AI Assistant
Transcript
00:23:630Paolo Guiotto: Okay, good morning.
00:28:920Paolo Guiotto: So, I left you to do… Some of the exercises… Oh, 574…
00:39:340Paolo Guiotto: Numbers 1, 3, 4, 5. We did the number 1.
00:45:120Paolo Guiotto: So… We do number 3.
00:50:790Paolo Guiotto: So, Exercise 5, 7… 4 number 3…
00:59:10Paolo Guiotto: We have to determine the volume of the set of points in R3, where X squared plus Y squared plus Z squared less than or equal than
01:10:630Paolo Guiotto: 16.
01:12:130Paolo Guiotto: And, X squared plus Y squared less equal than…
01:19:590Paolo Guiotto: Well, let's take this exercise also as the opportunity to do some figure.
01:27:270Paolo Guiotto: I told you that it's not always possible to have a reasonable picture of what is going on.
01:34:450Paolo Guiotto: But let me give you some information. Well, it is here that the first set is a set of points where the distance to the origin, which is the root of that quantity, is less or equal than the root of 164, so it's a sphere, okay?
01:52:660Paolo Guiotto: Well, the second one…
01:54:770Paolo Guiotto: it seems the exterior of a disc, but this is in the plane XY. You have to remind that here we are in space R3, so this should be always regarded as a condition for XYZ.
02:08:979Paolo Guiotto: Now, an interesting remark here is that
02:14:910Paolo Guiotto: You see that in both, this would be also important for the integration.
02:19:920Paolo Guiotto: In both conditions, it appears this quantity, X squared plus Y squared.
02:26:50Paolo Guiotto: Yes, in the first condition, there is also a plus X square, but there is not in the second one.
02:31:840Paolo Guiotto: Now, this quantity, X squared plus Y squared, is reminded from cylindrical coordinates.
02:38:420Paolo Guiotto: When you have a point, XYZ, huh?
02:44:240Paolo Guiotto: in R3.
02:48:840Paolo Guiotto: Now, this means that if you project down this point in the plane XY,
02:54:280Paolo Guiotto: The projection has coordinates X, Y, 0.
03:00:210Paolo Guiotto: So, the quantity X squared plus Y squared, or better, the root of that quantity, represents distance of that projection to the origin.
03:10:830Paolo Guiotto: So this is root of X squared plus Y squared.
03:16:390Paolo Guiotto: Which is, of course, the same of the distance between this point and the z-axis. This point here…
03:24:860Paolo Guiotto: on the z-axis is the point with abscess and dot unit equals 0, and the fourth equal to Z. When you do the distance between these two, this
03:34:880Paolo Guiotto: has length norm of, formally, XYZ minus 00Z, no? The distance between two points in R3. So this is the norm of vector XYZ0.
03:50:700Paolo Guiotto: And this is the root of X squared plus Y squared plus 0 squared, which is 0.
03:57:560Paolo Guiotto: So it is this one.
03:59:30Paolo Guiotto: So the quantity X squared plus Y squared root of X squared plus y squared is the distance between the point XYZ and the Z axis.
04:08:730Paolo Guiotto: Now, this quantity is invariant, is constant, when you rotate the point
04:15:450Paolo Guiotto: on a circle centered in the Z axis. So if you do a rotation around the z-axis, the distance to the z-axis does change.
04:26:280Paolo Guiotto: So, this means that, if you have just a point, let's say, that the point this one, which is in the set that verifies these two conditions.
04:36:820Paolo Guiotto: All points obtained by rotating that point
04:40:930Paolo Guiotto: around the z-axis are still in the set, you see?
04:45:170Paolo Guiotto: If, say, for just a tip… X spot.
04:50:390Paolo Guiotto: The two conditions are verified.
04:53:160Paolo Guiotto: The two positions will be verified for all points of this circle.
04:57:670Paolo Guiotto: Because the quantity X squared plus Y squared will be always the same.
05:02:520Paolo Guiotto: So, it doesn't change, and since you don't change Z, the first condition is verified, and the second one is verified as well. So, it means that…
05:12:800Paolo Guiotto: This is the remark.
05:18:210Paolo Guiotto: If… Effect.
05:23:130Paolo Guiotto: the… Depends.
05:28:810Paolo Guiotto: on X1.
05:32:10Paolo Guiotto: True.
05:36:310Paolo Guiotto: well, let's say formally would be the root of X squared plus Y squared, or… which is the same, X squared plus Y squared.
05:45:620Paolo Guiotto: You don't see, X or Y alone, but you see X squared plus Y squared.
05:53:130Paolo Guiotto: Then the set B, is invariant.
06:01:880Paolo Guiotto: by rotations.
06:08:20Paolo Guiotto: around.
06:11:290Paolo Guiotto: the Z axis.
06:14:350Paolo Guiotto: So if you want to draw the sector, there is a simple thing you could do.
06:20:10Paolo Guiotto: You take the z-axis.
06:22:200Paolo Guiotto: and choose one of the two other axes. For example, the y-axis. I choose the Y axis.
06:28:170Paolo Guiotto: Because of the figure anomaly. See, Z is the vertical axis, and XY are one original, but X would point out to the left. So, let's take this one, okay?
06:39:60Paolo Guiotto: So, in other words, the x-axis is pointing, perpendicular to that plane to U, okay?
06:50:440Paolo Guiotto: So, while you do the intersection of the domain D,
06:55:440Paolo Guiotto: with this plane, you get… well, you get something that cannot be like that, but… well, let's say that this is the intersection of D with the plane.
07:08:660Paolo Guiotto: ZY. Now, plane ZY can be described as the set of points where the X is zero, no? Because you are on the plane ZY when the X is 0.
07:21:80Paolo Guiotto: So, the intersection is X equals 0.
07:26:260Paolo Guiotto: Now, this is the section of the domain on the plane ZY. If you want now to reconstruct, let's add the
07:35:720Paolo Guiotto: the third axis, the x-axis. If you want to reconstruct the full domain, you have just to take this one and rotate around the z-axis.
07:47:480Paolo Guiotto: Okay? So, you should do something like this.
07:55:180Paolo Guiotto: Of course, it will be horrible, but let's make an easier figure. Imagine that in the set, in the plane YZ, you get a circle.
08:06:30Paolo Guiotto: If you add a third axis, X, and you rotate the circle around the Z axis.
08:13:190Paolo Guiotto: You get a donut.
08:17:160Paolo Guiotto: But in mathematics, this is called the Taurus.
08:21:800Paolo Guiotto: In fact, it's a donut.
08:23:970Paolo Guiotto: You'll see?
08:26:470Paolo Guiotto: While you don't see…
08:28:420Paolo Guiotto: Take the plane ZY. You have a disk, and imagine that you rotate around the Z-axis. The disk… if you cut the donut, what you get
08:38:330Paolo Guiotto: You got the disc, no?
08:41:240Paolo Guiotto: Sold.
08:43:409Paolo Guiotto: In particular, let's see in the case of our example. So, If the…
08:52:650Paolo Guiotto: is the set where X squared plus Y squared plus Z squared less equal than 16. And then we have X squared plus Y squared greater equal than…
09:07:640Paolo Guiotto: 4…
09:10:460Paolo Guiotto: Now, if we do the intersection of this T with the plane YZ, so we put X equals 0,
09:18:350Paolo Guiotto: we get points of type X, Y… no, sorry, X is 0. Y, Z, such that when you vanish X, you get Y squared.
09:31:790Paolo Guiotto: plus Z square is less or equal than 16.
09:36:240Paolo Guiotto: And, Y squared greater than or equal than,
09:41:570Paolo Guiotto: Let's see what… what this is.
09:43:740Paolo Guiotto: So we are in the plane YZ.
09:48:860Paolo Guiotto: The first,
09:51:100Paolo Guiotto: inequality defines a circle centered at the origin and radius 4. So, we have something like this.
10:08:740Paolo Guiotto: So this is 4, this is minus 4, this is plus 4, this is minus 4.
10:15:90Paolo Guiotto: And the condition defines everything contained in that list, so we may say.
10:21:710Paolo Guiotto: Maybe with the… I cannot change the color here.
10:28:00Paolo Guiotto: Well, let's use it this way.
10:32:30Paolo Guiotto: So… This is You don't see anything, buddy.
10:38:940Paolo Guiotto: Okay, let's take this one.
10:41:210Paolo Guiotto: Back to… So this one, the interior of the disc, Perhaps if I left.
10:49:550Paolo Guiotto: That color, it would have been.
10:52:160Paolo Guiotto: less aggressive.
10:55:980Paolo Guiotto: on this one.
10:59:10Paolo Guiotto: So this in gray is the disk.
11:04:250Paolo Guiotto: in plain YZ, Y squared plus Z squared less or equal than Z3. Then we have Y squared greater or equal than 4. Now, Y squared greater or equal than 4 means either Y is less than minus 2, or
11:21:890Paolo Guiotto: Y is greater than plus 2.
11:25:710Paolo Guiotto: So, it means that there is Y equal to It should be.
11:30:630Paolo Guiotto: At half of this.
11:33:280Paolo Guiotto: However, we don't need to be extremely precise. This is minus 2.
11:37:980Paolo Guiotto: So, we have to take the Y's, which are either greater than, plus 2, or less than. So, it means that if we take this vertical line.
11:48:140Paolo Guiotto: So we have to take these points.
11:50:920Paolo Guiotto: Those of this disk, Which are with Y less than minus 2 and greater than 2.
11:59:780Paolo Guiotto: So that thing in red is the, is the intersection of D with the plane X equals 0. So D intersection X equals 0. Now, we clean up everything, we do a second figure.
12:15:150Paolo Guiotto: Down here.
12:17:200Paolo Guiotto: During only the intersection. So we have said that we have, plus 2, plus 4,
12:24:370Paolo Guiotto: So we have, this… Piece of this here.
12:33:390Paolo Guiotto: And the semantic piece.
12:35:960Paolo Guiotto: On the other side.
12:45:620Paolo Guiotto: And now we have to imagine that these are the Y and Z axis.
12:51:620Paolo Guiotto: And this is what happens when we intersect
12:55:520Paolo Guiotto: our set with the YZ plane. So, to get the full set, we have to rotate around the z-axis this. When you do the rotation of this.
13:07:880Paolo Guiotto: You get a solid figure.
13:12:700Paolo Guiotto: Oh, this is not dashed, this one is dashed.
13:18:70Paolo Guiotto: Which is basically this sphere, so we see also this part here is solid. This sphere with a hole in the middle, a cylindrical hole.
13:28:710Paolo Guiotto: Okay.
13:29:920Paolo Guiotto: So that's the feature of the domain.
13:32:320Paolo Guiotto: Okay, now let's pass to the problem of…
13:36:180Paolo Guiotto: computing the volume, which is the integral of 1 dx, dy, dz.
13:43:580Paolo Guiotto: Now, we have to decide in which coordinates to use. Of course, we could use Cartesian coordinates, but the scope here is to use either polar or spherical or cylindrical coordinates. Now, if we look at the description, this is given by these two constraints.
14:02:250Paolo Guiotto: This last one was 16, and the other one.
14:05:520Paolo Guiotto: X squared plus Y squared.
14:10:40Paolo Guiotto: To greater equal than 4.
14:13:280Paolo Guiotto: So the first… the function is 1, and then we have DXTYZ. So clearly, the function does not change whatever is the system I use. It remains the constant 1. So we have to focus on the
14:27:230Paolo Guiotto: two conditions. The first one gets simplified if we use spherical coordinates, but not the second one.
14:35:940Paolo Guiotto: Because remember that the spherical coordinates are something like raw cos theta sine phi, and then there is Y, which is raw sine theta
14:49:170Paolo Guiotto: sine P.
14:51:110Paolo Guiotto: And then there is Z equal, cosine.
14:54:930Paolo Guiotto: Now, if we do X squared plus Y squared, we get raw square.
15:00:510Paolo Guiotto: Times cos squared plus sine squared theta, this is 1, but then we have sine phi.
15:06:550Paolo Guiotto: square.
15:07:580Paolo Guiotto: So that condition becomes here raw square sine P squared less or equal than 16. No, sorry, this one… sorry, this is raw square less or equal than 16. And the second one is raw square sine square.
15:27:110Paolo Guiotto: be greater or equal than 4.
15:30:30Paolo Guiotto: Which is not so bad, because we could say, well, the first one says raw between 0, remind that raw is positive, it's not raw between minus 4 and 4. But raw between 0 and 4…
15:45:190Paolo Guiotto: And then I could use the second one to provide a condition for phi, sine square phi greater than or equal then for my overall square.
15:56:760Paolo Guiotto: So, this would say, let's integrate first in phi and then in raw. If we could say, we should integrate for raw from 0 to 4.
16:06:920Paolo Guiotto: Well, there is also theta, but theta, you don't see theta, theta means between 0 and ti pi. So we first, well, let's say that.
16:16:680Paolo Guiotto: probably I'm doing a little bit faster. So if we do the change of variable, this becomes a row between 0 and 4, sine square P,
16:29:130Paolo Guiotto: greater than the… 4 over raw square and theta between 0 and 2 pi.
16:38:50Paolo Guiotto: of the function is one, the change of variable. In this case, this is the spherical. This is raw square sine phi, if I'm not wrong.
16:48:240Paolo Guiotto: DRA di phi di theta.
16:53:200Paolo Guiotto: So we may start to integrate first in theta, so we have a raw between 0 and 4.
17:01:140Paolo Guiotto: Science Square.
17:04:950Paolo Guiotto: feet… Great or equal than 4 over a square.
17:11:680Paolo Guiotto: Yeah, for example, you should notice that there could be potentially something which is implicit here.
17:18:550Paolo Guiotto: Because to have sine square feet greater than 4 over raw square.
17:23:810Paolo Guiotto: That 4 over raw square must be…
17:30:660Paolo Guiotto: But that's greater than zero, 4 divided by the positive. Now, there is another condition. Must be.
17:37:70Paolo Guiotto: must be less than 1. Otherwise, if it is greater than 1, you have no fee. Okay, so here there is something implicit, let's see. But, what does it mean is, it means less than 1, it means raw square.
17:51:40Paolo Guiotto: Great or equal, then, 4…
17:54:800Paolo Guiotto: So, since raw is between 0 and 4, this means, raw between 0 and 2.
18:05:250Paolo Guiotto: So this is, something implicit that you should put into, into this, this condition here. So that draw is actually between 0 and 2.
18:16:470Paolo Guiotto: No.
18:17:460Paolo Guiotto: Sorry.
18:18:530Paolo Guiotto: I did the opposite, inequality, rocked, waited.
18:23:630Paolo Guiotto: raw greater than 2, and so this, this means that, this condition is, correctly raw between 2 and 4, okay?
18:35:320Paolo Guiotto: If you write the raw from 0 to 4, that's wrong, because it takes about raw equals 0 would mean 4 divided 0 equals plus infinity. How can the sine phi greater than plus infinity? You understand that's not possible.
18:48:950Paolo Guiotto: Okay?
18:50:480Paolo Guiotto: So you have always to be careful with these things. So then, let's do first the integration in Titan.
18:58:900Paolo Guiotto: the square sine phi, D, theta, then we have D raw.
19:05:90Paolo Guiotto: D, 5. Now, this is a constant for the integration in theta, so it comes out we get 2 pi integral, so we say that, now this becomes raw between 2 and 4,
19:19:570Paolo Guiotto: And sine square phi greater or equal than for over Australia.
19:26:940Paolo Guiotto: Then we have raw square sine P.
19:31:380Paolo Guiotto: D raw B. And now, this is,
19:36:590Paolo Guiotto: the most complicated part of the exercise, because we have 2 pi, we do… the last integration will be in rho, 2 to 4, and then we have to integrate in phi. But here, the range is not written for phi, but through sine square phi.
19:53:760Paolo Guiotto: So, I should say. Now, remind that our fee is between… I'm showing this because this is not the most natural change of variable I should use here.
20:04:30Paolo Guiotto: It would be much better to use cylindrical coordinates, but I'm doing if I do not realize this. So, phi is between 0 and pi.
20:13:810Paolo Guiotto: So sine phi is positive, in particular. So the condition, sine square greater than 4 over raw square, means that sine phi, which is positive.
20:27:410Paolo Guiotto: So, 5 feet greater than the root of 4 over raw square, which is 2 over raw.
20:34:470Paolo Guiotto: or sine phi less than minus this, but this would be negative, so we have this. In terms of phi, this means…
20:45:410Paolo Guiotto: This means?
20:47:810Paolo Guiotto: But this could be a little bit to be careful, because,
20:52:930Paolo Guiotto: Yeah, before we write that sign, we have to be careful, because the phi is between 0 and pi. X sign is between minus pi half, pi half, so we have to be careful.
21:03:690Paolo Guiotto: So, let's think about… this is the interval zero to pile, where there is our fee, and we have sine, which is something like that, no?
21:13:440Paolo Guiotto: We have to understand what does it mean sine phi greater than this value, which is, we know, because of this condition, it is less than 1, okay? So it will be somewhere here, 2 over rho.
21:27:360Paolo Guiotto: So you see that it means phi between a value and a value. So the set of fees for which the sign is larger than 2 over raw is this interval. Now, what are these values?
21:42:450Paolo Guiotto: This one is, since here we are between 0 and pi half.
21:48:600Paolo Guiotto: The value phi, well, sine phi is 2 over rho, this value here, between 0 and pi f, this is really the arc sign
21:58:900Paolo Guiotto: Of 2 over raw.
22:02:280Paolo Guiotto: But what is this value?
22:05:80Paolo Guiotto: This is not the arc sign, because the arc sign yields a value between 0 and pi up.
22:10:640Paolo Guiotto: So this value here, of course, the figure is symmetric, so if this is… this phi means it is also this length here, this length is phi, so it should be the same you have here.
22:25:570Paolo Guiotto: So it means that this value is pi minus that blue segment, which is the ax sign.
22:33:720Paolo Guiotto: of 2 over O. As you can see, there are many technical problems by doing this way. So now I have the range for fee that I have to put here.
22:44:800Paolo Guiotto: Of 2 over rho and pi minus ars sine
22:51:450Paolo Guiotto: of 2 overall. This is the right range.
22:55:780Paolo Guiotto: Then we have the function, raw square sine field.
23:00:660Paolo Guiotto: the, the… this is in the feed, and then we have DRock.
23:06:770Paolo Guiotto: Now, we can carry this raw outside, so we get 2 pi.
23:12:00Paolo Guiotto: Integral from 2 to 4 of rho square. Then we have that integral, from arc sine…
23:20:700Paolo Guiotto: of 2 over rho, 2 pi minus X sine…
23:26:510Paolo Guiotto: of 2 of the raw, of sine… P, DP,
23:33:120Paolo Guiotto: You see, the problem is not the integration for the moment, but the parameterization of the integral is quite complicated. Now, this is the derivative with respect to phi of minus cos phi.
23:46:770Paolo Guiotto: So we have to do the evaluation of Myers cost P,
23:51:290Paolo Guiotto: between these two values, a sine…
23:55:670Paolo Guiotto: Of 2 over raw, and pi minus sign.
24:02:370Paolo Guiotto: of tools at all.
24:04:160Paolo Guiotto: And this is another thing which is a little bit…
24:06:980Paolo Guiotto: disturbing, because… let's start from this one. What is the cosine of the arc sign? It is not cosine of the arcos, but of the arc sign, so, cosine of arcsine
24:23:10Paolo Guiotto: of 2 over r is what?
24:28:880Paolo Guiotto: well.
24:29:900Paolo Guiotto: Be careful. It's a little bit, delicate to do this, because we are doing something like cost of an angle, we call this angle phi. Remind that that phi
24:42:330Paolo Guiotto: This would be important, is because it is an arc sine, this is between 0 and pi a half.
24:50:520Paolo Guiotto: In general, the angle at phi is between 0 and pi, but
24:54:540Paolo Guiotto: Since the value of x sign is between 0 and pi half, when the argument is positive, we have this. So, now what I use is the identity cos square phi plus sine square phi equal 1.
25:09:650Paolo Guiotto: This means cos square phi equals 1 minus sine squared phi, so cos phi is either plus or minus the root of 1 minus sine
25:21:480Paolo Guiotto: square phi. I'm doing this because I will have sine of phi is an arc sign, so sine of x sine, this will simplify.
25:30:880Paolo Guiotto: the argument. Now, plus or minus here.
25:35:480Paolo Guiotto: You have still… so you see how many times you have to be careful here? So it's… it's very… it's very likely that you do an error here somewhere, because there are too many details. You have to remind that, in this case.
25:50:420Paolo Guiotto: the angle phi is between 0 and pi half. Between 0 and pi half, this is cosine.
25:58:390Paolo Guiotto: 0 and pi half. This is cos P.
26:02:580Paolo Guiotto: So, it is positive. So, the right formula is cos phi equal root of 1 minus sine squared phi. This is not, of course, in general, but for that phi, it is the right formula. So, at the end, I can say that cos of arc sine
26:22:20Paolo Guiotto: T2 overall.
26:24:750Paolo Guiotto: It is equal to the root of 1 minus T squared of sine of arc sine
26:34:610Paolo Guiotto: of 2 over raw. Sine over sine are each the inverse of the other, so they kill, and they remain.
26:42:250Paolo Guiotto: 1 minus 2 over raw squared.
26:46:960Paolo Guiotto: Or, in other words, root of 1 minus 4 over raw sphere.
26:53:430Paolo Guiotto: We can also write, we do the common denominator, root of raw square.
26:59:710Paolo Guiotto: minus 4 divided the raw square, we can carry outside a row. I'm doing this because at the end here, I will have this factor, so I can simplify something. Now, in general, you should write a raw square
27:13:710Paolo Guiotto: minus 4 under the root, and then down here, the modulus of rho, because the root of rho squared is at the value of rho. But rho is positive here, so root of rho square minus 4 divided by raw. This is just to get the value of cosine, arc sine 2 over raw.
27:33:570Paolo Guiotto: What… we have to do also the value of cosine at pi minus that number.
27:40:30Paolo Guiotto: When we do costs of… Pi minus that number at sine.
27:48:290Paolo Guiotto: Of 2 overall? No.
27:51:360Paolo Guiotto: Determine what is cos of pi minus something, What is it?
28:00:690Paolo Guiotto: You see, now I'm fucked here, because I don't mind of this ponder.
28:07:130Paolo Guiotto: What these cars are…
28:14:70Paolo Guiotto: So we have, the addition formula for cost is cost, cost, minus x, right?
28:20:350Paolo Guiotto: So, cos pi is minus 1,
28:25:120Paolo Guiotto: Cos of minus arc sine…
28:30:170Paolo Guiotto: of 2 over… I'm using something which is not… Then I have minus sine sine, so sine pi, which is 0 times…
28:40:570Paolo Guiotto: So, I get it is, cos of minus is, even, so it is, like, minus cos of…
28:49:40Paolo Guiotto: arc sign, of, Q.
28:52:950Paolo Guiotto: Overall, so this is minus what we computed above.
28:57:480Paolo Guiotto: root of rho squared minus 4 divided rho. So, at the end, going back here to the evaluation, so we can take the minus outside, let's put the minus here.
29:10:370Paolo Guiotto: Then we have sod.
29:13:60Paolo Guiotto: The evaluation of cost, fee.
29:15:930Paolo Guiotto: Between phi equals R cos arc sign of 2 over rho, and phi equals phi minus arc sine
29:26:760Paolo Guiotto: etc. of 2 over raw. This is the final value, which is this one, it is minus root of raw square, minus 4 by the raw, minus the initial value, which is the same thing.
29:42:280Paolo Guiotto: So, minus root of raw square divided 4.
29:46:380Paolo Guiotto: minus 4 divided raw. So, minus 2 root of raw square minus 4 divided raw. So, this is the quantity we have to flag into this
29:57:700Paolo Guiotto: integral here. It is the value
30:01:220Paolo Guiotto: of this integral, apart for the minus we have outside. So, continuing it here… That's one star.
30:10:970Paolo Guiotto: It is up.
30:13:20Paolo Guiotto: one star.
30:14:810Paolo Guiotto: So we have minus 2 pi integral from 2 to 4. If I'm not wrong, there is a raw square. Then we have this quantity, minus 2, root of raw square minus 4 divided raw in the raw.
30:31:60Paolo Guiotto: We can simplify this, we can put the 2 out here, 4. We can simplify a raw square with a raw.
30:40:340Paolo Guiotto: And, at the end, we have to compute the integral from 2 to 4 of raw.
30:46:690Paolo Guiotto: Root of raw square minus 4, which is an easy integral, because if we look at this as a power, raw square minus 4 to explain 1 half.
30:57:880Paolo Guiotto: Here, you have more or less the derivative of, with respect to raw, of raw square minus 4 to the exponent 1 half plus 1, so 3 halves. In fact, if you do the derivative, you get 3 halves
31:11:580Paolo Guiotto: Rho squared minus 4 to the exponent 3 halves minus 1, which is 1 half, times the derivative of the argument, which is 2 rho.
31:20:90Paolo Guiotto: So, you see that if we put the 3 here, and so we give another 3 here, this becomes a derivative, and therefore the integral solves into an evaluation of rho square minus 4 tox from 3 halves.
31:36:660Paolo Guiotto: Between.
31:38:180Paolo Guiotto: 3 half, between raw equals 2 and draw equal 4.
31:46:440Paolo Guiotto: So, there is something wrong.
32:00:40Paolo Guiotto: I'm not, that is not an error. I'm… it's correct.
32:04:80Paolo Guiotto: So, 4 pi divided 3…
32:08:260Paolo Guiotto: When raw is 4, we get, raw square is 16, minus 4 is 12, 2 exponent 3 halves.
32:16:270Paolo Guiotto: Minus… when raw is 2, we get raw square is 4 minus 4 is 0, 0 to 3 halves.
32:22:520Paolo Guiotto: So, the conclusion is the value is 12, 3 halves, 4 pi divided 3.
32:31:350Paolo Guiotto: Now, let's see what happens. I will do this only once to illustrate to you the differences between spherical and cylindrical coordinates. And in this case.
32:42:820Paolo Guiotto: I guess that it would have been better to use cylinder. So what… If… we… used… cylindrical.
32:58:730Paolo Guiotto: coordinates.
33:02:510Paolo Guiotto: So, let's imagine now we discard all this long solution that is plenty of, of…
33:09:630Paolo Guiotto: perhaps, no, you see, there are lots of points where you have to be careful, and it's very easy, especially during an exam, for example, to do a mistake, okay? Because you cannot pretend to be 100% concentrated on these kind of details.
33:26:690Paolo Guiotto: So, let's return back to the beginning. So, we have to compute the integral on this domain, X squared plus Y squared plus Z squared, less than or equal 16.
33:36:770Paolo Guiotto: And the X squared plus Y squared greater than or equal than 4.
33:42:570Paolo Guiotto: DX, D, Y, DZ.
33:44:340Paolo Guiotto: The cylindrical coordinates we use are the standard one, those we have shown last time, so X equals rock cosine theta
33:54:920Paolo Guiotto: Fitta.
33:57:400Paolo Guiotto: and Z equals Z.
33:59:480Paolo Guiotto: These coordinates are the coordinates for which X squared plus Y squared becomes a unique variable raw square.
34:09:80Paolo Guiotto: What if I have, for example, what if I had here this, this, let's say, that suppose that instead of having X squared plus Y squared, I have X squared plus Z squared.
34:22:580Paolo Guiotto: Well, I have just to use the cylindrical coordinates for which X squared plus Z squared becomes raw squared. In this case, I would use X equals raw cosine theta, and Z is raw sine theta.
34:40:10Paolo Guiotto: And therefore, the third coordinate I do not change is Y.
34:45:30Paolo Guiotto: Now, since the letters are different, but the formulas are the same, it is clear that the modulus of the determinant of the Jacobian matrix of the phi minus 1, which is this one, it's clearly equal still to raw. Nothing changed. It's just the letter, which is different.
35:04:60Paolo Guiotto: But the formula are the same. So you have to be careful, because…
35:08:790Paolo Guiotto: There is not a unique system of cylindrical coordinates. This is the cylindrical coordinates with respect to the z-axis.
35:17:840Paolo Guiotto: The quantity X squared plus Y squared is the distance square of the distance to the z-axis. But if the figure is symmetric, because that figure written there would be invariant with respect to the rotations around the axis with the letter missing.
35:35:610Paolo Guiotto: So these are invariant with rotations with respect to the y-axis, okay? So in this case, you should use, as cylindrical coordinates those that are invariant with respect to the y-axis, and they are this one, okay? So you have to be flexible.
35:52:610Paolo Guiotto: So, okay, let's return to the original.
35:57:700Paolo Guiotto: notation. So, if we change variable.
36:00:910Paolo Guiotto: Now, we have… the domain in the new coordinates becomes… of course, we do not simplify completely the first condition. This becomes raw square plus Z squared, less or equal than 16.
36:15:920Paolo Guiotto: But the second condition becomes easier. Ross, where greater than 4,
36:21:460Paolo Guiotto: There is a third variable, which is theta, which is unconditioned, so this means theta between 0 and 2 pi.
36:31:60Paolo Guiotto: The function remains 1, and the change of variable is a factor raw.
36:37:350Paolo Guiotto: Okay.
36:38:470Paolo Guiotto: This is the new integration. So we can wash out the theta, so applying the reduction formula, we start integrating in theta, the function is independent of theta, and we finish with the integration across Z.
36:56:480Paolo Guiotto: So we have raw square plus Z square less or equal than 16, huh?
37:02:320Paolo Guiotto: And, raw square greater than or equal than 4.
37:07:980Paolo Guiotto: Okay, so this now is equal to 2 pi…
37:13:30Paolo Guiotto: And we have 2 pi times this integral rho squared plus z squared… less or equal to 16,
37:22:350Paolo Guiotto: And raw greater than or equal than 4.
37:25:760Paolo Guiotto: zero. This seems to be easy.
37:29:870Paolo Guiotto: that I want to remind… to you to remind of something here, because what is the set raw square plus n squared less or equal 16? I know that you give the wrong answer.
37:44:420Paolo Guiotto: What is this set?
37:47:310Paolo Guiotto: in the plane raw z. So you have a new plane, new variables, which are raw and Z.
37:53:620Paolo Guiotto: What is that set?
37:57:200Paolo Guiotto: A circle, right?
37:58:910Paolo Guiotto: No.
38:01:910Paolo Guiotto: are simple. Why? Because…
38:05:10Paolo Guiotto: raw is positive, so there is no negative values for raw. Raw is raw. So, the raw and Z are not in the full Cartesian plane, because this raw is still…
38:15:940Paolo Guiotto: raw cos theta, ras and theta is still a norm of something, it's positive. Remind that in these coordinates, raw is always positive. So, in fact, that set is not a full disk, but a half disk.
38:29:350Paolo Guiotto: It states, like, something like this, okay?
38:33:220Paolo Guiotto: Okay, now, we have to decide. We do integrate first in raw, or then in Z.
38:48:150Paolo Guiotto: Okay, well, let's say that this one means… at all?
38:55:70Paolo Guiotto: Yeah, there is no raw less than negative 2, okay?
38:59:80Paolo Guiotto: So, raw greater than 2.
39:03:280Paolo Guiotto: Since there is this one, I could look at a condition for Z, Z square less or equal than 16 minus raw square.
39:13:700Paolo Guiotto: So, tell me if this is correct. So, I said this is the integral from 2 to what?
39:21:170Paolo Guiotto: And Y4 when it comes from Paul.
39:28:640Paolo Guiotto: Yeah, you have to impose that this is positive, because it's larger than something which is positive. So there is something which is implicit here. It is raw square minus 16 minus raw square.
39:43:620Paolo Guiotto: must be positive, otherwise no Z.
39:48:10Paolo Guiotto: And since this positive means raw square are less than 16,
39:53:70Paolo Guiotto: Which is the old condition, raw less than 4.
39:57:910Paolo Guiotto: greater than 0, less than 4. So, if we put now together this one, row between 0 and 4, and this one, row greater than 2,
40:07:110Paolo Guiotto: you have to put 4 in that range. So this is indica from 2 to 4.
40:13:490Paolo Guiotto: And the second integration is now Z squared less or equal than 16 minus rho squared, so Z between
40:24:340Paolo Guiotto: Yeah, minus, because Z can be negative. Z is the old Z. So minus root of 16 minus raw square plus root of 16 minus raw square. Now, raw is the function, this is in Z, and then we have raw, and that's it.
40:41:300Paolo Guiotto: The raw comes out of this integral, so we have 2 pi integral 2 to 4.
40:48:450Paolo Guiotto: Yes, of course, you see there are also here some little danger, but much less than the previous version.
40:56:70Paolo Guiotto: So now we have to integrate from minus root of 16 minus raw square to plus the root of 16 minus raw square, just 1 in the Z, and this is the length of the interval, so it will yield 2 times root of 16 minus raw square.
41:14:510Paolo Guiotto: So we have 4 pi integral 2 to 4 of rho, and this is 16 minus raw squared 2X 1.
41:24:770Paolo Guiotto: out.
41:26:240Paolo Guiotto: Now, let's recognize that this is the derivative with respect to rho of 16 minus rho squared to the exponent 1 half plus 1, so 3 halves. In fact, if we do the derivative, we have 3 halves.
41:40:740Paolo Guiotto: 16 minus raw square times minus 2 raw. So, in fact, it's exactly
41:48:380Paolo Guiotto: more or less the same as before. We have minus 3 rho times to 1 half exponent, sorry, 16 minus rho squared exponent 1 half. So we put the minus 3 here, and we divide by minus 3 half, yeah.
42:03:930Paolo Guiotto: So, minus 4 pi divided 3, evaluation of 16 minus raw square to exponent 3 half between raw equal 2 and raw equal 4.
42:18:900Paolo Guiotto: Now, at row equals 4, we get 16 minus raw square, rho square is 16, so 0, so 0 to 3 half, minus a trifle 2, rho square is 4, 16 minus 4 is 12, 12 to exponent 3 half.
42:34:290Paolo Guiotto: And of course, we get the same result. So this evaluation is minus 12 to exponent 3 half, so with the minus out here, makes plus 12 to exponent 3 half.
42:46:710Paolo Guiotto: 4 pi over 3.
42:49:810Paolo Guiotto: I'd say that this is definitely easier than the other version.
42:54:770Paolo Guiotto: Less complicated calculations, less complications with the constraints, and so on, okay?
43:03:350Paolo Guiotto: So this would… would say that, as general rule.
43:07:420Paolo Guiotto: Whenever possible, you should use the cylindrical better than spherical, unless the domain is defined through
43:17:50Paolo Guiotto: directly the quantity X squared plus Y squared plus Z squared. In that case, if there is always that quantity repeating, you would use the spherical coordinates in such a way to simplify that quantity.
43:31:850Paolo Guiotto: Okay, do you want to take a 5-minute break?
43:36:50Paolo Guiotto: Okay.
43:44:510Paolo Guiotto: I want to illustrate, one more example.
43:50:540Paolo Guiotto: This is, this is a little bit, more complicated.
43:58:480Paolo Guiotto: We are always in the same kind of difficulty, but let's see, example…
44:07:330Paolo Guiotto: This is the example 5410, huh?
44:11:490Paolo Guiotto: It is to compute the volume of this atom.
44:16:30Paolo Guiotto: set of points, XYZ.
44:19:460Paolo Guiotto: In our 3… Such that we have these two conditions. Seems easy.
44:28:540Paolo Guiotto: And it is still the same type of center we have seen before. This is a sphere centering in the origin, radius R, here we assume R positive.
44:41:630Paolo Guiotto: And, we have this second condition, X minus, half squared plus Y squared.
44:50:880Paolo Guiotto: less or equal than R squared divided.
44:58:260Paolo Guiotto: Now, this second set, if I look in coordinates, XY is
45:06:370Paolo Guiotto: this second condition in coordinates XY, so forget for a second of Z, This is…
45:14:860Paolo Guiotto: The disc, you see the center is a0, radius, divided 2.
45:23:430Paolo Guiotto: And since there is also Z, and Z is, unconstrained, this means that you have to take XY in the disk, and whatever is Z. This is a cylinder.
45:32:420Paolo Guiotto: With the, with the circular section.
45:36:160Paolo Guiotto: And to be precise, if you want, we can do the figure here, it's a bit…
45:43:110Paolo Guiotto: So we go… let's start with the plain XY.
45:46:230Paolo Guiotto: Because in the plane XY, so if we put Z equals 0, the first becomes this center in the origin of radius R. So this is the section of the sphere.
45:59:710Paolo Guiotto: in the plane XY.
46:01:980Paolo Guiotto: of radius R, so this is R.
46:05:190Paolo Guiotto: This is us.
46:07:490Paolo Guiotto: Then we have a second circle, which is the circle centered at point R half, which is here, 0, with radius R half. So this is the second circle.
46:24:280Paolo Guiotto: Okay, so this is in the plane XY. What is X squared minus R half squared plus Y squared by less or equal R squared by the 4?
46:35:580Paolo Guiotto: Now we have the Z axis.
46:38:690Paolo Guiotto: And once we have visited axis, the first becomes a sphere, so we have an object like that.
46:51:70Paolo Guiotto: And this one is a cylinder.
46:54:540Paolo Guiotto: an infinite vertical cylinder with the axis parallel to the z-axis, but it is not the z-axis.
47:01:150Paolo Guiotto: And when we do the intersection with the sphere, what we get is something like this.
47:12:330Paolo Guiotto: Now here, it's a bit curved, because it is on the surface of the…
47:17:340Paolo Guiotto: Sphere, and something similar on this.
47:35:400Paolo Guiotto: It's not the… it's… But it's sort of, cylindrical… Part of the sphere.
47:46:310Paolo Guiotto: Okay, now, let's see what is the volume of this domain. So, it will be the integral on the domain itself of the function 1 dx dY.
47:58:390Paolo Guiotto: Visa.
48:00:550Paolo Guiotto: Now, the problem with this domain is that, of course, the first constraint, the function is 1, so whatever is the change of variable, it remains 1. So let's focus on the domain.
48:11:790Paolo Guiotto: We have a first condition, which is this one, for which it would be better to use a spherical coordinates, because in spherical coordinates, it becomes raw square less than r-square, so three variables.
48:22:870Paolo Guiotto: Shrinks down into one variable.
48:25:640Paolo Guiotto: But this is not good for these two.
48:29:740Paolo Guiotto: For these two, it could be better, for example, to use adapted, spherical, no, cylindic coordinates. What does it mean, adapted?
48:39:990Paolo Guiotto: I want, for example, that this one to be a raw square, so I have to do X minus R half equal raw cosine theta, Y equal raw sine theta, and Z equals Z. This would be an adapted
48:56:910Paolo Guiotto: System of, cylindrical coordinates.
49:00:470Paolo Guiotto: With this system, that condition, the second one, would become raw square less or equal than R squared divided 4. So, this would simplify the second, but probably would complicate the first one.
49:14:210Paolo Guiotto: Because if you put this into the first one, well, Z remains Z, but the X squared plus Y squared is no more equal raw square.
49:23:400Paolo Guiotto: So, you add in the first condition also theta, so in fact, the first condition remains conditioned in three variables, and it's more complicated.
49:31:860Paolo Guiotto: Now, the trick here is that we use the standard cylindrical coordinates. So, we do this by the assist… we change variable by using X equals raw cosine theta, y equals raw sine theta, and Z equals Z.
49:50:360Paolo Guiotto: Which is apparently not the best for each of these two concepts, because it does not simplify too much the first one, and apparently it does not simplify even the second one.
50:03:530Paolo Guiotto: But let's see what happens.
50:06:60Paolo Guiotto: So once we do the… the focus here is on the domain, not on the function. The function is 1, so you cannot simplify anymore this than 1, this function.
50:16:930Paolo Guiotto: Then we have the raw, which is the usual factor for the change of variables, the raw, theta, the Z. And now let's see what happens to the domain.
50:26:960Paolo Guiotto: So, let's say that point XYZ belongs to D, if and only if we have these two constants, X squared plus Y squared plus Z squared, less or equal than R squared.
50:42:690Paolo Guiotto: R is the parameter here.
50:44:790Paolo Guiotto: With this change of variable, X equals rho cos theta, Y equals rhos and theta z equals Z, this quantity here, X squared plus Y squared, becomes a raw square.
50:57:00Paolo Guiotto: So, Z remains Z, so this condition is just raw square plus Z squared less or equal than R squared.
51:06:300Paolo Guiotto: Let's give a look to the other condition. The other condition is X minus R half square plus Y squared less or equal than R squared divided 4.
51:17:120Paolo Guiotto: Now, it is convenient if we develop this square, because this will simplify a bit. There will be an X squared that goes together with Y squared, so I will expand this square, I get X squared minus the double product is minus
51:33:40Paolo Guiotto: RX, simply, plus R squared divided 4,
51:39:470Paolo Guiotto: plus Y squared less or equal than R squared divided 4.
51:45:400Paolo Guiotto: As you can see, we can cancel this, so it remains 0 on this side. And then we can put together X squared plus Y square.
51:55:580Paolo Guiotto: So these two together, a raw sphere.
52:00:970Paolo Guiotto: And for this one, there is nothing we can do except then replacing the value for X. So let's see what we get.
52:10:30Paolo Guiotto: So, X squared plus Y square yields raw square.
52:14:260Paolo Guiotto: minus RX. X is raw cosine theta.
52:20:400Paolo Guiotto: it is less or equal than zero. This is the condition.
52:25:30Paolo Guiotto: Okay? We can slightly simplify this, because raw is positive. Raw equals 0 is just the .00, so we can assume raw positive.
52:35:30Paolo Guiotto: Raw positive, this means raw minus R cosine theta is less equal than zero.
52:43:170Paolo Guiotto: Okay? So this is literally what happens when we…
52:47:670Paolo Guiotto: pass from Cartesian coordinates to polar coordinates, so we can say that, returning here, this
52:56:100Paolo Guiotto: has these conditions. Ros square plus z squared less or equal than R squared.
53:02:740Paolo Guiotto: This is the first one. The second one is raw.
53:07:330Paolo Guiotto: minus R cosine theta, less or equal than 0.
53:12:600Paolo Guiotto: Now, you see that this involves also theta.
53:15:550Paolo Guiotto: Okay, so there are all three variables represented here.
53:20:750Paolo Guiotto: So, for example, it is not the case where I don't see theta, so theta is… theta is in 0 to pi, but here there is also theta, so this is the parameterization. Okay, so, now, we have to decide… we have three integrations to do, which is the first one.
53:40:490Paolo Guiotto: That could be in one variable, and then we have two other variables remaining, or we could start with two variables.
53:49:170Paolo Guiotto: Now, let's decide what should be the first integration. If you look at these two conditions.
53:56:750Paolo Guiotto: What do you think?
54:03:90Paolo Guiotto: You say Zad.
54:09:750Paolo Guiotto: Yeah, the problem with, starting with Zed, to me is that, yes, I have apparently a condition, the first one.
54:18:410Paolo Guiotto: But then, the first one will imply a condition for rock, you see?
54:23:800Paolo Guiotto: Because this one, if you write Z squared less or equal than R squared minus raw square.
54:33:350Paolo Guiotto: Yeah, there could be… okay, let's see, let's explore this.
54:37:300Paolo Guiotto: Okay, I just copied the integration here. Rho square plus Z squared plus or equal than R squared, rho minus R cos theta.
54:48:140Paolo Guiotto: Plus or equal 0, then we have raw, zero.
54:53:970Paolo Guiotto: Okay, let's see. You proposed that we should start with Z. Z squared would be less or equal than R squared minus raw square.
55:04:660Paolo Guiotto: We noticed that because of this second condition here.
55:09:350Paolo Guiotto: We have also raw less or equal than R cosine theta.
55:14:440Paolo Guiotto: So, in particular.
55:16:640Paolo Guiotto: So, normally, if I do not look at that, I would say, okay, to have this, I would need that this quantity must be positive, right? So this would mean raw square less or equal than R squared, right? And since raw is positive, yeah.
55:36:120Paolo Guiotto: This means you're all less so equal than up.
55:38:930Paolo Guiotto: I would need this, I should…
55:41:160Paolo Guiotto: added this condition. But this condition here… Already implies that one.
55:49:130Paolo Guiotto: Okay? Because cosine will be less or equal than 1, so if you are less than R cos theta, you are definitely less than R, and therefore this is automatically verified. So, you don't need to add this condition, because it is already implied from that one.
56:08:00Paolo Guiotto: Okay, so let's see, this would mean that we can split this integration into integration for raw and Z. Now, Z, we said that we integrated for raw and theta. So what are the conditions for raw and theta, then?
56:24:610Paolo Guiotto: So, and then we have the integration in Z first, be raw. So the function is raw. We say that the Z varies, we can see this is Z between
56:41:450Paolo Guiotto: plus the root of R squared minus r squared, and minus the root of the same.
56:48:90Paolo Guiotto: So I have minus root of R squared.
56:51:950Paolo Guiotto: Minus raw square, and plus the root.
56:56:980Paolo Guiotto: of… no, this is… oh yeah, this is Zed, sorry, this is Zed.
57:02:670Paolo Guiotto: I was thinking that it was integration in Goh.
57:05:770Paolo Guiotto: this one, right? Now, the point is, what we have to write for the conditions for raw and theta.
57:14:300Paolo Guiotto: So…
57:15:320Paolo Guiotto: This condition has been already used to determine Z, and implicitly would require this one, but this one is already implied from this by this one, okay? So, that condition, this condition, is stronger than this, so I have to put that condition.
57:34:120Paolo Guiotto: raw less or equal, then R cos theta. This automatically implies raw S or equal then r, which makes sense for the second integration, and there is no other condition.
57:45:720Paolo Guiotto: Okay?
57:46:940Paolo Guiotto: But be careful, because here there is something implicit in that the conditions were less or equal than cos theta. Why? Because cos theta cannot be negative, right? Otherwise, there is no raw for that condition.
58:00:50Paolo Guiotto: But we will discuss in a moment. First, let's do the first integration, so we carry raw outside. We have an integral on the same set, rho less or equal than r cosine theta.
58:12:40Paolo Guiotto: rho, then it remains the integral of 1, so it is the length of the interval, and since the interval is symmetric, this is just 2 root of r squared minus rho squared. This is in zero.
58:27:70Paolo Guiotto: D, theta.
58:30:900Paolo Guiotto: And now… Here we have to decide if to integrate first in raw and then in theta, or
58:38:270Paolo Guiotto: Or vice versa.
58:41:730Paolo Guiotto: Apparently, if you look at this condition, you have raw less than, so it should be less than something that depends on theta. So this should say, well, let's start first integrating a row, and then we will integrate in theta.
58:55:90Paolo Guiotto: This is 2 raw root of r squared minus raw square. What are the range? So, raw is written there.
59:04:270Paolo Guiotto: must be less than r cos theta, so since raw is implicitly positive, because it's a positive number, so here you have to put 0, here you have to put R cosine theta, and what about theta?
59:18:520Paolo Guiotto: Theta is from 0 to 2 pi.
59:22:20Paolo Guiotto: Now, a little trick here.
59:25:20Paolo Guiotto: So, the range for theta… theta is the usual angle
59:30:400Paolo Guiotto: is from 0 to pi, and we need that this quantity be positive, so cosine be positive. Now, if you think to cosine, this is the…
59:41:280Paolo Guiotto: plot of cosine. When cosine is positive.
59:46:130Paolo Guiotto: here and here. What is this? This is 0 to pi half, and this is 3 half pi to 2 pi, okay? So, since we needed that raw less or equal than R cosine theta.
00:00:750Paolo Guiotto: is possible, or implies, that cosine theta
00:07:770Paolo Guiotto: be positive, so this means that theta is between 0 and pi half.
00:15:720Paolo Guiotto: Or theta between 3 halves.
00:18:890Paolo Guiotto: Pi, and 2 pi.
00:22:610Paolo Guiotto: But little trick.
00:24:950Paolo Guiotto: Now, so I… I should split this integral in two pieces. So the integral to 0, from 0 to pi half, plus the integral from 3 half pi to 2 pi, which is, of course, well possible.
00:40:240Paolo Guiotto: But a little tree.
00:42:310Paolo Guiotto: Now, since we are interested in the range of values of cosine theta when theta is in 0 to pi.
00:50:210Paolo Guiotto: 0 to pi is a period for the function, no? Then the function repeats.
00:54:740Paolo Guiotto: But, for example, also minus pi, 2 pi, is a period.
00:59:590Paolo Guiotto: And we have the same range of values for the… for the cosine, so this would be the plot of the cosine in that interval.
01:08:870Paolo Guiotto: And these values are minus, pi half, pi half, yeah.
01:13:240Paolo Guiotto: So, when theta is between minus pi and pi, it geometically means the same, because theta is that angle in the plane XY, so minus pi, 2 pi means still minus 180 degrees, 2 plus 180 degrees.
01:28:20Paolo Guiotto: In total, 360 degrees, so it's the same thing, but it's a little bit better, because with this choice, you can say that cosine theta is positive when theta is between minus pi half and plus pi half.
01:43:360Paolo Guiotto: So, you could also say that this integral can be written as a unique integral from minus pi half to plus pi half.
01:51:610Paolo Guiotto: It's just a trick.
01:54:210Paolo Guiotto: So, I prefer, of course, to use the trick. So, integral from minus 5 half
02:00:10Paolo Guiotto: to plus pi half of integral from 0 to r cos theta.
02:05:700Paolo Guiotto: of, two raw root of… As we are.
02:11:390Paolo Guiotto: Minus rho squared, this is the integral in raw, then we have a second expression in theta, and we are done.
02:17:850Paolo Guiotto: Now, this one is derivative, as usual. Let's take R squared minus raw squared, this is to exponent 1 half.
02:30:20Paolo Guiotto: So, this comes from doing the derivative of the same thing to the exponent 1 half plus 1, so 3 halves.
02:36:860Paolo Guiotto: If we do the derivative, we get 3 halves R squared minus rho squared to exponent 1 half times minus 2 rho. So, as you can see, the derivative is minus 3 rho, so the 2 is not useful. We put the minus 3 here and here, okay?
02:56:900Paolo Guiotto: So now we can say that it is minus 2 thirds, the second integration is from minus pi half to pi half.
03:04:290Paolo Guiotto: of…
03:05:390Paolo Guiotto: this thing, so R squared minus raw square to exponent 3 halves to be evaluated between rho equals 0 and rho equals r cosine theta, then this would be integrated in theta.
03:24:640Paolo Guiotto: When rho is r cos theta, we get r squared minus R squared cos…
03:31:850Paolo Guiotto: Square theta to exponent 3 half.
03:36:720Paolo Guiotto: We will see in a second that this simplifies. When rho is 0, we get R squared
03:43:710Paolo Guiotto: Two exponents, we have.
03:46:560Paolo Guiotto: So, if we factorize an R squared, so it becomes R squared to 3 halves.
03:52:700Paolo Guiotto: That multiplies 1 minus cos squared to 3 halves.
03:59:330Paolo Guiotto: But this is, sine square, right?
04:03:750Paolo Guiotto: So, we have minus 2 thirds
04:09:290Paolo Guiotto: Integral from minus pi half to pi half.
04:13:620Paolo Guiotto: Off.
04:14:560Paolo Guiotto: R squared to 3 half R is positive, it's the radius, this is R cubed.
04:20:00Paolo Guiotto: Times sine squared to 3 halves, be careful, because sine…
04:26:580Paolo Guiotto: It's not necessarily positive on that interval, so this becomes the absolute value sign.
04:32:190Paolo Guiotto: Theta to power 3.
04:34:440Paolo Guiotto: Now, because it is this to exponent 3 half, so it is sine squared, the root, if you want, of sine squared.
04:43:620Paolo Guiotto: theta to power 3. So this is modulus of sine theta cubed.
04:49:310Paolo Guiotto: This is, the first term, minus R cubed here.
04:54:400Paolo Guiotto: feedback.
04:56:520Paolo Guiotto: Okay, so first of all, let's take this R cube out and change the size, so we get 2 thirds R cube.
05:05:110Paolo Guiotto: Integral from minus pi half to pi half of 1 minus absolute value sine theta Cuba.
05:16:560Paolo Guiotto: Well, the first integral from minus pi half to pi half of 1 is the length of the interval, so this part here is just pi. So what remains is the integral from minus pi half to pi half of that module sine theta.
05:31:940Paolo Guiotto: Juba.
05:33:30Paolo Guiotto: But now…
05:34:390Paolo Guiotto: sine theta is odd, but with the absolute value, it becomes even. Cube remains even, so if you change theta with minus theta, you don't change that value. So you can say that this simplifies into 2 times the integral on half intervals.
05:51:790Paolo Guiotto: From 0 to pi half of modulus sine theta.
05:55:310Paolo Guiotto: Cuba.
05:56:580Paolo Guiotto: And in this case, now sine theta is positive, so this value is just sine theta.
06:02:740Paolo Guiotto: So, sine theta cube. So, at the end, we have to integrate this.
06:08:400Paolo Guiotto: Scienta Cuba.
06:10:800Paolo Guiotto: from 0 to pi half, which is an elementary integral. We do…
06:16:70Paolo Guiotto: Just for your pleasure, but it's elementary. Integral from 0 to pi half of sine cubed theta.
06:26:330Paolo Guiotto: We can do by parts.
06:30:300Paolo Guiotto: Because we… No. We write the sine square… Theta sine theta.
06:38:990Paolo Guiotto: We do this trick. This is 1 minus pose square theta.
06:45:190Paolo Guiotto: So we get the integral from 0 to pi half.
06:49:260Paolo Guiotto: We split this into integral sine theta minus the integral from 0 to pi r of cos square theta sine theta. We give the minus to the sine.
07:01:990Paolo Guiotto: In such a way that this is more or less derivative.
07:05:680Paolo Guiotto: So this is the derivative with respect to theta of minus cosine theta, so we get the evaluation of minus cos theta between 0 and pi half.
07:17:970Paolo Guiotto: By half year.
07:20:820Paolo Guiotto: While this guy should come from doing the derivative with respect to theta of cos theta cubed.
07:28:940Paolo Guiotto: Because when we do the derivative, we get 3 cos theta squared times the derivative of cos, which is minus sine theta.
07:37:710Paolo Guiotto: So we get… we need a factor of 3, so we put a 3 here, 1 third outside. We have 1 third the evaluation of cosine cubed theta between, again, 0 and pi half.
07:54:230Paolo Guiotto: The value of cosine at pi half is 0. So let's put the minus outside here. Minus. So the value at pi half is 0, minus the value at 0 is 1.
08:04:150Paolo Guiotto: Plus 1 3rd, again, the value at pi half is 0, so power 3 is 0, minus the value at 0 is 1 to power 3 is 1.
08:13:630Paolo Guiotto: So it is, 1 minus 1 third… 2 tent.
08:21:310Paolo Guiotto: Okay, so this is the value of…
08:27:300Paolo Guiotto: Of what?
08:28:960Paolo Guiotto: I forgot what I'm computing here.
08:35:160Paolo Guiotto: Yeah, but they already have the… so it's the integral of sine cubed should be correct.
08:42:200Paolo Guiotto: Okay, so the integral of sine cubed comes to third…
08:46:830Paolo Guiotto: So, it means that… so let's put equal here, and let's face a conclusion.
08:54:990Paolo Guiotto: So we have… We have, at the beginning, two-third our cube.
09:05:200Paolo Guiotto: Then we have a first integral, which is equal to pi, minus…
09:12:790Paolo Guiotto: A second integral, which is the integral of sine cube. We said it's 2 times the integral from 0 pi half of sine cube. We computed this one, which is 2 thirds, so we have to take 2 times this, so 2 times 2 thirds.
09:32:40Paolo Guiotto: So, at the end, it is 2 thirds R cubed pi minus forward to it.
09:39:580Paolo Guiotto: And this is… V.
09:43:560Paolo Guiotto: There may be errors, but I can say that this is more or less developed.
09:49:470Paolo Guiotto: Okay.
09:52:700Paolo Guiotto: Good.
09:54:760Paolo Guiotto: So, finish the exercise left, of… finish… 574…
10:07:590Paolo Guiotto: So, it remains to do the numbers 4, which is very similar to the number 3 we have done before, and number 5.
10:15:980Paolo Guiotto: If you want to try… Well, there is no number 6, sir. Number 6, it's well possible.
10:24:340Paolo Guiotto: If you want to…
10:26:900Paolo Guiotto: to suicide. Try to do number 7. It's a bit, let's say 2 star plus.
10:34:840Paolo Guiotto: It's a bit, technical. I just give you the… the little hint that use… cylindrical.
10:47:840Paolo Guiotto: coordinates.
10:49:240Paolo Guiotto: But you will get created, too.
10:51:830Paolo Guiotto: Describe the domain. Of course, the function is 1, because it's a volume. Do also D575…
11:04:270Paolo Guiotto: Number 4… 5 and 6, huh?
11:12:60Paolo Guiotto: These are with the integrations in three variables.
11:17:860Paolo Guiotto: Okay. Now,
11:21:160Paolo Guiotto: let's return back to the change of variable formula, because we have seen the application of change of variable to very special change of variables. So,
11:32:80Paolo Guiotto: D… Change off.
11:37:420Paolo Guiotto: variable formula.
11:43:20Paolo Guiotto: Says… But…
11:47:120Paolo Guiotto: If you want to compute an integral on a domain D of a function f of a certain array X,
11:54:400Paolo Guiotto: This is the same of the internet, and you want to introduce a new variable.
11:59:70Paolo Guiotto: Y equal to a certain function P . Now, to be a change of variable, we need that this function ph be invertible, so you can express X in function of Y.
12:17:540Paolo Guiotto: And, here, and the two, phi and phi minus 1, both differentiable.
12:26:510Paolo Guiotto: So this integration is transformed into an integration in Y,
12:31:460Paolo Guiotto: Y varies in the image set through the map fee of the initial domain V, so this is the range for Y. The function f is written in the new variable Y, so this means that we don't have to replace the letter X with the letter Y, but
12:48:420Paolo Guiotto: you have to replace whatever is X with respect to Y,
12:54:460Paolo Guiotto: formula. Then there is the modulus of the determinant of the Jacobiometrics of phi minus 1, Wow.
13:02:920Paolo Guiotto: DY.
13:05:570Paolo Guiotto: And we have seen how this formula works in very particular cases of change of value.
13:11:990Paolo Guiotto: I want to illustrate some examples of different change of variables. Now, I told you when we have seen this formula for the first time that the use of this formula
13:28:200Paolo Guiotto: is a little bit more complicated, not just because there is a multiple integral, but because when you have to introduce a change of variable, normally,
13:40:40Paolo Guiotto: In one variable, this is quite clear what you have to do, because one simplifies, so you introduce a new variable, which is a certain expression of the initial variable, to simplify.
13:51:610Paolo Guiotto: Here, there are several variables, so introducing a change of variable is never a very standard operation. So that's why sometimes it is guided, for example, as in the exercise.
14:06:310Paolo Guiotto: 5… 7… 6, huh?
14:10:140Paolo Guiotto: So let's take the number 1.
14:13:130Paolo Guiotto: Integral on D of function XY, DXDY,
14:18:540Paolo Guiotto: where domain D is the set of points XY, in R3.
14:27:830Paolo Guiotto: Such that there are these conditions.
14:30:760Paolo Guiotto: X times Y is between 1 and 3.
14:35:50Paolo Guiotto: And X is less or equal than Y, which is less or equal than 3X.
14:43:00Paolo Guiotto: And it says, use… Change.
14:48:570Paolo Guiotto: of valuable.
14:50:330Paolo Guiotto: where you define new variables, UV, equal.
14:56:790Paolo Guiotto: One is XY, And V is Y over X.
15:04:180Paolo Guiotto: So it is giving the assignment where this is what we, in the previous notation, called the Y variable, the new variable. So it is, you see, UV is a function of X and Y.
15:20:850Paolo Guiotto: So it means that this time, the change… and differently from polar, spherical, or cylindrical coordinates, where the change of variable is given in this format.
15:32:600Paolo Guiotto: You have the Cartesian function of the polar coordinates. This is how it's given in… opposed to organize.
15:42:40Paolo Guiotto: In this case, we have a change of alcohol where we have new coordinate functions of the initial ones, which are X and Y.
15:52:790Paolo Guiotto: Okay?
15:53:990Paolo Guiotto: So let's see how it works.
15:57:240Paolo Guiotto: But basically, there is nothing really new, because we have seen for this particular change of variables, like the polar coordinates, the spherical and the cylindrical coordinates.
16:07:970Paolo Guiotto: How it works, so it's the same here, but let's see, more or less, let's see how it works exactly.
16:16:670Paolo Guiotto: First of all, let's have a look of the domain, huh?
16:21:00Paolo Guiotto: this case, It is possible.
16:25:420Paolo Guiotto: You see that we have a domain where…
16:30:680Paolo Guiotto: We are in the plane XY.
16:33:90Paolo Guiotto: We have this condition first, X times Y between 1 and 3.
16:38:880Paolo Guiotto: How we could represent this first condition.
16:46:270Paolo Guiotto: Now, I do also this for other purposes, because it may happen to have
16:52:570Paolo Guiotto: draw sets like that. So, I know that most of people will write something like this, Y between 1 over X and 3 over X?
17:05:590Paolo Guiotto: But is that correct?
17:12:90Paolo Guiotto: Yeah, that's the point. For X greater than 0, it is like that. For X negative.
17:18:570Paolo Guiotto: Well, for X equals 0, I cannot… I cannot write this because I… I have to divide by 0, but you see that if you put X equals 0 there, you get 0 between 1 and 3, and that's 4, so X equals 0 is not in that condition, so we can consider just X positive and X negative.
17:36:820Paolo Guiotto: For X negative, this is the right condition, and basically the two are flipped, so I will have below 3 over X, and above 1 over X.
17:48:350Paolo Guiotto: So these are hyperbolas.
17:50:590Paolo Guiotto: So for example, in the first quarter, when X is positive, 1 over X, say that it is like this.
17:59:480Paolo Guiotto: It is like this.
18:01:930Paolo Guiotto: This is 1 over X. 3 over X is 3 times this, so it would be bigger, will be above.
18:11:170Paolo Guiotto: So this is… 3 over Xa.
18:14:660Paolo Guiotto: And we have to take whatever Y is between the two lines, so the region between these two hyperbolas.
18:22:780Paolo Guiotto: Is, the… is what is described by this.
18:28:300Paolo Guiotto: On the other side, we have the opposite, because now 3 over X is more negative than 1 over X, because 1 over X is negative, so 3 times is more negative, let's say. So this is 3 over X.
18:44:550Paolo Guiotto: This is 1 over X.
18:48:50Paolo Guiotto: And the region is still what is between these two lines, so this is the region, okay? So we have this here.
19:02:240Paolo Guiotto: Y minus? Why do you see a minus here?
19:11:110Paolo Guiotto: You don't have to think that's a number.
19:14:90Paolo Guiotto: To the negatives must have a minus in front, okay?
19:18:350Paolo Guiotto: then, okay, this is part of the domain, because it's the first condition, then we have a second condition, which is why between…
19:27:800Paolo Guiotto: X and 3X. This is a little bit simpler, because it says Y above the line Y equal X, so Y equals X is this one.
19:42:830Paolo Guiotto: So you must be above that line, but below 3X. 3x is similar, it passed to the origin, but with the higher slope.
19:52:860Paolo Guiotto: Now, reminder, this is Y equal X, and this is Y equal 3X.
19:58:670Paolo Guiotto: So when you can be between these two, above Y equal X, below Y equals 3X.
20:11:660Paolo Guiotto: Hello, so the question is, is the region
20:15:360Paolo Guiotto: Y between above X, close the X, all this. What is this?
20:20:900Paolo Guiotto: two lines?
20:28:60Paolo Guiotto: Yeah, of course, it's wrong.
20:31:290Paolo Guiotto: But, so, is this the region?
20:37:440Paolo Guiotto: plus this…
20:44:340Paolo Guiotto: this one. Y between… above X, below 3X, is the green region
21:00:400Paolo Guiotto: No.
21:03:190Paolo Guiotto: It's only the other part.
21:05:480Paolo Guiotto: Why?
21:11:530Paolo Guiotto: Because this is X, so this is Y equal X, right? This is Y equal 3X. How can you be above Y equal X and below 3X on that side?
21:26:360Paolo Guiotto: You see, Y must be above.
21:30:30Paolo Guiotto: and below 3X. Above X, below 3X. If you are above Y for X, you are here.
21:38:30Paolo Guiotto: But if you are below 3X, you are here, so you do not be between the two.
21:44:790Paolo Guiotto: This would be Y above the X below X, okay?
21:50:130Paolo Guiotto: So now, if we do the intersection, we see that what remains is this, region here.
22:01:150Paolo Guiotto: And this is the integration domain D.
22:04:670Paolo Guiotto: Okay?
22:07:370Paolo Guiotto: Okay, so, yes, we have 5 minutes, we can do. Okay, now, to, use the change of coordinates.
22:16:780Paolo Guiotto: What we have to do.
22:18:600Paolo Guiotto: We have to pass the integration in the initial domain on the… where to determine what is the new domain.
22:26:130Paolo Guiotto: in coordinates, if we want UVE, how the function becomes in U coordinates, and what is the determinant of phi minus 1.
22:36:480Paolo Guiotto: Okay, so the first point, let's start to understand what is phi of D. Phi of D is what is the set of UV,
22:48:670Paolo Guiotto: is a set of UV, such that UV is a fee of XY.
22:58:880Paolo Guiotto: Now, UV is… according to the definition, it's X times Y, X, or Y over X.
23:08:580Paolo Guiotto: X or Y or Wild X?
23:17:170Paolo Guiotto: Now, remind that point XY belongs to D,
23:21:640Paolo Guiotto: If and only if the two conditions are X times Y is between 1 and 3,
23:29:50Paolo Guiotto: And, X is less frequent than Y, which is less frequent than 3X.
23:37:140Paolo Guiotto: Now, how… what can be said about UV? As you can see, this is just UV.
23:44:450Paolo Guiotto: So, in terms of… it's… I'm doing the same thing we did with the polar coordinates. We now try to revive these conditions in terms of U and V. So, the first one becomes U between 1 and 3.
23:57:960Paolo Guiotto: What about the second?
24:00:110Paolo Guiotto: Well, for the second, you see that since V is this, I would like to divide by X. Can I do that? Well, you have to be careful, because first of all, X cannot be 0. And second, you have to be careful, because X positive, X negative could change the order. But…
24:18:830Paolo Guiotto: This is useful to have done this figure, because in this figure, we see that when points XY are indeed
24:26:550Paolo Guiotto: Necessarily, the X is positive. There cannot be X negative. So, when X is in V, these two conditions together implies that X must be positive, and therefore, from this one, I get that Y over X must be between 1 and 3.
24:43:770Paolo Guiotto: I'm authorized to divide by X, because for those points, remind that XY is indeed… is not everywhere, it's in D, and indeed, the point must have an X positive, so that condition becomes Y over X between 1 and 3, and this means V between
25:01:640Paolo Guiotto: One… and 3.
25:05:580Paolo Guiotto: So, what does it mean, this?
25:07:450Paolo Guiotto: XY belongs to D, if and only if.
25:11:330Paolo Guiotto: the pair UV, which is phi of XY,
25:17:230Paolo Guiotto: belongs to that set, which is just a product. Interval 1, 3, Cartesian product, to interval 1, 3.
25:28:280Paolo Guiotto: And this means exactly, if and only if UV belongs to this set, this set is the image in the map V of the domain D.
25:39:210Paolo Guiotto: Okay, so let's start composing the integral. So…
25:44:120Paolo Guiotto: Our integral for the function f, the function was X times Y, right?
25:49:550Paolo Guiotto: X times Y on V, becomes the integral on that domain, so U between 1 and 3.
25:59:900Paolo Guiotto: V between 1 and 3.
26:03:470Paolo Guiotto: The function is this one. You don't need to, to invert it for the moment to compute C minus 1, because incidentally, XY is just U. So the function in the new variables will be just U. Then, what I need to compute is the modulus of the determinant of P-1 prime.
26:23:760Paolo Guiotto: UV.
26:26:200Paolo Guiotto: Do you, do you?
26:28:790Paolo Guiotto: So let's see what is this. We need to…
26:32:430Paolo Guiotto: do the inverse operation. So we know that UV
26:39:550Paolo Guiotto: is P of XY means that U is…
26:47:780Paolo Guiotto: X times Y, and V is Y over X.
26:53:00Paolo Guiotto: the inverse mapper, let's say XY,
26:57:630Paolo Guiotto: is a phi minus 1 of UV, if and only if I have something like X equals something, Y equals something. So I have to see what corresponds to XY in terms of UV.
27:11:290Paolo Guiotto: How could I do? Well, you have to extract to solve this system for X and Y. So you can see that, for example, if I multiply these two conditions, I get U times V equal what?
27:24:210Paolo Guiotto: XY times Y over X, the X disappears, so this is Y squared.
27:31:200Paolo Guiotto: Okay, so I get why here, because since these are positive.
27:36:270Paolo Guiotto: U and V are between 1 and 3, etc.
27:39:310Paolo Guiotto: And Y must be positive. At the end, the unit possibility is Y equals the root of U.
27:47:330Paolo Guiotto: About X, well, now, for example, I can use one of these two equations and say that X is equal to
27:54:860Paolo Guiotto: U of Y, but Y is the root of UV, so I have U divided the root of UV,
28:03:930Paolo Guiotto: If you want a better expression, you can put under the unique rotor, becomes root of u divided D.
28:13:930Paolo Guiotto: So, I have that X is equal to phi XY if, and only if, X.
28:25:370Paolo Guiotto: is equal to… root of U over V, and Y is root of UV.
28:34:790Paolo Guiotto: Now, we should compute the Jacobian matrix, but since time is over, you try to finish So, compute.
28:47:780Paolo Guiotto: the determinant of P minus 1 prime.
28:54:680Paolo Guiotto: And… Finish.
28:58:780Paolo Guiotto: Tomorrow, of course, we will see the end of these exercises.
29:07:200Paolo Guiotto: And the due also the remaining of 576.
29:18:270Paolo Guiotto: There is no class tomorrow.