Figuring out the boundaries of integration. Created by Sal Khan.
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- Can you do a lesson on doing triple integrals in cylindrical and/or spherical coordinates? Or can anyone show me a good resource that explains those coordinate systems well?(114 votes)
- The "Solid of a Revolution" videos are the Calc I way of doing it. I'm not sure if Khan has any videos covering this, but "Paul's Online Math Notes" is an excellent resource, and it does cover cylindrical and spherical coordinates. Here's the URL: http://tutorial.math.lamar.edu/Classes/CalcIII/CalcIII.aspx . It's under the "Multiple Integrals" chapter.(32 votes)
- My question is at10:10of the video, when setting the bounds for the x and then y planes. I understand the bottom bounds being 0. What I don't understand is why the upper bound of x is the line x=3-y/2 and why the upper bound of y IS NOT also that line, only in terms of y (y=6-2x). To me it seems like both upper boundaries would either be the line or their simpler numbers, like we gave y at the end.
Thanks for your help, I always get mixed up using boundaries!(31 votes)
- Sal drew a little triangle in the bottom right of the screen which is the projection of the volume on to the x-y plane. Sal also drew a horizontal rectangular "bar" with height dy. The area of the triangle is the sum of all of the little bars "stacked" on top of each other. Each bar runs from 0 to the end of the line in the x direction, so these are the x boundaries. And the bars are stacked from 0 to 6 in the y direction, so these are the y boundaries. Hope this helps.(22 votes)
- Towards the very end, Why isn't upper y boundary y= -2x + 6? I'm not sure why its just 6..(8 votes)
- Every time you integrate, you are essentially removing a dimension from your problem as you approach the answer. The first integral removes the z dimension so z=0 the rest of the way through because you integrated with respect to z, which is why you use the 2 dimensional projection in the xy plane to get the upper bound of x. Once you integrate with respect to x, you have finished with that dimension and x=0 from then on. Finally, now that z=0 and x=0, you can plug those into the original surface equation to get your upper y bound of 6. Then, you do your final one-dimensional integral to solve the problem.(43 votes)
- Can your x integral be from 0 to 3?(7 votes)
- Only if you integrate that last and you use y=6-2x as the upper bound for y. That boundary must be used at least once and your last integration must only have constants.(17 votes)
- did anyone try to evaluate this? i got 40.5 kg but i'm probably wrong(5 votes)
- Why z=2 is that green triangle surface?(3 votes)
PLEASE do more on tripple integrals! I love your videos, but its a bummer you stop here with this trivial example. As someone else menioned in a comment years ago, can you continue into clindrical and sphereical coordinat systems?? Paraboloids, and frustrums, and more complex surfaces?(6 votes)
- Why do we put the variable bounds in the innermost integral?(3 votes)
- because if we use it in the ¿outermost integral? we will have in the final result a function, and we can´t have this, we need to have a fixed number...(5 votes)
- At11:11, if we want the area between z = 2 and 2x + 3z + y = 6, why do we integrate from x = 3 - y/2 to x= 0? Wouldn't we integrate from x = 3 to x = 3 - y/2 since x = 3 - y/2 is the lower bound?(3 votes)
- Very good observation. I followed in the footsteps of those who solved this hideous problem, without realizing this, and now I know my solution is meaningless since the limits are mixed up. I would recommend mentioning this as an error in the video so that others may be able to avoid similar meaningless work.
Thank you for your attention to detail and precision!(3 votes)
- At9:21, i don't understand why you used the x-y plane instead of the x-z plane. I am not very good at visualizing 3D diagrams, so could you please show me why?(2 votes)
- Because if your integration order takes care of Z first, i.e. dzdxdy, then once you find your Z limits in the first integral, then you are done with Z altogether, the next step to solve the triple integral is to project into the remaining variables' plane, in other words, project into the x-y plane. You would only project your solid into the x-z plane if you integrated with the order dydxdz or dydzdx, in other words, integrating y first.(5 votes)
Let's now do another triple integral, and in this one I won't actually evaluate the triple integral. But what we'll do is we'll define the triple integral. We're going to something similar that we did in the second video where we figured out the mass using a density function. But what I want to do in this video is show you how to set the boundaries when the figure is a little bit more complicated. And if we have time we'll try to do it where we change the order of integration. So let's say I have the surface -- let me just make something up -- 2x plus 3z plus y is equal to 6. Let's draw that surface. It looks something like this. This will be my x-axis. This is going to be my z-axis. This is going to be my y-axis. Draw them out. x, y and z. And I care about the surface in the kind of positive octant, right, because when you're dealing with three-dimensionals we have, instead of four quadrants we have eight octants. But we want the octant where all x, y and z is positive, which is the one I drew here. So let's see, let me draw some -- what is the x-intercept? When y and z are 0, so we'll write here, that's the x-intercept. 2x is equal to 6, so x is equal to 3. So 1, 2, 3. So that's the x-intercept. The y-intercept when x and z are 0 are on the y-axis, so y will be equal to 6. so we have 1, 2, 3, 4, 5, 6 is the y-intercept. Then finally the z-intercept when x and y are 0 we're on the z-axis. 3z will be equal to 6. So z is equal to 1, 2. So the figure that I care about will look something like this -- it's be this inclined surface. It will look something like that. In this positive octant. So this is the surface defined by this function. Let's say that I care about the volume, and I'm going to make it a little bit more complicated. We could say oh, well let's find the volume between the surface and the xy plane. But I'm going to make it a little bit more complicated. Let's say the volume above this surface, and the surface z is equal to 2. So the volume we care about is going to look something like this. Let me see if I can pull off drawing this. If we go up 2 here -- let me draw the top in a different color, let me draw the top in green. So this is along the zy plane. And then the other edge is going to look something like this. Let me make sure I can draw it -- this is the hardest part. We'll go up 2 here, and this is along the zx plane. And we'd have another line connecting these two. So this green triangle, this is part of the plane z is equal to 2. The volume we care about is the volume between this top green plane and this tilted plane defined by 2x plus 3z plus y is equal to 6. So this area in between. Let me see if I can make it a little bit clearer. Because the visualization, as I say, is often the hardest part. So we'd have kind of a front wall here, and then the back wall would be this wall back here, and then there'd be another wall here. And then the base of it, the base I'll do in magenta will be this plane. So the base is that plane -- that's the bottom part. Anyway, I don't know if I should have made it that messy because we're going to have to draw dv's and d volumes on it. But anyway, let's try our best. So, if we're going to figure out the volume -- and actually, since we're doing a triple integral and we want to show that we have to do the triple integral, instead of doing a volume, let's do the mass of something of variable density. So let's say the density in this volume that we care about, the density function is a function of x, y and z. It can be anything. That's not the point of what I'm trying to teach here. But I'll just define something. Let's say it's x squared yz. Our focus is really just to set up the integrals. So the first thing I like to do is I visualize -- what we're going to do is we're going to set up a little cube in the volume under consideration. So if I had a -- let me do it in a bold color so that you can see it -- so if I have a cube -- maybe I'll do it in brown, it's not as bold but it's different enough from the other colors. So if I had a little cube here in the volume under consideration, that's a little cube -- you consider that dv. The volume of that cube is kind of a volume differential. And that is equal to dx -- no, sorry, this is dy. Let me do this in yellow, or green even better. So dy, which is this. dy times dx, dx times dz. That's the volume of that little cube. And if we wanted to know the mass of that cube, we would multiply the density function at that point times this dv. So the mass, you could call it d -- I don't know, dm. The mass differential is going to be equal to that times that. So x squared y z times this. dy, dx, and dz. And we normally switch this order around, depending on what we're going to integrate with respect to first so we don't get confused. So let's try to do this. Let's try to set up this integral. So let's do it traditionally. The last couple of triple integrals we did we integrated with respect to z first. So let's do that. So we're going to integrate with respect to z first. so we're going to take this cube and we're going to sum up all of the cubes in the z-axis. So going up and down first, right? So if we do that, what is the bottom boundary? So when you sum up up and down, these cubes are going to turn to columns, right? So what is the bottom of the column, the bottom bound? What's the surface? It's the surface defined right here. So, if we want that bottom bound defined in terms of z, we just have to solve this in terms of z. So let's subtract. So what do we get. If we want this defined in terms of z, we get 3z is equal to 6 minus 2x minus y. Or z is equal 2 minus 2/3x minus y over 3. This is the same thing as that. But when we're talking about z, explicitly defining a z, this is how we get it, algebraically manipulated. So the bottom boundary -- and you can visualize it, right? The bottom of these columns are going to go up and down. We're going to add up all the columns in up and down direction, right? You can imagine summing them. The bottom boundary is going to be this surface. z is equal to 2 minus 2/3x minus y over 3. And then what's the upper bound? Well, the top of the column is going to be this green plane, and what did we say the green plane was? It was z is equal to 2. That's this plane, this surface right here. Z is equal to 2. And, of course, what is the volume of that column? Well, it's going to be the density function, x squared yz times the volume differential, but we're integrating with respect to z first. Let me write dz there. I don't know, let's say we want to integrate with respect to -- I don't know, we want to integrate with respect to x for next. In the last couple of videos, I integrated with respect to y next. So let's do x just to show you it really doesn't matter. So we're going to integrate with respect to x. So, now we have these columns, right? When we integrate with respect to z, we get the volume of each of these columns wher the top boundary is that plane. Let's see if I can draw it decently. The top boundary is that plane. The bottom boundary is this surface. Now we want to integrate with respect to x. So we're going to add up all of the dx's. So what is the bottom boundary for the x's? Well, this surface is defined all the way to -- the volume under question is defined all the way until x is equal to 0. And if you get confused, and it's not that difficult to get confused when you're imagining these three-dimensional things, say you know what, we already integrated with respect to z. The two variables I have left are x and y. Let me draw the projection of our volume onto the xy plane, and what does that look like? So I will do that. Because that actually does help simplify things. So if we twist it, if we take this y and flip it out like that, and x like that we'll get in kind of the traditional way that we learned when we first learned algebra. The xy-axis. So this is x, this is y. And this point is what? Or this point? What is that? That's x is equal to 3. So it's 1, 2, 3. That's x is equal to 3. And this point right here is y is equal to 6. So 1, 2, 3, 4, 5, 6. So on the xy-axis, kind of the domain -- you can view it that -- looks something like that. So one way to think about it is we've figured out if these columns -- we've integrated up/down or along the z-axis. But when you view it looking straight down onto it, you're looking on the xy plane, each of our columns are going to look like this where the column's going to pop out out of your screen in the z direction. But the base of each column is going to dx like that, and then dy up and down, right? So we decided to integrate with respect to x next. So we're going to add up each of those columns in the x direction, in the horizontal direction. So the question was what is the bottom boundary? What is the lower bound in the x direction? Well, it's x is equal to 0. If there was a line here, then it would be that line probably as a function of y, or definitely as a function of y. So our bottom bound here is x is equal to 0. What's our top bound? I realize I'm already pushing. Well, our top bound is this relation, but it has to be in terms of x, right? So what is this relation. So, you could view it as kind of saying well, if z is equal to 0, what is this line? What is this line right here? So z is equal to 0. We have 2x plus y is equal to 6. We want the relationship in terms of x. So we get 2x is equal to 6 minus y where x is equal to 3 minus y over 2. And then finally we're going to integrate with respect to y. And this is the easy part. So we've integrated up and down to get a column. These are the bases of the column, so we've integrated in the x direction. Now we just have to go up and down with respect to y, or in the xy plane with respect to y. So what is the y bottom boundary? Well, it's 0. y is equal to 0. And the top boundary is y is equal to 6. And there you have it. We have set up the integral and now it's just a matter of chugging through it mechanically. But I've run out of time and I don't want this video to get rejected. So I'll leave you there. See you in the next video.