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So here, we have r2, we're going to evaluate it at 1 half, because this is at time is equal to 1 half second, and this is going to be equal to 2i, this isn't dependent at all on time, so 2i plus 8 times the time, so time right here is 1 half, so 8 times 1 half is 4, so plus 4j. So what does this look like? The instantaneous derivative here, and this is the derivative, we have to be very clear. So 2i, let me draw some more, so 2i maybe gets us about that far, plus 4j will get us up to right around there, plus 4j is that vector, so when you add those two heads to tails, you get this thing. You get something that looks like that. I didn't draw it as neatly as I would like to, but let's notice something. Both of these vectors are going in the exact same direction, they're both tangential to the path, to our curve, but this vector is going, its length, its magnitude, is much larger than this vector's magnitude, and that makes sense because I hinted at it when we first talked about these vector-valued position functions and their derivatives, is that the length, you can kind of view it as the speed.
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Vector valued function derivative example Multivariable Calculus Khan Academy.mp3
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So 2i, let me draw some more, so 2i maybe gets us about that far, plus 4j will get us up to right around there, plus 4j is that vector, so when you add those two heads to tails, you get this thing. You get something that looks like that. I didn't draw it as neatly as I would like to, but let's notice something. Both of these vectors are going in the exact same direction, they're both tangential to the path, to our curve, but this vector is going, its length, its magnitude, is much larger than this vector's magnitude, and that makes sense because I hinted at it when we first talked about these vector-valued position functions and their derivatives, is that the length, you can kind of view it as the speed. The length is equal to the speed. If you imagine time, t being time, and these parametrizations are representing a dot moving along these curves, so in this case, the particle only takes a second to go there, so at this point in its path, it's moving much faster than this particle is. If you think about it, this vector right here, if you imagine this is a position vector, this is velocity.
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Vector valued function derivative example Multivariable Calculus Khan Academy.mp3
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Both of these vectors are going in the exact same direction, they're both tangential to the path, to our curve, but this vector is going, its length, its magnitude, is much larger than this vector's magnitude, and that makes sense because I hinted at it when we first talked about these vector-valued position functions and their derivatives, is that the length, you can kind of view it as the speed. The length is equal to the speed. If you imagine time, t being time, and these parametrizations are representing a dot moving along these curves, so in this case, the particle only takes a second to go there, so at this point in its path, it's moving much faster than this particle is. If you think about it, this vector right here, if you imagine this is a position vector, this is velocity. This right here is velocity. Velocity is speed plus direction. Speed is just how fast are you going, velocity is how fast you're going in what direction.
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Vector valued function derivative example Multivariable Calculus Khan Academy.mp3
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If you think about it, this vector right here, if you imagine this is a position vector, this is velocity. This right here is velocity. Velocity is speed plus direction. Speed is just how fast are you going, velocity is how fast you're going in what direction. I'm going this fast, and you could calculate it using Pythagorean theorem, but I just want to give you the intuition right here. I'm going that fast in this direction. Here I'm going this fast.
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Vector valued function derivative example Multivariable Calculus Khan Academy.mp3
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Speed is just how fast are you going, velocity is how fast you're going in what direction. I'm going this fast, and you could calculate it using Pythagorean theorem, but I just want to give you the intuition right here. I'm going that fast in this direction. Here I'm going this fast. I'm going even faster. That's my magnitude, but I'm still going in the same direction. Hopefully you have a gut feeling now what the derivative of these position vectors really are.
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Vector valued function derivative example Multivariable Calculus Khan Academy.mp3
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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. Looks something like this. This will be my x-axis. It's going to be my z-axis.
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Triple integrals 3 Double and triple integrals Multivariable Calculus Khan Academy.mp3
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Let's draw that surface. Looks something like this. This will be my x-axis. It's going to be my z-axis. That's going to be my y-axis. Draw them out. x, y, and z.
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Triple integrals 3 Double and triple integrals Multivariable Calculus Khan Academy.mp3
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It's going to be my z-axis. That's 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 dimensions, 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.
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Triple integrals 3 Double and triple integrals Multivariable Calculus Khan Academy.mp3
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x, y, and z. And I care about the surface in the kind of positive octant, right, because when you're dealing with three dimensions, 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're right here, that's the x-intercept. 2x is equal to 6, so x is equal to 3.
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Triple integrals 3 Double and triple integrals Multivariable Calculus Khan Academy.mp3
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Let me draw some. What is the x-intercept? When y and z are 0, so we're right 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, we're on the y-axis, so y will be equal to 6.
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Triple integrals 3 Double and triple integrals Multivariable Calculus Khan Academy.mp3
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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, we're on the y-axis, so y will be equal to 6. So we have 1, 2, 3, 4, 5, 6. So y-intercept. And then finally, the z-intercept.
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Triple integrals 3 Double and triple integrals Multivariable Calculus Khan Academy.mp3
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The y-intercept, when x and z are 0, we're on the y-axis, so y will be equal to 6. So we have 1, 2, 3, 4, 5, 6. So y-intercept. And 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.
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Triple integrals 3 Double and triple integrals Multivariable Calculus Khan Academy.mp3
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And 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'll be this inclined surface. It will look something like that.
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Triple integrals 3 Double and triple integrals Multivariable Calculus Khan Academy.mp3
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So z is equal to 1, 2. So the figure that I care about will look something like this. It'll be this inclined surface. It will look something like that. And this positive octant. So this is the surface defined by this function. Let's say that I care about the volume.
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Triple integrals 3 Double and triple integrals Multivariable Calculus Khan Academy.mp3
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It will look something like that. And 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, what's the volume between the surface and the xy-plane? But I'm going to make it a little bit more complicated.
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Triple integrals 3 Double and triple integrals Multivariable Calculus Khan Academy.mp3
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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, what's 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 essentially, the surface z, so the volume we care about is going to look something like this. Let me see if I can pull off drawing this.
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Triple integrals 3 Double and triple integrals Multivariable Calculus Khan Academy.mp3
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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 essentially, the surface z, so the volume we care about is going to look something like this. Let me see if I can pull off drawing this. So if we go up 2 here. So if we go up 2 here. And then we have, let me draw the top in a different color.
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Triple integrals 3 Double and triple integrals Multivariable Calculus Khan Academy.mp3
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Let me see if I can pull off drawing this. So if we go up 2 here. So if we go up 2 here. And then we have, 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.
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Triple integrals 3 Double and triple integrals Multivariable Calculus Khan Academy.mp3
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And then we have, 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.
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Triple integrals 3 Double and triple integrals Multivariable Calculus Khan Academy.mp3
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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 then 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.
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Triple integrals 3 Double and triple integrals Multivariable Calculus Khan Academy.mp3
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We'll go up 2 here. And then 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. But what we can do is, what 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.
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Triple integrals 3 Double and triple integrals Multivariable Calculus Khan Academy.mp3
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So this green triangle, this is part of the plane z is equal to 2. But what we can do is, what 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.
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Triple integrals 3 Double and triple integrals Multivariable Calculus Khan Academy.mp3
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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.
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Triple integrals 3 Double and triple integrals Multivariable Calculus Khan Academy.mp3
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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.
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Triple integrals 3 Double and triple integrals Multivariable Calculus Khan Academy.mp3
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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 use a triple integral, instead of just 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, it's a function of x, y, and z. It can be anything.
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Triple integrals 3 Double and triple integrals Multivariable Calculus Khan Academy.mp3
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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 use a triple integral, instead of just 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, it's 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, y, z.
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Triple integrals 3 Double and triple integrals Multivariable Calculus Khan Academy.mp3
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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, y, z. Our focus here 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.
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Triple integrals 3 Double and triple integrals Multivariable Calculus Khan Academy.mp3
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Let's say it's x squared, y, z. Our focus here 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.
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Triple integrals 3 Double and triple integrals Multivariable Calculus Khan Academy.mp3
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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 could 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.
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Triple integrals 3 Double and triple integrals Multivariable Calculus Khan Academy.mp3
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So if I had a little cube here in the volume under consideration, that's a little cube. You could 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 length, dy times dx, dx times dz. That's the volume of that little cube.
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Triple integrals 3 Double and triple integrals Multivariable Calculus Khan Academy.mp3
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No, sorry, this is dy. Let me do this in yellow or green even better. So dy, which is this length, 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 dm. The mass differential is going to be equal to that times that.
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Triple integrals 3 Double and triple integrals Multivariable Calculus Khan Academy.mp3
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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 dm. The mass differential is going to be equal to that times that. So x squared yz times this. dy dx 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.
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Triple integrals 3 Double and triple integrals Multivariable Calculus Khan Academy.mp3
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The mass differential is going to be equal to that times that. So x squared yz times this. dy dx 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.
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Triple integrals 3 Double and triple integrals Multivariable Calculus Khan Academy.mp3
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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 way the last couple 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.
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Triple integrals 3 Double and triple integrals Multivariable Calculus Khan Academy.mp3
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So let's do it traditionally. The way the last couple 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 them up, we're going to sum up all of the cubes in the z axis, so going up and down first. 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.
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Triple integrals 3 Double and triple integrals Multivariable Calculus Khan Academy.mp3
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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 them up, we're going to sum up all of the cubes in the z axis, so going up and down first. 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. So what is the bottom of the column, the bottom bound? Well, it's the surface. It's the surface defined right here.
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Triple integrals 3 Double and triple integrals Multivariable Calculus Khan Academy.mp3
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So when you sum up, up and down, these cubes are going to turn to columns. So what is the bottom of the column, the bottom bound? Well, it'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?
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Triple integrals 3 Double and triple integrals Multivariable Calculus Khan Academy.mp3
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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 to 2 minus 2 thirds x 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. We just algebraically manipulate it.
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Triple integrals 3 Double and triple integrals Multivariable Calculus Khan Academy.mp3
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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 to 2 minus 2 thirds x 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. We just algebraically manipulate it. So the bottom boundary, and you visualize it, right? The bottom of these columns, because we're going to go up and down, we're going to add up all the columns in an up and down direction, right? You can imagine summing them.
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Triple integrals 3 Double and triple integrals Multivariable Calculus Khan Academy.mp3
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We just algebraically manipulate it. So the bottom boundary, and you visualize it, right? The bottom of these columns, because we're going to go up and down, we're going to add up all the columns in an 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 thirds x minus y over 3. And then what's the upper bound?
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Triple integrals 3 Double and triple integrals Multivariable Calculus Khan Academy.mp3
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You can imagine summing them. The bottom boundary is going to be this surface. z is equal to 2 minus 2 thirds x 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 that green plane was? Well, it's z is equal to 2.
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Triple integrals 3 Double and triple integrals Multivariable Calculus Khan Academy.mp3
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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 that green plane was? Well, it's z is equal to 2. And that's this plane, this surface right here. z is equal to 2. And of course, what is the volume of that column?
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Triple integrals 3 Double and triple integrals Multivariable Calculus Khan Academy.mp3
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Well, it's z is equal to 2. And 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 y z times the volume differential, but we're integrating with respect to z first. Let me write dz there. And let's say we want to integrate with respect to x for next.
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Triple integrals 3 Double and triple integrals Multivariable Calculus Khan Academy.mp3
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And of course, what is the volume of that column? Well, it's going to be the density function, x squared y z times the volume differential, but we're integrating with respect to z first. Let me write dz there. And let's say we want to integrate with respect to x for next. In the last couple videos, I integrated with respect to y next. Well, let's do x, just to show you it really doesn't matter. So we're going to integrate with respect to x.
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Triple integrals 3 Double and triple integrals Multivariable Calculus Khan Academy.mp3
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And let's say we want to integrate with respect to x for next. In the last couple videos, I integrated with respect to y next. Well, 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, where the top boundary is that plane. Let's see if I can draw it decently.
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Triple integrals 3 Double and triple integrals Multivariable Calculus Khan Academy.mp3
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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, where 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. And now we want to integrate with respect to x.
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Triple integrals 3 Double and triple integrals Multivariable Calculus Khan Academy.mp3
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Let's see if I can draw it decently. The top boundary is that plane. The bottom boundary is this surface. And 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 x.
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Triple integrals 3 Double and triple integrals Multivariable Calculus Khan Academy.mp3
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And 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 x. 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've already integrated with respect to z.
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Triple integrals 3 Double and triple integrals Multivariable Calculus Khan Academy.mp3
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Well, this surface is defined all the way to x. 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've 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?
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Triple integrals 3 Double and triple integrals Multivariable Calculus Khan Academy.mp3
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We've 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, so if we take this y and flip it out like that, x like that, we'll get it in the traditional way that we learned when we first learned algebra. The xy axis.
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Triple integrals 3 Double and triple integrals Multivariable Calculus Khan Academy.mp3
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And what does that look like? So I will do that, because that actually does help simplify things. So if we twist it, so if we take this y and flip it out like that, x like that, we'll get it in 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.
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Triple integrals 3 Double and triple integrals Multivariable Calculus Khan Academy.mp3
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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.
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Triple integrals 3 Double and triple integrals Multivariable Calculus Khan Academy.mp3
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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.
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Triple integrals 3 Double and triple integrals Multivariable Calculus Khan Academy.mp3
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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. And so the xy axis, kind of the domain, you could view it that way, looks something like that. And 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 base of the column, where the column's going to pop out of your screen in the z direction, but the base of each column is going to be dx like that, and then dy up and down.
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Triple integrals 3 Double and triple integrals Multivariable Calculus Khan Academy.mp3
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So 1, 2, 3, 4, 5, 6. And so the xy axis, kind of the domain, you could view it that way, looks something like that. And 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 base of the column, where the column's going to pop out of your screen in the z direction, but the base of each column is going to be dx like that, and then dy up and down. 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?
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Triple integrals 3 Double and triple integrals Multivariable Calculus Khan Academy.mp3
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Where the base of the column, where the column's going to pop out of your screen in the z direction, but the base of each column is going to be dx like that, and then dy up and down. 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.
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Triple integrals 3 Double and triple integrals Multivariable Calculus Khan Academy.mp3
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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. And what's our top bound? I realize I'm already pushing it.
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Triple integrals 3 Double and triple integrals Multivariable Calculus Khan Academy.mp3
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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. And what's our top bound? I realize I'm already pushing it. Well, our top bound is this relation, but it has to be in terms of x. 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?
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Triple integrals 3 Double and triple integrals Multivariable Calculus Khan Academy.mp3
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I realize I'm already pushing it. Well, our top bound is this relation, but it has to be in terms of x. 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 if z is equal to 0, we have 2x plus y is equal to 6. We want the relationship in terms of x.
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Triple integrals 3 Double and triple integrals Multivariable Calculus Khan Academy.mp3
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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 if 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, or x is equal to 3 minus y over 2. 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 figured out, we've integrated up and down to get a column.
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Triple integrals 3 Double and triple integrals Multivariable Calculus Khan Academy.mp3
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We want the relationship in terms of x. So we get 2x is equal to 6 minus y, or x is equal to 3 minus y over 2. 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 figured out, we've integrated up and down to get a column. These are the bases of the column, and 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?
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Triple integrals 3 Double and triple integrals Multivariable Calculus Khan Academy.mp3
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So we've figured out, we've integrated up and down to get a column. These are the bases of the column, and 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.
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Triple integrals 3 Double and triple integrals Multivariable Calculus Khan Academy.mp3
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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.
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Triple integrals 3 Double and triple integrals Multivariable Calculus Khan Academy.mp3
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We've taken partial derivatives of non-vector valued functions before, where we only vary one of the variables. We only take it with respect to one variable. You hold the other one constant. We're going to do the exact same thing here. And we've taken regular derivatives of vector valued functions in the past, and those just ended up being the regular derivative of each of the terms. And we're going to see it's going to be the same thing here with the partial derivative. So let's define the partial derivative of r with respect to s. And everything I do with respect to s, you could just swap it with t, and you're going to get the same exact result.
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Partial derivatives of vector-valued functions Multivariable Calculus Khan Academy.mp3
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We're going to do the exact same thing here. And we've taken regular derivatives of vector valued functions in the past, and those just ended up being the regular derivative of each of the terms. And we're going to see it's going to be the same thing here with the partial derivative. So let's define the partial derivative of r with respect to s. And everything I do with respect to s, you could just swap it with t, and you're going to get the same exact result. So I'm going to define it as being equal to the limit as delta s approaches 0 of r of s plus delta s. We're only finding the limit with respect to a change in s comma t. We're holding t, you can imagine, constant for a given t, minus r of s and t, all of that over delta s. Now, if you do a little bit of algebra here, you literally, r of s plus delta s comma t, that's the same thing as x of s plus delta st, i, plus y of s plus delta st, j, plus z, all of that minus this thing. If you do a little bit of algebra with that, and if you don't believe me, try it out, this is going to be equal to the limit of delta s approaching 0, and I'm going to write it small because it can take up a lot of space, of x of s plus delta s comma t, minus x of s and t. I think you know where I'm going, this is all a little bit monotonous to write it all out, but never hurts. Times s, or divided by delta s, times i, and then I'll do it in different colors, so it's less monotonous, plus y.
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Partial derivatives of vector-valued functions Multivariable Calculus Khan Academy.mp3
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So let's define the partial derivative of r with respect to s. And everything I do with respect to s, you could just swap it with t, and you're going to get the same exact result. So I'm going to define it as being equal to the limit as delta s approaches 0 of r of s plus delta s. We're only finding the limit with respect to a change in s comma t. We're holding t, you can imagine, constant for a given t, minus r of s and t, all of that over delta s. Now, if you do a little bit of algebra here, you literally, r of s plus delta s comma t, that's the same thing as x of s plus delta st, i, plus y of s plus delta st, j, plus z, all of that minus this thing. If you do a little bit of algebra with that, and if you don't believe me, try it out, this is going to be equal to the limit of delta s approaching 0, and I'm going to write it small because it can take up a lot of space, of x of s plus delta s comma t, minus x of s and t. I think you know where I'm going, this is all a little bit monotonous to write it all out, but never hurts. Times s, or divided by delta s, times i, and then I'll do it in different colors, so it's less monotonous, plus y. This limit as delta approaches 0 applies to every term I'm writing out here. y of s plus delta s comma t, minus y of s comma t, all of that over delta s times j. And then finally, plus z of s plus delta s comma t, minus z of s and t, all of that over delta s times the z unit vector, k. And this all comes out of this definition.
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Partial derivatives of vector-valued functions Multivariable Calculus Khan Academy.mp3
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Times s, or divided by delta s, times i, and then I'll do it in different colors, so it's less monotonous, plus y. This limit as delta approaches 0 applies to every term I'm writing out here. y of s plus delta s comma t, minus y of s comma t, all of that over delta s times j. And then finally, plus z of s plus delta s comma t, minus z of s and t, all of that over delta s times the z unit vector, k. And this all comes out of this definition. If you literally just put s plus delta s in place for s, you evaluate all this, do a little algebra, you're going to get this exact same thing. And this hopefully pops out at you as, gee, we're just taking the partial derivative of each of these functions with respect to s. And these functions right here, this x of s and t, this is a non-vector valued function. This y, this is also a non-vector valued function.
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Partial derivatives of vector-valued functions Multivariable Calculus Khan Academy.mp3
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And then finally, plus z of s plus delta s comma t, minus z of s and t, all of that over delta s times the z unit vector, k. And this all comes out of this definition. If you literally just put s plus delta s in place for s, you evaluate all this, do a little algebra, you're going to get this exact same thing. And this hopefully pops out at you as, gee, we're just taking the partial derivative of each of these functions with respect to s. And these functions right here, this x of s and t, this is a non-vector valued function. This y, this is also a non-vector valued function. z is also a non-vector valued function. When you put them all together, it becomes a vector valued function, because we're multiplying the first one times a vector, the second one times another vector, the third one times another vector. But independently, these functions are non-vector valued.
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Partial derivatives of vector-valued functions Multivariable Calculus Khan Academy.mp3
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This y, this is also a non-vector valued function. z is also a non-vector valued function. When you put them all together, it becomes a vector valued function, because we're multiplying the first one times a vector, the second one times another vector, the third one times another vector. But independently, these functions are non-vector valued. So this is just the definition of the regular partial derivatives, where you're taking the limit as delta s approaches 0 in each of these cases. So this is the exact same thing. This is equal to, this is the exact same thing as the partial derivative of x with respect to s times i, plus the partial derivative of y with respect to s times j, plus the partial derivative of z with respect to s times k. I'm going to do one more thing here, and this is pseudo-mathy.
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Partial derivatives of vector-valued functions Multivariable Calculus Khan Academy.mp3
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But independently, these functions are non-vector valued. So this is just the definition of the regular partial derivatives, where you're taking the limit as delta s approaches 0 in each of these cases. So this is the exact same thing. This is equal to, this is the exact same thing as the partial derivative of x with respect to s times i, plus the partial derivative of y with respect to s times j, plus the partial derivative of z with respect to s times k. I'm going to do one more thing here, and this is pseudo-mathy. But it's going to come out, the whole reason I'm even doing this video is it's going to give us some good tools or in our toolkit for the videos that I'm about to do on surface integrals. So I'm going to do one thing here that's a little pseudo-mathy. And that's really because differentials are these things that are very hard to define rigorously.
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Partial derivatives of vector-valued functions Multivariable Calculus Khan Academy.mp3
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This is equal to, this is the exact same thing as the partial derivative of x with respect to s times i, plus the partial derivative of y with respect to s times j, plus the partial derivative of z with respect to s times k. I'm going to do one more thing here, and this is pseudo-mathy. But it's going to come out, the whole reason I'm even doing this video is it's going to give us some good tools or in our toolkit for the videos that I'm about to do on surface integrals. So I'm going to do one thing here that's a little pseudo-mathy. And that's really because differentials are these things that are very hard to define rigorously. But I think it'll give you the intuition of what's going on. So this thing right here, I'm going to say this is also equal to, and you're not going to see this in any math textbook. And hardcore mathematicians are going to kind of cringe when they see me do this.
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Partial derivatives of vector-valued functions Multivariable Calculus Khan Academy.mp3
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And that's really because differentials are these things that are very hard to define rigorously. But I think it'll give you the intuition of what's going on. So this thing right here, I'm going to say this is also equal to, and you're not going to see this in any math textbook. And hardcore mathematicians are going to kind of cringe when they see me do this. But I like to do it because I think it'll give you the intuition on what's going on when we take our surface integrals. So I'm going to say that this whole thing right here, that that is equal to r of s plus the differential of s, a super small change in s, t minus r of s and t, all of that over that same smooth, super small change in s. So hopefully you understand at least why I view things this way. When I take the limit as delta s approaches 0, these delta s's are going to get super duper duper small.
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Partial derivatives of vector-valued functions Multivariable Calculus Khan Academy.mp3
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And hardcore mathematicians are going to kind of cringe when they see me do this. But I like to do it because I think it'll give you the intuition on what's going on when we take our surface integrals. So I'm going to say that this whole thing right here, that that is equal to r of s plus the differential of s, a super small change in s, t minus r of s and t, all of that over that same smooth, super small change in s. So hopefully you understand at least why I view things this way. When I take the limit as delta s approaches 0, these delta s's are going to get super duper duper small. And in my head, that's how I imagine differentials. When someone writes the derivative of y with respect to x, and let's say that they say that is 2. And we've done a little bit of math with differentials before.
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Partial derivatives of vector-valued functions Multivariable Calculus Khan Academy.mp3
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When I take the limit as delta s approaches 0, these delta s's are going to get super duper duper small. And in my head, that's how I imagine differentials. When someone writes the derivative of y with respect to x, and let's say that they say that is 2. And we've done a little bit of math with differentials before. You can imagine multiplying both sides by dx, and you could get dy is equal to 2 dx. We've done this throughout calculus. The way I imagine it is super small change in y. Infinitely small change in y is equal to 2 times, you can imagine an equally small change in x.
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Partial derivatives of vector-valued functions Multivariable Calculus Khan Academy.mp3
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And we've done a little bit of math with differentials before. You can imagine multiplying both sides by dx, and you could get dy is equal to 2 dx. We've done this throughout calculus. The way I imagine it is super small change in y. Infinitely small change in y is equal to 2 times, you can imagine an equally small change in x. So if you have a super small change in x, your change in y is going to be still super small, but it's going to be 2 times that. I guess that's the best way to view it. But in general, I view differentials as super small changes in a variable.
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Partial derivatives of vector-valued functions Multivariable Calculus Khan Academy.mp3
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The way I imagine it is super small change in y. Infinitely small change in y is equal to 2 times, you can imagine an equally small change in x. So if you have a super small change in x, your change in y is going to be still super small, but it's going to be 2 times that. I guess that's the best way to view it. But in general, I view differentials as super small changes in a variable. So with that out of the way, and me explaining to you that many mathematicians would cringe at what I just wrote, hopefully this gives you a little, you know, this isn't like some crazy thing I did. I'm just saying, oh, delta s is delta s approaches 0. I kind of imagine that as ds.
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Partial derivatives of vector-valued functions Multivariable Calculus Khan Academy.mp3
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But in general, I view differentials as super small changes in a variable. So with that out of the way, and me explaining to you that many mathematicians would cringe at what I just wrote, hopefully this gives you a little, you know, this isn't like some crazy thing I did. I'm just saying, oh, delta s is delta s approaches 0. I kind of imagine that as ds. And the whole reason I did that is if you take this side and that side and multiply both sides times this differential ds, then what happens? The left hand side, you get the partial of r with respect to s is equal to this, oh, let me, times ds. I'll do ds in maybe pink.
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Partial derivatives of vector-valued functions Multivariable Calculus Khan Academy.mp3
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I kind of imagine that as ds. And the whole reason I did that is if you take this side and that side and multiply both sides times this differential ds, then what happens? The left hand side, you get the partial of r with respect to s is equal to this, oh, let me, times ds. I'll do ds in maybe pink. Times ds, this is just a regular differential, super small change in s. This was kind of a partial with respect to s. That's going to be equal to, well, if you multiply this side of the equation times ds, this guy's going to disappear. So it's going to be r of s plus our super small change in s. t minus r of s and t. And let me put a little square around this. This is going to be valuable for us in the next video.
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Partial derivatives of vector-valued functions Multivariable Calculus Khan Academy.mp3
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I'll do ds in maybe pink. Times ds, this is just a regular differential, super small change in s. This was kind of a partial with respect to s. That's going to be equal to, well, if you multiply this side of the equation times ds, this guy's going to disappear. So it's going to be r of s plus our super small change in s. t minus r of s and t. And let me put a little square around this. This is going to be valuable for us in the next video. We're going to actually think about what this means and how to visualize this on a surface. As you can imagine, this is a vector right here, right? You have two vector valued functions.
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Partial derivatives of vector-valued functions Multivariable Calculus Khan Academy.mp3
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This is going to be valuable for us in the next video. We're going to actually think about what this means and how to visualize this on a surface. As you can imagine, this is a vector right here, right? You have two vector valued functions. You're taking the difference. And we're going to visualize it in the next video. It's going to really help us with surface integrals.
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Partial derivatives of vector-valued functions Multivariable Calculus Khan Academy.mp3
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You have two vector valued functions. You're taking the difference. And we're going to visualize it in the next video. It's going to really help us with surface integrals. By the same exact logic, we can do everything we did here with s, we can do it with t as well. So we can define the partial. I'll draw a little.
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Partial derivatives of vector-valued functions Multivariable Calculus Khan Academy.mp3
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It's going to really help us with surface integrals. By the same exact logic, we can do everything we did here with s, we can do it with t as well. So we can define the partial. I'll draw a little. I can define the partial of r with respect. Let me do it in a different color, completely different color. It's orange.
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Partial derivatives of vector-valued functions Multivariable Calculus Khan Academy.mp3
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I'll draw a little. I can define the partial of r with respect. Let me do it in a different color, completely different color. It's orange. The partial of r with respect to t. The definition is just right here. The limit as delta t approaches 0 of r of s t plus delta t minus r of s and t. In this situation, we're holding the s. You can imagine it constant. We're finding its change in t. All of that over delta t. And the same thing falls out.
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Partial derivatives of vector-valued functions Multivariable Calculus Khan Academy.mp3
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It's orange. The partial of r with respect to t. The definition is just right here. The limit as delta t approaches 0 of r of s t plus delta t minus r of s and t. In this situation, we're holding the s. You can imagine it constant. We're finding its change in t. All of that over delta t. And the same thing falls out. This is equal to the partial of x with respect to t, i, plus y with respect to t, j, plus z with respect to t, k. Same exact thing. You just kind of swap the s's and the t's. And by that same logic, you'd have the same result but in terms of t. If you do this pseudo mathy thing that I did up here, then you would get the partial of r with respect to t times a super small change in t, dt, or t differential, you could imagine, is equal to r of s t plus dt minus r of s and t. So let's box these two guys away.
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Partial derivatives of vector-valued functions Multivariable Calculus Khan Academy.mp3
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We're finding its change in t. All of that over delta t. And the same thing falls out. This is equal to the partial of x with respect to t, i, plus y with respect to t, j, plus z with respect to t, k. Same exact thing. You just kind of swap the s's and the t's. And by that same logic, you'd have the same result but in terms of t. If you do this pseudo mathy thing that I did up here, then you would get the partial of r with respect to t times a super small change in t, dt, or t differential, you could imagine, is equal to r of s t plus dt minus r of s and t. So let's box these two guys away. And in the next video, we're going to actually visualize what these mean. And sometimes when you kind of do a bunch of silly math like this, you're always like, hey, whatever it's all about. Remember, all I did is I said, what does it mean to take the derivative of this with respect to s or t, played around with it a little bit.
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Partial derivatives of vector-valued functions Multivariable Calculus Khan Academy.mp3
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In all of the double integrals we've done so far, the boundaries on x and y were fixed. Now we'll see what happens when the boundaries on x and y are variable. So let's say I have the same surface. And I'm not going to draw it the way it looks. I'll just kind of draw it figuratively. But the problem we're actually going to do is z. And this is the exact same one we've been doing all along.
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Double integrals 5 Double and triple integrals Multivariable Calculus Khan Academy.mp3
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And I'm not going to draw it the way it looks. I'll just kind of draw it figuratively. But the problem we're actually going to do is z. And this is the exact same one we've been doing all along. But the point of here isn't to show you how to integrate. The point of here is to show you how to visualize and think about these problems. And frankly, in double integral problems, the hardest part is figuring out the boundaries.
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Double integrals 5 Double and triple integrals Multivariable Calculus Khan Academy.mp3
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And this is the exact same one we've been doing all along. But the point of here isn't to show you how to integrate. The point of here is to show you how to visualize and think about these problems. And frankly, in double integral problems, the hardest part is figuring out the boundaries. Once you do that, the integration is pretty straightforward. It's really not any harder than single variable integration. So let's say that's our surface.
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Double integrals 5 Double and triple integrals Multivariable Calculus Khan Academy.mp3
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And frankly, in double integral problems, the hardest part is figuring out the boundaries. Once you do that, the integration is pretty straightforward. It's really not any harder than single variable integration. So let's say that's our surface. z is equal to xy squared. And let me draw the axes again. So that's my x-axis.
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Double integrals 5 Double and triple integrals Multivariable Calculus Khan Academy.mp3
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So let's say that's our surface. z is equal to xy squared. And let me draw the axes again. So that's my x-axis. That's my z-axis. That's my y-axis, x, y, and z. And you saw what this graph looked like several videos ago.
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Double integrals 5 Double and triple integrals Multivariable Calculus Khan Academy.mp3
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So that's my x-axis. That's my z-axis. That's my y-axis, x, y, and z. And you saw what this graph looked like several videos ago. I took out the whole grapher and we rotated and things. I'm not going to draw the graph the way it looks. I'm just going to draw it fairly abstractly as just an abstract surface.
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Double integrals 5 Double and triple integrals Multivariable Calculus Khan Academy.mp3
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And you saw what this graph looked like several videos ago. I took out the whole grapher and we rotated and things. I'm not going to draw the graph the way it looks. I'm just going to draw it fairly abstractly as just an abstract surface. Because the point here is really to figure out the boundaries of integration. Before I actually even draw the surface, I'm going to draw the boundaries. So the first time we did this problem, we said, OK, x goes from 0 to 2, y goes from 0 to 1, and then we figured out the volume above that bounded domain.
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Double integrals 5 Double and triple integrals Multivariable Calculus Khan Academy.mp3
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I'm just going to draw it fairly abstractly as just an abstract surface. Because the point here is really to figure out the boundaries of integration. Before I actually even draw the surface, I'm going to draw the boundaries. So the first time we did this problem, we said, OK, x goes from 0 to 2, y goes from 0 to 1, and then we figured out the volume above that bounded domain. Now let's do something else. Let's figure out, let's say that x goes from 0 to 1. So x goes from 0 to 1.
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Double integrals 5 Double and triple integrals Multivariable Calculus Khan Academy.mp3
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So the first time we did this problem, we said, OK, x goes from 0 to 2, y goes from 0 to 1, and then we figured out the volume above that bounded domain. Now let's do something else. Let's figure out, let's say that x goes from 0 to 1. So x goes from 0 to 1. And let's say that the volume that we want to figure out under the surface, it's not from a fixed y to an upper bound y. Or the kind of, well, I'll show you. It's actually a curve.
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Double integrals 5 Double and triple integrals Multivariable Calculus Khan Academy.mp3
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So x goes from 0 to 1. And let's say that the volume that we want to figure out under the surface, it's not from a fixed y to an upper bound y. Or the kind of, well, I'll show you. It's actually a curve. So this is all on the xy-plane, everything I'm drawing here. And this curve, we could view it two ways. We could say y is a function of x, y is equal to x squared.
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Double integrals 5 Double and triple integrals Multivariable Calculus Khan Academy.mp3
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It's actually a curve. So this is all on the xy-plane, everything I'm drawing here. And this curve, we could view it two ways. We could say y is a function of x, y is equal to x squared. Or we could write x is equal to square root of y. We don't have to write plus or minus or anything like that, because we're in the first quadrant. So this is the area above which we want to figure out the volume.
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Double integrals 5 Double and triple integrals Multivariable Calculus Khan Academy.mp3
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We could say y is a function of x, y is equal to x squared. Or we could write x is equal to square root of y. We don't have to write plus or minus or anything like that, because we're in the first quadrant. So this is the area above which we want to figure out the volume. Let me, yeah, it doesn't hurt to color it in, just so we can really hone in on what we care about. So that's the area above which we want to figure out the volume. Or you could kind of say that's our bounded domain.
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Double integrals 5 Double and triple integrals Multivariable Calculus Khan Academy.mp3
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So this is the area above which we want to figure out the volume. Let me, yeah, it doesn't hurt to color it in, just so we can really hone in on what we care about. So that's the area above which we want to figure out the volume. Or you could kind of say that's our bounded domain. And so x goes from 0 to 1, and then this point is going to be what? That point's going to be 1 comma 1, right? 1 is equal to 1 squared, 1 is equal to square root of 1.
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Double integrals 5 Double and triple integrals Multivariable Calculus Khan Academy.mp3
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Or you could kind of say that's our bounded domain. And so x goes from 0 to 1, and then this point is going to be what? That point's going to be 1 comma 1, right? 1 is equal to 1 squared, 1 is equal to square root of 1. So this point is y is equal to 1. That's y is equal to 1. And then I'm not going to draw this surface exactly.
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Double integrals 5 Double and triple integrals Multivariable Calculus Khan Academy.mp3
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1 is equal to 1 squared, 1 is equal to square root of 1. So this point is y is equal to 1. That's y is equal to 1. And then I'm not going to draw this surface exactly. I'm just trying to give you a sense of what the volume of the figure we're trying to calculate is. So the top of, if this is just some arbitrary surface, let me do it in a different color. So this is the top.
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Double integrals 5 Double and triple integrals Multivariable Calculus Khan Academy.mp3
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