Learn About Triple Integrals - Calculus 3 Video

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Learn About Triple Integrals - Calculus 3

Hi and welcome to this section of the Calculus III Tutor volume number two, and in this section of the class, we are going to continue our discussion of Calculus III marching our way on through to the end of what most of you guys will cover in your Calculus class. We are going to talk about the topic of triple integrals. Let me just pause right here and say that a lot of what I am covering here and through the remainder of the class is going to assume that you are already have been exposed to the material covered in the Calculus III Tutor volume one, so if you do not already have that, you probably should go ahead and get it especially with the triple integrals we are doing now, because although I am going to start from the beginning and walk you through it, I am going to sort of assume that you have been exposed to me introducing the double integrals to you which is covered in volume one that is going to be directly applicable to what we are doing here. So what we are going to do in this class is we are going to continue on with the triple integrals in this section and we are going to march through the different types of triple integration and on into the vector fields and stuff that we will do in here in a little bit. So what is a triple integral? It is one of those things that you will look at the first time, kind of like a double integral and you are like my goodness, we did double integrals, that was different and now we are doing three integrals, why would you ever do that? We are going take it one step at a time and tell you why we were doing that. First, let us go ahead and recall the double integral that we did just briefly which at the tail end of the volume one of the Calculus III, okay. The double integral recall is just if you can visualize it, you have your X and your Y and your Z, right and so that is your coordinate system and you have a function that was sort of “hovers”. It is a function X and Y, function of two variables and I use my hand here because I am sort of indicating that it is a function that is sort of this mountain or valley or something that is kind of hovering your XY plans, it is kind of hovering there. There is always a region R that we were talking about when we are doing our double integrals, because the function that I am talking about is a function of X and Y. It is a function of two variables. So when we try to integrate those functions, we have to integrate it over two directions, X and Y because that is what the functions define the terms of X and Y. So you define this region R in the plane. It could be any shape, it could be an ellipse, it could be anything. The way we set the integrals up we could use any arbitrary shape in the XY plane to integrate this function over. And what we were really doing in that section is we were adding up the contribution of the function as it relates to the region R of the XY plane. Basically, we were integrating that the “height” so to speak of that function over the XY plane, over any arbitrary shape that we wanted to choose in there. If I could summarize it in one sentence okay; the double integral was integrating a function of two variables, X and Y and we were integrating it in two directions, X and Y because we were integrating a function of two variables over two directions. It makes sense, right? Two variables, two directions, okay. And when we did it in polar coordinates, it was really no different instead of a function of X and Y, it was a function of R and theta which is again two variables. So when we did our integration over those polar functions of R and theta, we did not integrate over X and Y because that is Cartesian. We integrated over dr and dtheta, so it looks different but it is the same thing. You are integrating over two variables because you have a function of two variables, so to sort of summarize that just really briefly because it is going to be directly applicable to what we are doing here, just recall for me that the double integral looks something like this and we sort of, in the beginning anyway before we actually solved any problems, we said we were integrating over a region R which is that region in the XY plane down here that I am actually integrating this function over. We are integrating the function of two variables, X and Y, so that is why we have X and Y there and the differentials that we were integrating over are dx and dy. Okay so this was sort of are general form, you have a function of two variables over X and Y, we integrate them over those two directions basically and the dx and dy and the limits of integration that we had define the region R here in the XY plane that we integrated this function over; that is what we were doing. In integration, do not forget it is just a shorthand way of adding up an infinite number of things so you are adding up the contribution of its function over that region X and Y. Now when we actually boiled it down to where the rubber meets the road, the way we actually broke this down, is we again we had the two integral signs, we had the function of X and Y, right? And we had the dy and dx, let us say, because we said before and we are going to change this, just to be consistent dy and dx. You can flip the order of integration. We talked about that but just to give you sort of an example, what we said was we were going to integrate, let me go and change colors here just to make it clear. Since we are integrating over Y at first, Y was some function of G1 of X up to some function of G2 of X and we said we are going to integrate from X=A to X=B. So real briefly, I mean not too really belabor it too much but if this is X and Y, so we are looking down on the XY plane and our function here that we are integrating is kind of hovering over the XY plane because it is a function of X and Y then what we are really saying here is; is since we were integrating along Y first, we are integrating along this direction first, we had some function here and we called it G1, and we integrated up to some function here G2, integrating from G1 to G2 of X, these are just functions of X. And then me integrated from A over to B, and the combination of all these stuff from G1 to G2 which are functions and from A to B along X, define this region R that is what we are integrating over, so by choosing G1 and G2, these functions in our limits of integration, we can sort of define any arbitrary shape in the XY plane that we care about and we can integrate this function over this region R. That is what we were doing. So we were going to make a direct extension to what we were doing when we were talking about the triple integrals, that is why I am giving you a little review here. So to recap, for double integrals, they are integrating a function of two variables, X and Y and because it is a function of two variables, we are integrating it over X and Y in the XY plane. We are integrating over two directions basically. Now for the triple integral, what we are going to do is we were going to integrate a function of three variables, X, Y, and Z, and because there is three variables involved, we have to integrate in three directions and so it is going to be integrating over dx, dy, and dz, and instead of integrating over a surface area that this function is hovering over, we are actually going to integrate this function over a volume region dv. So instead of seeing integrals with da at the end, which is kind of what we had here, I guess I erased it but this could be, this da could be dy and dx, you can sort of replace it with da. A lot of times, you will see it in your textbook, an area or unit, okay. So it would integrating over some area in the XY plane, what we will be doing is we will be integrating this function of three variables over a volume. So in order to get the dimension of volume, we have to add dx, dy, and dz. So burn that in your head, that is really important and we were going to work some problems here. You are going to see they are not going to be very different from double integrals, I mean really, it is just tedious. It is the same thing. We get into the next section where we will be doing the triple integrals and the cylindrical coordinates and then even beyond that, the triple integrals and the spherical coordinates, they sound so complicated and so hard but just please erase your anxiety over it because what you are going to find is it is exactly the same thing. It is just using a different coordinate system to define the volume that you are integrating over. So what we are going to do now is draw a picture because I like to draw pictures and I hope that it makes it clear. So what we are going to do is we are going to draw a little picture of what this triple integral thing is and then we are going to start working some problems and you are going to find out that they are really not a big deal. So, let us go ahead and start by drawing the thing that we always draw which is your coordinate axis and I would encourage you guys to do this yourself for every problem if you can, I mean you do not have to but get used to drawing this because it really does help visualize things. So here is your X, Y, and Z, it is a three dimensional system. Now what we are going to do is I am going to draw and define for you the following. I am going to draw several things up here and I am going to label a few things and you are probably going to look and say what is he talking about? Let me just sort of get it up here and define it for you. What I am going to do is I am going to define some arbitrary volume which is going to be a volume V or a volume D actually that we are going to talk about here. So let me go ahead and define, this is sort of like a cylindrical thing that is why I have the straight sides here but the tops and the bottom can be sort of any arbitrary shape. You see that is why I have got lumps here; lumps here on the top and lumps here on the bottom. This can be very smooth, it could be hills and valleys on the top or whatever. It is some arbitrary thing on the top. It is some arbitrary thing on the bottom. So what we are doing here is we are defining what the top here connecting down to the bottom which are arbitrary kind of shapes, this is a volume D. Now why do they call it volume D? Why do they not call it volume V? Most of the books call it volume D because if you start talking about V here then when you get into actual integration where you are talking about dv, it can get a little bit confusing with V and dv. So the volume is D, it is just a letter, do not let it worry you, okay. So here is the volume D, and this volume D if you were to shine a flashlight on top and project its shadow so to speak down here into the XY plain, it is going to cast a shadow, right. So this shadow actually, I am going to draw in a different color. I am going to draw in this color. I am going to do my best to kind of show you that what we have here is maybe not the best representation but what I am trying to show you here is that this shadow cast down here, and this shadow is the very familiar region R that we have been talking about from the other sections. So this is the region R; when you look at these books, the region R is always talking about the XY plane. What something looks like in the XY plane, the volume D is actually the entire volume that you are encompassing here. Now what we are going to end up doing is I am going to have a function here; a function of three dimensional space, so before we were talking about functions of two variables, you had your XY plane and every point in the XY plane, you plug in X and Y and you get a third value, you can call it Z, right. And so, then you can plot that and it will sort of hover over the XY plane. The function of three variables, you really cannot draw it on a board because when you think about what you are doing in this room for instance, at every point X, Y, and Z, at every single point my finger is at, is a value, okay. And that value could be let us say I am measuring the temperature, I am measuring the temperature here, and here, and here, and here, and here, and here, so at every point I have a different value. Now that value is a function of X, Y, and Z, right. Because everywhere I move my finger, I have a different temperature so I could construct a function of X, Y and Z that could predict the temperature at different points in space; that would be a function of three variables. Now how do you draw that? How do you actually draw that on a board? I mean you really cannot because you would have to have basically at every point X, Y, and Z, you have a different value of temperature, and that is really hard to draw as a surface, so you really cannot draw it too well but you can sort of visualize what is happening. At every single point inside of this volume element that we have sort of created here, at every point there is a different value and if it helps you to think about it, think about this value as a function that is like a temperature or something, it could be the pressure at every point in space. It could be the magnetic fields strength or something like this; it could be anything that is a function of the space, the three functions of space. So that function that we are talking about here on the board is going to be a function of X, Y, and Z. That function kind of permeates space at every point of X, Y, and Z, whatever my function is defined to be, I calculate a value at that point and what I am going to do is I am going to integrate this function over this volume. So this function is everywhere, it is all the dimensions going into the board coming out of the board, left and right, it is different values that function calculates a different value at every single point in space but I do not care about every point in space, I care about adding up the contribution of the function only in this volume element that I have defined here. That is what integration is. And in real life, you might be integrating, later on we will do a lot more thus you might be integrating the electric fields strength over a volume or you could be integrating at any number of things over a volume, that would actually mean something. Like you might be integrating the electric field and it might relate to the amount of charge you have inside of that volume or any number of physical applications but you are integrating an actual three dimensional over three dimensions. So in order to pull this off, we have to add a couple of more things to our drawing. This upper boundary and this lower boundary are arbitrary surfaces so to speak because this is sort of a surface on the top and a surface on the bottom and they can be arbitrary, that is why I have got these squiggly lines here, okay. So the top one or let us go to the bottom one first. The bottom one, we are going to say, let me go ahead and erase this and put this is going to be volume D, nothing has changed. This bottom surface, we are going to call it Z is equal to some function, I am going to call it function F1 of X and Y, let me finish here and I am going to show you what I am doing here, so do not worry if this does not make. The top surface, so to speak that caps the top end of this can is going to be some other function F2, that is why there is a two there because it is a different a function, it is also a function of X and Y. So what am I saying here? This surface, so to speak on the top is sort of like my hand it is cupping it right. It can have hills and valleys, and divots and depressions, and mountains and everything else here and it is going to be a function of two variables; a function of X and Y. So at any point X and Y, that is how I am defining this guy here because this function of X and Y is going to give me this surface that is going to sit right on the top. I have a different function of X and Y that defines the bottom surface here. Again it is kind of cupping it down like this. At any point XY down here, I am defining the surface here. I will plug in X and Y and I calculate the function, a number and I am going to call it Z and that is going to be the Z value here in this direction that is going to define the actual shape of these boundaries, so here I have everything I need to actually proceed. I have got some function on the bottom of X and Y of two variables because it is a surface down here, I have got a different function on the top that governs what the top looks likes. It is also a different surface and they are just connected by vertical lines here and I form by doing this a volume D. Now if I take a flashlight and shine down like this; the shadow of this whole thing is going to define a region R here. That is just sort of giving a new continuity from the last section before, but what we are going to do is we are going to take a function of three variables, X, Y, and Z which exist everywhere. I can plug a value of X, Y, and Z in and get a number back that is my function and I am going to integrate this function over this volume element D; that is what this entire section is about. So, how do we do it? We are going to work way up to it slowly so we are not going to just, I am not going to throw you into the grease too much but conceptually when you think about it, we need three integral signs. We know that because we are integrating in three directions and we are going to integrate not over a region R like what we were doing in the section on double integrals but we are going to integrate over this volume element D because this volume element D is what we have defined right here. So when we do the integration, however, we set up our limits of integration must correspond to this actual volume element. So in integrating over the volume D, we are integrating a function that we just talked about; X, Y, and Z. And before getting too technical, we are just going to put dv here. So if you remember back to your double integrals from the last DVD that we talked about, it was usually written as double integral over a region R of a function of two variables da, which is this da is an area and it corresponds to the region R. Now da got busted into dy and dx because these two directions give you an area. What you are doing is, you see it is not too complicated, you are just having a function of three variables and instead of an area, you are integrating over a volume and we have an extra integral sign here and so that is just sort of given right like that. So what we are going to do here is continue on with what we know to be true here and I am going to write down a few more things and we are going to arrive at a very useful thing to actually do our integration. Now let us go ahead and write this a little bit further. Let me say that this is then going to be equal to; let me just write it down and show you. Okay, so do not worry about this here yet. I am going to put a big parenthesis here and I am going to put my other integral sign inside and I am going to integrate from F1 of X and Y to F2 of X and Y. I am just integrate my function X, Y, and Z, and then over here, what I am going to have is dz, now I am actually going to change this parenthetical, I am going to change colors, I am going make it hopefully super easy to understand. And then over here, I am going to have da, all right. Now let me explain where I am doing here, I am changing from this form to this form and I am show it slowly unraveling things to you. Eventually we are going to get down to even a more explicit form here. All right, what am I doing here? This is a triple integral over three directions here. The same function now, I mean integrating this function. What I am doing is I am you giving you a little more detail about what this first integral is. What I am going to end up doing, first is integrating along dz. I am going to integrate along dz, now look what happens if I actually do this; if I integrate this function along dz, where Z is the actual variable I am integrating over this thing here, okay. Then what I am going to get in the answer is a function of X and Y, because this is a definite integral, if I integrate this as a function of Z and then I evaluate the limits of integration which are they say here they are functions of X and Y, so whatever I get in the end when I evaluate the limits of integration, this thing returns a function of X and Y. This inner thing here when I integrate this function over Dz and then take and evaluate the limits of integration, when you evaluate the limits of integration, you take this function you plug into whatever you had for Z here after you do the integration and the same thing with the bottom thing, you are not going to have any Zs leftover, you are only going to have Xs and Ys, so once you evaluate this integral, the only thing you are going to have left after you plug in the limits of integration after you have done the integration is a function of X and Y. So this whole thing is a function of X and Y. Now we already learned how to do double integrals over functions of X and Y; da. We did that in the last DVD at the very end and it is called the double integral. You have done that all along. The only thing we are doing here is restarting one of the functions of three variables and we are integrating it along the third variable first, dz and we were going to get our answer and then we were going to take that and we were going to integrate it over the region R, that I conveniently wrote in your drawing down here just for reference. So what we are going to effectively do is we are going take the region, the function that permeates all of space here in three directions. We are going to first integrate it along Z, okay that is what we are doing first, but in order to integrate along Z, we have to integrate from the lower surface which is a function of X and Y to the upper surface which is a function of X and Y. That is why the limit of integration are these two functions, so we were going to integrate in this direction first and then after we get the answer, we are going to integrate whatever is leftover along Y first and then along X and we are going to be integrating it over the region R because that is the projection of the entire volume elements. So all you are really doing, I mean I am trying to explain things in several different ways to hopefully make it easy but really all you are doing is, you are going to integrate over Z first and then Y and then X. Once you do that first integration along Z, you are going to get back the function of X and Y, which you already know how to integrate because you have been doing double integrals over X and Y in the last section on double integrals. So that is all I am trying to tell you here okay. So let me go ahead and expand, now that you know what this is, let me go ahead and write the whole thing in its full glory to show you what you are actually going to be doing. What are you actually going to be doing in your problems so if R and I am going to go ahead and switch colors here; if the region R in the XY plane is vertically simple, then we are going to have a certain form and if it is horizontally simple, we will have another form. What does vertically simple mean? Just to refresh your memory, if this is X and this is Y, vertically simple just means and you can go back and look in the section on double integrals that your region R is defined like this; this is R okay and this is the function G2 and this is the function down here G1, okay and what it means is this is vertically simple because you can take a little pencil and you can pierce the first function and pierce the second function all in one time. Horizontally simple is going to be if your region R is defined by two little functions that kind of go vertically like this and then your regions define like that and so going horizontally, you pierce the function one time each going like this, so basically, the vertically simple regions are what you are going to pretty much run into with most regularity in your classes, I am just sort of giving you the full spectrum in case you do have a problem like that you can solve. So if this looks foreign to you, go back to your double integrals and this is all explained in detailed here but this is a vertically simple region in the XY plane when the two functions that you are bounding by are sorts of one on top of another and they are kind of arranged like this, you integrate along Y first and then along X, that is vertically simple. So, let us go ahead and write down the form of this triple integral if we were going to do this thing and it is a vertically simple region. What we are going to say is the triple integral, over the volume D, okay of the function X, Y, and Z, over the volume dv which is what we are talking about here, it is a volume element you are integrating over, you have a ton of little volume element inside here and you are integrating over every little volume element that you can sort of infantesimally have infinite number of little volume elements inside of this big volume D. We are integrating over all of them. What you are going to have here, this is going to be equal to obviously, it is going to be three integration signs so let me go ahead and put the three integration signs there, and it is going to be function of X, Y, and Z, nothing surprising, you are integrating the function and when there is nothing surprising here, you are integrating along Z first, and then along Y, and then along X. So you are integrating into three directions, so what are the limits of integration going to be? We just said up here that the first integral that we do, the inside integral as along Z, so we are integrating from this surface down here to the surface up here so we are integrating between this two functions here, so I am going to write these limits of integration in a different thing so we are going to integrate along F1 which is a function of two variables up to F2 which is a function of two variables. These are surfaces, the top surface and the bottom surface. Now whatever we get as a result of that we are going to integrate along dx and dy which is down here and the XY plane, we have been doing that all along. So we are going to integrate along Y first, so that means we are going to integrate between the function G1 of X up to G2 of X. And then along X, we are going to integrate from the numbers A to B. So if you sort of travel back to this little scribble drawing that I have here, once you integrate along the surface from F1 up to F2 which are the surfaces here, it is on the top and the bottom of the cylinder, you get that, then to finish integrating, you got to go along Y and then along X, so if you are going to integrate along Y whatever this function is along Y then you integrate from the lower function up to the upper function of X. Lower function of X to the upper function of X and then to finish sweeping the area, you would go from A to B, so this would be A and this would be B, that is for a vertically simple region, so that is a lot of talking but I really want you to absorb this because most of the time you are going to have vertically simple regions, most of the time. So all you are doing, is you are taking this function, you are integrating along Z, then you plug in these limits of integration, you take that result, then you integrate it along Y and you plug in these limits of integration, which is going to be integrating along Y from this function to this function. And then you integrate from A to B like this along X. Now I will just say it real quickly that what you are doing here by having the three limits of integration; dz, dy, and dx, and having all these limits of integration here, what you are doing is, by choosing those cleverly, you are defining uniquely what this volume really looks like, you see. Because you have defined the surface Z from here to here and then you have defined what it looks like in the Y direction by those other two functions there, and then from going from A to B in the X direction, you have done it the other way, so you have X, and Y, and Z, going Z this way, then along Y, and then along X, putting all of these together is what actually specifies this volume that you are integrating over. You need all three directions to do that. So this is what you are going to use all the time, almost all the time, I am going to write down one more thing only because you will see it in your Calculus textbook and I want you to sort of know what it means even if you do not use it very much, so I am going to write that down here. If R is horizontally simple, then the integral will be very, very similar but not quite the same so if it is going to be done like this then it will be triple integrals obviously. You will be integrating this function of X, Y, and Z. You will always integrate along dz first and the only difference is, if you remember back to the double integrals when you had the difference between vertically and horizontally simple, the only real difference is you switch the order here. So here you will be integrating along the X first and along the Y. We flip the order. Now the inner integral is going to be the same exact limit of integration is before. You integrate from that lower surface to the upper surface, so it is going to be F1 of X and Y up to F2 of X and Y. That is what this limit of integration corresponding to Z is then when you integrate along X, this is the only real difference here, and you are going to be integrating from H1 of Y to H2 of Y and then here is going to be integrating from the number C to the number D. Now just refresh your memory a little bit here, just so you are not totally lost with this. If it is a horizontally simple region, all that means is that when you are looking down in the XY plane, instead of having your functions like this one of top of another and then from one number to another number and you have the functions like this, your functions are sort of laying like this so if you put a horizontal line, they only pierce the function one time. So what are you going to end up doing is that you can kind of tilt your head to side and look at it like this, Y is the independent variable here and X is the dependent variable, so this function right here is actually H2 of Y and this is H1 of Y. This function is a function of Y and this is another function of Y. And then if you are going to complete things here, you integrate from some number here and some number here and this number is called C and this number is called D. They are just numbers so instead of having your functions oriented like this, where they are vertically simple, you kind of tilt your head and you are orienting like this so they are horizontally simple. So what it all really amounts to is that if you are doing something where you have horizontally simple region, you integrate along Z first and then you integrate along dx and you have to integrate from this function up to this function which are now functions of Y because you kind of turn your head over to the side looking at it like this. Y is the independent variable here so they are functions of Y integrating along X and then to finish it out, you would integrate along Y from C to D which is what we were doing here from C to D. But I do not want you to lose track of the big, big picture because this is a lot of theory and what you will see it in a minute is when we actually erase all of this and go work some problems, it is not going to be that difficult because you are going to have a function here. All you are going to do is you are going to integrate along Z, plug in your limits of integration, take the result, integrate it along Y, plug the limits of integration, and then you will integrate along X and plug in those limits of integration and you will get the answers. So, there is really not going to be anything more than this going on. It is just that the theory behind it, if you were told to integrate over some specific boundary then you will have to know how to set it up. You have some function here down to the bottom governing this thing. You have a function up here governing what the top looks like and then you would obviously have to choose what boundary you are going to integrate this thing over in the XY plane as well, and then putting all of that together, top, bottom, and then whatever you define in the XY plane, that is going to define what this actual volume D is. And so most of the time, you are going to set it up like this, integrating along Z from that function F1 to F2, then taking the result integrating along Y and then integrating along X and plugging in your limits of integration here, so let us go ahead and work some example problems and I think when we do that, you will be a lot more comfortable with triple integrals. Okay, so our first problem of the day is going to be a triple integral. The function is actually going to be y-x.z. We are going to integrate along dz first and then along dy and then along dx and the limits of integration are from two to four, negative one to one, and zero to three. So look at what we have here. We are integrating a function of three variables; X, and Y and Z, we are integrating along three directions and then these are just simply the limits of integration along Z, the limits of integration along Y, the limits of integration along X, so by doing all of this, you are defining what that volume looks like. Because these limits of integration really are not complicated, you see this looks a whole lot simpler than what we were looking at on the previous screen because these limits of integration are just numbers, this volume region that we were actually integrating over is just a cube. It is literally just going from Z to Z, Y to Y, and X to X. So you are actually just forming a little cube there. If these were more complicated functions up here, then maybe instead of going from Z to Z up here, you would have some weird looking function on the bottom and some weird looking function on the top and you would define a more complicated looking volume, but when you just have numbers in the limits integration, you are really just integrating over simpler shapes and simple little constant values so it is going to form a cube in this case. So mathematically speaking, what do we do? What I do when I write this down on my paper is I literally write a curly brace like this and I work on the inner integral first and I get the answer and then I do the next one and then I do the next one, so that is how we are going to do it here. When you were integrating along Z, from here on out by the way, nothing is different from your double integrals that we have already done in the last volume one so if you have not look at that, please go look at it now, you have already gotten a lot of practice with what we are going to do here, it is just going to have an extra integral sign. So what we are doing is we are integrating along Z so that means Z is what we are integrating over. Anything else X or Y anything else is just a constant. You must get to the point where you can look at these things and say, I am integrating along Z, X and Y are just numbers. They could be negative 10 or negative 54, I treat them exactly as a constant when I am doing the integration. So what this is going to equal, since Y is a constant, the answer to this integral is Y times Z minus X over two Z squared. Now make sure you understand that okay. This is a number. It is like two. So if I am integrating over dz and this is like two over this is just a constant and it comes down and then you have your variable. That is a simple integral from Calculus I, and if X is a constant because it is, we are integrating over Z. This is a constant and it just comes out, one half Z squared. It is just a regular polynomial integral that you have been doing all of your life. You evaluate this from two to four because those are the limits of integration along Z. So you see it is really not that bad, you just have to know what to hold constant, so when you do this, when you are plugging in your limits of integration, you integrated along Z, so the limits of integration must be plugged in only into the Zs because that is what you integrated over. Everything else is just a constant so the way this will work out here, actually let me go and write it a little bit down below here. What you are going to have here when you evaluate this definite integral here is you will have Y times four okay minus X over two times four squared, I just plugged four into the Z squared and four into the Z minus, and I am going to do the same thing with the two. It is going to be Y times two minus X over two and it is going two squared. All I did was take the upper limit, plug into wherever Z is and the lower limit, plug into wherever Z is, and I just simplify and let us see what we get. Okay, so over here, let us not get too crazy. Let us not try to do too many things at once so let us just rewrite this as four Y minus four times four is going to be 16. Sixteen divided by two is going to give you eight so you are going to have eight X here. You are going to have a negative sign here but I would rather just go ahead and do it. Let us go and do it this way; negative two Y, and you have two Y here, negative times negative gives me a positive here and then over here, I am going to have two squared is four, four divided by two is two, so you have two X and it is positive because of this. So collecting all of my terms, what I have here is I am going to have negative six X plus two Y. Now notice that when I integrated this inner function here, I integrated this function and I plugged in the limits of integration into Zs and I got a function of X and Y back. Now remember, I took great care to tell you that that would always happens just to give you a little bit of the theory behind what is going on here. You integrate along Z and you plug into values of Z because that is what you integrated over. You are always going to get a function of X and Y back so when you take this function and you integrate it along dx and dy, this now reduces to a double integral that you have done in the last couple of sections. Integrating a functions of X and Y over X and Y so now you see, you have just done another step here and now it reduces to a problem you have already done before. So you are actually getting some extra help and practice in doing double integrals as well because what we are going to do now, I am going to switch colors as I am going to take this, okay and now I am going to integrate it like this, so what I am going to do is just to make it totally clear to you, I am going to rewrite everything. Zero to three, negative one to one, the function is negative six X plus two Y and I am going rename that dy and Dx. I have just taken the answer that I am integrating along dy and dx and I am going to do exactly the same thing that I did before. I am going to draw a little curly brace and I am going to evaluate that inner integral first. Now here I am integrating along dy so Y is what I am integrating over. Anything else is constant including this X. So to actually do this integration, you are going to have negative six XY because this is a constant so it is a number, it comes down. You have your Y because that is how you integrate simple integrals like that. We have the number and this is just going to be a Y. It is a polynomial integral, plus I am going to have two over two Y squared because this is a two. It is a constant, one half Y squared that is just a regular old Plain Jane integral from Calculus I, evaluate this from negative one to one. So what I am going to actually have in the end here, plugging in here, I am going to actually plug in the top limit of the integration into the Ys and the bottom number into the Ys because I have integrated along Y so when I evaluate the limits, I need to plug them into only the Ys because that is what they integrated over, so it will have is negative six X times Y which is one, plus two over two, I am just going to say is a one, I am not going to worry about it, plugging in here is going to be one squared, I am going to write it down just for completeness. So I do not run out of space I am going to subtract off, this is plugging in the top. I am going to subtract off plugging in the bottom. I am going to have negative six X times negative one because that goes into here. Notice I kept the negative here, negative six X times this and then I am going to have plus again two over two is one so I am not going to worry about it. Plugging a negative here, I am going to have negative one squared, that is going to be squared. So all I need to do is evaluate what that gives me here. So going up here negative six X times one is negative six X plus one squared is plus one. I have a negative here so negative times negative is a positive but then I have another negative one here so this is going to be negative six X and then I have a negative distributed over here and negative one squared is positive one but it is times negative so I am going to have a negative one here. I think you can convince yourself that that is what happened. Now the positive one and the negative one just add zero, so all I am going to have after all of that is done is negative 12X just adding these two things together, negative 12X. Now what do I do, going back up to the top of the page you can see that I have done the integral along dz. Now I have done the integral along dy. Now I need to do the integral along dx, so it is real simple matter to take this and finish the integral. So now I am going to integrate from zero to three, okay, the integral is negative 12X dx. This is a super simple integral. It is negative 12 over two X squared, just a regular polynomial integral evaluated from zero to three. So negative 12 over two is negative six, and then on the inside, I am going to plug in the top value which is three squared minus and then the bottom is zero squared. All I did is I evaluated the outside. I kept it out and then I just evaluate the limits of integration on the inside, so what you are going to have is negative six times; three times three is nine, and so the answer is going to be negative 54. Nine times six gives me 54 and that is the answer so let us go ahead and recap because I want to make sure and all the problems are going to follow this general form. Integrating over three directions, a function of three variables again along three directions; dz, dy, and dx. First we integrate along Z so Z is what we are going to integrate over, everything else is constant, we got this as a result. We plugged in the limits of integration into the variable Z, we got this, we simplified it down to this expression in X and Y. So we have effectively eliminated the Z dependents because we integrated along Z. We know what the Z contribution is. Now we are left with X and Y. We take that value and we integrate along X and Y using the original limits of integration. Now we integrate along Y first, X is now constant, so he behaves like a constant. We get this as an answer and we plugged in the limits of integration into the Y variables which we got here. We simplified all this down. We got a very simple answer, negative 12X. We take this and we integrate it along X using the original limits of integration and do everything just as before integrating a Calculus I integral then the limits of integration gives you a negative 54. So what you have done here is you are integrating this function of three variables and you have defined a volume by picking the limits of integration like this. What you are doing is you are really looking at this function and the values of this function along this region, this volume V that we have define and we are sort of adding up the contributions that every point in space governed by this volume that we have defined here. So every problem is going to be like this, I mean this is it, this is the section. All we are going to do now is just some more problems to sort of get you some practice in doing it but it is exactly the same thing. You integrated along Z first and then Y and then X and then you know, that is going to be basically the form of all of these problems.

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Hi and welcome to this section of the Calculus III Tutor volume number two, and in this section of the class, we are going to continue our discussion of Calculus III marching our way on through to the end of what most of you guys will cover in your Calculus class. We are going to talk about the topic of triple integrals. Let me just pause right here and say that a lot of what I am covering here and through the remainder of the class is going to assume that you are already have been exposed to the material covered in the Calculus III Tutor volume one, so if you do not already have that, you probably should go ahead and get it especially with the triple integrals we are doing now, because although I am going to start from the beginning and walk you through it, I am going to sort of assume that you have been exposed to me introducing the double integrals to you which is covered in volume one that is going to be directly applicable to what we are doing here.

So what we are going to do in this class is we are going to continue on with the triple integrals in this section and we are going to march through the different types of triple integration and on into the vector fields and stuff that we will do in here in a little bit.

So what is a triple integral? It is one of those things that you will look at the first time, kind of like a double integral and you are like my goodness, we did double integrals, that was different and now we are doing three integrals, why would you ever do that? We are going take it one step at a time and tell you why we were doing that.

First, let us go ahead and recall the double integral that we did just briefly which at the tail end of the volume one of the Calculus III, okay. The double integral recall is just if you can visualize it, you have your X and your Y and your Z, right and so that is your coordinate system and you have a function that was sort of “hovers”. It is a function X and Y, function of two variables and I use my hand here because I am sort of indicating that it is a function that is sort of this mountain or valley or something that is kind of hovering your XY plans, it is kind of hovering there. There is always a region R that we were talking about when we are doing our double integrals, because the function that I am talking about is a function of X and Y. It is a function of two variables. So when we try to integrate those functions, we have to integrate it over two directions, X and Y because that is what the functions define the terms of X and Y.

So you define this region R in the plane. It could be any shape, it could be an ellipse, it could be anything. The way we set the integrals up we could use any arbitrary shape in the XY plane to integrate this function over. And what we were really doing in that section is we were adding up the contribution of the function as it relates to the region R of the XY plane. Basically, we were integrating that the “height” so to speak of that function over the XY plane, over any arbitrary shape that we wanted to choose in there.

If I could summarize it in one sentence okay; the double integral was integrating a function of two variables, X and Y and we were integrating it in two directions, X and Y because we were integrating a function of two variables over two directions. It makes sense, right? Two variables, two directions, okay. And when we did it in polar coordinates, it was really no different instead of a function of X and Y, it was a function of R and theta which is again two variables.

So when we did our integration over those polar functions of R and theta, we did not integrate over X and Y because that is Cartesian. We integrated over dr and dtheta, so it looks different but it is the same thing. You are integrating over two variables because you have a function of two variables, so to sort of summarize that just really briefly because it is going to be directly applicable to what we are doing here, just recall for me that the double integral looks something like this and we sort of, in the beginning anyway before we actually solved any problems, we said we were integrating over a region R which is that region in the XY plane down here that I am actually integrating this function over.

We are integrating the function of two variables, X and Y, so that is why we have X and Y there and the differentials that we were integrating over are dx and dy. Okay so this was sort of are general form, you have a function of two variables over X and Y, we integrate them over those two directions basically and the dx and dy and the limits of integration that we had define the region R here in the XY plane that we integrated this function over; that is what we were doing.

In integration, do not forget it is just a shorthand way of adding up an infinite number of things so you are adding up the contribution of its function over that region X and Y. Now when we actually boiled it down to where the rubber meets the road, the way we actually broke this down, is we again we had the two integral signs, we had the function of X and Y, right? And we had the dy and dx, let us say, because we said before and we are going to change this, just to be consistent dy and dx. You can flip the order of integration. We talked about that but just to give you sort of an example, what we said was we were going to integrate, let me go and change colors here just to make it clear. Since we are integrating over Y at first, Y was some function of G1 of X up to some function of G2 of X and we said we are going to integrate from X=A to X=B.

So real briefly, I mean not too really belabor it too much but if this is X and Y, so we are looking down on the XY plane and our function here that we are integrating is kind of hovering over the XY plane because it is a function of X and Y then what we are really saying here is; is since we were integrating along Y first, we are integrating along this direction first, we had some function here and we called it G1, and we integrated up to some function here G2, integrating from G1 to G2 of X, these are just functions of X.

And then me integrated from A over to B, and the combination of all these stuff from G1 to G2 which are functions and from A to B along X, define this region R that is what we are integrating over, so by choosing G1 and G2, these functions in our limits of integration, we can sort of define any arbitrary shape in the XY plane that we care about and we can integrate this function over this region R. That is what we were doing.

So we were going to make a direct extension to what we were doing when we were talking about the triple integrals, that is why I am giving you a little review here. So to recap, for double integrals, they are integrating a function of two variables, X and Y and because it is a function of two variables, we are integrating it over X and Y in the XY plane. We are integrating over two directions basically.

Now for the triple integral, what we are going to do is we were going to integrate a function of three variables, X, Y, and Z, and because there is three variables involved, we have to integrate in three directions and so it is going to be integrating over dx, dy, and dz, and instead of integrating over a surface area that this function is hovering over, we are actually going to integrate this function over a volume region dv.

So instead of seeing integrals with da at the end, which is kind of what we had here, I guess I erased it but this could be, this da could be dy and dx, you can sort of replace it with da. A lot of times, you will see it in your textbook, an area or unit, okay. So it would integrating over some area in the XY plane, what we will be doing is we will be integrating this function of three variables over a volume. So in order to get the dimension of volume, we have to add dx, dy, and dz.

So burn that in your head, that is really important and we were going to work some problems here. You are going to see they are not going to be very different from double integrals, I mean really, it is just tedious. It is the same thing. We get into the next section where we will be doing the triple integrals and the cylindrical coordinates and then even beyond that, the triple integrals and the spherical coordinates, they sound so complicated and so hard but just please erase your anxiety over it because what you are going to find is it is exactly the same thing. It is just using a different coordinate system to define the volume that you are integrating over.

So what we are going to do now is draw a picture because I like to draw pictures and I hope that it makes it clear. So what we are going to do is we are going to draw a little picture of what this triple integral thing is and then we are going to start working some problems and you are going to find out that they are really not a big deal.

So, let us go ahead and start by drawing the thing that we always draw which is your coordinate axis and I would encourage you guys to do this yourself for every problem if you can, I mean you do not have to but get used to drawing this because it really does help visualize things.

So here is your X, Y, and Z, it is a three dimensional system. Now what we are going to do is I am going to draw and define for you the following. I am going to draw several things up here and I am going to label a few things and you are probably going to look and say what is he talking about? Let me just sort of get it up here and define it for you.

What I am going to do is I am going to define some arbitrary volume which is going to be a volume V or a volume D actually that we are going to talk about here. So let me go ahead and define, this is sort of like a cylindrical thing that is why I have the straight sides here but the tops and the bottom can be sort of any arbitrary shape. You see that is why I have got lumps here; lumps here on the top and lumps here on the bottom. This can be very smooth, it could be hills and valleys on the top or whatever. It is some arbitrary thing on the top. It is some arbitrary thing on the bottom. So what we are doing here is we are defining what the top here connecting down to the bottom which are arbitrary kind of shapes, this is a volume D.

Now why do they call it volume D? Why do they not call it volume V? Most of the books call it volume D because if you start talking about V here then when you get into actual integration where you are talking about dv, it can get a little bit confusing with V and dv. So the volume is D, it is just a letter, do not let it worry you, okay.

So here is the volume D, and this volume D if you were to shine a flashlight on top and project its shadow so to speak down here into the XY plain, it is going to cast a shadow, right. So this shadow actually, I am going to draw in a different color. I am going to draw in this color. I am going to do my best to kind of show you that what we have here is maybe not the best representation but what I am trying to show you here is that this shadow cast down here, and this shadow is the very familiar region R that we have been talking about from the other sections.

So this is the region R; when you look at these books, the region R is always talking about the XY plane. What something looks like in the XY plane, the volume D is actually the entire volume that you are encompassing here.

Now what we are going to end up doing is I am going to have a function here; a function of three dimensional space, so before we were talking about functions of two variables, you had your XY plane and every point in the XY plane, you plug in X and Y and you get a third value, you can call it Z, right. And so, then you can plot that and it will sort of hover over the XY plane. The function of three variables, you really cannot draw it on a board because when you think about what you are doing in this room for instance, at every point X, Y, and Z, at every single point my finger is at, is a value, okay. And that value could be let us say I am measuring the temperature, I am measuring the temperature here, and here, and here, and here, and here, and here, so at every point I have a different value.

Now that value is a function of X, Y, and Z, right. Because everywhere I move my finger, I have a different temperature so I could construct a function of X, Y and Z that could predict the temperature at different points in space; that would be a function of three variables. Now how do you draw that? How do you actually draw that on a board? I mean you really cannot because you would have to have basically at every point X, Y, and Z, you have a different value of temperature, and that is really hard to draw as a surface, so you really cannot draw it too well but you can sort of visualize what is happening. At every single point inside of this volume element that we have sort of created here, at every point there is a different value and if it helps you to think about it, think about this value as a function that is like a temperature or something, it could be the pressure at every point in space. It could be the magnetic fields strength or something like this; it could be anything that is a function of the space, the three functions of space.

So that function that we are talking about here on the board is going to be a function of X, Y, and Z. That function kind of permeates space at every point of X, Y, and Z, whatever my function is defined to be, I calculate a value at that point and what I am going to do is I am going to integrate this function over this volume.

So this function is everywhere, it is all the dimensions going into the board coming out of the board, left and right, it is different values that function calculates a different value at every single point in space but I do not care about every point in space, I care about adding up the contribution of the function only in this volume element that I have defined here. That is what integration is. And in real life, you might be integrating, later on we will do a lot more thus you might be integrating the electric fields strength over a volume or you could be integrating at any number of things over a volume, that would actually mean something. Like you might be integrating the electric field and it might relate to the amount of charge you have inside of that volume or any number of physical applications but you are integrating an actual three dimensional over three dimensions.

So in order to pull this off, we have to add a couple of more things to our drawing. This upper boundary and this lower boundary are arbitrary surfaces so to speak because this is sort of a surface on the top and a surface on the bottom and they can be arbitrary, that is why I have got these squiggly lines here, okay. So the top one or let us go to the bottom one first. The bottom one, we are going to say, let me go ahead and erase this and put this is going to be volume D, nothing has changed. This bottom surface, we are going to call it Z is equal to some function, I am going to call it function F1 of X and Y, let me finish here and I am going to show you what I am doing here, so do not worry if this does not make. The top surface, so to speak that caps the top end of this can is going to be some other function F2, that is why there is a two there because it is a different a function, it is also a function of X and Y.

So what am I saying here? This surface, so to speak on the top is sort of like my hand it is cupping it right. It can have hills and valleys, and divots and depressions, and mountains and everything else here and it is going to be a function of two variables; a function of X and Y. So at any point X and Y, that is how I am defining this guy here because this function of X and Y is going to give me this surface that is going to sit right on the top. I have a different function of X and Y that defines the bottom surface here. Again it is kind of cupping it down like this. At any point XY down here, I am defining the surface here. I will plug in X and Y and I calculate the function, a number and I am going to call it Z and that is going to be the Z value here in this direction that is going to define the actual shape of these boundaries, so here I have everything I need to actually proceed. I have got some function on the bottom of X and Y of two variables because it is a surface down here, I have got a different function on the top that governs what the top looks likes. It is also a different surface and they are just connected by vertical lines here and I form by doing this a volume D.

Now if I take a flashlight and shine down like this; the shadow of this whole thing is going to define a region R here. That is just sort of giving a new continuity from the last section before, but what we are going to do is we are going to take a function of three variables, X, Y, and Z which exist everywhere. I can plug a value of X, Y, and Z in and get a number back that is my function and I am going to integrate this function over this volume element D; that is what this entire section is about.

So, how do we do it? We are going to work way up to it slowly so we are not going to just, I am not going to throw you into the grease too much but conceptually when you think about it, we need three integral signs. We know that because we are integrating in three directions and we are going to integrate not over a region R like what we were doing in the section on double integrals but we are going to integrate over this volume element D because this volume element D is what we have defined right here. So when we do the integration, however, we set up our limits of integration must correspond to this actual volume element. So in integrating over the volume D, we are integrating a function that we just talked about; X, Y, and Z. And before getting too technical, we are just going to put dv here.

So if you remember back to your double integrals from the last DVD that we talked about, it was usually written as double integral over a region R of a function of two variables da, which is this da is an area and it corresponds to the region R. Now da got busted into dy and dx because these two directions give you an area. What you are doing is, you see it is not too complicated, you are just having a function of three variables and instead of an area, you are integrating over a volume and we have an extra integral sign here and so that is just sort of given right like that.

So what we are going to do here is continue on with what we know to be true here and I am going to write down a few more things and we are going to arrive at a very useful thing to actually do our integration.

Now let us go ahead and write this a little bit further. Let me say that this is then going to be equal to; let me just write it down and show you. Okay, so do not worry about this here yet. I am going to put a big parenthesis here and I am going to put my other integral sign inside and I am going to integrate from F1 of X and Y to F2 of X and Y. I am just integrate my function X, Y, and Z, and then over here, what I am going to have is dz, now I am actually going to change this parenthetical, I am going to change colors, I am going make it hopefully super easy to understand.

And then over here, I am going to have da, all right. Now let me explain where I am doing here, I am changing from this form to this form and I am show it slowly unraveling things to you. Eventually we are going to get down to even a more explicit form here. All right, what am I doing here? This is a triple integral over three directions here. The same function now, I mean integrating this function. What I am doing is I am you giving you a little more detail about what this first integral is.

What I am going to end up doing, first is integrating along dz. I am going to integrate along dz, now look what happens if I actually do this; if I integrate this function along dz, where Z is the actual variable I am integrating over this thing here, okay. Then what I am going to get in the answer is a function of X and Y, because this is a definite integral, if I integrate this as a function of Z and then I evaluate the limits of integration which are they say here they are functions of X and Y, so whatever I get in the end when I evaluate the limits of integration, this thing returns a function of X and Y. This inner thing here when I integrate this function over Dz and then take and evaluate the limits of integration, when you evaluate the limits of integration, you take this function you plug into whatever you had for Z here after you do the integration and the same thing with the bottom thing, you are not going to have any Zs leftover, you are only going to have Xs and Ys, so once you evaluate this integral, the only thing you are going to have left after you plug in the limits of integration after you have done the integration is a function of X and Y.

So this whole thing is a function of X and Y. Now we already learned how to do double integrals over functions of X and Y; da. We did that in the last DVD at the very end and it is called the double integral. You have done that all along. The only thing we are doing here is restarting one of the functions of three variables and we are integrating it along the third variable first, dz and we were going to get our answer and then we were going to take that and we were going to integrate it over the region R, that I conveniently wrote in your drawing down here just for reference. So what we are going to effectively do is we are going take the region, the function that permeates all of space here in three directions. We are going to first integrate it along Z, okay that is what we are doing first, but in order to integrate along Z, we have to integrate from the lower surface which is a function of X and Y to the upper surface which is a function of X and Y. That is why the limit of integration are these two functions, so we were going to integrate in this direction first and then after we get the answer, we are going to integrate whatever is leftover along Y first and then along X and we are going to be integrating it over the region R because that is the projection of the entire volume elements. So all you are really doing, I mean I am trying to explain things in several different ways to hopefully make it easy but really all you are doing is, you are going to integrate over Z first and then Y and then X.

Once you do that first integration along Z, you are going to get back the function of X and Y, which you already know how to integrate because you have been doing double integrals over X and Y in the last section on double integrals. So that is all I am trying to tell you here okay. So let me go ahead and expand, now that you know what this is, let me go ahead and write the whole thing in its full glory to show you what you are actually going to be doing. What are you actually going to be doing in your problems so if R and I am going to go ahead and switch colors here; if the region R in the XY plane is vertically simple, then we are going to have a certain form and if it is horizontally simple, we will have another form.

What does vertically simple mean? Just to refresh your memory, if this is X and this is Y, vertically simple just means and you can go back and look in the section on double integrals that your region R is defined like this; this is R okay and this is the function G2 and this is the function down here G1, okay and what it means is this is vertically simple because you can take a little pencil and you can pierce the first function and pierce the second function all in one time. Horizontally simple is going to be if your region R is defined by two little functions that kind of go vertically like this and then your regions define like that and so going horizontally, you pierce the function one time each going like this, so basically, the vertically simple regions are what you are going to pretty much run into with most regularity in your classes, I am just sort of giving you the full spectrum in case you do have a problem like that you can solve.

So if this looks foreign to you, go back to your double integrals and this is all explained in detailed here but this is a vertically simple region in the XY plane when the two functions that you are bounding by are sorts of one on top of another and they are kind of arranged like this, you integrate along Y first and then along X, that is vertically simple.

So, let us go ahead and write down the form of this triple integral if we were going to do this thing and it is a vertically simple region. What we are going to say is the triple integral, over the volume D, okay of the function X, Y, and Z, over the volume dv which is what we are talking about here, it is a volume element you are integrating over, you have a ton of little volume element inside here and you are integrating over every little volume element that you can sort of infantesimally have infinite number of little volume elements inside of this big volume D. We are integrating over all of them. What you are going to have here, this is going to be equal to obviously, it is going to be three integration signs so let me go ahead and put the three integration signs there, and it is going to be function of X, Y, and Z, nothing surprising, you are integrating the function and when there is nothing surprising here, you are integrating along Z first, and then along Y, and then along X.

So you are integrating into three directions, so what are the limits of integration going to be? We just said up here that the first integral that we do, the inside integral as along Z, so we are integrating from this surface down here to the surface up here so we are integrating between this two functions here, so I am going to write these limits of integration in a different thing so we are going to integrate along F1 which is a function of two variables up to F2 which is a function of two variables. These are surfaces, the top surface and the bottom surface. Now whatever we get as a result of that we are going to integrate along dx and dy which is down here and the XY plane, we have been doing that all along. So we are going to integrate along Y first, so that means we are going to integrate between the function G1 of X up to G2 of X. And then along X, we are going to integrate from the numbers A to B.

So if you sort of travel back to this little scribble drawing that I have here, once you integrate along the surface from F1 up to F2 which are the surfaces here, it is on the top and the bottom of the cylinder, you get that, then to finish integrating, you got to go along Y and then along X, so if you are going to integrate along Y whatever this function is along Y then you integrate from the lower function up to the upper function of X. Lower function of X to the upper function of X and then to finish sweeping the area, you would go from A to B, so this would be A and this would be B, that is for a vertically simple region, so that is a lot of talking but I really want you to absorb this because most of the time you are going to have vertically simple regions, most of the time.

So all you are doing, is you are taking this function, you are integrating along Z, then you plug in these limits of integration, you take that result, then you integrate it along Y and you plug in these limits of integration, which is going to be integrating along Y from this function to this function. And then you integrate from A to B like this along X.

Now I will just say it real quickly that what you are doing here by having the three limits of integration; dz, dy, and dx, and having all these limits of integration here, what you are doing is, by choosing those cleverly, you are defining uniquely what this volume really looks like, you see. Because you have defined the surface Z from here to here and then you have defined what it looks like in the Y direction by those other two functions there, and then from going from A to B in the X direction, you have done it the other way, so you have X, and Y, and Z, going Z this way, then along Y, and then along X, putting all of these together is what actually specifies this volume that you are integrating over. You need all three directions to do that.

So this is what you are going to use all the time, almost all the time, I am going to write down one more thing only because you will see it in your Calculus textbook and I want you to sort of know what it means even if you do not use it very much, so I am going to write that down here.

If R is horizontally simple, then the integral will be very, very similar but not quite the same so if it is going to be done like this then it will be triple integrals obviously. You will be integrating this function of X, Y, and Z. You will always integrate along dz first and the only difference is, if you remember back to the double integrals when you had the difference between vertically and horizontally simple, the only real difference is you switch the order here. So here you will be integrating along the X first and along the Y. We flip the order.

Now the inner integral is going to be the same exact limit of integration is before. You integrate from that lower surface to the upper surface, so it is going to be F1 of X and Y up to F2 of X and Y. That is what this limit of integration corresponding to Z is then when you integrate along X, this is the only real difference here, and you are going to be integrating from H1 of Y to H2 of Y and then here is going to be integrating from the number C to the number D. Now just refresh your memory a little bit here, just so you are not totally lost with this. If it is a horizontally simple region, all that means is that when you are looking down in the XY plane, instead of having your functions like this one of top of another and then from one number to another number and you have the functions like this, your functions are sort of laying like this so if you put a horizontal line, they only pierce the function one time.

So what are you going to end up doing is that you can kind of tilt your head to side and look at it like this, Y is the independent variable here and X is the dependent variable, so this function right here is actually H2 of Y and this is H1 of Y. This function is a function of Y and this is another function of Y. And then if you are going to complete things here, you integrate from some number here and some number here and this number is called C and this number is called D. They are just numbers so instead of having your functions oriented like this, where they are vertically simple, you kind of tilt your head and you are orienting like this so they are horizontally simple.

So what it all really amounts to is that if you are doing something where you have horizontally simple region, you integrate along Z first and then you integrate along dx and you have to integrate from this function up to this function which are now functions of Y because you kind of turn your head over to the side looking at it like this. Y is the independent variable here so they are functions of Y integrating along X and then to finish it out, you would integrate along Y from C to D which is what we were doing here from C to D.

But I do not want you to lose track of the big, big picture because this is a lot of theory and what you will see it in a minute is when we actually erase all of this and go work some problems, it is not going to be that difficult because you are going to have a function here. All you are going to do is you are going to integrate along Z, plug in your limits of integration, take the result, integrate it along Y, plug the limits of integration, and then you will integrate along X and plug in those limits of integration and you will get the answers.

So, there is really not going to be anything more than this going on. It is just that the theory behind it, if you were told to integrate over some specific boundary then you will have to know how to set it up. You have some function here down to the bottom governing this thing. You have a function up here governing what the top looks like and then you would obviously have to choose what boundary you are going to integrate this thing over in the XY plane as well, and then putting all of that together, top, bottom, and then whatever you define in the XY plane, that is going to define what this actual volume D is.

And so most of the time, you are going to set it up like this, integrating along Z from that function F1 to F2, then taking the result integrating along Y and then integrating along X and plugging in your limits of integration here, so let us go ahead and work some example problems and I think when we do that, you will be a lot more comfortable with triple integrals.

Okay, so our first problem of the day is going to be a triple integral. The function is actually going to be y-x.z. We are going to integrate along dz first and then along dy and then along dx and the limits of integration are from two to four, negative one to one, and zero to three. So look at what we have here. We are integrating a function of three variables; X, and Y and Z, we are integrating along three directions and then these are just simply the limits of integration along Z, the limits of integration along Y, the limits of integration along X, so by doing all of this, you are defining what that volume looks like.

Because these limits of integration really are not complicated, you see this looks a whole lot simpler than what we were looking at on the previous screen because these limits of integration are just numbers, this volume region that we were actually integrating over is just a cube. It is literally just going from Z to Z, Y to Y, and X to X. So you are actually just forming a little cube there. If these were more complicated functions up here, then maybe instead of going from Z to Z up here, you would have some weird looking function on the bottom and some weird looking function on the top and you would define a more complicated looking volume, but when you just have numbers in the limits integration, you are really just integrating over simpler shapes and simple little constant values so it is going to form a cube in this case.

So mathematically speaking, what do we do? What I do when I write this down on my paper is I literally write a curly brace like this and I work on the inner integral first and I get the answer and then I do the next one and then I do the next one, so that is how we are going to do it here. When you were integrating along Z, from here on out by the way, nothing is different from your double integrals that we have already done in the last volume one so if you have not look at that, please go look at it now, you have already gotten a lot of practice with what we are going to do here, it is just going to have an extra integral sign.

So what we are doing is we are integrating along Z so that means Z is what we are integrating over. Anything else X or Y anything else is just a constant. You must get to the point where you can look at these things and say, I am integrating along Z, X and Y are just numbers. They could be negative 10 or negative 54, I treat them exactly as a constant when I am doing the integration.

So what this is going to equal, since Y is a constant, the answer to this integral is Y times Z minus X over two Z squared. Now make sure you understand that okay. This is a number. It is like two. So if I am integrating over dz and this is like two over this is just a constant and it comes down and then you have your variable. That is a simple integral from Calculus I, and if X is a constant because it is, we are integrating over Z. This is a constant and it just comes out, one half Z squared. It is just a regular polynomial integral that you have been doing all of your life. You evaluate this from two to four because those are the limits of integration along Z.

So you see it is really not that bad, you just have to know what to hold constant, so when you do this, when you are plugging in your limits of integration, you integrated along Z, so the limits of integration must be plugged in only into the Zs because that is what you integrated over. Everything else is just a constant so the way this will work out here, actually let me go and write it a little bit down below here. What you are going to have here when you evaluate this definite integral here is you will have Y times four okay minus X over two times four squared, I just plugged four into the Z squared and four into the Z minus, and I am going to do the same thing with the two. It is going to be Y times two minus X over two and it is going two squared.

All I did was take the upper limit, plug into wherever Z is and the lower limit, plug into wherever Z is, and I just simplify and let us see what we get. Okay, so over here, let us not get too crazy. Let us not try to do too many things at once so let us just rewrite this as four Y minus four times four is going to be 16. Sixteen divided by two is going to give you eight so you are going to have eight X here. You are going to have a negative sign here but I would rather just go ahead and do it. Let us go and do it this way; negative two Y, and you have two Y here, negative times negative gives me a positive here and then over here, I am going to have two squared is four, four divided by two is two, so you have two X and it is positive because of this.

So collecting all of my terms, what I have here is I am going to have negative six X plus two Y. Now notice that when I integrated this inner function here, I integrated this function and I plugged in the limits of integration into Zs and I got a function of X and Y back. Now remember, I took great care to tell you that that would always happens just to give you a little bit of the theory behind what is going on here. You integrate along Z and you plug into values of Z because that is what you integrated over.

You are always going to get a function of X and Y back so when you take this function and you integrate it along dx and dy, this now reduces to a double integral that you have done in the last couple of sections. Integrating a functions of X and Y over X and Y so now you see, you have just done another step here and now it reduces to a problem you have already done before. So you are actually getting some extra help and practice in doing double integrals as well because what we are going to do now, I am going to switch colors as I am going to take this, okay and now I am going to integrate it like this, so what I am going to do is just to make it totally clear to you, I am going to rewrite everything. Zero to three, negative one to one, the function is negative six X plus two Y and I am going rename that dy and Dx. I have just taken the answer that I am integrating along dy and dx and I am going to do exactly the same thing that I did before.

I am going to draw a little curly brace and I am going to evaluate that inner integral first. Now here I am integrating along dy so Y is what I am integrating over. Anything else is constant including this X. So to actually do this integration, you are going to have negative six XY because this is a constant so it is a number, it comes down. You have your Y because that is how you integrate simple integrals like that. We have the number and this is just going to be a Y. It is a polynomial integral, plus I am going to have two over two Y squared because this is a two. It is a constant, one half Y squared that is just a regular old Plain Jane integral from Calculus I, evaluate this from negative one to one.

So what I am going to actually have in the end here, plugging in here, I am going to actually plug in the top limit of the integration into the Ys and the bottom number into the Ys because I have integrated along Y so when I evaluate the limits, I need to plug them into only the Ys because that is what they integrated over, so it will have is negative six X times Y which is one, plus two over two, I am just going to say is a one, I am not going to worry about it, plugging in here is going to be one squared, I am going to write it down just for completeness. So I do not run out of space I am going to subtract off, this is plugging in the top. I am going to subtract off plugging in the bottom. I am going to have negative six X times negative one because that goes into here. Notice I kept the negative here, negative six X times this and then I am going to have plus again two over two is one so I am not going to worry about it. Plugging a negative here, I am going to have negative one squared, that is going to be squared.

So all I need to do is evaluate what that gives me here. So going up here negative six X times one is negative six X plus one squared is plus one. I have a negative here so negative times negative is a positive but then I have another negative one here so this is going to be negative six X and then I have a negative distributed over here and negative one squared is positive one but it is times negative so I am going to have a negative one here. I think you can convince yourself that that is what happened. Now the positive one and the negative one just add zero, so all I am going to have after all of that is done is negative 12X just adding these two things together, negative 12X. Now what do I do, going back up to the top of the page you can see that I have done the integral along dz. Now I have done the integral along dy. Now I need to do the integral along dx, so it is real simple matter to take this and finish the integral.

So now I am going to integrate from zero to three, okay, the integral is negative 12X dx. This is a super simple integral. It is negative 12 over two X squared, just a regular polynomial integral evaluated from zero to three. So negative 12 over two is negative six, and then on the inside, I am going to plug in the top value which is three squared minus and then the bottom is zero squared. All I did is I evaluated the outside. I kept it out and then I just evaluate the limits of integration on the inside, so what you are going to have is negative six times; three times three is nine, and so the answer is going to be negative 54. Nine times six gives me 54 and that is the answer so let us go ahead and recap because I want to make sure and all the problems are going to follow this general form. Integrating over three directions, a function of three variables again along three directions; dz, dy, and dx. First we integrate along Z so Z is what we are going to integrate over, everything else is constant, we got this as a result. We plugged in the limits of integration into the variable Z, we got this, we simplified it down to this expression in X and Y. So we have effectively eliminated the Z dependents because we integrated along Z. We know what the Z contribution is. Now we are left with X and Y.

We take that value and we integrate along X and Y using the original limits of integration. Now we integrate along Y first, X is now constant, so he behaves like a constant. We get this as an answer and we plugged in the limits of integration into the Y variables which we got here. We simplified all this down. We got a very simple answer, negative 12X. We take this and we integrate it along X using the original limits of integration and do everything just as before integrating a Calculus I integral then the limits of integration gives you a negative 54.

So what you have done here is you are integrating this function of three variables and you have defined a volume by picking the limits of integration like this. What you are doing is you are really looking at this function and the values of this function along this region, this volume V that we have define and we are sort of adding up the contributions that every point in space governed by this volume that we have defined here.

So every problem is going to be like this, I mean this is it, this is the section. All we are going to do now is just some more problems to sort of get you some practice in doing it but it is exactly the same thing. You integrated along Z first and then Y and then X and then you know, that is going to be basically the form of all of these problems.

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