terms of raw. Otherwise, we should transform this into a condition for raw. So, since it is written, and for the function it's indifferent, what is the variable.
33:35:120Paolo Guiotto: So, we start, let's say, with Z, and we finish with R. So, by the way, this will make this outside.
33:42:990Paolo Guiotto: Now, what are the parameterizations I have to put there?
33:49:50Paolo Guiotto: What is the range for Zed?
33:51:790Paolo Guiotto: That's easy.
33:54:260Paolo Guiotto: Here we are, from raw square, definitely, to 18 minus raw square. And what is the rate of raw?
34:09:340Paolo Guiotto: 0 to 18.
34:14:350Paolo Guiotto: Think about when do you have Zed?
34:19:370Paolo Guiotto: I don't want the answer. I want to let you read the condition you have to discuss.
34:24:930Paolo Guiotto: So you have Z here.
34:28:480Paolo Guiotto: when raw square is less than 18 minus raw square, otherwise you don't have any Z.
34:36:80Paolo Guiotto: Okay? So, this means that 2 rows square is less than…
34:41:570Paolo Guiotto: 2 rho square is less than 18, so rho square less than 9, and this means raw between 0, it is positive.
34:52:210Paolo Guiotto: and 3.
34:53:570Paolo Guiotto: So that's the condition.
34:56:100Paolo Guiotto: Okay, so 0 and 3. And now it's easy, no? I just do the first step, you finish. 0 and 3 of raw, then we have that integral, which is just the integral of 1, so you do the difference of the two endpoints, so 18…
35:15:700Paolo Guiotto: Minus 2 raw square, it seems.
35:21:550Paolo Guiotto: So now we have 18 raw minus 2 raw cubes.
35:27:730Paolo Guiotto: So, this is 2 pi.
35:31:220Paolo Guiotto: The first one is 18. The integral of rho is raw square of a 2 between 0 and 3.
35:38:610Paolo Guiotto: Minus 2 raw cube is raw power 4 divided 4 between 0 and 3.
35:44:930Paolo Guiotto: And this leads to battle.
35:46:900Paolo Guiotto: Okay?
35:50:170Paolo Guiotto: Okay, have a nice weekend.
35:53:920Paolo Guiotto: Please do these exercises, it's very important, okay? 574.
35:59:990Paolo Guiotto: You have to struggle a bit,
36:04:570Paolo Guiotto: Okay, let's stop recording.