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You could kind of view it the area under the curve times a dx, so we have some depth here. So it's times dx. And if we want to figure out the entire volume under this surface, between the surface and the xy plane, given this constraint to our domain, we just integrate from x is equal to 0 to 2. All right. So let's think about it. This area, this area in green here that we started with, that should be a function of x. We held x constant, but depending on which x you pick, this area is going to change. | Double integrals 3 Double and triple integrals Multivariable Calculus Khan Academy.mp3 |
All right. So let's think about it. This area, this area in green here that we started with, that should be a function of x. We held x constant, but depending on which x you pick, this area is going to change. So when we evaluate this magenta inner integral with respect to y, we should get a function of x. And then when we evaluate the whole thing, we'll get our volume. So let's do it. | Double integrals 3 Double and triple integrals Multivariable Calculus Khan Academy.mp3 |
We held x constant, but depending on which x you pick, this area is going to change. So when we evaluate this magenta inner integral with respect to y, we should get a function of x. And then when we evaluate the whole thing, we'll get our volume. So let's do it. Let's evaluate this inner integral. Hold x constant. What's the antiderivative of y squared? | Double integrals 3 Double and triple integrals Multivariable Calculus Khan Academy.mp3 |
So let's do it. Let's evaluate this inner integral. Hold x constant. What's the antiderivative of y squared? It's y to the third over 3. So it's y to the third over 3. The x is a constant, right? | Double integrals 3 Double and triple integrals Multivariable Calculus Khan Academy.mp3 |
What's the antiderivative of y squared? It's y to the third over 3. So it's y to the third over 3. The x is a constant, right? And we're going to evaluate that at 1 and at 0. And the outer integral is still with respect to x dx. This is equal to, let's see, when you evaluate y is equal to 1, you get 1 to the third. | Double integrals 3 Double and triple integrals Multivariable Calculus Khan Academy.mp3 |
The x is a constant, right? And we're going to evaluate that at 1 and at 0. And the outer integral is still with respect to x dx. This is equal to, let's see, when you evaluate y is equal to 1, you get 1 to the third. That's 1. So it's x over 3 minus, when y is 0, then that whole thing just becomes 0, right? So this purple expression is just x over 3. | Double integrals 3 Double and triple integrals Multivariable Calculus Khan Academy.mp3 |
This is equal to, let's see, when you evaluate y is equal to 1, you get 1 to the third. That's 1. So it's x over 3 minus, when y is 0, then that whole thing just becomes 0, right? So this purple expression is just x over 3. And then we still have the outside integral from 0 to 2 dx. So given what x we have, the area of this green surface, that was where we started, given any given x, that area, I wanted something with some contrast. This area is x over 3, depending on which x you pick. | Double integrals 3 Double and triple integrals Multivariable Calculus Khan Academy.mp3 |
So this purple expression is just x over 3. And then we still have the outside integral from 0 to 2 dx. So given what x we have, the area of this green surface, that was where we started, given any given x, that area, I wanted something with some contrast. This area is x over 3, depending on which x you pick. If x is 1, this area right here is 1 third, right? But now we're going to integrate underneath the entire surface and get our volume. And like I said, when you integrate it, it's a function of x. | Double integrals 3 Double and triple integrals Multivariable Calculus Khan Academy.mp3 |
This area is x over 3, depending on which x you pick. If x is 1, this area right here is 1 third, right? But now we're going to integrate underneath the entire surface and get our volume. And like I said, when you integrate it, it's a function of x. So let's do that. And this is just plain old vanilla standard integral. So what's the antiderivative of x? | Double integrals 3 Double and triple integrals Multivariable Calculus Khan Academy.mp3 |
And like I said, when you integrate it, it's a function of x. So let's do that. And this is just plain old vanilla standard integral. So what's the antiderivative of x? It's x squared over 2. We have a 1 third there. So it equals x squared over 2 times 3. | Double integrals 3 Double and triple integrals Multivariable Calculus Khan Academy.mp3 |
So what's the antiderivative of x? It's x squared over 2. We have a 1 third there. So it equals x squared over 2 times 3. So x squared over 6. We're going to evaluate it at 2 and at 0. 2 squared over 6 is 4 6 minus 0 over 6, which is equal to 0, equals 4 6. | Double integrals 3 Double and triple integrals Multivariable Calculus Khan Academy.mp3 |
So it equals x squared over 2 times 3. So x squared over 6. We're going to evaluate it at 2 and at 0. 2 squared over 6 is 4 6 minus 0 over 6, which is equal to 0, equals 4 6. What is 4 6? Well, that's just the same thing as 2 thirds. So the volume under the surface is 2 thirds. | Double integrals 3 Double and triple integrals Multivariable Calculus Khan Academy.mp3 |
2 squared over 6 is 4 6 minus 0 over 6, which is equal to 0, equals 4 6. What is 4 6? Well, that's just the same thing as 2 thirds. So the volume under the surface is 2 thirds. And if you watched the previous video, you will appreciate the fact that when we integrated the other way around, when we did it with respect to x first and then y, we got the exact same answer. So the universe is in proper working order. And I've surprisingly actually finished this video with extra time. | Double integrals 3 Double and triple integrals Multivariable Calculus Khan Academy.mp3 |
So the volume under the surface is 2 thirds. And if you watched the previous video, you will appreciate the fact that when we integrated the other way around, when we did it with respect to x first and then y, we got the exact same answer. So the universe is in proper working order. And I've surprisingly actually finished this video with extra time. So for fun, we can just spin this graph and just appreciate the fact that we have figured out the volume between this surface, xy squared, and the xy plane. Pretty neat. Anyway, I'll see you in the next video. | Double integrals 3 Double and triple integrals Multivariable Calculus Khan Academy.mp3 |
So this is F dot dr, where our path is this boundary right over here, this path C. We had expressed it in terms of a line integral around this boundary, around path C1, which is the boundary of region R. And the reason why this is going to be valuable to us is now we can directly apply Green's theorem to this to turn this into a double integral over this region right over here, the region that it is actually bounding. So let's actually do that. We're just applying Green's theorem here. So Green's theorem. Actually, let me box this off right over here. This was just something that we wrote to remind ourselves in the last video. So let me just box it off right over here. | Stokes' theorem proof part 7 Multivariable Calculus Khan Academy.mp3 |
So Green's theorem. Actually, let me box this off right over here. This was just something that we wrote to remind ourselves in the last video. So let me just box it off right over here. When you apply Green's theorem, you get that this is going to be the exact same thing as the double integral over the region that C1 bounds, that region R that's in our xy-plane, of the partial of this with respect to x. So the partial with respect to x of this business. So I'll do that in that same green color. | Stokes' theorem proof part 7 Multivariable Calculus Khan Academy.mp3 |
So let me just box it off right over here. When you apply Green's theorem, you get that this is going to be the exact same thing as the double integral over the region that C1 bounds, that region R that's in our xy-plane, of the partial of this with respect to x. So the partial with respect to x of this business. So I'll do that in that same green color. q plus r times the partial of z with respect to y minus the partial with respect to y of p plus r times the partial of z with respect to x. p plus r times the partial of z with respect to x. And then dA, a little differential of our region. And so let's actually calculate this. | Stokes' theorem proof part 7 Multivariable Calculus Khan Academy.mp3 |
So I'll do that in that same green color. q plus r times the partial of z with respect to y minus the partial with respect to y of p plus r times the partial of z with respect to x. p plus r times the partial of z with respect to x. And then dA, a little differential of our region. And so let's actually calculate this. You take the partials of each of these expressions. And we'll see that if we expand them and then simplify, we'll get something very similar, actually, hopefully identical to this right over here. And that'll show that this line integral for the special case is the same thing as this surface integral, which will prove Stokes' theorem for this special case. | Stokes' theorem proof part 7 Multivariable Calculus Khan Academy.mp3 |
And so let's actually calculate this. You take the partials of each of these expressions. And we'll see that if we expand them and then simplify, we'll get something very similar, actually, hopefully identical to this right over here. And that'll show that this line integral for the special case is the same thing as this surface integral, which will prove Stokes' theorem for this special case. So let's do it. So let's just apply the partial derivative operator. So first, we want to take the partial derivative with respect to q. | Stokes' theorem proof part 7 Multivariable Calculus Khan Academy.mp3 |
And that'll show that this line integral for the special case is the same thing as this surface integral, which will prove Stokes' theorem for this special case. So let's do it. So let's just apply the partial derivative operator. So first, we want to take the partial derivative with respect to q. And we need to remind ourselves, and we did this way up here, when we first thought about it, we saw that q, well, p, q, and r, they're each functions of x, y, and z. So we're assuming that's how they're represented. And if z was not a function of x, then if we were to take the partial of q with respect to x, we would just write it as a partial of q with respect to x. | Stokes' theorem proof part 7 Multivariable Calculus Khan Academy.mp3 |
So first, we want to take the partial derivative with respect to q. And we need to remind ourselves, and we did this way up here, when we first thought about it, we saw that q, well, p, q, and r, they're each functions of x, y, and z. So we're assuming that's how they're represented. And if z was not a function of x, then if we were to take the partial of q with respect to x, we would just write it as a partial of q with respect to x. But we know that we've assumed that z is itself a function of x and y. So if we're taking the partial with respect to x over here, we have to think, first, well, how can q directly change with respect to x? And then how can it change due to something else changing due to x? | Stokes' theorem proof part 7 Multivariable Calculus Khan Academy.mp3 |
And if z was not a function of x, then if we were to take the partial of q with respect to x, we would just write it as a partial of q with respect to x. But we know that we've assumed that z is itself a function of x and y. So if we're taking the partial with respect to x over here, we have to think, first, well, how can q directly change with respect to x? And then how can it change due to something else changing due to x? And so that other thing that could change due to x is z. y is independent of x. But z is a function of x. So let's keep that in mind. | Stokes' theorem proof part 7 Multivariable Calculus Khan Academy.mp3 |
And then how can it change due to something else changing due to x? And so that other thing that could change due to x is z. y is independent of x. But z is a function of x. So let's keep that in mind. So we're really going to do the multivariable chain rule. So when we try to take the derivative of this part, we're taking, essentially, of this whole function with respect to x, we have to think about how will q change directly with respect to x. And to that, we have to add how q could change due to changes in other variables due to changes in x. | Stokes' theorem proof part 7 Multivariable Calculus Khan Academy.mp3 |
So let's keep that in mind. So we're really going to do the multivariable chain rule. So when we try to take the derivative of this part, we're taking, essentially, of this whole function with respect to x, we have to think about how will q change directly with respect to x. And to that, we have to add how q could change due to changes in other variables due to changes in x. And the other only variable that q is a function of that could change due to a change in x is z. So q could also change to z because gz has changed due to x. So operated here is the partial with respect to x plus the partial of q with respect to z times a partial of z with respect to x. | Stokes' theorem proof part 7 Multivariable Calculus Khan Academy.mp3 |
And to that, we have to add how q could change due to changes in other variables due to changes in x. And the other only variable that q is a function of that could change due to a change in x is z. So q could also change to z because gz has changed due to x. So operated here is the partial with respect to x plus the partial of q with respect to z times a partial of z with respect to x. If we rewrote q so that z was substituted with x's and y's because it is a function of x and y's, then we would just have to write this first term right over here. But we're assuming this is expressed as a function of x, y, and z. And z itself is a function of x. | Stokes' theorem proof part 7 Multivariable Calculus Khan Academy.mp3 |
So operated here is the partial with respect to x plus the partial of q with respect to z times a partial of z with respect to x. If we rewrote q so that z was substituted with x's and y's because it is a function of x and y's, then we would just have to write this first term right over here. But we're assuming this is expressed as a function of x, y, and z. And z itself is a function of x. And so that's why we had to use the multivariable chain rule. Now let's move to the next part. And both of these might have some x's in them. | Stokes' theorem proof part 7 Multivariable Calculus Khan Academy.mp3 |
And z itself is a function of x. And so that's why we had to use the multivariable chain rule. Now let's move to the next part. And both of these might have some x's in them. So we have to use the product rule right over here. So first we can take the derivative of r with respect to x and then multiply that times z sub y. Then we have to take the derivative of z sub y with respect to x and multiply that times r. So this is going to be plus. | Stokes' theorem proof part 7 Multivariable Calculus Khan Academy.mp3 |
And both of these might have some x's in them. So we have to use the product rule right over here. So first we can take the derivative of r with respect to x and then multiply that times z sub y. Then we have to take the derivative of z sub y with respect to x and multiply that times r. So this is going to be plus. So if we take the derivative of this with respect to x, same exact logic. r can change directly due to x. And it can change due to y. | Stokes' theorem proof part 7 Multivariable Calculus Khan Academy.mp3 |
Then we have to take the derivative of z sub y with respect to x and multiply that times r. So this is going to be plus. So if we take the derivative of this with respect to x, same exact logic. r can change directly due to x. And it can change due to y. And it can change due to z. And multiply that times how z could change due to x. Once again, you could use this as the multivariable chain rule in action here. | Stokes' theorem proof part 7 Multivariable Calculus Khan Academy.mp3 |
And it can change due to y. And it can change due to z. And multiply that times how z could change due to x. Once again, you could use this as the multivariable chain rule in action here. But of course, we take the derivative of the first term times the second term. Do the second term in magenta. Times the second term. | Stokes' theorem proof part 7 Multivariable Calculus Khan Academy.mp3 |
Once again, you could use this as the multivariable chain rule in action here. But of course, we take the derivative of the first term times the second term. Do the second term in magenta. Times the second term. So the partial of z with respect to y. Plus the derivative of the second term, which is the partial of z with respect to y. And then taking the partial of that with respect to x, which we could just write as that, times the first term. | Stokes' theorem proof part 7 Multivariable Calculus Khan Academy.mp3 |
Times the second term. So the partial of z with respect to y. Plus the derivative of the second term, which is the partial of z with respect to y. And then taking the partial of that with respect to x, which we could just write as that, times the first term. Times our r. So that's the partial with respect to x of all of this business right over here. And then we need to subtract the partial of this with respect to y. And we're going to use the exact same logic. | Stokes' theorem proof part 7 Multivariable Calculus Khan Academy.mp3 |
And then taking the partial of that with respect to x, which we could just write as that, times the first term. Times our r. So that's the partial with respect to x of all of this business right over here. And then we need to subtract the partial of this with respect to y. And we're going to use the exact same logic. So then we're going to subtract. And I'll put it in parentheses like that. So p could change directly due to y. I will circle p. Let me do it with a color that I haven't used yet. | Stokes' theorem proof part 7 Multivariable Calculus Khan Academy.mp3 |
And we're going to use the exact same logic. So then we're going to subtract. And I'll put it in parentheses like that. So p could change directly due to y. I will circle p. Let me do it with a color that I haven't used yet. p could change directly due to y. So this is the partial of p with respect to y. But it could also change due to z changing because of y. | Stokes' theorem proof part 7 Multivariable Calculus Khan Academy.mp3 |
So p could change directly due to y. I will circle p. Let me do it with a color that I haven't used yet. p could change directly due to y. So this is the partial of p with respect to y. But it could also change due to z changing because of y. So plus the partial with respect to z times the partial of z with respect to y. Plus, and I'll do r maybe in that same color, plus the derivative of r. Well, we already figured that. But actually now it's with respect to x, not with respect to y. I have to be careful. | Stokes' theorem proof part 7 Multivariable Calculus Khan Academy.mp3 |
But it could also change due to z changing because of y. So plus the partial with respect to z times the partial of z with respect to y. Plus, and I'll do r maybe in that same color, plus the derivative of r. Well, we already figured that. But actually now it's with respect to x, not with respect to y. I have to be careful. So it's going to be the partial of r with respect to y plus the partial of r with respect to z times the partial of z with respect to y times z sub x. Plus now we take the derivative of the second term times the first term. The derivative of the partial of z with respect to x then with respect to y is going to be z. | Stokes' theorem proof part 7 Multivariable Calculus Khan Academy.mp3 |
But actually now it's with respect to x, not with respect to y. I have to be careful. So it's going to be the partial of r with respect to y plus the partial of r with respect to z times the partial of z with respect to y times z sub x. Plus now we take the derivative of the second term times the first term. The derivative of the partial of z with respect to x then with respect to y is going to be z. So the partial of z with respect to x and we're going to take the partial of that with respect and then we're going to multiply that times r. Then we're going to multiply that times r. Now let's see if we can expand this out and hopefully things simplify. Just a reminder, I'm just working on the inside of this double integral. I'll rewrite the outside, the double integral and the dA once I get this all cleaned up a little bit. | Stokes' theorem proof part 7 Multivariable Calculus Khan Academy.mp3 |
The derivative of the partial of z with respect to x then with respect to y is going to be z. So the partial of z with respect to x and we're going to take the partial of that with respect and then we're going to multiply that times r. Then we're going to multiply that times r. Now let's see if we can expand this out and hopefully things simplify. Just a reminder, I'm just working on the inside of this double integral. I'll rewrite the outside, the double integral and the dA once I get this all cleaned up a little bit. Let me rewrite it a little bit. This is equal to the partial of q. I'll try to color code the same way. This is just algebra at this point. | Stokes' theorem proof part 7 Multivariable Calculus Khan Academy.mp3 |
I'll rewrite the outside, the double integral and the dA once I get this all cleaned up a little bit. Let me rewrite it a little bit. This is equal to the partial of q. I'll try to color code the same way. This is just algebra at this point. The partial of q with respect to x plus the partial of q with respect to z times the partial of z with respect to x plus the partial of r with respect to x times the partial of z with respect to y plus the partial of r with respect to z times the partial of z with respect to x times the partial of z with respect to y and then we have this term right over here, which I'll just do in purple, plus the partial of z with respect to y and then with respect to x times r. Now we're going to subtract all of this business right over here. So minus the partial of p with respect to y minus the partial of p with respect to z times the partial of z with respect to y and then we're going to subtract from that minus the partial of r with respect to y times the partial of z with respect to x minus the partial of r where we need to go. The partial of r with respect to z times the partial of z with respect to y times the partial of z with respect to x. | Stokes' theorem proof part 7 Multivariable Calculus Khan Academy.mp3 |
This is just algebra at this point. The partial of q with respect to x plus the partial of q with respect to z times the partial of z with respect to x plus the partial of r with respect to x times the partial of z with respect to y plus the partial of r with respect to z times the partial of z with respect to x times the partial of z with respect to y and then we have this term right over here, which I'll just do in purple, plus the partial of z with respect to y and then with respect to x times r. Now we're going to subtract all of this business right over here. So minus the partial of p with respect to y minus the partial of p with respect to z times the partial of z with respect to y and then we're going to subtract from that minus the partial of r with respect to y times the partial of z with respect to x minus the partial of r where we need to go. The partial of r with respect to z times the partial of z with respect to y times the partial of z with respect to x. And then finally, this term right over here minus this, because we have that negative out there, minus the partial of z with respect to x and then with respect to y, r. Now let's see if we can simplify things. So the first thing, this and this look to be the same. We just can commute the order in which we actually multiply. | Stokes' theorem proof part 7 Multivariable Calculus Khan Academy.mp3 |
The partial of r with respect to z times the partial of z with respect to y times the partial of z with respect to x. And then finally, this term right over here minus this, because we have that negative out there, minus the partial of z with respect to x and then with respect to y, r. Now let's see if we can simplify things. So the first thing, this and this look to be the same. We just can commute the order in which we actually multiply. But these are the exact same terms. So that is going to cancel out with that. And because we assumed that the second, because we assumed way up here that we have continuous second derivatives of the function z, z is a function of x and y, that that is equal to that, we can now say that these two right over here are going to be the negatives of each other or that they are going to cancel out. | Stokes' theorem proof part 7 Multivariable Calculus Khan Academy.mp3 |
We just can commute the order in which we actually multiply. But these are the exact same terms. So that is going to cancel out with that. And because we assumed that the second, because we assumed way up here that we have continuous second derivatives of the function z, z is a function of x and y, that that is equal to that, we can now say that these two right over here are going to be the negatives of each other or that they are going to cancel out. And so this simplified things a good bit. And let's see if I can write it in a way, let me group terms in a way that might start to make sense. And actually, I'm going to try to see if I can make them similar to this. | Stokes' theorem proof part 7 Multivariable Calculus Khan Academy.mp3 |
And because we assumed that the second, because we assumed way up here that we have continuous second derivatives of the function z, z is a function of x and y, that that is equal to that, we can now say that these two right over here are going to be the negatives of each other or that they are going to cancel out. And so this simplified things a good bit. And let's see if I can write it in a way, let me group terms in a way that might start to make sense. And actually, I'm going to try to see if I can make them similar to this. So I have all the terms that have a z sub x and the z sub y and then the rest of them. So z sub x I'll do in blue. So you have the terms that have a z sub x in it. | Stokes' theorem proof part 7 Multivariable Calculus Khan Academy.mp3 |
And actually, I'm going to try to see if I can make them similar to this. So I have all the terms that have a z sub x and the z sub y and then the rest of them. So z sub x I'll do in blue. So you have the terms that have a z sub x in it. So you have this term right over here and this term right over here. We can factor out the z sub x and we get the partial of z with respect to x times the partial of q with respect to z minus the partial of r with respect to y. And then let's do the, and I want to get the colors the same way too. | Stokes' theorem proof part 7 Multivariable Calculus Khan Academy.mp3 |
So you have the terms that have a z sub x in it. So you have this term right over here and this term right over here. We can factor out the z sub x and we get the partial of z with respect to x times the partial of q with respect to z minus the partial of r with respect to y. And then let's do the, and I want to get the colors the same way too. I did yellow next. Plus I have all the terms with the partial of z with respect to y. So it's that term and that term right over here. | Stokes' theorem proof part 7 Multivariable Calculus Khan Academy.mp3 |
And then let's do the, and I want to get the colors the same way too. I did yellow next. Plus I have all the terms with the partial of z with respect to y. So it's that term and that term right over here. So it becomes plus the partial of z with respect to y times the partial of r with respect to x minus the partial of p with respect to z. And then we have these last two terms. And I used the color green up there. | Stokes' theorem proof part 7 Multivariable Calculus Khan Academy.mp3 |
So it's that term and that term right over here. So it becomes plus the partial of z with respect to y times the partial of r with respect to x minus the partial of p with respect to z. And then we have these last two terms. And I used the color green up there. So I'll use the color green again. So for these two terms, I'll just write plus the partial of q with respect to x minus the partial of p with respect to y. So our double integrals has kind of, I can't really say simplified, but we can rewrite it like this. | Stokes' theorem proof part 7 Multivariable Calculus Khan Academy.mp3 |
And I used the color green up there. So I'll use the color green again. So for these two terms, I'll just write plus the partial of q with respect to x minus the partial of p with respect to y. So our double integrals has kind of, I can't really say simplified, but we can rewrite it like this. And we don't want to forget this was all because this was all a simplification of our double integral over the region dA. This is what we have been able to, using Green's theorem and the multivariable chain rule and whatever else, we've been able to say that that line integral around the boundary of our surface is the same thing as this. And now we can compare that to what our surface integral was. | Stokes' theorem proof part 7 Multivariable Calculus Khan Academy.mp3 |
So our double integrals has kind of, I can't really say simplified, but we can rewrite it like this. And we don't want to forget this was all because this was all a simplification of our double integral over the region dA. This is what we have been able to, using Green's theorem and the multivariable chain rule and whatever else, we've been able to say that that line integral around the boundary of our surface is the same thing as this. And now we can compare that to what our surface integral was. Let's see if I have space. So copy, and then let me see if I have space up here to paste it. Well, it doesn't look like I actually have much space to paste it, although I'll try anyway. | Stokes' theorem proof part 7 Multivariable Calculus Khan Academy.mp3 |
And now we can compare that to what our surface integral was. Let's see if I have space. So copy, and then let me see if I have space up here to paste it. Well, it doesn't look like I actually have much space to paste it, although I'll try anyway. So if I paste it, you see that they are identical. They are identical. Our line integral is identical to this. | Stokes' theorem proof part 7 Multivariable Calculus Khan Academy.mp3 |
Well, it doesn't look like I actually have much space to paste it, although I'll try anyway. So if I paste it, you see that they are identical. They are identical. Our line integral is identical to this. We get the exact same thing. So our line integral, f dot dr around this pat c simplified to this, and our surface integral simplified to this. So using the assumptions we had, they both simplified to the same thing. | Stokes' theorem proof part 7 Multivariable Calculus Khan Academy.mp3 |
And to do that, I will assume that our surface can be parameterized by the position vector function, r. And r is a function of two parameters. It's a function of u, and it is a function of v. You give me a u and a v, and that will essentially specify a point on this two-dimensional surface right over here. It could be bent, so it kind of exists in three-dimensional space. But a u and a v will specify a given point on this surface. Now, let's think about what the directions of r, the partial of r with respect to u looks like, and what the direction of the partial of r with respect to v looks like. So let's say that we're at some point u, v. So for some u, v, if you find the position vector, it takes us to that point on the surface right over there. Now, let's say that we increment u just a little bit. | Constructing a unit normal vector to a surface Multivariable Calculus Khan Academy.mp3 |
But a u and a v will specify a given point on this surface. Now, let's think about what the directions of r, the partial of r with respect to u looks like, and what the direction of the partial of r with respect to v looks like. So let's say that we're at some point u, v. So for some u, v, if you find the position vector, it takes us to that point on the surface right over there. Now, let's say that we increment u just a little bit. And as we increment u just a little bit, we're going to get to another point on our surface. And let's say that other point on the surface is right over there. So what would this r sub u vector look like? | Constructing a unit normal vector to a surface Multivariable Calculus Khan Academy.mp3 |
Now, let's say that we increment u just a little bit. And as we increment u just a little bit, we're going to get to another point on our surface. And let's say that other point on the surface is right over there. So what would this r sub u vector look like? Well, its magnitude is essentially going to be dependent on how fast it's happening, how fast we're moving towards that point. But its direction is going to be in that direction. It's going to be towards that point. | Constructing a unit normal vector to a surface Multivariable Calculus Khan Academy.mp3 |
So what would this r sub u vector look like? Well, its magnitude is essentially going to be dependent on how fast it's happening, how fast we're moving towards that point. But its direction is going to be in that direction. It's going to be towards that point. It's going to be along the surface. We're going from one point of the surface to another. It's essentially going to be tangent to the surface at that point. | Constructing a unit normal vector to a surface Multivariable Calculus Khan Academy.mp3 |
It's going to be towards that point. It's going to be along the surface. We're going from one point of the surface to another. It's essentially going to be tangent to the surface at that point. And I could draw it a little bit bigger. It would look something like that, r sub u. So I've just zoomed in right over here. | Constructing a unit normal vector to a surface Multivariable Calculus Khan Academy.mp3 |
It's essentially going to be tangent to the surface at that point. And I could draw it a little bit bigger. It would look something like that, r sub u. So I've just zoomed in right over here. Now, let's go back to this point. And now, let's make v a little bit bigger. Now, let's say if we make v a little bit bigger, we go to this point right over here. | Constructing a unit normal vector to a surface Multivariable Calculus Khan Academy.mp3 |
So I've just zoomed in right over here. Now, let's go back to this point. And now, let's make v a little bit bigger. Now, let's say if we make v a little bit bigger, we go to this point right over here. So then our position vector r would point to this point. And so what would r sub v look like? Well, its magnitude, once again, would be dependent on how fast we're going there. | Constructing a unit normal vector to a surface Multivariable Calculus Khan Academy.mp3 |
Now, let's say if we make v a little bit bigger, we go to this point right over here. So then our position vector r would point to this point. And so what would r sub v look like? Well, its magnitude, once again, would be dependent on how fast we're going there. But the direction is what's interesting. The direction would also be tangential to the surface. We're going from one point on the surface to another as we change v. So r sub v might look something like that. | Constructing a unit normal vector to a surface Multivariable Calculus Khan Academy.mp3 |
Well, its magnitude, once again, would be dependent on how fast we're going there. But the direction is what's interesting. The direction would also be tangential to the surface. We're going from one point on the surface to another as we change v. So r sub v might look something like that. And these two aren't necessarily perpendicular to each other. In fact, the way I drew them, they're not perpendicular. So r sub v is looking like this. | Constructing a unit normal vector to a surface Multivariable Calculus Khan Academy.mp3 |
We're going from one point on the surface to another as we change v. So r sub v might look something like that. And these two aren't necessarily perpendicular to each other. In fact, the way I drew them, they're not perpendicular. So r sub v is looking like this. But they're both tangential to the plane. They're both essentially telling us right at that point, what is the tangent, what is the slope in the u direction or in the v direction? Now, when you have two vectors that are tangential to the plane and they're not the same vector, these are actually already specifying, these are already kind of determining a plane. | Constructing a unit normal vector to a surface Multivariable Calculus Khan Academy.mp3 |
So r sub v is looking like this. But they're both tangential to the plane. They're both essentially telling us right at that point, what is the tangent, what is the slope in the u direction or in the v direction? Now, when you have two vectors that are tangential to the plane and they're not the same vector, these are actually already specifying, these are already kind of determining a plane. And so you can imagine a plane that looks something like this. If you took linear combinations of these two things, you would get a plane that both of these would lie on. Now, we've done this before, but I'll revisit it. | Constructing a unit normal vector to a surface Multivariable Calculus Khan Academy.mp3 |
Now, when you have two vectors that are tangential to the plane and they're not the same vector, these are actually already specifying, these are already kind of determining a plane. And so you can imagine a plane that looks something like this. If you took linear combinations of these two things, you would get a plane that both of these would lie on. Now, we've done this before, but I'll revisit it. What happens when I take the cross product of r sub u and r sub v? What happens when I take the cross product? Well, first, this is going to give us another vector. | Constructing a unit normal vector to a surface Multivariable Calculus Khan Academy.mp3 |
Now, we've done this before, but I'll revisit it. What happens when I take the cross product of r sub u and r sub v? What happens when I take the cross product? Well, first, this is going to give us another vector. It's going to give us a vector that is perpendicular to both, to r sub u and r sub v. Or another way to think about it is this plane, when you take the cross product, this plane is essentially a tangential plane to the surface. And if something's going to be perpendicular to both of these characters, it's going to have to be normal to them, or it's definitely going to be perpendicular to both of them, but it's going to be normal to this plane, which is essentially going to be perpendicular to the surface itself. So this right over here is going to be a normal vector. | Constructing a unit normal vector to a surface Multivariable Calculus Khan Academy.mp3 |
Well, first, this is going to give us another vector. It's going to give us a vector that is perpendicular to both, to r sub u and r sub v. Or another way to think about it is this plane, when you take the cross product, this plane is essentially a tangential plane to the surface. And if something's going to be perpendicular to both of these characters, it's going to have to be normal to them, or it's definitely going to be perpendicular to both of them, but it's going to be normal to this plane, which is essentially going to be perpendicular to the surface itself. So this right over here is going to be a normal vector. Well, let me just write it this way. A normal vector. I'm not saying the normal vector, because you could have different normal vectors of different magnitudes. | Constructing a unit normal vector to a surface Multivariable Calculus Khan Academy.mp3 |
So this right over here is going to be a normal vector. Well, let me just write it this way. A normal vector. I'm not saying the normal vector, because you could have different normal vectors of different magnitudes. This is a normal vector when you take the cross product. And we can even think about what direction it's pointing in. And so when you have something cross something else, the easiest way I remember how to do it is you point your right thumb in the direction of the first vector. | Constructing a unit normal vector to a surface Multivariable Calculus Khan Academy.mp3 |
I'm not saying the normal vector, because you could have different normal vectors of different magnitudes. This is a normal vector when you take the cross product. And we can even think about what direction it's pointing in. And so when you have something cross something else, the easiest way I remember how to do it is you point your right thumb in the direction of the first vector. So in this case, r sub u. So let me see if I can draw this. I'm literally looking at my hand and trying to draw it. | Constructing a unit normal vector to a surface Multivariable Calculus Khan Academy.mp3 |
And so when you have something cross something else, the easiest way I remember how to do it is you point your right thumb in the direction of the first vector. So in this case, r sub u. So let me see if I can draw this. I'm literally looking at my hand and trying to draw it. So you put your right thumb, so this is a right hand rule essentially, in the direction of the first vector. And then you put your index finger in the direction of the second vector right over here. So this is the second vector. | Constructing a unit normal vector to a surface Multivariable Calculus Khan Academy.mp3 |
I'm literally looking at my hand and trying to draw it. So you put your right thumb, so this is a right hand rule essentially, in the direction of the first vector. And then you put your index finger in the direction of the second vector right over here. So this is the second vector. So that's the direction of my index finger. So my index finger is going to look something like that. And then you bend your middle finger inward, and that'll tell you the direction of the cross product. | Constructing a unit normal vector to a surface Multivariable Calculus Khan Academy.mp3 |
So this is the second vector. So that's the direction of my index finger. So my index finger is going to look something like that. And then you bend your middle finger inward, and that'll tell you the direction of the cross product. So if I bend my middle finger inward, it will look something like this. And of course, my other two fingers are just going to be folded in like that, and they're not really relevant. But my other two fingers and my hand looks like that. | Constructing a unit normal vector to a surface Multivariable Calculus Khan Academy.mp3 |
And then you bend your middle finger inward, and that'll tell you the direction of the cross product. So if I bend my middle finger inward, it will look something like this. And of course, my other two fingers are just going to be folded in like that, and they're not really relevant. But my other two fingers and my hand looks like that. And so that tells us the direction. The direction is going to be like that. It's going to be upward facing. | Constructing a unit normal vector to a surface Multivariable Calculus Khan Academy.mp3 |
But my other two fingers and my hand looks like that. And so that tells us the direction. The direction is going to be like that. It's going to be upward facing. That's important because you're two normal vectors. Or there's two directions of normalcy, I guess you could say. One is going out like that, outwards. | Constructing a unit normal vector to a surface Multivariable Calculus Khan Academy.mp3 |
It's going to be upward facing. That's important because you're two normal vectors. Or there's two directions of normalcy, I guess you could say. One is going out like that, outwards. Or I guess in the upward direction, one would be going downwards, or going, I guess you could say, into the surface. But the way I've set it up right now, this would be going outwards. It would be a normal vector to the surface. | Constructing a unit normal vector to a surface Multivariable Calculus Khan Academy.mp3 |
One is going out like that, outwards. Or I guess in the upward direction, one would be going downwards, or going, I guess you could say, into the surface. But the way I've set it up right now, this would be going outwards. It would be a normal vector to the surface. Now, in order to go from a normal vector to the unit normal vector, we just have to normalize this. We just have to divide this by its magnitude. So now we have our drum roll, the unit vector. | Constructing a unit normal vector to a surface Multivariable Calculus Khan Academy.mp3 |
It would be a normal vector to the surface. Now, in order to go from a normal vector to the unit normal vector, we just have to normalize this. We just have to divide this by its magnitude. So now we have our drum roll, the unit vector. And it's going to essentially be a function of u and v. You give me a u or v, and I'll give you that unit normal vector. It is going to be equal to the partial of r with respect to u crossed with the partial of r with respect to v. That just now, that gives us a normal vector, but it hasn't been normalized. So we want to divide by the magnitude of the exact same thing, r sub u crossed with r sub v. And we're done. | Constructing a unit normal vector to a surface Multivariable Calculus Khan Academy.mp3 |
So in the last video, I described how to interpret three-dimensional graphs. And I have another three-dimensional graph here. It's a very bumpy guy. And this happens to be the graph of the function f of x, y is equal to cosine of x multiplied by the sine of y. And you know, I could also say like that this graph represents, I could also say that this graph represents z is equal to that whole value because we think about the output of the function as the z-coordinate of each point. And what I want to do here is describe how you can interpret the relationship between this graph and these functions that you know by taking slices of it. So for example, let's say that we took a slice with this plane here. | Interpreting graphs with slices Multivariable calculus Khan Academy.mp3 |
And this happens to be the graph of the function f of x, y is equal to cosine of x multiplied by the sine of y. And you know, I could also say like that this graph represents, I could also say that this graph represents z is equal to that whole value because we think about the output of the function as the z-coordinate of each point. And what I want to do here is describe how you can interpret the relationship between this graph and these functions that you know by taking slices of it. So for example, let's say that we took a slice with this plane here. And what this plane here is is it represents the value x equals zero. And you can kind of see that because this is the x-axis. So when you're at zero on the x-axis, you know, you pass through the origin. | Interpreting graphs with slices Multivariable calculus Khan Academy.mp3 |
So for example, let's say that we took a slice with this plane here. And what this plane here is is it represents the value x equals zero. And you can kind of see that because this is the x-axis. So when you're at zero on the x-axis, you know, you pass through the origin. And then the values of y and z can go freely. So you end up with this, this plane. And let's say you want to just consider where this cuts through the graph, okay? | Interpreting graphs with slices Multivariable calculus Khan Academy.mp3 |
So when you're at zero on the x-axis, you know, you pass through the origin. And then the values of y and z can go freely. So you end up with this, this plane. And let's say you want to just consider where this cuts through the graph, okay? So we'll limit our graph just down to the point where it cuts it. And I'm gonna draw a little red line over that spot. Now what you might notice here, that red line looks like a sinusoidal wave. | Interpreting graphs with slices Multivariable calculus Khan Academy.mp3 |
And let's say you want to just consider where this cuts through the graph, okay? So we'll limit our graph just down to the point where it cuts it. And I'm gonna draw a little red line over that spot. Now what you might notice here, that red line looks like a sinusoidal wave. In fact, it looks exactly like the sine function itself. You know, it passes through the origin. It starts by going up. | Interpreting graphs with slices Multivariable calculus Khan Academy.mp3 |
Now what you might notice here, that red line looks like a sinusoidal wave. In fact, it looks exactly like the sine function itself. You know, it passes through the origin. It starts by going up. And this makes sense if we start to plug things into the original form here. Because if you take, you know, if you take f and you plug in x equals zero, but then we still let y range freely, what it means is you're looking at cosine of zero multiplied by sine of y. Sine of y. | Interpreting graphs with slices Multivariable calculus Khan Academy.mp3 |
It starts by going up. And this makes sense if we start to plug things into the original form here. Because if you take, you know, if you take f and you plug in x equals zero, but then we still let y range freely, what it means is you're looking at cosine of zero multiplied by sine of y. Sine of y. And what is cosine of zero? Cosine of zero evaluates to one. So this whole function should look just like sine of y, in that when we let y run freely, the output, which is still represented by the z-coordinate, will give us this graph that's just a normal two-dimensional graph that we're probably familiar with. | Interpreting graphs with slices Multivariable calculus Khan Academy.mp3 |
Sine of y. And what is cosine of zero? Cosine of zero evaluates to one. So this whole function should look just like sine of y, in that when we let y run freely, the output, which is still represented by the z-coordinate, will give us this graph that's just a normal two-dimensional graph that we're probably familiar with. And let's try this at a different point. Let's see what would happen if instead of plugging in x equals zero, let's imagine that we plugged in y equals zero. And this time, before I graph it and before I show everything that goes on, let's just try to figure out purely from the formula here what it's gonna look like when we plug in y equals zero. | Interpreting graphs with slices Multivariable calculus Khan Academy.mp3 |
So this whole function should look just like sine of y, in that when we let y run freely, the output, which is still represented by the z-coordinate, will give us this graph that's just a normal two-dimensional graph that we're probably familiar with. And let's try this at a different point. Let's see what would happen if instead of plugging in x equals zero, let's imagine that we plugged in y equals zero. And this time, before I graph it and before I show everything that goes on, let's just try to figure out purely from the formula here what it's gonna look like when we plug in y equals zero. So now I'm gonna write over on the other side, we have fx will still run freely, y is gonna be fixed at zero, and what this means is we have cosine of x, so maybe you expect to see something that looks kinda like a cosine graph, and then sine of zero. Except, what is sine of zero? Sine of zero cancels out and just becomes zero, which, multiplied by cosine of x, means everything cancels out and becomes zero. | Interpreting graphs with slices Multivariable calculus Khan Academy.mp3 |
And this time, before I graph it and before I show everything that goes on, let's just try to figure out purely from the formula here what it's gonna look like when we plug in y equals zero. So now I'm gonna write over on the other side, we have fx will still run freely, y is gonna be fixed at zero, and what this means is we have cosine of x, so maybe you expect to see something that looks kinda like a cosine graph, and then sine of zero. Except, what is sine of zero? Sine of zero cancels out and just becomes zero, which, multiplied by cosine of x, means everything cancels out and becomes zero. So what you'd expect is that this is gonna look like a constant function that's constantly equal to zero. And let's see if that's what we get. So I'm gonna slice it with y equals zero here, and you notice we look at the y-axis, we see when it's zero, and x and z both run freely. | Interpreting graphs with slices Multivariable calculus Khan Academy.mp3 |
Sine of zero cancels out and just becomes zero, which, multiplied by cosine of x, means everything cancels out and becomes zero. So what you'd expect is that this is gonna look like a constant function that's constantly equal to zero. And let's see if that's what we get. So I'm gonna slice it with y equals zero here, and you notice we look at the y-axis, we see when it's zero, and x and z both run freely. I'm gonna chop off my graph at that point, and indeed, it chops it just at this straight line, the straight line that goes right along the x-axis. But let's say that we did a different constant value of y. Rather than y equals zero, so rather than y equals zero, and we'll erase all of this, let's say that I cut things at some other value. | Interpreting graphs with slices Multivariable calculus Khan Academy.mp3 |
So I'm gonna slice it with y equals zero here, and you notice we look at the y-axis, we see when it's zero, and x and z both run freely. I'm gonna chop off my graph at that point, and indeed, it chops it just at this straight line, the straight line that goes right along the x-axis. But let's say that we did a different constant value of y. Rather than y equals zero, so rather than y equals zero, and we'll erase all of this, let's say that I cut things at some other value. So in this case, what I've chosen is y is equal to pi halves, and it looks kinda like we've got a wave here, and it looks like a cosine wave. And you can probably see where this is going. This is when x is running freely, and if we start to imagine plugging this in, I'll just actually write it out. | Interpreting graphs with slices Multivariable calculus Khan Academy.mp3 |
Rather than y equals zero, so rather than y equals zero, and we'll erase all of this, let's say that I cut things at some other value. So in this case, what I've chosen is y is equal to pi halves, and it looks kinda like we've got a wave here, and it looks like a cosine wave. And you can probably see where this is going. This is when x is running freely, and if we start to imagine plugging this in, I'll just actually write it out. We've got cosine of x, and then y is held at a constant, sine of pi halves. Sine of pi halves, this just always equals one, so we could replace this with one, which means the function as a whole should look like cosine x. So again, the multivariable function, we've frozen y, and we're letting x range freely, and it ends up looking like a cosine function. | Interpreting graphs with slices Multivariable calculus Khan Academy.mp3 |
This is when x is running freely, and if we start to imagine plugging this in, I'll just actually write it out. We've got cosine of x, and then y is held at a constant, sine of pi halves. Sine of pi halves, this just always equals one, so we could replace this with one, which means the function as a whole should look like cosine x. So again, the multivariable function, we've frozen y, and we're letting x range freely, and it ends up looking like a cosine function. And I think a really good way to understand a given three-dimensional graph when you see it, so let's say you look back at the original graph, and we don't have anything going on. Get rid of that little line. So you've got this graph, and it looks wavy and bumpy and a little bit hard to understand at first, but if you just think in terms of holding one variable constant, it boils down always into a normal two-dimensional graph. | Interpreting graphs with slices Multivariable calculus Khan Academy.mp3 |
In the last video, we had this rectangle, and we used a triple integral to figure out its volume. And I know you were probably thinking, well, I could have just used my basic geometry to multiply the height times the width times the depth. And that's true, because this was a constant-valued function. And then even once we evaluated, once we integrated with respect to z, we ended up with a double integral, which is exactly what you would have done in the last several videos, when we just learned the volume under a surface. But then we added a twist at the end of the video. We said, fine, you could have figured out the volume within this rectangular domain, I guess, very straightforward using things you already knew. But what if our goal was not to figure out the volume? | Triple integrals 2 Double and triple integrals Multivariable Calculus Khan Academy.mp3 |
And then even once we evaluated, once we integrated with respect to z, we ended up with a double integral, which is exactly what you would have done in the last several videos, when we just learned the volume under a surface. But then we added a twist at the end of the video. We said, fine, you could have figured out the volume within this rectangular domain, I guess, very straightforward using things you already knew. But what if our goal was not to figure out the volume? Our goal was to figure out the mass of this volume. And even more, the mass, the material that we're taking the volume of, whether it's a volume of gas or a volume of some solid, that its density is not constant. So now the mass becomes kind of interesting to calculate. | Triple integrals 2 Double and triple integrals Multivariable Calculus Khan Academy.mp3 |
But what if our goal was not to figure out the volume? Our goal was to figure out the mass of this volume. And even more, the mass, the material that we're taking the volume of, whether it's a volume of gas or a volume of some solid, that its density is not constant. So now the mass becomes kind of interesting to calculate. And so we defined a density function. And rho, this p-looking thing with a curvy bottom, that gives us the density at any given point. At the end of the last video, we said, well, what is mass? | Triple integrals 2 Double and triple integrals Multivariable Calculus Khan Academy.mp3 |
So now the mass becomes kind of interesting to calculate. And so we defined a density function. And rho, this p-looking thing with a curvy bottom, that gives us the density at any given point. At the end of the last video, we said, well, what is mass? Mass is just density times volume. Or you could view it another way. Density is the same thing as mass divided by volume. | Triple integrals 2 Double and triple integrals Multivariable Calculus Khan Academy.mp3 |
At the end of the last video, we said, well, what is mass? Mass is just density times volume. Or you could view it another way. Density is the same thing as mass divided by volume. So the mass around a very, very small point, and we call that d mass, the differential of the mass, is going to equal the density at that point, or the rough density at exactly that point, times the volume differential around that point. Times the volume of this little small cube. And then we, as we saw it on the last video, if you're using rectangular coordinates, this volume differential could just be the x distance times the y distance times the z distance. | Triple integrals 2 Double and triple integrals Multivariable Calculus Khan Academy.mp3 |
Density is the same thing as mass divided by volume. So the mass around a very, very small point, and we call that d mass, the differential of the mass, is going to equal the density at that point, or the rough density at exactly that point, times the volume differential around that point. Times the volume of this little small cube. And then we, as we saw it on the last video, if you're using rectangular coordinates, this volume differential could just be the x distance times the y distance times the z distance. So what if we wanted, the density was, that our density function is defined to be x, y, and z, and we wanted to figure out the mass of this volume. And let's say that our x, y, and z coordinates, their values, let's say they're in meters, and let's say this density is in kilograms per meter cubed. So our answer is going to be in kilograms, if that was the case. | Triple integrals 2 Double and triple integrals Multivariable Calculus Khan Academy.mp3 |
And then we, as we saw it on the last video, if you're using rectangular coordinates, this volume differential could just be the x distance times the y distance times the z distance. So what if we wanted, the density was, that our density function is defined to be x, y, and z, and we wanted to figure out the mass of this volume. And let's say that our x, y, and z coordinates, their values, let's say they're in meters, and let's say this density is in kilograms per meter cubed. So our answer is going to be in kilograms, if that was the case. And those are kind of the traditional SI units. So let's figure out the mass of this variably dense volume. So all we do is we have the same integral up here. | Triple integrals 2 Double and triple integrals Multivariable Calculus Khan Academy.mp3 |
So our answer is going to be in kilograms, if that was the case. And those are kind of the traditional SI units. So let's figure out the mass of this variably dense volume. So all we do is we have the same integral up here. So the differential of mass is going to be this value. So let's write that down. So it's, let me, it is x, I want to make sure I don't run a space, x, y, z times, and I'm going to integrate with respect to dz first. | Triple integrals 2 Double and triple integrals Multivariable Calculus Khan Academy.mp3 |
So all we do is we have the same integral up here. So the differential of mass is going to be this value. So let's write that down. So it's, let me, it is x, I want to make sure I don't run a space, x, y, z times, and I'm going to integrate with respect to dz first. But you could do that, you could actually switch the order, maybe we'll do that in the next video. We'll do dz first, then we'll do dy, then we'll do dx. And then we just define, so this is, once again, this is just the mass at any small differential of volume. | Triple integrals 2 Double and triple integrals Multivariable Calculus Khan Academy.mp3 |
So it's, let me, it is x, I want to make sure I don't run a space, x, y, z times, and I'm going to integrate with respect to dz first. But you could do that, you could actually switch the order, maybe we'll do that in the next video. We'll do dz first, then we'll do dy, then we'll do dx. And then we just define, so this is, once again, this is just the mass at any small differential of volume. And if we integrate with z first, we said z goes from what, the boundaries on z were 0 to 2, the boundaries on y were 0 to 4, and the boundaries on x, x went from 0 to 3. And how do we evaluate this? Well, what is the anti-derivative of, we're integrating with respect to z first, so what's the anti-derivative of x, y, z with respect to z? | Triple integrals 2 Double and triple integrals Multivariable Calculus Khan Academy.mp3 |
And then we just define, so this is, once again, this is just the mass at any small differential of volume. And if we integrate with z first, we said z goes from what, the boundaries on z were 0 to 2, the boundaries on y were 0 to 4, and the boundaries on x, x went from 0 to 3. And how do we evaluate this? Well, what is the anti-derivative of, we're integrating with respect to z first, so what's the anti-derivative of x, y, z with respect to z? Well, it's, let's see, this is just a constant, so it'll be x, y, z squared over 2. Right? Yeah, that's right. | Triple integrals 2 Double and triple integrals Multivariable Calculus Khan Academy.mp3 |
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