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So in this case, the line integral along c1 of F dot dr is going to be equal to the line integral of c2, over the path c2 of F dot dr. If we have a potential in a region, and we're maybe everywhere, then the line integral between any two points is independent of the path. That's the neat thing about a conservative field. Now what I want to do in this video is do a little bit of an extension of the takeaway of the last video. And it's actually a pretty important extension. It might already be obvious to you. I've already written this here.
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Closed curve line integrals of conservative vector fields Multivariable Calculus Khan Academy.mp3
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Now what I want to do in this video is do a little bit of an extension of the takeaway of the last video. And it's actually a pretty important extension. It might already be obvious to you. I've already written this here. I could rearrange this equation a little bit. So let me do it. So let me rearrange this.
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Closed curve line integrals of conservative vector fields Multivariable Calculus Khan Academy.mp3
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I've already written this here. I could rearrange this equation a little bit. So let me do it. So let me rearrange this. I'll just rewrite this in orange. So the line integral on path c1 dot dr minus, I'll just subtract this from both sides, minus the line integral c2 of F dot dr is going to be equal to 0. All I did is I took this takeaway from the last video, and I subtracted this from both sides.
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Closed curve line integrals of conservative vector fields Multivariable Calculus Khan Academy.mp3
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So let me rearrange this. I'll just rewrite this in orange. So the line integral on path c1 dot dr minus, I'll just subtract this from both sides, minus the line integral c2 of F dot dr is going to be equal to 0. All I did is I took this takeaway from the last video, and I subtracted this from both sides. Now, we learned several videos ago that if we're dealing with a line integral of a vector field, not a scalar field, with a vector field, the direction of the path is important. We learned that the line integral over, say, c2 of F dot dr is equal to the negative of the line integral of minus c2 of F dot dr, where we denoted minus c2 as the same path as c2, but just in the opposite direction. So for example, minus c2, I would write like this.
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Closed curve line integrals of conservative vector fields Multivariable Calculus Khan Academy.mp3
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All I did is I took this takeaway from the last video, and I subtracted this from both sides. Now, we learned several videos ago that if we're dealing with a line integral of a vector field, not a scalar field, with a vector field, the direction of the path is important. We learned that the line integral over, say, c2 of F dot dr is equal to the negative of the line integral of minus c2 of F dot dr, where we denoted minus c2 as the same path as c2, but just in the opposite direction. So for example, minus c2, I would write like this. So let me do it in a different color. So let's say this is minus c2. It would be a path just like c2.
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Closed curve line integrals of conservative vector fields Multivariable Calculus Khan Academy.mp3
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So for example, minus c2, I would write like this. So let me do it in a different color. So let's say this is minus c2. It would be a path just like c2. I'm going to call this minus c2, but instead of going in that direction, I'm now going to go in that direction. So ignore the old c2 arrows. We're now starting from there and coming back here.
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Closed curve line integrals of conservative vector fields Multivariable Calculus Khan Academy.mp3
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It would be a path just like c2. I'm going to call this minus c2, but instead of going in that direction, I'm now going to go in that direction. So ignore the old c2 arrows. We're now starting from there and coming back here. So this is minus c2. Or we could put the minus on the other side, and we could say that the negative of the c2 line integral along the path of c2 of F dot dr is equal to the line integral over the reverse path of F dot dr. All I did is I switched the negative on the other side, multiplied both sides by negative 1. So let's replace in this equation, we have the minus of the c2 path.
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Closed curve line integrals of conservative vector fields Multivariable Calculus Khan Academy.mp3
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We're now starting from there and coming back here. So this is minus c2. Or we could put the minus on the other side, and we could say that the negative of the c2 line integral along the path of c2 of F dot dr is equal to the line integral over the reverse path of F dot dr. All I did is I switched the negative on the other side, multiplied both sides by negative 1. So let's replace in this equation, we have the minus of the c2 path. We have that right there. So we could just replace this with this right there. So let me do that.
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Closed curve line integrals of conservative vector fields Multivariable Calculus Khan Academy.mp3
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So let's replace in this equation, we have the minus of the c2 path. We have that right there. So we could just replace this with this right there. So let me do that. So I'll write this first part first. So the integral along the curve c1 of F dot dr, instead of minus the line integral along c2, I'm going to say plus the integral along minus c2. This, we've established, is the same thing as this.
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Closed curve line integrals of conservative vector fields Multivariable Calculus Khan Academy.mp3
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So let me do that. So I'll write this first part first. So the integral along the curve c1 of F dot dr, instead of minus the line integral along c2, I'm going to say plus the integral along minus c2. This, we've established, is the same thing as this. The negative of this curve, or the line integral along this path, is the same thing as the line integral, the positive of the line integral along the reverse path. So we'll say plus the line integral of minus c2 of F dot dr is equal to 0. Now there's something interesting.
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Closed curve line integrals of conservative vector fields Multivariable Calculus Khan Academy.mp3
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This, we've established, is the same thing as this. The negative of this curve, or the line integral along this path, is the same thing as the line integral, the positive of the line integral along the reverse path. So we'll say plus the line integral of minus c2 of F dot dr is equal to 0. Now there's something interesting. Let's look at what the combination of the path of c1 and minus c2 is. c1 starts over here. Let me get a nice, vibrant color.
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Closed curve line integrals of conservative vector fields Multivariable Calculus Khan Academy.mp3
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Now there's something interesting. Let's look at what the combination of the path of c1 and minus c2 is. c1 starts over here. Let me get a nice, vibrant color. c1 starts over here at this point. It moves from this point along this curve c1 and ends up at this point. And then we do the minus c2.
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Closed curve line integrals of conservative vector fields Multivariable Calculus Khan Academy.mp3
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Let me get a nice, vibrant color. c1 starts over here at this point. It moves from this point along this curve c1 and ends up at this point. And then we do the minus c2. Minus c2 starts at this point and then goes and comes back to the original point. It completes a loop. So this is a closed line integral.
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Closed curve line integrals of conservative vector fields Multivariable Calculus Khan Academy.mp3
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And then we do the minus c2. Minus c2 starts at this point and then goes and comes back to the original point. It completes a loop. So this is a closed line integral. So if you combine this, we could rewrite this. Remember, this is just a loop. By reversing this, instead of having two guys starting here and going there, I now can start here, go all the way there, and then come all the way back on this reverse path of c2.
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Closed curve line integrals of conservative vector fields Multivariable Calculus Khan Academy.mp3
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So this is a closed line integral. So if you combine this, we could rewrite this. Remember, this is just a loop. By reversing this, instead of having two guys starting here and going there, I now can start here, go all the way there, and then come all the way back on this reverse path of c2. So this is equivalent to a closed line integral. So that is the same thing as an integral along a closed path, and we could call a closed path maybe c1 plus minus c2 if we wanted to be particular about the closed path. But this could be, I drew c1 and c2 or minus c2 arbitrarily.
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Closed curve line integrals of conservative vector fields Multivariable Calculus Khan Academy.mp3
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By reversing this, instead of having two guys starting here and going there, I now can start here, go all the way there, and then come all the way back on this reverse path of c2. So this is equivalent to a closed line integral. So that is the same thing as an integral along a closed path, and we could call a closed path maybe c1 plus minus c2 if we wanted to be particular about the closed path. But this could be, I drew c1 and c2 or minus c2 arbitrarily. This could be any closed path where our vector field F has a potential or where it is the gradient of a scalar field or where it is conservative. And so this can be written as a closed path of c1 plus the reverse of c2 of F dot dr. That's just a rewriting of that. And so that's going to be equal to 0.
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Closed curve line integrals of conservative vector fields Multivariable Calculus Khan Academy.mp3
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But this could be, I drew c1 and c2 or minus c2 arbitrarily. This could be any closed path where our vector field F has a potential or where it is the gradient of a scalar field or where it is conservative. And so this can be written as a closed path of c1 plus the reverse of c2 of F dot dr. That's just a rewriting of that. And so that's going to be equal to 0. And this is our takeaway for this video. This is, you can view it as a corollary. It's kind of a low-hanging conclusion that you can make after this conclusion.
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Closed curve line integrals of conservative vector fields Multivariable Calculus Khan Academy.mp3
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And so that's going to be equal to 0. And this is our takeaway for this video. This is, you can view it as a corollary. It's kind of a low-hanging conclusion that you can make after this conclusion. So now we know that if we have a vector field that's the gradient of a scalar field in some region or maybe over the entire xy-plane, and this is called the potential of F. This is a potential function. Oftentimes it'll be the negative of it, but it's easy to mess with negatives. But if we have a vector field that is the gradient of a scalar field, we call that vector field conservative.
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Closed curve line integrals of conservative vector fields Multivariable Calculus Khan Academy.mp3
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It's kind of a low-hanging conclusion that you can make after this conclusion. So now we know that if we have a vector field that's the gradient of a scalar field in some region or maybe over the entire xy-plane, and this is called the potential of F. This is a potential function. Oftentimes it'll be the negative of it, but it's easy to mess with negatives. But if we have a vector field that is the gradient of a scalar field, we call that vector field conservative. That tells us that at any point in the region where this is valid, the line integral from one point to another is independent of the path. That's what we got from the last video. And because of that, a closed-loop line integral or a closed-line integral, so if we take some other place, if we take any other closed-line integral or we take the line integral of the vector field on any closed loop, it will become 0 because it is path independent.
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Closed curve line integrals of conservative vector fields Multivariable Calculus Khan Academy.mp3
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And we ended up working out through some nice geometrical reasoning that we need to solve this system of equations, so there's nothing left to do but to just solve the system of equations. We'll start with this first one at the top, see what we can simplify. We notice there's an x term in each one, so we'll go ahead and cancel those out, which is basically a way of saying we're assuming that x is not zero, and we can kind of return to that to see if x equals zero could be a solution. So maybe we'll kind of write that down. We're assuming x is not equal to zero in order to cancel out, and we kind of can revisit whether that could give us another possible solution later, but that will be two times y is equal to lambda times two. And from here, the twos can cancel out, no worries about two equaling zero. And we know that y equals lambda.
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Finishing the intro lagrange multiplier example.mp3
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So maybe we'll kind of write that down. We're assuming x is not equal to zero in order to cancel out, and we kind of can revisit whether that could give us another possible solution later, but that will be two times y is equal to lambda times two. And from here, the twos can cancel out, no worries about two equaling zero. And we know that y equals lambda. So that's a nice simplified form for this equation. And for this next equation, we can use what we just found, that y is equal to lambda, to replace the lambda that we see. And instead, if I replace this with a y, what I'm gonna get is that x squared is equal to y times two times y.
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Finishing the intro lagrange multiplier example.mp3
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And we know that y equals lambda. So that's a nice simplified form for this equation. And for this next equation, we can use what we just found, that y is equal to lambda, to replace the lambda that we see. And instead, if I replace this with a y, what I'm gonna get is that x squared is equal to y times two times y. So that's two times y squared. And I'll leave it in that form because I see that in the next equation, I see an x squared, I see a y squared, so it might be nice to be able to plug this guy right into it. So in that next equation, x squared, I'm gonna go ahead and replace that with y squared.
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Finishing the intro lagrange multiplier example.mp3
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And instead, if I replace this with a y, what I'm gonna get is that x squared is equal to y times two times y. So that's two times y squared. And I'll leave it in that form because I see that in the next equation, I see an x squared, I see a y squared, so it might be nice to be able to plug this guy right into it. So in that next equation, x squared, I'm gonna go ahead and replace that with y squared. So that's two y squared plus y squared equals one. And then from there, simplifies to three times y squared equals one, which in turn means y squared is equal to 1 3rd. And so y is equal to plus or minus the square root of 1 3rd.
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Finishing the intro lagrange multiplier example.mp3
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So in that next equation, x squared, I'm gonna go ahead and replace that with y squared. So that's two y squared plus y squared equals one. And then from there, simplifies to three times y squared equals one, which in turn means y squared is equal to 1 3rd. And so y is equal to plus or minus the square root of 1 3rd. Great, so this gives us y. And I'll go ahead and put a box around that, that we have found what y must be. Now if y squared is equal to 1 3rd, then when we look up here and we say, hmm, two times y squared, that's gonna be the same thing as two times 1 3rd.
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Finishing the intro lagrange multiplier example.mp3
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And so y is equal to plus or minus the square root of 1 3rd. Great, so this gives us y. And I'll go ahead and put a box around that, that we have found what y must be. Now if y squared is equal to 1 3rd, then when we look up here and we say, hmm, two times y squared, that's gonna be the same thing as two times 1 3rd. So two times 1 3rd. So if x squared is equal to 2 3rds, what that implies is that x is equal to plus or minus the square root of 2 3rds. And then there we go, that's another one of the solutions.
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Finishing the intro lagrange multiplier example.mp3
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Now if y squared is equal to 1 3rd, then when we look up here and we say, hmm, two times y squared, that's gonna be the same thing as two times 1 3rd. So two times 1 3rd. So if x squared is equal to 2 3rds, what that implies is that x is equal to plus or minus the square root of 2 3rds. And then there we go, that's another one of the solutions. And I could write down what lambda is, right? And we could, I mean, in this case it's easy because y equals lambda. But all we really want in their final form are x and y, since that's gonna give us the answer to the original constraint problem.
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Finishing the intro lagrange multiplier example.mp3
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And then there we go, that's another one of the solutions. And I could write down what lambda is, right? And we could, I mean, in this case it's easy because y equals lambda. But all we really want in their final form are x and y, since that's gonna give us the answer to the original constraint problem. So this, this gives us what we want. And we just have that pesky little possibility that x equals zero to address. And for that, we can take a look and say, if x equals zero, you know, let's go through the possibility that maybe that's one of the, one of the constraint solutions.
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Finishing the intro lagrange multiplier example.mp3
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But all we really want in their final form are x and y, since that's gonna give us the answer to the original constraint problem. So this, this gives us what we want. And we just have that pesky little possibility that x equals zero to address. And for that, we can take a look and say, if x equals zero, you know, let's go through the possibility that maybe that's one of the, one of the constraint solutions. Well, in this equation, that would make sense, since two times zero would equal zero. In this equation, that would mean that we're setting zero equal to lambda times two times y. Well, since lambda equals y, that would mean that for this side to equal zero, y would have to equal zero.
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Finishing the intro lagrange multiplier example.mp3
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And for that, we can take a look and say, if x equals zero, you know, let's go through the possibility that maybe that's one of the, one of the constraint solutions. Well, in this equation, that would make sense, since two times zero would equal zero. In this equation, that would mean that we're setting zero equal to lambda times two times y. Well, since lambda equals y, that would mean that for this side to equal zero, y would have to equal zero. So, so evidently, you know, if it was the case that x equals zero, that would have to imply from the second equation that y equals zero. But if x and y both equal zero, this constraint can't be satisfied. So none of this is possible, so we never even had to worry about this to start with.
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Finishing the intro lagrange multiplier example.mp3
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Well, since lambda equals y, that would mean that for this side to equal zero, y would have to equal zero. So, so evidently, you know, if it was the case that x equals zero, that would have to imply from the second equation that y equals zero. But if x and y both equal zero, this constraint can't be satisfied. So none of this is possible, so we never even had to worry about this to start with. But it's something you do need to check, just every time you're dividing by a variable, you're basically assuming that it's not equal to zero. So this right here gives us four possible solutions, four possible values for x and y that satisfy this constraint and which potentially maximize this. And remember, when I say potentially maximize, the whole idea of this Lagrange multiplier is that we were looking for where there's a point of tangency between the contour lines.
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Finishing the intro lagrange multiplier example.mp3
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So none of this is possible, so we never even had to worry about this to start with. But it's something you do need to check, just every time you're dividing by a variable, you're basically assuming that it's not equal to zero. So this right here gives us four possible solutions, four possible values for x and y that satisfy this constraint and which potentially maximize this. And remember, when I say potentially maximize, the whole idea of this Lagrange multiplier is that we were looking for where there's a point of tangency between the contour lines. So just to make it explicit, the four points that we're dealing with, here, I'll write them all here. So x could be the square root of 2 3rds, square root of 2 3rds, and y could be the positive square root of 1 3rd. And then we can basically just toggle, you know, maybe x is the negative square root of 2 3rds, and y is still the positive square root of 1 3rd.
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Finishing the intro lagrange multiplier example.mp3
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And remember, when I say potentially maximize, the whole idea of this Lagrange multiplier is that we were looking for where there's a point of tangency between the contour lines. So just to make it explicit, the four points that we're dealing with, here, I'll write them all here. So x could be the square root of 2 3rds, square root of 2 3rds, and y could be the positive square root of 1 3rd. And then we can basically just toggle, you know, maybe x is the negative square root of 2 3rds, and y is still the positive square root of 1 3rd. Or maybe x is the positive square root of 2 3rds, and y is the negative square root of 1 3rd. Kind of monotonous, but just getting all of the different possibilities on the table here. x is negative square root of 2 3rds, and then y is positive, no, negative, that's the last one, square root of 1 3rd.
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Finishing the intro lagrange multiplier example.mp3
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And then we can basically just toggle, you know, maybe x is the negative square root of 2 3rds, and y is still the positive square root of 1 3rd. Or maybe x is the positive square root of 2 3rds, and y is the negative square root of 1 3rd. Kind of monotonous, but just getting all of the different possibilities on the table here. x is negative square root of 2 3rds, and then y is positive, no, negative, that's the last one, square root of 1 3rd. So these are the four points where the contour lines are tangent. And to find which one of these maximizes our function, here, let's go ahead and write down our function again. It gets easy to forget.
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Finishing the intro lagrange multiplier example.mp3
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x is negative square root of 2 3rds, and then y is positive, no, negative, that's the last one, square root of 1 3rd. So these are the four points where the contour lines are tangent. And to find which one of these maximizes our function, here, let's go ahead and write down our function again. It gets easy to forget. So the whole thing we're doing is maximizing f of x, y equals x squared times y. So let me just put that down again. We're looking at f of x, y is equal to x squared times y.
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Finishing the intro lagrange multiplier example.mp3
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It gets easy to forget. So the whole thing we're doing is maximizing f of x, y equals x squared times y. So let me just put that down again. We're looking at f of x, y is equal to x squared times y. So we could just plug these values in and see which one of them is actually greatest. And the first thing to observe is x squared is always gonna be positive. So if I plug in a negative value for y, or if I plug in either this guy here, or this guy here, where the value for y is negative, the entire function would be negative.
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Finishing the intro lagrange multiplier example.mp3
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We're looking at f of x, y is equal to x squared times y. So we could just plug these values in and see which one of them is actually greatest. And the first thing to observe is x squared is always gonna be positive. So if I plug in a negative value for y, or if I plug in either this guy here, or this guy here, where the value for y is negative, the entire function would be negative. So I'm just gonna say that neither of these can be the maximum, because it'll be some positive number, some x squared times a negative. Whereas I know that these guys are gonna produce a positive number. And specifically, if we plug in, if we plug in f of, let's say, this top one, square root of 2 3rds, square root of 1 3rd, well, x squared is gonna be 2 3rds, and then y is square root of 1 3rd.
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Finishing the intro lagrange multiplier example.mp3
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So if I plug in a negative value for y, or if I plug in either this guy here, or this guy here, where the value for y is negative, the entire function would be negative. So I'm just gonna say that neither of these can be the maximum, because it'll be some positive number, some x squared times a negative. Whereas I know that these guys are gonna produce a positive number. And specifically, if we plug in, if we plug in f of, let's say, this top one, square root of 2 3rds, square root of 1 3rd, well, x squared is gonna be 2 3rds, and then y is square root of 1 3rd. And in fact, that's gonna be the same as what we get plugging in this other value. So either one of these maximizes the function. It's got two different maximizing points, and each one of them has a maximum value of 2 3rds times the square root of 1 3rd.
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Finishing the intro lagrange multiplier example.mp3
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And specifically, if we plug in, if we plug in f of, let's say, this top one, square root of 2 3rds, square root of 1 3rd, well, x squared is gonna be 2 3rds, and then y is square root of 1 3rd. And in fact, that's gonna be the same as what we get plugging in this other value. So either one of these maximizes the function. It's got two different maximizing points, and each one of them has a maximum value of 2 3rds times the square root of 1 3rd. And that's the final answer. But I do wanna emphasize that the takeaway here is not the specific algebra that you work out going towards the end, but it's the whole idea of this Lagrange multiplier technique, to find the gradient of one function, find the gradient of the constraining function, and then set them proportional to each other. That's the key takeaway.
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Finishing the intro lagrange multiplier example.mp3
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It's got two different maximizing points, and each one of them has a maximum value of 2 3rds times the square root of 1 3rd. And that's the final answer. But I do wanna emphasize that the takeaway here is not the specific algebra that you work out going towards the end, but it's the whole idea of this Lagrange multiplier technique, to find the gradient of one function, find the gradient of the constraining function, and then set them proportional to each other. That's the key takeaway. And then the rest of it is just making sure that we check our work and go through the minute details, which is important. It has its place. And coming up, I'll go through a few more examples.
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Finishing the intro lagrange multiplier example.mp3
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And in order to take a surface integral, we had to find the partial of our parameterization with respect to s and the partial with respect to t. And now we're ready to take the cross product, and then we can take the magnitude of the cross product. And then we can actually take this double integral and figure out the surface area. So let's just do it step by step. Here we can take the cross product, which is not a non-hairy operation. This is why you don't see many surface integrals actually get done or many examples done. So let's take the cross product of these two fellows. So the partial of r with respect to s crossed with, in magenta, the partial of r with respect to t. This will be a little bit of a review of cross products for you.
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Example of calculating a surface integral part 2 Multivariable Calculus Khan Academy.mp3
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Here we can take the cross product, which is not a non-hairy operation. This is why you don't see many surface integrals actually get done or many examples done. So let's take the cross product of these two fellows. So the partial of r with respect to s crossed with, in magenta, the partial of r with respect to t. This will be a little bit of a review of cross products for you. You might remember this is going to be equal to the determinant. The determinant, I'm going to write the unit vectors up here of the first rows i, j, and k. And then the next two rows are going to be the components of these guys. So let me copy and paste them.
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Example of calculating a surface integral part 2 Multivariable Calculus Khan Academy.mp3
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So the partial of r with respect to s crossed with, in magenta, the partial of r with respect to t. This will be a little bit of a review of cross products for you. You might remember this is going to be equal to the determinant. The determinant, I'm going to write the unit vectors up here of the first rows i, j, and k. And then the next two rows are going to be the components of these guys. So let me copy and paste them. You have that right there. Let me copy and paste. Put that guy right there.
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Example of calculating a surface integral part 2 Multivariable Calculus Khan Academy.mp3
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So let me copy and paste them. You have that right there. Let me copy and paste. Put that guy right there. Then you have this fellow right there. Copy and paste. Put them right there.
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Example of calculating a surface integral part 2 Multivariable Calculus Khan Academy.mp3
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Put that guy right there. Then you have this fellow right there. Copy and paste. Put them right there. And then you got this guy right here. This will save us some time. Copy and paste.
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Example of calculating a surface integral part 2 Multivariable Calculus Khan Academy.mp3
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Put them right there. And then you got this guy right here. This will save us some time. Copy and paste. Put them right there. Then the last row is going to be this guy's components. Copy and paste.
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Example of calculating a surface integral part 2 Multivariable Calculus Khan Academy.mp3
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Copy and paste. Put them right there. Then the last row is going to be this guy's components. Copy and paste. Put them right here. Almost done. This guy.
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Example of calculating a surface integral part 2 Multivariable Calculus Khan Academy.mp3
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Copy and paste. Put them right here. Almost done. This guy. Copy and paste. Put them right there. Make sure we know that these are separate terms.
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Example of calculating a surface integral part 2 Multivariable Calculus Khan Academy.mp3
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This guy. Copy and paste. Put them right there. Make sure we know that these are separate terms. And finally, we don't have to copy and paste it, but just since we did it for all of the other terms, I'll do it for that zero as well. So the cross product of these is literally the determinant of this matrix right here. It's this determinant.
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Example of calculating a surface integral part 2 Multivariable Calculus Khan Academy.mp3
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Make sure we know that these are separate terms. And finally, we don't have to copy and paste it, but just since we did it for all of the other terms, I'll do it for that zero as well. So the cross product of these is literally the determinant of this matrix right here. It's this determinant. It's that determinant. And so, just as a bit of a refresher of taking determinants, this is going to be i times the sub-determinant right here, if you cross out this column and that row. It's going to be equal to i.
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Example of calculating a surface integral part 2 Multivariable Calculus Khan Academy.mp3
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It's this determinant. It's that determinant. And so, just as a bit of a refresher of taking determinants, this is going to be i times the sub-determinant right here, if you cross out this column and that row. It's going to be equal to i. You're not used to seeing the unit vector written first, but we can switch the order later. Times i times the sub-matrix right here, if you cross out this column and that row. So it's going to be this term times zero, which is just zero, minus this term times that term.
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Example of calculating a surface integral part 2 Multivariable Calculus Khan Academy.mp3
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It's going to be equal to i. You're not used to seeing the unit vector written first, but we can switch the order later. Times i times the sub-matrix right here, if you cross out this column and that row. So it's going to be this term times zero, which is just zero, minus this term times that term. So minus this term times this term. The negative signs are going to cancel out, so this will be positive. So it would be i times this term times this term without a negative sign right there.
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Example of calculating a surface integral part 2 Multivariable Calculus Khan Academy.mp3
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So it's going to be this term times zero, which is just zero, minus this term times that term. So minus this term times this term. The negative signs are going to cancel out, so this will be positive. So it would be i times this term times this term without a negative sign right there. So i times this term, which is a cosine of s. It's really that term times that term minus that term times that term, but the negatives cancel out. That times that is zero, so that's how we can do it. So it's a cosine of s times b plus a cosine of s. I'll switch to the same color, sine of t. So we've got our i term for the cross product.
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Example of calculating a surface integral part 2 Multivariable Calculus Khan Academy.mp3
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So it would be i times this term times this term without a negative sign right there. So i times this term, which is a cosine of s. It's really that term times that term minus that term times that term, but the negatives cancel out. That times that is zero, so that's how we can do it. So it's a cosine of s times b plus a cosine of s. I'll switch to the same color, sine of t. So we've got our i term for the cross product. Now it's going to be minus j. You remember when you take the determinant, you have to checkerboard of switching signs. So now it's going to be minus j times, so you cross out that row and that column, and it's going to be this term times this term, which is just zero, minus this term times this term.
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Example of calculating a surface integral part 2 Multivariable Calculus Khan Academy.mp3
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So it's a cosine of s times b plus a cosine of s. I'll switch to the same color, sine of t. So we've got our i term for the cross product. Now it's going to be minus j. You remember when you take the determinant, you have to checkerboard of switching signs. So now it's going to be minus j times, so you cross out that row and that column, and it's going to be this term times this term, which is just zero, minus this term times this term. And once again, when you have, oh sorry, you cross out this column and that row, so it's going to be that guy times that guy minus this guy times this guy. So it's going to be minus this guy times this guy, so it's going to be, let me do it in yellow, so the negative times negative that guy, b plus a cosine of s cosine of t times this guy, a cosine of s. We'll clean it up in a little bit. Immediately, oh we'll clean this up.
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Example of calculating a surface integral part 2 Multivariable Calculus Khan Academy.mp3
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So now it's going to be minus j times, so you cross out that row and that column, and it's going to be this term times this term, which is just zero, minus this term times this term. And once again, when you have, oh sorry, you cross out this column and that row, so it's going to be that guy times that guy minus this guy times this guy. So it's going to be minus this guy times this guy, so it's going to be, let me do it in yellow, so the negative times negative that guy, b plus a cosine of s cosine of t times this guy, a cosine of s. We'll clean it up in a little bit. Immediately, oh we'll clean this up. You see this negative and that negative will cancel out. We're just multiplying everything. And then finally, the k term.
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Example of calculating a surface integral part 2 Multivariable Calculus Khan Academy.mp3
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Immediately, oh we'll clean this up. You see this negative and that negative will cancel out. We're just multiplying everything. And then finally, the k term. So plus, I'll go to the next line, plus k times, cross out that row, that column, it's going to be that times that minus that times that. So that looks like a kind of a beastly thing, but I think if we take step by step it shouldn't be too bad. So that times that, the negatives are going to cancel out.
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Example of calculating a surface integral part 2 Multivariable Calculus Khan Academy.mp3
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And then finally, the k term. So plus, I'll go to the next line, plus k times, cross out that row, that column, it's going to be that times that minus that times that. So that looks like a kind of a beastly thing, but I think if we take step by step it shouldn't be too bad. So that times that, the negatives are going to cancel out. So we have this term right here is going to be a sine of t sine of s, and then this term right here is b plus a cosine of s sine of t, so that's that times that and the negatives cancelled out, that's why I didn't put any negatives here, minus this times this. So this times this is going to be a negative number, but if you take the negative of it, it's going to be a positive value. So it's going to be plus that a cosine of t sine of s times that, times b plus a cosine of s cosine of t. Now you see why you don't see many examples of surface integrals being done.
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Example of calculating a surface integral part 2 Multivariable Calculus Khan Academy.mp3
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So that times that, the negatives are going to cancel out. So we have this term right here is going to be a sine of t sine of s, and then this term right here is b plus a cosine of s sine of t, so that's that times that and the negatives cancelled out, that's why I didn't put any negatives here, minus this times this. So this times this is going to be a negative number, but if you take the negative of it, it's going to be a positive value. So it's going to be plus that a cosine of t sine of s times that, times b plus a cosine of s cosine of t. Now you see why you don't see many examples of surface integrals being done. Let's see if we can clean this up a little bit, especially if we can clean up this last term a bit. So let's see what we can do to simplify it. So our first term, so let's just multiply it out, I guess is the easiest way to do it.
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Example of calculating a surface integral part 2 Multivariable Calculus Khan Academy.mp3
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So it's going to be plus that a cosine of t sine of s times that, times b plus a cosine of s cosine of t. Now you see why you don't see many examples of surface integrals being done. Let's see if we can clean this up a little bit, especially if we can clean up this last term a bit. So let's see what we can do to simplify it. So our first term, so let's just multiply it out, I guess is the easiest way to do it. Actually the easiest first step would just be to factor out the b plus a cosine of s, because that's in every term, b plus a cosine of s, b plus a cosine of s, b plus a cosine of s, b plus a cosine of s. So let's just factor that out. This whole crazy thing can be written as b plus a cosine of s. So we factored it out times, I'll put in maybe some brackets here so you know it multiplies times every component. So the i component, we need to factor this guy out, is going to be a cosine of s sine of t. Let me write it in green.
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Example of calculating a surface integral part 2 Multivariable Calculus Khan Academy.mp3
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So our first term, so let's just multiply it out, I guess is the easiest way to do it. Actually the easiest first step would just be to factor out the b plus a cosine of s, because that's in every term, b plus a cosine of s, b plus a cosine of s, b plus a cosine of s, b plus a cosine of s. So let's just factor that out. This whole crazy thing can be written as b plus a cosine of s. So we factored it out times, I'll put in maybe some brackets here so you know it multiplies times every component. So the i component, we need to factor this guy out, is going to be a cosine of s sine of t. Let me write it in green. It's going to be a cosine of s sine of t. You're not used to seeing the i before it, so I'm going to write the i here. And then plus, we're factoring this guy out, so you're just going to be left with cosine of t, a cosine of s. Or we could write it as a cosine of s cosine of t. That's that right there, just putting it in the same order as that, times the unit vector j. And then when we factored this guy out, so we're not going to see that or that anymore.
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Example of calculating a surface integral part 2 Multivariable Calculus Khan Academy.mp3
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So the i component, we need to factor this guy out, is going to be a cosine of s sine of t. Let me write it in green. It's going to be a cosine of s sine of t. You're not used to seeing the i before it, so I'm going to write the i here. And then plus, we're factoring this guy out, so you're just going to be left with cosine of t, a cosine of s. Or we could write it as a cosine of s cosine of t. That's that right there, just putting it in the same order as that, times the unit vector j. And then when we factored this guy out, so we're not going to see that or that anymore. When we factor that out, we can multiply this out, and what do we get? So in green I'll write again. So if you multiply sine of t times this thing over here, because that's all that we have left after we factor out this thing, we get a sine of s sine squared of t. We have sine of t times sine of t. Sine squared of t, so that's that over there, plus what do we have over here?
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Example of calculating a surface integral part 2 Multivariable Calculus Khan Academy.mp3
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And then when we factored this guy out, so we're not going to see that or that anymore. When we factor that out, we can multiply this out, and what do we get? So in green I'll write again. So if you multiply sine of t times this thing over here, because that's all that we have left after we factor out this thing, we get a sine of s sine squared of t. We have sine of t times sine of t. Sine squared of t, so that's that over there, plus what do we have over here? We have a sine of s times cosine squared of t. A sine of s times cosine squared of t, and all of that times the k unit vector. All of that times the k unit vector. And so things are looking a little bit more simplified, but you might see something jump out at you.
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Example of calculating a surface integral part 2 Multivariable Calculus Khan Academy.mp3
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So if you multiply sine of t times this thing over here, because that's all that we have left after we factor out this thing, we get a sine of s sine squared of t. We have sine of t times sine of t. Sine squared of t, so that's that over there, plus what do we have over here? We have a sine of s times cosine squared of t. A sine of s times cosine squared of t, and all of that times the k unit vector. All of that times the k unit vector. And so things are looking a little bit more simplified, but you might see something jump out at you. You have a sine squared and a cosine squared. So somehow if I can just make that just sine squared plus cosine squared of t, those will simplify to 1, and we can. And this term right here, if we just focus on that term, and this is all kind of algebraic manipulation, if we just focus on that term, this term right here can be rewritten as a sine of s, if we factor that out, times sine squared of t plus cosine squared of t times our unit vector k. I just factored out an a sine of s from both of these terms.
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Example of calculating a surface integral part 2 Multivariable Calculus Khan Academy.mp3
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And so things are looking a little bit more simplified, but you might see something jump out at you. You have a sine squared and a cosine squared. So somehow if I can just make that just sine squared plus cosine squared of t, those will simplify to 1, and we can. And this term right here, if we just focus on that term, and this is all kind of algebraic manipulation, if we just focus on that term, this term right here can be rewritten as a sine of s, if we factor that out, times sine squared of t plus cosine squared of t times our unit vector k. I just factored out an a sine of s from both of these terms. And this is our most fundamental trig identity from the unit circle. This is equal to 1. So this last term simplifies to a sine of s times k. So, so far, we've gotten pretty far.
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Example of calculating a surface integral part 2 Multivariable Calculus Khan Academy.mp3
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And this term right here, if we just focus on that term, and this is all kind of algebraic manipulation, if we just focus on that term, this term right here can be rewritten as a sine of s, if we factor that out, times sine squared of t plus cosine squared of t times our unit vector k. I just factored out an a sine of s from both of these terms. And this is our most fundamental trig identity from the unit circle. This is equal to 1. So this last term simplifies to a sine of s times k. So, so far, we've gotten pretty far. We were able to figure out the cross product of these two, I guess, partial derivatives of the vector value to our original parameterization there. We were able to figure out what this thing, this thing right here, before we take the magnitude of it, it translates to this thing right here. Let me rewrite it.
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Example of calculating a surface integral part 2 Multivariable Calculus Khan Academy.mp3
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So this last term simplifies to a sine of s times k. So, so far, we've gotten pretty far. We were able to figure out the cross product of these two, I guess, partial derivatives of the vector value to our original parameterization there. We were able to figure out what this thing, this thing right here, before we take the magnitude of it, it translates to this thing right here. Let me rewrite it. Well, I don't need to rewrite it. You know that. Well, I'll rewrite it.
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Example of calculating a surface integral part 2 Multivariable Calculus Khan Academy.mp3
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Let me rewrite it. Well, I don't need to rewrite it. You know that. Well, I'll rewrite it. So that's equal to, I'll rewrite it neatly, and we'll use this in the next video, b plus a cosine of s times, open bracket, a cosine of s sine of t times i plus, switch back to the blue, plus a cosine of s cosine of t times j plus, switch back to the blue, this thing, plus, this simplified nicely, a sine of s times k, times the unit vector k. This, this right here is this expression right there. And I'll finish this video since I'm already over 10 minutes. And in the next video, we're going to take the magnitude of it, and then if we have time, actually take this double integral, and we'll all be done.
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Example of calculating a surface integral part 2 Multivariable Calculus Khan Academy.mp3
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In the video on parametric surfaces, I give you guys this function here. It's a very complicated looking function. It's got a two-dimensional input and a three-dimensional output. And I talked about how you can think about it as drawing a surface in three-dimensional space. And that one came out to be the surface of a donut, which we also call a torus. So what I want to do here is talk about how you might think of this as a transformation. And first, let me just get straight what the input space here is.
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Transformations, part 3 Multivariable calculus Khan Academy.mp3
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And I talked about how you can think about it as drawing a surface in three-dimensional space. And that one came out to be the surface of a donut, which we also call a torus. So what I want to do here is talk about how you might think of this as a transformation. And first, let me just get straight what the input space here is. So the input space, you could think about it as the entire t-s plane. Right, we might draw this as the entire t-axis and the s-axis, and just everything here and see where it maps. But you can actually go to just a small subset of that.
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Transformations, part 3 Multivariable calculus Khan Academy.mp3
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And first, let me just get straight what the input space here is. So the input space, you could think about it as the entire t-s plane. Right, we might draw this as the entire t-axis and the s-axis, and just everything here and see where it maps. But you can actually go to just a small subset of that. So if you limit yourself to t going between zero, so between zero and let's say 2 pi, and then similarly with s going from zero up to 2 pi, and you imagine what, you know, that would be sort of a square region, just limiting yourself to that, you're actually going to get all of the points that you need to draw the torus. And the basic reason for that is that as t ranges from zero to 2 pi, cosine of t goes over its full range before it starts becoming periodic. Sine of t does the same.
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Transformations, part 3 Multivariable calculus Khan Academy.mp3
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But you can actually go to just a small subset of that. So if you limit yourself to t going between zero, so between zero and let's say 2 pi, and then similarly with s going from zero up to 2 pi, and you imagine what, you know, that would be sort of a square region, just limiting yourself to that, you're actually going to get all of the points that you need to draw the torus. And the basic reason for that is that as t ranges from zero to 2 pi, cosine of t goes over its full range before it starts becoming periodic. Sine of t does the same. And same deal with s. If you let s range from zero to 2 pi, that covers a full period of cosine, a full period of sine. So you'll get no new information by going elsewhere. So what we can do is think about this portion of the t-s plane kind of as living inside three-dimensional space.
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Transformations, part 3 Multivariable calculus Khan Academy.mp3
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Sine of t does the same. And same deal with s. If you let s range from zero to 2 pi, that covers a full period of cosine, a full period of sine. So you'll get no new information by going elsewhere. So what we can do is think about this portion of the t-s plane kind of as living inside three-dimensional space. This is sort of cheating, but it's a little bit easier to do this than to imagine, you know, moving from some separate area into the space. At the very least for the animation efforts, it's easier to just start it off in 3D. So what I'm thinking about here, this square is representing that t-s plane.
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Transformations, part 3 Multivariable calculus Khan Academy.mp3
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So what we can do is think about this portion of the t-s plane kind of as living inside three-dimensional space. This is sort of cheating, but it's a little bit easier to do this than to imagine, you know, moving from some separate area into the space. At the very least for the animation efforts, it's easier to just start it off in 3D. So what I'm thinking about here, this square is representing that t-s plane. And for this function, which is taking all of the points in this square as its input and outputs a point in three-dimensional space, you can think about how those points move to their corresponding output points. Okay, so I'll show that again. We start off with our t-s plane here.
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Transformations, part 3 Multivariable calculus Khan Academy.mp3
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So what I'm thinking about here, this square is representing that t-s plane. And for this function, which is taking all of the points in this square as its input and outputs a point in three-dimensional space, you can think about how those points move to their corresponding output points. Okay, so I'll show that again. We start off with our t-s plane here. And then whatever your input point is, if you were to follow it, and you were to follow it through this whole transformation, the place where it lands would be the corresponding output of this function. And one thing I should mention is all of the interpolating values as you go in between these don't really matter. A function is really a very static thing.
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Transformations, part 3 Multivariable calculus Khan Academy.mp3
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We start off with our t-s plane here. And then whatever your input point is, if you were to follow it, and you were to follow it through this whole transformation, the place where it lands would be the corresponding output of this function. And one thing I should mention is all of the interpolating values as you go in between these don't really matter. A function is really a very static thing. There's just an input and there's an output. And if I'm thinking in terms of a transformation actually moving it, there's a little bit of magic sauce that has to go into making an animation do this. And in this case, I kind of put it into two different phases to sort of roll up one side and roll up the other.
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Transformations, part 3 Multivariable calculus Khan Academy.mp3
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A function is really a very static thing. There's just an input and there's an output. And if I'm thinking in terms of a transformation actually moving it, there's a little bit of magic sauce that has to go into making an animation do this. And in this case, I kind of put it into two different phases to sort of roll up one side and roll up the other. It doesn't really matter. But the general idea of starting with a square and somehow warping that, however you do choose to warp it, is actually a pretty powerful thought. And as we get into multivariable calculus and you start thinking a little bit more deeply about surfaces, I think it really helps if you, you know, you think about what a slight little movement over here on your input space would look like.
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Transformations, part 3 Multivariable calculus Khan Academy.mp3
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And in this case, I kind of put it into two different phases to sort of roll up one side and roll up the other. It doesn't really matter. But the general idea of starting with a square and somehow warping that, however you do choose to warp it, is actually a pretty powerful thought. And as we get into multivariable calculus and you start thinking a little bit more deeply about surfaces, I think it really helps if you, you know, you think about what a slight little movement over here on your input space would look like. What happens to that tiny little movement or that tiny little traversal? What it looks like if you did that same movement somewhere on the output space. And you'll get lots of chances to wrap your mind about this and engage with the idea.
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Transformations, part 3 Multivariable calculus Khan Academy.mp3
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So in this video, I'm gonna introduce vector fields. Now these are a concept that come up all the time in multivariable calculus, and that's probably because they come up all the time in physics. You know, it comes up with fluid flow, with electrodynamics. You see them all over the place. And what a vector field is, is it's pretty much a way of visualizing functions that have the same number of dimensions in their input as in their output. So here I'm gonna write a function that's got a two-dimensional input, x and y, and then its output is gonna be a two-dimensional vector. And each of the components will somehow depend on x and y. I'll make the first one y cubed minus nine y.
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Vector fields, introduction Multivariable calculus Khan Academy.mp3
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You see them all over the place. And what a vector field is, is it's pretty much a way of visualizing functions that have the same number of dimensions in their input as in their output. So here I'm gonna write a function that's got a two-dimensional input, x and y, and then its output is gonna be a two-dimensional vector. And each of the components will somehow depend on x and y. I'll make the first one y cubed minus nine y. And then the second component, the y component of the output, will be x cubed minus nine x. I made them symmetric here, looking kind of similar. They don't have to be. I'm just kind of a sucker for symmetry.
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Vector fields, introduction Multivariable calculus Khan Academy.mp3
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And each of the components will somehow depend on x and y. I'll make the first one y cubed minus nine y. And then the second component, the y component of the output, will be x cubed minus nine x. I made them symmetric here, looking kind of similar. They don't have to be. I'm just kind of a sucker for symmetry. So if you imagine trying to visualize a function like this with, I don't know, like a graph, it would be really hard because you have two dimensions in the input, two dimensions in the output, so you'd have to somehow visualize this thing in four dimensions. So instead what we do, we look only in the input space. So that means we look only in the xy plane.
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Vector fields, introduction Multivariable calculus Khan Academy.mp3
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I'm just kind of a sucker for symmetry. So if you imagine trying to visualize a function like this with, I don't know, like a graph, it would be really hard because you have two dimensions in the input, two dimensions in the output, so you'd have to somehow visualize this thing in four dimensions. So instead what we do, we look only in the input space. So that means we look only in the xy plane. So I'll draw these coordinate axes, and just mark it up. This here is our x-axis. This here is our y-axis.
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Vector fields, introduction Multivariable calculus Khan Academy.mp3
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So that means we look only in the xy plane. So I'll draw these coordinate axes, and just mark it up. This here is our x-axis. This here is our y-axis. And for each individual input point, like let's say one, two. So let's say we go to one, two. I'm gonna consider the vector that it outputs and attach that vector to the point.
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Vector fields, introduction Multivariable calculus Khan Academy.mp3
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This here is our y-axis. And for each individual input point, like let's say one, two. So let's say we go to one, two. I'm gonna consider the vector that it outputs and attach that vector to the point. So let's walk through an example of what I mean by that. So if we actually evaluate f at one, two, x is equal to one, y is equal to two, so we plug in two cubed, whoops, two cubed minus nine times two up here in the x component, and then one cubed minus nine times y, nine times one, excuse me, down in the y component, two cubed is eight, nine times two is 18, so eight minus 18 is negative 10, negative 10, and then one cubed is one, nine times one is nine, so one minus nine is negative eight. Now first, imagine that this was, if we just drew this vector where we count, starting from the origin, negative one, two, three, four, five, six, seven, eight, nine, 10, so it's gonna have this as its x component, and then negative eight, one, two, three, four, five, six, seven, we're gonna actually go off the screen.
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Vector fields, introduction Multivariable calculus Khan Academy.mp3
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I'm gonna consider the vector that it outputs and attach that vector to the point. So let's walk through an example of what I mean by that. So if we actually evaluate f at one, two, x is equal to one, y is equal to two, so we plug in two cubed, whoops, two cubed minus nine times two up here in the x component, and then one cubed minus nine times y, nine times one, excuse me, down in the y component, two cubed is eight, nine times two is 18, so eight minus 18 is negative 10, negative 10, and then one cubed is one, nine times one is nine, so one minus nine is negative eight. Now first, imagine that this was, if we just drew this vector where we count, starting from the origin, negative one, two, three, four, five, six, seven, eight, nine, 10, so it's gonna have this as its x component, and then negative eight, one, two, three, four, five, six, seven, we're gonna actually go off the screen. It's a very, very large vector, so it's gonna be something here, and it ends up having to go off the screen. But the nice thing about vectors, it doesn't matter where they start, so instead we can start it here, and we still want it to have that negative 10 x component, and then negative eight, so negative one, two, three, four, five, six, seven, eight, negative eight as its y component there. So this is a really big vector, and a plan with the vector field is to do this at not just one, two, but at a whole bunch of different points and see what vectors attach to them, and if we drew them all according to their size, this would be a real mess.
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Vector fields, introduction Multivariable calculus Khan Academy.mp3
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Now first, imagine that this was, if we just drew this vector where we count, starting from the origin, negative one, two, three, four, five, six, seven, eight, nine, 10, so it's gonna have this as its x component, and then negative eight, one, two, three, four, five, six, seven, we're gonna actually go off the screen. It's a very, very large vector, so it's gonna be something here, and it ends up having to go off the screen. But the nice thing about vectors, it doesn't matter where they start, so instead we can start it here, and we still want it to have that negative 10 x component, and then negative eight, so negative one, two, three, four, five, six, seven, eight, negative eight as its y component there. So this is a really big vector, and a plan with the vector field is to do this at not just one, two, but at a whole bunch of different points and see what vectors attach to them, and if we drew them all according to their size, this would be a real mess. There'd be markings all over the place, and you'd have, you know, this one might have some huge vector attached to it, and this one would have some huge vector attached to it, and it would get really, really messy. But instead what we do, so I'm gonna just clear up the board here, we scale them down, this is common, you'll scale them down so that you're kind of lying about what the vectors themselves are, but you get a much better feel for what each thing corresponds to. And another thing about this drawing that's not entirely faithful to the original function that we have is that all of these vectors are the same length.
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Vector fields, introduction Multivariable calculus Khan Academy.mp3
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So this is a really big vector, and a plan with the vector field is to do this at not just one, two, but at a whole bunch of different points and see what vectors attach to them, and if we drew them all according to their size, this would be a real mess. There'd be markings all over the place, and you'd have, you know, this one might have some huge vector attached to it, and this one would have some huge vector attached to it, and it would get really, really messy. But instead what we do, so I'm gonna just clear up the board here, we scale them down, this is common, you'll scale them down so that you're kind of lying about what the vectors themselves are, but you get a much better feel for what each thing corresponds to. And another thing about this drawing that's not entirely faithful to the original function that we have is that all of these vectors are the same length. You know, I made this one just kind of the same unit, this one the same unit, and over here, they all just have the same length, even though in reality, the length of the vectors output by this function can be wildly different. This is kind of common practice when vector fields are drawn or when some kind of software is drawing them for you. So there are ways of getting around this.
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Vector fields, introduction Multivariable calculus Khan Academy.mp3
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And another thing about this drawing that's not entirely faithful to the original function that we have is that all of these vectors are the same length. You know, I made this one just kind of the same unit, this one the same unit, and over here, they all just have the same length, even though in reality, the length of the vectors output by this function can be wildly different. This is kind of common practice when vector fields are drawn or when some kind of software is drawing them for you. So there are ways of getting around this. One way is to just use colors with your vectors, so I'll switch over to a different vector field here, and here, color is used to kind of give a hint of length. So it still looks organized because all of them have the same lengths, but the difference is that red and warmer colors are supposed to indicate this is a very long vector somehow, and then blue would indicate that it's very short. Another thing you could do is scale them to be roughly proportional to what they should be.
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Vector fields, introduction Multivariable calculus Khan Academy.mp3
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So there are ways of getting around this. One way is to just use colors with your vectors, so I'll switch over to a different vector field here, and here, color is used to kind of give a hint of length. So it still looks organized because all of them have the same lengths, but the difference is that red and warmer colors are supposed to indicate this is a very long vector somehow, and then blue would indicate that it's very short. Another thing you could do is scale them to be roughly proportional to what they should be. So notice all the blue vectors scaled way down to basically be zero. Red vectors kind of stayed the same size. And even though in reality, this might be representing a function where the true vector here should be really long or the true vector here should be kind of medium length, it's still common for people to just shrink them down so it's a reasonable thing to view.
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Vector fields, introduction Multivariable calculus Khan Academy.mp3
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So I have this region, this simple solid right over here. It's x can go between negative 1 and 1. z, this kind of arch part right over here, is going to be a function of x. That's the upper bound on z. The lower bound on z is just 0. And then y can go anywhere between 0, and then it's bounded here by this plane where we can express y as a function of z. y is 2 minus z along this plane right over here. And we're given this crazy vector field. It has natural logs and tangents in it.
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Divergence theorem example 1 Divergence theorem Multivariable Calculus Khan Academy.mp3
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The lower bound on z is just 0. And then y can go anywhere between 0, and then it's bounded here by this plane where we can express y as a function of z. y is 2 minus z along this plane right over here. And we're given this crazy vector field. It has natural logs and tangents in it. And we're asked to evaluate the surface integral, or I should say the flux of our vector field across the boundary of this region, across the surface of this region right over here. And surface integrals are messy as is, especially when you have a crazy vector field like this. But you could imagine that there might be a way to simplify this, perhaps using the divergence theorem.
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Divergence theorem example 1 Divergence theorem Multivariable Calculus Khan Academy.mp3
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It has natural logs and tangents in it. And we're asked to evaluate the surface integral, or I should say the flux of our vector field across the boundary of this region, across the surface of this region right over here. And surface integrals are messy as is, especially when you have a crazy vector field like this. But you could imagine that there might be a way to simplify this, perhaps using the divergence theorem. The divergence theorem tells us that the flux across the boundary of this simple solid region is going to be the same thing as the triple integral over the volume of it, or I'll just call it over the region, of the divergence of f, dv, where dv is some combination of dx, dy, dz, the divergence times each little cubic volume, infinitesimal cubic volume. So times dv. So let's see if this simplifies things a bit.
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Divergence theorem example 1 Divergence theorem Multivariable Calculus Khan Academy.mp3
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But you could imagine that there might be a way to simplify this, perhaps using the divergence theorem. The divergence theorem tells us that the flux across the boundary of this simple solid region is going to be the same thing as the triple integral over the volume of it, or I'll just call it over the region, of the divergence of f, dv, where dv is some combination of dx, dy, dz, the divergence times each little cubic volume, infinitesimal cubic volume. So times dv. So let's see if this simplifies things a bit. So let's calculate the divergence of f first. So the divergence of f is going to be the partial of the x component, or the partial of the, you could say, the i component, or the x component with respect to x. Well, the derivative of this with respect to x is just x.
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Divergence theorem example 1 Divergence theorem Multivariable Calculus Khan Academy.mp3
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So let's see if this simplifies things a bit. So let's calculate the divergence of f first. So the divergence of f is going to be the partial of the x component, or the partial of the, you could say, the i component, or the x component with respect to x. Well, the derivative of this with respect to x is just x. Derivative of this with respect to x, luckily, is just 0. This is a constant in terms of x. Now let's go over here, the partial of this with respect to y.
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Divergence theorem example 1 Divergence theorem Multivariable Calculus Khan Academy.mp3
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Well, the derivative of this with respect to x is just x. Derivative of this with respect to x, luckily, is just 0. This is a constant in terms of x. Now let's go over here, the partial of this with respect to y. The partial of this with respect to y is just x. And then this is just a constant in terms of y, so it's just going to be 0 when you take the derivative with respect to y. And then finally, the partial of this with respect to z.
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Divergence theorem example 1 Divergence theorem Multivariable Calculus Khan Academy.mp3
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Now let's go over here, the partial of this with respect to y. The partial of this with respect to y is just x. And then this is just a constant in terms of y, so it's just going to be 0 when you take the derivative with respect to y. And then finally, the partial of this with respect to z. Well, this is just a constant in terms of z. It doesn't change when z changes, so the partial with respect to z is just going to be 0 here. And so taking the divergence really, really, really simplified things, the divergence of f simplified down to 2x.
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Divergence theorem example 1 Divergence theorem Multivariable Calculus Khan Academy.mp3
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And then finally, the partial of this with respect to z. Well, this is just a constant in terms of z. It doesn't change when z changes, so the partial with respect to z is just going to be 0 here. And so taking the divergence really, really, really simplified things, the divergence of f simplified down to 2x. So now we can restate the flux across the surface as a triple integral of 2x. So let me just write 2x here. And let's think about the ordering.
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Divergence theorem example 1 Divergence theorem Multivariable Calculus Khan Academy.mp3
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And so taking the divergence really, really, really simplified things, the divergence of f simplified down to 2x. So now we can restate the flux across the surface as a triple integral of 2x. So let me just write 2x here. And let's think about the ordering. So y can go between 0 and this plane that is a function of z. So let's write that down. So y is bounded below by 0 and above by this plane 2 minus z.
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Divergence theorem example 1 Divergence theorem Multivariable Calculus Khan Academy.mp3
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And let's think about the ordering. So y can go between 0 and this plane that is a function of z. So let's write that down. So y is bounded below by 0 and above by this plane 2 minus z. And z is bounded below by 0 and above by these parabolas of 1 minus x squared. And then x is bounded below by negative 1 and bounded above by 1. So negative 1 is less than or equal to x is less than or equal to 1.
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Divergence theorem example 1 Divergence theorem Multivariable Calculus Khan Academy.mp3
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So y is bounded below by 0 and above by this plane 2 minus z. And z is bounded below by 0 and above by these parabolas of 1 minus x squared. And then x is bounded below by negative 1 and bounded above by 1. So negative 1 is less than or equal to x is less than or equal to 1. And so this is probably a good order of integration. We can integrate with respect to y first, and then we'll get a function of z. Then we can integrate with respect to z, and we'll get a function of x.
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Divergence theorem example 1 Divergence theorem Multivariable Calculus Khan Academy.mp3
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So negative 1 is less than or equal to x is less than or equal to 1. And so this is probably a good order of integration. We can integrate with respect to y first, and then we'll get a function of z. Then we can integrate with respect to z, and we'll get a function of x. And then we can integrate with respect to x. So let's do it in that order. So first we'll integrate with respect to y.
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Divergence theorem example 1 Divergence theorem Multivariable Calculus Khan Academy.mp3
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