 So, last time we talked about finding the integral over some region of some number of variables right this way, right that way. So, yes and no. So, next week's exam will have an integral, at least one, probably one. So, yes and no. If you understood everything we did to this point completely, 100%, and sometimes you're willing to work hard, you can now go to sleep and wake up in two weeks and it'll be okay. But, this material is sort of kind of on the exam if you want it to be. We have some function picking up, so this is sort of a, so dv, dimensional volume element, dx1 dx2 sum to dxn is some subset of a three-dimensional volume. I have a clue what I'm talking about, right? So, for example, so this takes r2 of this graph, so I want the volume at 2, it's going to give me a chunk, and y goes from 0 to 1, this chunk, this x goes from 1 to 2. Even worse, 1 to 2 goes from 1 to 2. And that gives me a volume and you just do the integral at a lot of levels. Right? Nobody needs me to actually do this calculation. I just integrate first the x's, then I get a function in y, then I integrate that function in there. And we talked at some length, which I guess I'll do some more in a minute, of parameterizing the thing where the region is not a rectangle. Right here I'm integrating x goes from 1 to 2, y goes from 0 to 1 over this rectangle, and things are easy, but if my boundary has some kind of curved edge, then I have to worry about this right. So I talked at some length last time about that. By the way, how many people watch the videos? Anybody? So last time I set the video up, it was all good to go. Somebody asked me a question, I talked to them, and I went and talked to class, I neglected to push on button, so there's no video for last time. I forgot to push on. This time it's on. Let me check. Yes, there's still a red dot, it's still on. So rather than doing more examples of that, I want to move along a little bit. My notes are completely backwards. No, they aren't. So one thing that we do in a single variable, say you have the integral of x times the square root of x plus 1 dx from, I don't know, number one. Yes, then the way we would make the substitution u equals x plus 1, so du is dx, that's a 1. Doesn't look like a 1, but it is. And here, x is u minus 1, and so this integral would become when x is 0, u is 1, when x is 1, u is 2, x is u minus 1, x plus 1 is u, so du is dx. So we can do this integral now because it's easy, because this is the integral from 1 to 2 of u to the 3 half, so 1 half, du is, or integrated, is 2 thirds u minus u to the 5 halfs, 2 fifths, that's a 5. And 3 halves, 2 thirds, 3 halves from 1 to 2 is summing up. 5 halves, 2 thirds, 2 to the 3 halves, 2 fifths, it's summing up. So we can do that, okay, fine. Now if we have more than one variable, maybe things are a little different. Now in fact, before I go to more than one variable, I'm going to talk about, so I want to change coordinates here. If I have a multiple integral, suppose I had it set up this way, I can do the same business. Right here, there's no problem. So here I could still do this part by making the substitution u is x plus 1, blah, blah, blah. I can still do that integral, no problem. Maybe what I want to do instead, so if you recall, so this is the same. Several classes ago, we talked about functions in polar coordinates, cylindrical coordinates, spherical coordinates, and maybe our region is more naturally described in one of these other coordinate systems. So for example, suppose that I want to integrate over a circle. So suppose I just want to integrate over a circle like this. So I'm going to describe, this is the same circle, and suppose that I have some function. So suppose I want to find plus y squared over, so with, and again I'm doing an easy one, but let's say it's this one, x and y inside the circle. Set this up, so one way I could do this is I could set it up. Some integral, some integral, x squared plus y squared dx dy. And now finding these boundaries, let me actually set it up this way. It's a little bit tricky, it's not hard. It's a little bit tricky, so we're thinking about this circle, radius 2, which is y. So the circle is x squared plus y squared is 4. And so y will be 4 minus x squared on the top part. y will be minus 4, minus square root of 4 minus x squared. And I guess I really am setting this up dy dx. So if I take, so my x values are going to range from minus 2 to 2, and as x ranges from minus 2 to 2, a particular, here's a particular x, the y values will go from this curve to that curve from minus square root of 4 minus x squared to plus square root of 4 minus x squared. Right, this is what we did last time. And so now this is just an integral that you just did. So we integrate the y's and we get, let me do the first step. I'm not going to keep going, minus 2 to 2. When I integrate the y's, I get y x squared plus y cubed over 3, evaluated from equals minus 4 minus x squared, square root of 4 minus x squared. I plug in, I get an integral in x, there's a dx somewhere. I plug in, I get an integral in x, blah blah blah. Anybody confused about how to do this? So I'm assuming everybody can do this. I'm not going to. But instead of that, I don't want to set it up this way. This function is particularly nice in polar coordinates. This region is particularly nice in polar coordinates. So if I want to think in polar coordinates, I want to set everything up and just do everything natively in polar coordinates. So in polar coordinates, my function plus y squared is just r squared. And so my circle of x, y is r squared. So this is some other function. My region, my domain, is just, goes from 0 to 2. Also much simpler. Yeah? What is the function of r and theta? OK. So let me say this a little more. Then what is the third angle? No, this is plane. I'll do that. OK. Right? This is polar coordinates. So I'm putting in r and theta and out comes something. Out comes a number. Right? So my graph sits over this circle. So we describe my input variables in r and theta rather than in x and y. And my output z is x squared plus y squared. So if I put in x and y, out comes x squared plus y squared. But that's the same as if I tell you r and theta, out comes r squared. And it doesn't depend on theta because this thing is circularly, it's got a circular symmetry. It's a surface of revolution. So I mean you could do this problem already because you could do it using cylindrical shelves or whatever stuff you know how to do. But I'm using this as an example to try and illustrate a more general fact. So you could do this, this directly, not a problem. But I'm thinking I want to make the substitution that x, y goes to r cosine theta, r sine theta. And how do I have to adjust the integral to make that? That's really the question that I'm trying to make. I'm making a substitution just not x and not just substituting x and substituting y separately. I'm substituting variables that depend on both of them. And we need to account for that. Another way to say this is I'm trying to just set everything up in polar coordinates from the start because everything is much simpler if I just think of it in polar coordinates. It is not. So this is not the answer. The integral as r goes from 0 to 2 and theta goes from 0 to 2 pi of r squared dr d theta. This is wrong. You might think it is, but it's wrong because I didn't account for the dx dy changing to dr d theta. I didn't account for the factor. This is a bad example because du is dx. But if du is not dx, I have to account for du when I change variables. Similarly, this is wrong because it doesn't represent the same thing as this integral. So I want to explain what it should be. Is this clear to people why this is wrong? So let's go back to the beginning. Well, sort of the beginning. And think about setting up this integral to represent this volume natively in terms of r and theta. So what am I doing? I'm constructing little pillars that reach up to the graph but my pillars, the cross section of the pillars is not a rectangle. In this case, my little area form is a rectangle and the area of this is dx dy. But in the polar case, my little area thing, the cross section of my slice is a sector of angle d theta and with dr. And its area is not dr d theta because if it's du close here, these guys have the same... Well, I guess these guys, the black guy here and this guy here have the same dr d theta but the area is quite different. So we have to account for the fact that the area changes with r. And so the area of this thing, well, dr is still about the same but this length is not d theta but r d theta. This length here, however much theta increased, times how far away you are. And so this area is just about r d theta. This is the length this direction times dr. This is my area. Anybody confused by this and would like to admit it? I'm seeing some faces that are like this. If I take a little sector like this, a little polar rectangle, then its area is about... It's width is more formally. I can chop it up, blah, blah, blah. But I think is this clear enough for people? So that means that the correct thing in this case, let's write it a little more formally, we're going to integrate over the circle of my function dx dy, maybe it shouldn't be an algebra because I don't just say f. Or f of x, y, I don't care. f of my variable is dA. If I'm doing this in polar, then my changing area will be integral over the circle that's actually the disc. It's the inside of the circle. My function, and then dA is r, I should write r d theta dr, but it sort of rolls off the tongue easier to say r dr d theta. Because it's like r dr dr. So if you go lying in a r dr, people don't know that you're lying, right? That's an old cartoon, the 60s. But anyway, I'm just going to watch it a lot. Anyway, so r dr d theta is our area form. How the area, a little slice of r's and theta's gives us our area. So that means that this integral represented as a double integral in polar would be r cubed dr d theta. And r goes from 0 to 2. And theta goes all the way around, so that will be 0 to 2. And now this is an easy integral. And since it's easy, I guess I can even do it. So let me do it here. So this guy is r to the fourth over 4, evaluated from 0 to 2, integral to the other two. This is just a pi. If you do this ugly integral here, which will come up with nasty square roots and all sorts of ugly stuff, you should also get a pi. Sort of surprising that there's some pi sitting in here, but we're probably going to have to do a trig substitution and all sorts of icky stuff. This integral is not fun. This looks easy. I want to also do coordinate systems that we discussed. So now let me come back to Tommy's question. You might encounter an integral like this on the exam, which you can do in principle in x and y. But it's a heck of a lot easier if you do it in polar. You don't have to know about polar to be able to do this, but it will make your life easier. So that's really the answer to your question. Yes, it is. No, it isn't. So I will not ask a question like, do this integral in cylindrical coordinates. But I may ask an integral that's easier to do in cylindrical coordinates. We can do that same one. So now let's let me do cylindrical coordinates first. So now if we think in terms of cylindrical coordinates, so now I have some function of three variables, density over a spatial region or average temperature or something that depends on three variables. And my region that I'm integrating over could be just a q. This is my area form. If I could integrate this thing cylindrical like this in spherical coordinates, so let me remind you, which I shouldn't have to remind you too much because it is on the floor if it's doing this time, that we can describe something in spherical coordinates. No, I'm doing cylindrical, I'm sorry. Where we're doing polar coordinates in x, y plane and then we go up to some height. So I describe my points by polar coordinates instead of x and y and z in the height. And it's not very surprising, so if I'm imagining that I'm trying to describe some three-dimensional region in cylindrical coordinates, then a little instead of a box like this being my natural, tiny little infinitesimal element of volume, I'm going to think of, well, I move my theta's a little bit. That'll give me some sector. I'm trying to draw it in the base. Some sector like this. So I should have dotted it. The perspective's a bit off. My natural block of volume, I'm going to let, and I guess this curves. I'm going to let theta vary. So this width is d theta and this distance is dr. Yes, it'll be r to theta. And this is dz. But of course this length depends on how far out I am. So in fact, it's r. So if I let theta, like before, r dr d theta, in some sense I don't need to even talk about cylindrical coordinates in this case, because I can think of my cylindrical object. So if I wanted to, you know, the change of variables for cylindrical is x, y, z, r cosine theta, r sine theta, and z is just left alone. I can think of if I have some integral expressed in cylindrical coordinates, I can think of that as just x and y are expressed in polar and z is just z. So there's nothing sort of new there. Okay? On the other hand, an example. Yes? How are we going to be that the unit of rows will give us the same volume as this one? I'm sorry? How do we know that the unit of rows will give us the same volume? So one has to actually prove that this change of variables works this way, which I'll do a little more generally. I won't do it prove actually. I'll just state the theorem. Okay. So think about it. We can think of it in two ways. In one way, I can redo the whole theory. So if you remember, let me draw the picture in three dimensions because it's a little hard to draw in two dimensions. I can draw it in three dimensions. So I'm putting in x, y, z. So I'm integrating over some three-dimensional, this is some three-dimensional blob. And on this three-dimensional blob, if I'm integrating x, y, z, then I calculate, I take a little tiny cube and the function is constant. As the cube goes to zero, so it will be constant. Right? So that is my integral over my blob. I thought you meant over all the cubes. No, no, no. Over one cube is constant. So my integral over the cube of the blob of f of x, y, z, d volume will be the sum over all the little cubes. I'll sort of inter-value the little blobs as long as the cubes, as long as the cube will eventually go to zero. It still works. Yeah. Okay. So I do it over my cubes and I add up the volume of the cube times f of x star, y star, z star, where x, y, z are in the cube. So here with volume of the cube, let's call it sub i. So xi, yj, zk, so ijk is in the cube. And then I take it as the size of the cubes. That's how I define when we're integrating x, y, z. And so if this exists, then this converges to whatever the range is of x, y, z, dx, dy, dz. But I could have, instead of using little cubes, I could have used some other shape. And the other shape that I could use, so for example, if I want to describe my little cubes, not as cubes, but as little blocks of cheese. So any shape that can just fill up an object will work. Provided that I can continuously fill up the object. So you can have a more general definition of an integral. Absolutely. I did. Yes. So it doesn't matter. I mean, these are common coordinate systems. So if I want some other coordinate system and I can describe them in terms of, so you know, imagine some MC Escher kind of situation where you've titled the plane by the dragon. You just have to somehow describe the area of the little dragon depending on where it is. Fine. Get the same thing. Okay, not the same formula, but the same answer. So if I want to integrate over in dragon coordinates, that's fine. Yeah. I'm just going to show you where you're getting this. Do you need a pie slice or whatever? I understand the r kind of thing is r. But how does that tell you the area of the plane? Okay. So this is not exactly the thing. Yeah. It's just, it's this. So this is actually kind of a lie. If the thing, this is here is in. So the r here and r here are very close to the same. So this width dr is tiny. So this width, this r value and this r value are pretty much the same. So this length is rd theta for the outer. And this length is rd theta for the inner r. But if r, if these two r's, so this is r in and r out, and if r in and r out are very close together, then r in d theta is very close to r out d theta. And since I'm taking the limit as they come to one, it's okay to pretend that they are exactly the same. And similarly, but here, the difference between r in and r out doesn't depend on theta at all. So this distance here is just dr. And the distance between this z and this z doesn't depend on theta or r at all. So it's just a z. And so this, the volume of this little guy has to be corrected by a factor of r. If I'm taking a little guy who is far out, he will be bigger than a little guy who is closer. Now, yeah. So I am saying this rather informally. The book is a little more formal. I'm being a little more informal to try and give you an idea of what's going on. I'm being sort of mathematically sloppy. But I'm being sloppy on purpose because I think it's easier to understand the slide. It doesn't mean I don't know how to do it in a formal way. Oh, yeah, they have to tie it. So you have to have one of those arbitrary things. The dragons have to fit together. That's why I said it's an M.C.S. your picture. Right? You know, you have these M.C.S. your pictures with little fish and all the fish fit together. You can do lots of them. It doesn't, I mean, you know, anything, I mean, we can integrate in terms of anything that we can tile or play with that we can describe that thing. But, of course, usually you're not going to use those dragons. I'll say how to do that a little more generally. So cylindrical coordinates, polar coordinates, they're really the same thing. I've just got some extra dimensions thrown in there. We do get something different if we think in terms of spherical coordinates, or we ask spherical coordinates. Those would be spherical coordinates, I guess. So in spherical coordinates, we measure, we take with the x-axis and call that theta. And then that gives us a plane here. And then we take within that plane some angle from the vertical and then along that line we go some distance and the book calls it r, but I'm going to call it rho. And so we can describe a point in space here by theta, which is the angle to the x-z plane, phi, which is the angle. We talked about this a couple of classes ago and this is x, and I'll wait a minute. This is x, and this could be x. This is x, that's y, and z. And I want to point at that projection. So in terms of theta, I'm going to turn 45 degrees and now I'm at that projector. Of course we'll do pi over 4, but my theta is pi over 4. My phi is I put my arms straight up and I go down until I'm pointing at the projector. My angle is about, I don't know, it looks like about 60 degrees. So this is pi over 3, thank you. So this is about pi over 3. And then rho looks like about 12 feet, 18 feet. You want any meters? Okay, it's about 5 meters. Maybe it's not that long. Okay, I'm not very good at judging this. It's about 3.6259 meters. Anyway, it's that far. So phi is this angle. And if I turn, I get all of the things that I see in this cone. So this is the angle from the vertical. This is the angle that just fit around and rho is how far out we go. I mean, and this is a good coordinate system for things which have some kind of spherical symmetry. My volume element in this, probably not going to be able to draw it, but it's a, so imagine a big sphere and I'm going to take a little chunk of it, a little chunk of a sphere in all the dimensions. This is supposed to be this angle here. Let's go back to the origin. And this distance here is d rho. My phi looks like a zero. So maybe I should draw it again. So I have that little thing. And now it's a little more complicated to describe what this is. So let's start with the phi. So if I take a little piece here where phi varies, then this height, well, this one's easy. This is d rho. The thickness of this little piece of the sphere, think of it as a thick sphere, and I cut out a square piece, right? I took a sphere and I took a little square on it, and then I carved it out. My big sphere had some thickness. So that's what I got. Something like that. That's where I was trying to draw there. Okay, so the thickness is easy. It's d rho. That's not going to change. So I want the volume of this thing. The change in phi, when we have the same problem, same issue, we need to know how far out we are. So this is going to be rho d phi. That is the link here is going to be the same as the link here on the same sphere. So this height is going to be rho d phi. You might think initially that this width is going to be rho d theta, but it won't be. Well, I will certainly have a rho d theta, but it's not really rho, because I have to project it down to the plane. When I project down to the plane to know how theta changes, I'm going to get a factor of sine phi. So it's really rho sine phi. This is the projection down to the plane where theta is polar, sine phi, but this link is rho sine phi. So that means that my volume, so the answer should be four-thirds high. Maybe how about radius r? So the answer, we already know, this is something that you should know in sixth grade, something like that. Ninth grade, somewhere in there where this comes out. I don't know, somewhere in there. You get it in first grade and then again in tenth. But now we could, of course, so again, I'm doing a stupid problem so that we can see that it works rather than doing a hard problem and saying, yeah, sure, we got so it must be true. I could, you know, crank this up and, you know, let the density vary and load. So that means I'm going to integrate. So, of course, I could set this up in x, y, z coordinates. That means that x squared plus y squared plus z squared will be less than r squared. Blah, blah, blah. I could also set it up in cylindrical coordinates. I could set it up in lots of coordinates, but it's sort of trivial in spherical coordinates. So in spherical coordinates, well, theta goes all the way around. So I'm going to do d theta last. And then phi goes from the top to the bottom. Zero to the top here. Oh, yeah, it's from the top to the bottom. Right, zero to the top. So phi goes from the top to the bottom and r goes from the origin out to the edge of this field. But I, so, one, that's my function, and I didn't leave enough room, so I'll write it again. So d phi is last, d theta is last, d phi is next, and then my radius rho goes from zero to r, but then I need the volume factor, which is this rho squared sine phi. What? And I need this rho squared sine phi, d stuff in whatever order I want. Since I wrote them in this order, I guess I write them in this order. It's okay? Yeah. I'm just curious. If you want to write stuff to a rectangle, would you be divided by d? Jacobian? Yeah, d rho, sine theta. Now we have to switch variables. Yeah. So if you wanted this in rectangular, then this is going to be a triple integral of messy stuff. So let me just write mess here. Right? It's going to be a triple integral of a mess. X goes from, only my triple integral of a mess of 1 dx dy dz. But these bounds are going to be they're going to involve square roots and stuff. Ultimately this is like, is the volume factor the Jacobian, or is it the regular rho? Talk about Jacobian again. This is the Jacobian. You know the word Jacobian, that's the Jacobian of the Jacobian. Yeah. Okay, so I'm imagining that I have a function that I'm going to add up to a little sphere, a spherey cube. I'm going to chop up my sphere into little cubes that aren't really cubical. They're, I don't know, they're sectors, cross sectors. The one is the value of the function. It doesn't matter. So you can think of this as another way to say that, is that the volume of the sphere is the same as the mass of the sphere so you can imagine that I'm calculating the mass of the sphere but the sphere is uniform that is made of something with a uniform density and the density is one. So then the volume is the mass or the same. Is the thing that I'm integrating over the volume. So if I were to draw a picture of this I would need to draw it in four dimensions where I have three dimensions. It's a sphere and my fourth dimension is one. So just like if I want to calculate so I'll put this on hold for a second and I want to calculate of a cylinder if I take a cylinder of height one then the area of the circle and the volume of the cylinder are the same. I'll leave this area of light instead adding a dimension and integrating over the circle V is my volume filled and P is some point inside this sphere. So I'm just saying think of the sphere consider all the points in the sphere and at every point I assign the value one to my function. So then when I add them all up I should get the volume of the sphere. So I don't want to do this messy rectangular one this one's easier. So if I integrate this guy D row cubed over three evaluated from zero to r which gives me r cubed over three that's the thirds and the fourth thirds sine phi d phi d theta I'm going to actually integrate away the d theta first so I don't have to keep writing it so actually let's do it this way since theta and phi don't depend I can switch the order without a change and so now if I integrate d theta I get the integral from pi of two pi r cubed over three sine phi over three as I go from zero to pi um now I need a place to write more just right here so that you write the sine I get the cosine if the sine of phi is going to become cosine if phi evaluated from zero to pi oh from zero to pi which is two I would say zero to pi is four okay so then I get two pi r cubed cosine phi evaluated from zero to pi which is two pi r cubed over three minus a minus two pi r cubed which is four thirds not forty three four thirds so so uh yeah probably oh yeah but I evaluate you guys it is minus cosine c cosine of pi is minus one minus minus one is a plus one minus a plus one still a plus one so it is a cosine but it didn't matter because I did the order backwards too so two wrongs didn't make it right and we can do that okay so magically we know how to compute the volume of the sphere yay we can pass a trade um so but you know we can see right here before third pi r cubed but this is maybe so this integral is pretty easy in spherical coordinates now if we wanted to change the problem to imagine that you know you're going to make your sphere out of something that gets later as you move out then instead of integrating one we would integrate the density you know density function yeah so I'm still don't want to get we do the it's not going to go too high right you lost it here yeah from here how we describe a sphere a spherical coordinates my mouth isn't saying what my brain is thinking so in spherical coordinates I want to describe the sphere of radius two goes out to the edge that phi vary phi can go from zero down to so that'll give me and now I'm going to let the semicircle revolve the whole spin this semicircle so that'll be two pi so theta here will change by two pi so phi changes from zero to pi and theta changes from zero to two pi all the way around notice we didn't let phi go from zero to two pi because then I'd get the circle twice so that's why that's why my bounds are theta goes from zero to two pi phi goes from zero to pi I started the other way around but I wanted to get rid of the easy stuff first so I changed the order going from here to here because it made my life easier okay so so fun so now let's look at this in a slightly more general setting what are we really doing here I mean of course we could just memorize and now you can do integrals in polar, cylindrical, and spherical but suppose you have some weird coordinate system you know, diagonal coordinate system so what are we really doing we're describing a change of variables so just like when you make the substitution u x in terms of u when you make a u substitution you have to describe how u changes when I change x a little bit so in one variable let's again think back to one variable in some sense there's nothing new in this glass for quite a while so when I make the substitution when I transform f of x to some g of u how a little change you can do the same sort of thing here where I'm thinking of finding three dimensions but it doesn't really matter how many dimensions I have this is my one volume object and then I have some transformation which is my change of coordinates that gives me my new shape here they start with my x, y, z and then my change of variables gives me u, v, v really small scale assuming that my change of coordinates is nice function then this is going to transform by the derivative Pt so if I write my change of variables and I calculate the derivative well it's not really the derivative it's the determinant of the derivative if I calculate the determinant of the derivative then that will tell me how a little cube transforms to some other thing the thing is called the Jacobian the determinant actually gives it a formula so what I want to do now is check for example if I compute the determinant in spherical coordinates let's start with polar I get rho squared sine theta as the determinant of that change of variables let's do either polar or cylindrical it's right there on the board so this is my change of variables for transformation I want to calculate dT which will be a 3 by 3 matrix but a fairly easy one so dT will be my matrix of partials so I take r cosine theta I have 3 variables and I take the derivative with respect to r, theta, and z so the derivative of r cosine theta with respect to cosine theta and now I take the derivative with respect to theta and I get minus r sine theta and now with respect to z I get 0 do r sine theta I get sine theta derivative of sine theta is r cosine theta derivative with respect to z is 0 derivative of z with respect to r is 0 derivative of z with respect to z zero through z with respect to theta, zero through z with respect to z1. So this is my derivative matrix of this function t which takes x, y, z into our cosine theta. Okay. I want to compute this determinant, so the Jacobian is the determinant, and rather than that, let me expand it this way instead. So I'm going to expand again about maybe this row. Alright, so it will be plus minus a plus here. So I get zero from expanding from taking the determinant of this part, zero times this, plus zero times this, plus one times the determinant of this matrix. Let me write it out. It is r cosine squared theta plus r back to the r out, r dr d theta. My Jacobian, so the thing that I integrate by, my volume form is r dr d theta. Correct. So if I have, again if I'm integrating by little dragons that tile space, if I have something that tells me how to transform a cube into a fire, if I have a transformation from a cube into a dragon, then the derivative of that transformation, take the determinant and give me the way I need to change my variables to describe something in dragon form. Okay. So what does the determinant, what does this mean? What does this dt mean? This means at a very small scale, if I move theta a little bit and I move r some other little bit and I move z, well some little bit doesn't matter how I move z, then how much does the function change? The change of variables function change. So if I describe a little change over a cube where I move r, the volume of my cylindrical cube change, it changes by this. This is a linear transformation. If I'm just doing, multiplying by 2 in every direction, then my determinant matrix is 2, 2, 2. So if I'm doing x, y, z goes to 2x, 2y, 2z, then this is my derivative matrix. And if I increase x a little bit, z a little bit, y a little bit, everything changes by a factor of 8. My volumes will blow up by a factor of 8. Here it's more complicated, but the determinant measures how much my volumes change as I move around in infinitesimal. Okay. Okay. You can do it the other direction, right? You can get 1 over square root of x. Yeah, if you want. It's more complicated, but yes. You can also just calculate this. If we know how xy is, suppose I want to change now from polar to rectangle. I can write down the transformation. So I was going to do the other one. No, no, no. Let me continue in this vein because I think it may be reasonable. Suppose I want to go the other way. r, theta, z goes to, so what is this? Well, if I know z, it goes to z. That's easy. If I know theta, it's the arctangent. And r is the square root, so that's my change of variables going the other way. Okay. It's not fun. I mean, it's doable, sure. But if there's square roots, there's arctans. Ew. But if you do it, you'll get a matrix. You take its determinant. Okie dokie. We're good. But dt is a linear map. Determinant is not zero. That is, we're away from the origin. I can take it in. Let's call this s and this t because we already have t on board. So s is the transformation from cylindrical to rectangle. So the determinant of ds, well, s is the inverse of t as long as the function is invertible. So that means that this is 1 over the determinant of dt, right? Because this is a matrix. Its inverse is the inverse matrix. So let's write it in one more step. This is the determinant of ds inverse dt because s is t inverse. But the derivative of the inverse matrix is the inverse of the derivative matrix. So this is just going to be 1 over the determinant of dt, which is 1 over r. But I want it in x, y, and c. So that's 1 over square root x squared equals y squared. So this is an easier calculation. If you want to do this long calculation, you should get this answer. It might be easier to look at the inverse transformation and figure out how it relates to the forward transformation. So the Jacobian of this transformation from cylindrical to rectangular is going to be this. That's not an 11 integral. Okay? I have two minutes. Yeah. So if you do the calculation for the spherical one, let me not do the calculation for the spherical one. You will get a big matrix with lots of cosines and sines. When you start calculating everything out, use the Pythagorean theorem a couple of times and blah, blah, blah, blah, blah, blah, blah, blah, and then outfalls square root x squared. The calculation looks like this. So let's read it. There you go. In one minute, the general theorem, taking some domain, which is in Rn, some change of variables here, continuously differentiable, be smooth pieces. So the edge of u, the boundary of u needs to be smooth bits. And it needs to land in, so I want my area, the boundary is inside the domain, the interior of the domain. In other words, everything's nice in the region where I'm looking. Then t is 1 to 1, not 1 to e, 1 to 1 on the region that I'm integrating. That's one thing I need. And another thing I need is that the determinant, the dt, is not 0 on the region of interest, or I guess t of r, actually. Provided f is all nice, then if I integrate f of, let's call it x vector over the region r, let's call it dv. And this is going to be, I just need to adjust by the region r in the new coordinates. And I'll do f in the new coordinates, adjust by the Jacobian, the determinant of t part, the du that you need in one variable. It's just the determinant of the derivative of the substitution.