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And this is neat, because using Stokes' theorem in this special case where we're dealing with a flattened out surface in the xy plane, in this situation, this just boiled down to Green's theorem. This thing right over here just boiled down to Green's theorem. So we see that Green's theorem is really just a special case. Let me write theorem a little bit neater. We see that Green's theorem is really just a special case of Stokes' theorem where our surface is flattened out and it's in the xy plane. So that should make us feel pretty good. Although we still have not proven Stokes' theorem.
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Green's and Stokes' theorem relationship Multivariable Calculus Khan Academy.mp3
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Let me write theorem a little bit neater. We see that Green's theorem is really just a special case of Stokes' theorem where our surface is flattened out and it's in the xy plane. So that should make us feel pretty good. Although we still have not proven Stokes' theorem. But the one thing that I do like about this is seeing that Green's theorem and Stokes' theorem is consistent is now it starts to make sense of this right over here. When we first learned Green's theorem, we're like, what's going on over here? But now this is telling us this is just taking the curl in this region along this surface.
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Green's and Stokes' theorem relationship Multivariable Calculus Khan Academy.mp3
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And then the base of this building, or the contour of its walls, was defined by the path where we have a circle of radius 2 along here. Then we go down along the y-axis, and then we take another left, and we go along the x-axis, and that was our building. And in the last video, we figured out this first wall's surface area. In fact, you can think of it, our original problem is we wanted to figure out the line integral along the closed path, so it was a closed line integral, along the closed path c of f of x, y, and we're always multiplying f of x, y times a little small distance of our path, ds. We're writing this in the most abstract way possible. And what we saw in the last video is the easiest way to do this is to break this up into multiple paths, or into multiple problems. So you can imagine this whole contour, this whole path we call c, but we could call this part that we figured out in the last video c1.
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Line integral example 2 (part 2) Multivariable Calculus Khan Academy.mp3
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In fact, you can think of it, our original problem is we wanted to figure out the line integral along the closed path, so it was a closed line integral, along the closed path c of f of x, y, and we're always multiplying f of x, y times a little small distance of our path, ds. We're writing this in the most abstract way possible. And what we saw in the last video is the easiest way to do this is to break this up into multiple paths, or into multiple problems. So you can imagine this whole contour, this whole path we call c, but we could call this part that we figured out in the last video c1. This part we can call, let me make a pointer, c2. And then this point right here, c3. So we could redefine, or we can break up this line integral, this closed line integral, into three non-closed line integrals.
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Line integral example 2 (part 2) Multivariable Calculus Khan Academy.mp3
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So you can imagine this whole contour, this whole path we call c, but we could call this part that we figured out in the last video c1. This part we can call, let me make a pointer, c2. And then this point right here, c3. So we could redefine, or we can break up this line integral, this closed line integral, into three non-closed line integrals. This will be equal to the line integral along the path c1 of f of x, y, ds, plus the line integral along c2 of f of x, y, ds, plus the line integral, and you might have guessed it, along c3 of f of x, y, ds. And in the last video, we got as far as figuring out this first part, this first curvy wall right here. Its surface area, we figured out, was 4 plus 2 pi.
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Line integral example 2 (part 2) Multivariable Calculus Khan Academy.mp3
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So we could redefine, or we can break up this line integral, this closed line integral, into three non-closed line integrals. This will be equal to the line integral along the path c1 of f of x, y, ds, plus the line integral along c2 of f of x, y, ds, plus the line integral, and you might have guessed it, along c3 of f of x, y, ds. And in the last video, we got as far as figuring out this first part, this first curvy wall right here. Its surface area, we figured out, was 4 plus 2 pi. Now we've got to figure out the other two parts. So let's do c2, let's do this line integral next. And in order to do it, we need to do another parametrization of x and y.
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Line integral example 2 (part 2) Multivariable Calculus Khan Academy.mp3
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Its surface area, we figured out, was 4 plus 2 pi. Now we've got to figure out the other two parts. So let's do c2, let's do this line integral next. And in order to do it, we need to do another parametrization of x and y. It's going to be different than what we did for this part. We're no longer along this circle, we're just along the y-axis. So as long as we're there, x is definitely going to be equal to 0.
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Line integral example 2 (part 2) Multivariable Calculus Khan Academy.mp3
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And in order to do it, we need to do another parametrization of x and y. It's going to be different than what we did for this part. We're no longer along this circle, we're just along the y-axis. So as long as we're there, x is definitely going to be equal to 0. So that's my parametrization. x is equal to 0. So if we're along the y-axis, x is definitely equal to 0.
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Line integral example 2 (part 2) Multivariable Calculus Khan Academy.mp3
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So as long as we're there, x is definitely going to be equal to 0. So that's my parametrization. x is equal to 0. So if we're along the y-axis, x is definitely equal to 0. And then y, we could say it starts off at y is equal to 2. Maybe we'll say y is equal to 2 minus t, for t is between 0, t is greater than or equal to 0, less than or equal to 2. And that should work.
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Line integral example 2 (part 2) Multivariable Calculus Khan Academy.mp3
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So if we're along the y-axis, x is definitely equal to 0. And then y, we could say it starts off at y is equal to 2. Maybe we'll say y is equal to 2 minus t, for t is between 0, t is greater than or equal to 0, less than or equal to 2. And that should work. When t is equal to 0, we're at this point right there. And then as t increases towards 2, we move down the y-axis. And eventually, when t is equal to 2, we're at that point right there.
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Line integral example 2 (part 2) Multivariable Calculus Khan Academy.mp3
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And that should work. When t is equal to 0, we're at this point right there. And then as t increases towards 2, we move down the y-axis. And eventually, when t is equal to 2, we're at that point right there. So that's our parametrization. And so let's evaluate this line. And actually, we could do our derivatives too, if we like.
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Line integral example 2 (part 2) Multivariable Calculus Khan Academy.mp3
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And eventually, when t is equal to 2, we're at that point right there. So that's our parametrization. And so let's evaluate this line. And actually, we could do our derivatives too, if we like. What's the derivative? Let me write it right here. What's dx dt?
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Line integral example 2 (part 2) Multivariable Calculus Khan Academy.mp3
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And actually, we could do our derivatives too, if we like. What's the derivative? Let me write it right here. What's dx dt? Pretty straightforward. Derivative of 0 is 0. And dy dt is equal to the derivative of this.
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Line integral example 2 (part 2) Multivariable Calculus Khan Academy.mp3
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What's dx dt? Pretty straightforward. Derivative of 0 is 0. And dy dt is equal to the derivative of this. It's just minus 1. 2 minus t, derivative of minus t is just minus 1. And so let's just break it up.
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Line integral example 2 (part 2) Multivariable Calculus Khan Academy.mp3
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And dy dt is equal to the derivative of this. It's just minus 1. 2 minus t, derivative of minus t is just minus 1. And so let's just break it up. So we have this thing right here. So we have the integral along C2. But instead of writing C2, I'll leave C2 there.
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Line integral example 2 (part 2) Multivariable Calculus Khan Academy.mp3
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And so let's just break it up. So we have this thing right here. So we have the integral along C2. But instead of writing C2, I'll leave C2 there. But we'll say we're going from t is equal to 0 to 2 of f of xy. So f of xy is this thing right here, is x plus y squared. And then times ds.
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Line integral example 2 (part 2) Multivariable Calculus Khan Academy.mp3
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But instead of writing C2, I'll leave C2 there. But we'll say we're going from t is equal to 0 to 2 of f of xy. So f of xy is this thing right here, is x plus y squared. And then times ds. ds times ds. And we know from the last several videos, ds can be rewritten as the square root of dx dt squared. So 0 squared plus dy dt squared.
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Line integral example 2 (part 2) Multivariable Calculus Khan Academy.mp3
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And then times ds. ds times ds. And we know from the last several videos, ds can be rewritten as the square root of dx dt squared. So 0 squared plus dy dt squared. So minus 1 squared is 1. All of that times dt. And obviously, this is pretty nice.
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Line integral example 2 (part 2) Multivariable Calculus Khan Academy.mp3
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So 0 squared plus dy dt squared. So minus 1 squared is 1. All of that times dt. And obviously, this is pretty nice. It gets very nice and clean. This is 0 plus 1 square root. This just becomes 1.
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Line integral example 2 (part 2) Multivariable Calculus Khan Academy.mp3
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And obviously, this is pretty nice. It gets very nice and clean. This is 0 plus 1 square root. This just becomes 1. And then what is x? x, if we write it in terms of our parametrization, is always going to be equal to 0. And then y squared is going to be 2 minus t squared.
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Line integral example 2 (part 2) Multivariable Calculus Khan Academy.mp3
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This just becomes 1. And then what is x? x, if we write it in terms of our parametrization, is always going to be equal to 0. And then y squared is going to be 2 minus t squared. So this is going to be 2 minus t squared. So this whole crazy thing simplified to, we're going to go from t is equal to 0 to t is equal to 2. The x disappears in our parametrization.
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Line integral example 2 (part 2) Multivariable Calculus Khan Academy.mp3
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And then y squared is going to be 2 minus t squared. So this is going to be 2 minus t squared. So this whole crazy thing simplified to, we're going to go from t is equal to 0 to t is equal to 2. The x disappears in our parametrization. x just stays 0 regardless of what t is. And then you have y squared. But y is the same thing as 2 minus t. So 2 minus t squared.
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Line integral example 2 (part 2) Multivariable Calculus Khan Academy.mp3
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The x disappears in our parametrization. x just stays 0 regardless of what t is. And then you have y squared. But y is the same thing as 2 minus t. So 2 minus t squared. And then you have your dt sitting out there. This is pretty straightforward. I always find it easier when you're finding an antiderivative of this.
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Line integral example 2 (part 2) Multivariable Calculus Khan Academy.mp3
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But y is the same thing as 2 minus t. So 2 minus t squared. And then you have your dt sitting out there. This is pretty straightforward. I always find it easier when you're finding an antiderivative of this. Although you can do this in your head. I like to just actually multiply out this binomial. So this is going to be equal to the antiderivative from t is equal to 0 to t is equal to 2 of 4 minus 4t plus t squared.
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Line integral example 2 (part 2) Multivariable Calculus Khan Academy.mp3
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I always find it easier when you're finding an antiderivative of this. Although you can do this in your head. I like to just actually multiply out this binomial. So this is going to be equal to the antiderivative from t is equal to 0 to t is equal to 2 of 4 minus 4t plus t squared. Just like that. dt. And this is pretty straightforward.
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Line integral example 2 (part 2) Multivariable Calculus Khan Academy.mp3
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So this is going to be equal to the antiderivative from t is equal to 0 to t is equal to 2 of 4 minus 4t plus t squared. Just like that. dt. And this is pretty straightforward. This is going to be the antiderivative of this is 4t minus 2t squared. When you take the derivative, 2 times minus 2 is minus 4t. And then you have plus 1 third t to the third.
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Line integral example 2 (part 2) Multivariable Calculus Khan Academy.mp3
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And this is pretty straightforward. This is going to be the antiderivative of this is 4t minus 2t squared. When you take the derivative, 2 times minus 2 is minus 4t. And then you have plus 1 third t to the third. These are just simple antiderivatives. And we need to evaluate it from 0 to 2. And so let's evaluate it at 2.
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Line integral example 2 (part 2) Multivariable Calculus Khan Academy.mp3
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And then you have plus 1 third t to the third. These are just simple antiderivatives. And we need to evaluate it from 0 to 2. And so let's evaluate it at 2. 4 times 2 is 8. Let me pick a new color. 4 times 2 is 8 minus 2 times 2 squared.
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Line integral example 2 (part 2) Multivariable Calculus Khan Academy.mp3
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And so let's evaluate it at 2. 4 times 2 is 8. Let me pick a new color. 4 times 2 is 8 minus 2 times 2 squared. So 2 times 4. So minus 8 plus 1 third times 2 to the third power. So 1 third times 8.
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Line integral example 2 (part 2) Multivariable Calculus Khan Academy.mp3
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4 times 2 is 8 minus 2 times 2 squared. So 2 times 4. So minus 8 plus 1 third times 2 to the third power. So 1 third times 8. So these cancel out. We have 8 minus 8. And we just have 8 thirds.
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Line integral example 2 (part 2) Multivariable Calculus Khan Academy.mp3
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So 1 third times 8. So these cancel out. We have 8 minus 8. And we just have 8 thirds. So this just becomes 8 thirds. And then we have to put a 0 in. Minus 0 evaluated here.
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Line integral example 2 (part 2) Multivariable Calculus Khan Academy.mp3
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And we just have 8 thirds. So this just becomes 8 thirds. And then we have to put a 0 in. Minus 0 evaluated here. But this is just going to be 0. We have 4 times 0, 2 times 0. All of these are going to be 0.
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Line integral example 2 (part 2) Multivariable Calculus Khan Academy.mp3
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Minus 0 evaluated here. But this is just going to be 0. We have 4 times 0, 2 times 0. All of these are going to be 0. So minus 0. So just like that, we found our surface area of our second wall. So this turned out being this right here is 8 thirds.
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Line integral example 2 (part 2) Multivariable Calculus Khan Academy.mp3
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All of these are going to be 0. So minus 0. So just like that, we found our surface area of our second wall. So this turned out being this right here is 8 thirds. And now we have our last wall. And then we could just add them up. So we have our last wall.
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Line integral example 2 (part 2) Multivariable Calculus Khan Academy.mp3
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So this turned out being this right here is 8 thirds. And now we have our last wall. And then we could just add them up. So we have our last wall. We're looking. So let's just do another parametrization. I want to have the graph there.
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Line integral example 2 (part 2) Multivariable Calculus Khan Academy.mp3
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So we have our last wall. We're looking. So let's just do another parametrization. I want to have the graph there. Well, maybe I can paste it again. Edit. So there's the graph again.
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Line integral example 2 (part 2) Multivariable Calculus Khan Academy.mp3
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I want to have the graph there. Well, maybe I can paste it again. Edit. So there's the graph again. And now we're going to do our last wall. So our last wall is this one right here, which is we could write it. This was C3.
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Line integral example 2 (part 2) Multivariable Calculus Khan Academy.mp3
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So there's the graph again. And now we're going to do our last wall. So our last wall is this one right here, which is we could write it. This was C3. Let me switch colors here. So this is C. We're going to go along contour C3 of f of x, y, ds, which is the same thing as let's do a parametrization. Along this curve, if we just say x is equal to t, very straightforward, for t is greater than or equal to 0, less than or equal to 2.
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Line integral example 2 (part 2) Multivariable Calculus Khan Academy.mp3
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This was C3. Let me switch colors here. So this is C. We're going to go along contour C3 of f of x, y, ds, which is the same thing as let's do a parametrization. Along this curve, if we just say x is equal to t, very straightforward, for t is greater than or equal to 0, less than or equal to 2. And then this whole time that we're along the x-axis, y is going to be equal to 0. That's a pretty straightforward parametrization. So this is going to be equal to.
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Line integral example 2 (part 2) Multivariable Calculus Khan Academy.mp3
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Along this curve, if we just say x is equal to t, very straightforward, for t is greater than or equal to 0, less than or equal to 2. And then this whole time that we're along the x-axis, y is going to be equal to 0. That's a pretty straightforward parametrization. So this is going to be equal to. We're going to go from t is equal to 0 to t is equal to 2 of f of x, y, which is I'll write in terms of x right now, x and y, x plus y squared, times ds. Now, what is dx? Well, let me write ds right here, times ds.
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Line integral example 2 (part 2) Multivariable Calculus Khan Academy.mp3
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So this is going to be equal to. We're going to go from t is equal to 0 to t is equal to 2 of f of x, y, which is I'll write in terms of x right now, x and y, x plus y squared, times ds. Now, what is dx? Well, let me write ds right here, times ds. That's what we're dealing with. Now, we know what ds is. ds is equal to the square root of dx dt squared plus dy dt squared times dt.
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Line integral example 2 (part 2) Multivariable Calculus Khan Academy.mp3
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Well, let me write ds right here, times ds. That's what we're dealing with. Now, we know what ds is. ds is equal to the square root of dx dt squared plus dy dt squared times dt. We proved that in the first video. Or we didn't rigorously prove it, but we got the sense of why this is true. And what's the derivative of x with respect to t?
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Line integral example 2 (part 2) Multivariable Calculus Khan Academy.mp3
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ds is equal to the square root of dx dt squared plus dy dt squared times dt. We proved that in the first video. Or we didn't rigorously prove it, but we got the sense of why this is true. And what's the derivative of x with respect to t? Well, it's just 1. So this is just going to be a 1. 1 squared, same thing.
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Line integral example 2 (part 2) Multivariable Calculus Khan Academy.mp3
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And what's the derivative of x with respect to t? Well, it's just 1. So this is just going to be a 1. 1 squared, same thing. And derivative of y with respect to t is 0. So this is a 0. 1 plus 0 is 1.
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Line integral example 2 (part 2) Multivariable Calculus Khan Academy.mp3
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1 squared, same thing. And derivative of y with respect to t is 0. So this is a 0. 1 plus 0 is 1. Square root of 1 is 1. So this thing just becomes dt. ds is going to be equal to dt in this case.
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Line integral example 2 (part 2) Multivariable Calculus Khan Academy.mp3
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1 plus 0 is 1. Square root of 1 is 1. So this thing just becomes dt. ds is going to be equal to dt in this case. So this just becomes a dt. And then our x is going to be equal to a t. That's our definition of our parametrization. And y is 0, so we can ignore it.
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Line integral example 2 (part 2) Multivariable Calculus Khan Academy.mp3
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ds is going to be equal to dt in this case. So this just becomes a dt. And then our x is going to be equal to a t. That's our definition of our parametrization. And y is 0, so we can ignore it. So this was a super simple integral. So this simplified down to, we're going to go from 0 to 2 of t dt, which is equal to the anti derivative of t is just 1 half t squared. And we're going to go 0 to 2, which is equal to 1 half times 2 squared.
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Line integral example 2 (part 2) Multivariable Calculus Khan Academy.mp3
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And y is 0, so we can ignore it. So this was a super simple integral. So this simplified down to, we're going to go from 0 to 2 of t dt, which is equal to the anti derivative of t is just 1 half t squared. And we're going to go 0 to 2, which is equal to 1 half times 2 squared. 2 squared is 4 times 1 half is 2. And then minus 1 half times 0 squared. So this third wall's area right there is just 2, pretty straightforward.
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Line integral example 2 (part 2) Multivariable Calculus Khan Academy.mp3
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And we're going to go 0 to 2, which is equal to 1 half times 2 squared. 2 squared is 4 times 1 half is 2. And then minus 1 half times 0 squared. So this third wall's area right there is just 2, pretty straightforward. So that right there, the area there is just 2. And so to answer our question, what was this line integral evaluated over this closed path of f of x, y? Well, we just add up these numbers.
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Line integral example 2 (part 2) Multivariable Calculus Khan Academy.mp3
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So this third wall's area right there is just 2, pretty straightforward. So that right there, the area there is just 2. And so to answer our question, what was this line integral evaluated over this closed path of f of x, y? Well, we just add up these numbers. We have 4 plus 2 pi plus 8 thirds plus 2. Well, what is this? 8 thirds is the same thing as 2 and 2 thirds.
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Line integral example 2 (part 2) Multivariable Calculus Khan Academy.mp3
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Well, we just add up these numbers. We have 4 plus 2 pi plus 8 thirds plus 2. Well, what is this? 8 thirds is the same thing as 2 and 2 thirds. So we have 4 plus 2 and 2 thirds is 6 and 2 thirds plus another 2 is 8 and 2 thirds. So this whole thing becomes 8 and 2 thirds, if you write it as a mixed number, plus 2 pi. And we're done.
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Line integral example 2 (part 2) Multivariable Calculus Khan Academy.mp3
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Draw my y-axis, that is my x-axis. Let's say the path looks like this. It looks something like this. It's the same one we had in the last video. It might not look exactly like it. Let me see what I did in the last video. It looked like that in the last video, but close enough.
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Green's theorem proof (part 2) Multivariable Calculus Khan Academy.mp3
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It's the same one we had in the last video. It might not look exactly like it. Let me see what I did in the last video. It looked like that in the last video, but close enough. Let's say we're dealing with the exact same curve as the last video. We could call that curve C. In the last video, we dealt with a vector field that only had vectors in the i direction. Let's deal with another vector field that only has vectors in the j direction, or the vertical direction.
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Green's theorem proof (part 2) Multivariable Calculus Khan Academy.mp3
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It looked like that in the last video, but close enough. Let's say we're dealing with the exact same curve as the last video. We could call that curve C. In the last video, we dealt with a vector field that only had vectors in the i direction. Let's deal with another vector field that only has vectors in the j direction, or the vertical direction. Let's say that Q, the vector field Q of x, y, is equal to capital Q of x, y times j. We are going to concern ourselves with the closed line integral around the path C of Q dot dr. We've seen it already. dr can be rewritten as dx times i plus dy times j.
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Green's theorem proof (part 2) Multivariable Calculus Khan Academy.mp3
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Let's deal with another vector field that only has vectors in the j direction, or the vertical direction. Let's say that Q, the vector field Q of x, y, is equal to capital Q of x, y times j. We are going to concern ourselves with the closed line integral around the path C of Q dot dr. We've seen it already. dr can be rewritten as dx times i plus dy times j. If we were to take the dot product of these two, this line integral is going to be the exact same thing. This is going to be the same thing as the closed line integral over C of Q dot dr. Q only has a j component. If you take its zero i, so zero times dx, that's just zero.
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Green's theorem proof (part 2) Multivariable Calculus Khan Academy.mp3
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dr can be rewritten as dx times i plus dy times j. If we were to take the dot product of these two, this line integral is going to be the exact same thing. This is going to be the same thing as the closed line integral over C of Q dot dr. Q only has a j component. If you take its zero i, so zero times dx, that's just zero. Then you're going to have Q x, y times dy. It had no i component. This is just going to be Q of x and y. Q of x and y times dy.
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Green's theorem proof (part 2) Multivariable Calculus Khan Academy.mp3
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If you take its zero i, so zero times dx, that's just zero. Then you're going to have Q x, y times dy. It had no i component. This is just going to be Q of x and y. Q of x and y times dy. That's the dot product. There was no i component. That's why we lose the dx.
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Green's theorem proof (part 2) Multivariable Calculus Khan Academy.mp3
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This is just going to be Q of x and y. Q of x and y times dy. That's the dot product. There was no i component. That's why we lose the dx. Let's see if there's any way that we can solve this line integral without having to resort to a third parameter, t. Just like we did in the last video. It will be almost identical. We're just dealing with y's now instead of x's.
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Green's theorem proof (part 2) Multivariable Calculus Khan Academy.mp3
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That's why we lose the dx. Let's see if there's any way that we can solve this line integral without having to resort to a third parameter, t. Just like we did in the last video. It will be almost identical. We're just dealing with y's now instead of x's. What we can do is say, what's our minimum y and our maximum y? Our minimum y, let's say it's right here. Let's call that a.
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Green's theorem proof (part 2) Multivariable Calculus Khan Academy.mp3
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We're just dealing with y's now instead of x's. What we can do is say, what's our minimum y and our maximum y? Our minimum y, let's say it's right here. Let's call that a. Let's say our maximum y that we attain is right over there. Let's call that b. Just like in the last, I forgot to tell you the direction of the curve.
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Green's theorem proof (part 2) Multivariable Calculus Khan Academy.mp3
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Let's call that a. Let's say our maximum y that we attain is right over there. Let's call that b. Just like in the last, I forgot to tell you the direction of the curve. This is the same path as last time. We're going in a counterclockwise direction. It's the exact same curve, exact same path.
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Green's theorem proof (part 2) Multivariable Calculus Khan Academy.mp3
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Just like in the last, I forgot to tell you the direction of the curve. This is the same path as last time. We're going in a counterclockwise direction. It's the exact same curve, exact same path. We're going in that direction. In the last video, we broke it up into two functions of x's. Two y's, a function of x.
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Green's theorem proof (part 2) Multivariable Calculus Khan Academy.mp3
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It's the exact same curve, exact same path. We're going in that direction. In the last video, we broke it up into two functions of x's. Two y's, a function of x. Now we want to deal with y's. Let's break it up into two functions of y. If we break this path into two paths, those are kind of our extreme points.
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Green's theorem proof (part 2) Multivariable Calculus Khan Academy.mp3
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Two y's, a function of x. Now we want to deal with y's. Let's break it up into two functions of y. If we break this path into two paths, those are kind of our extreme points. Let's call this path right here. Let's call that path right there. Let's call that y.
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Green's theorem proof (part 2) Multivariable Calculus Khan Academy.mp3
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If we break this path into two paths, those are kind of our extreme points. Let's call this path right here. Let's call that path right there. Let's call that y. Let's call that x of. Or along this path, x is equal to, I could just write this path 2 or I'll call it c2. We could say it's x is equal to x2 of y.
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Green's theorem proof (part 2) Multivariable Calculus Khan Academy.mp3
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Let's call that y. Let's call that x of. Or along this path, x is equal to, I could just write this path 2 or I'll call it c2. We could say it's x is equal to x2 of y. That's that path. Then the first path, it doesn't have to be the first path, depending where you start. You can start anywhere.
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Green's theorem proof (part 2) Multivariable Calculus Khan Academy.mp3
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We could say it's x is equal to x2 of y. That's that path. Then the first path, it doesn't have to be the first path, depending where you start. You can start anywhere. Let's say this one in magenta. We'll call that path 1. We could say that that's defined as x is equal to x1 of y.
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Green's theorem proof (part 2) Multivariable Calculus Khan Academy.mp3
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You can start anywhere. Let's say this one in magenta. We'll call that path 1. We could say that that's defined as x is equal to x1 of y. It's a little confusing when you have x as a function of y, but it's really completely analogous to what we did in the last video. We're literally just swapping x's and y's. We're now expressing x as a function of y instead of y as a function of x.
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Green's theorem proof (part 2) Multivariable Calculus Khan Academy.mp3
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We could say that that's defined as x is equal to x1 of y. It's a little confusing when you have x as a function of y, but it's really completely analogous to what we did in the last video. We're literally just swapping x's and y's. We're now expressing x as a function of y instead of y as a function of x. We have these two curves. You can imagine just flipping it, and we're doing the exact same thing that we did in the last video. It's just now in terms of y.
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Green's theorem proof (part 2) Multivariable Calculus Khan Academy.mp3
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We're now expressing x as a function of y instead of y as a function of x. We have these two curves. You can imagine just flipping it, and we're doing the exact same thing that we did in the last video. It's just now in terms of y. If you look at it this way, this line integral can be rewritten. This could be rewritten as being equal to the integral. Let's just do c2 first.
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Green's theorem proof (part 2) Multivariable Calculus Khan Academy.mp3
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It's just now in terms of y. If you look at it this way, this line integral can be rewritten. This could be rewritten as being equal to the integral. Let's just do c2 first. This is the integral from b to a. We start at b, and we go to a. We're coming back from a high y to a low y.
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Green's theorem proof (part 2) Multivariable Calculus Khan Academy.mp3
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Let's just do c2 first. This is the integral from b to a. We start at b, and we go to a. We're coming back from a high y to a low y. The integral from b to a of q of, instead of having an x there, we know along this curve right here, x is equal to, we want everything in terms of y. Here, x is equal to x2 of y. q of x2 of y, maybe I'm using too many colors here, but I think you get the idea. dy.
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Green's theorem proof (part 2) Multivariable Calculus Khan Academy.mp3
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We're coming back from a high y to a low y. The integral from b to a of q of, instead of having an x there, we know along this curve right here, x is equal to, we want everything in terms of y. Here, x is equal to x2 of y. q of x2 of y, maybe I'm using too many colors here, but I think you get the idea. dy. This is the part of the line integral just over this left-hand curve. Then we're going to add to that the line integral, or really just the regular integral now, from y is equal to a to y is equal to b of q of, and instead of x being equal to x2, now x is equal to x1 of y. It's equal to this curve, this other function.
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Green's theorem proof (part 2) Multivariable Calculus Khan Academy.mp3
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dy. This is the part of the line integral just over this left-hand curve. Then we're going to add to that the line integral, or really just the regular integral now, from y is equal to a to y is equal to b of q of, and instead of x being equal to x2, now x is equal to x1 of y. It's equal to this curve, this other function. x1 of y, y, dy. We can do exactly what we did in the previous video. We don't like the larger number on the bottom, so let's swap these two around.
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Green's theorem proof (part 2) Multivariable Calculus Khan Academy.mp3
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It's equal to this curve, this other function. x1 of y, y, dy. We can do exactly what we did in the previous video. We don't like the larger number on the bottom, so let's swap these two around. If you swap these two, if you make this into an a and this into a b, that makes it the negative of the integral. When you swap the two, change the direction. This is exactly what we did in the last video, so hopefully it's nothing too fancy.
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Green's theorem proof (part 2) Multivariable Calculus Khan Academy.mp3
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We don't like the larger number on the bottom, so let's swap these two around. If you swap these two, if you make this into an a and this into a b, that makes it the negative of the integral. When you swap the two, change the direction. This is exactly what we did in the last video, so hopefully it's nothing too fancy. Now that we have the same boundaries of integration, these two definite integrals, we can just write them as one definite integral. This is going to be equal to the integral from a to b. I'll write this one first, since it's positive. q of x1 of y, y minus this one.
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Green's theorem proof (part 2) Multivariable Calculus Khan Academy.mp3
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This is exactly what we did in the last video, so hopefully it's nothing too fancy. Now that we have the same boundaries of integration, these two definite integrals, we can just write them as one definite integral. This is going to be equal to the integral from a to b. I'll write this one first, since it's positive. q of x1 of y, y minus this one. We have the minus sign here. Minus q of x2 of y and y, dy. Let me do that in that neutral color.
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Green's theorem proof (part 2) Multivariable Calculus Khan Academy.mp3
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q of x1 of y, y minus this one. We have the minus sign here. Minus q of x2 of y and y, dy. Let me do that in that neutral color. dy, that's multiplied by all of these things. I distributed out the dy. I think you get the idea.
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Green's theorem proof (part 2) Multivariable Calculus Khan Academy.mp3
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Let me do that in that neutral color. dy, that's multiplied by all of these things. I distributed out the dy. I think you get the idea. This is identical to what we did in the last video. This could be rewritten as this is equal to the integral from a to b of, and inside of the integral we're evaluating the function q of xy from the boundaries, the upper boundary, where the upper boundary is going to be from x is equal to x1 of y and the lower boundary is x is equal to x2 of y. All the x's, we substitute it with that and then we get some expression.
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Green's theorem proof (part 2) Multivariable Calculus Khan Academy.mp3
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I think you get the idea. This is identical to what we did in the last video. This could be rewritten as this is equal to the integral from a to b of, and inside of the integral we're evaluating the function q of xy from the boundaries, the upper boundary, where the upper boundary is going to be from x is equal to x1 of y and the lower boundary is x is equal to x2 of y. All the x's, we substitute it with that and then we get some expression. From that, we subtract this with x, substitute it as x2 of y. That's exactly what we did, just like I said in the last video. We're going in the reverse direction that we normally go in definite integrals.
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Green's theorem proof (part 2) Multivariable Calculus Khan Academy.mp3
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All the x's, we substitute it with that and then we get some expression. From that, we subtract this with x, substitute it as x2 of y. That's exactly what we did, just like I said in the last video. We're going in the reverse direction that we normally go in definite integrals. We normally get to this and then the next step is we get this. Now we're going in the reverse direction, but it's all the same difference. All of that times dy.
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Green's theorem proof (part 2) Multivariable Calculus Khan Academy.mp3
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We're going in the reverse direction that we normally go in definite integrals. We normally get to this and then the next step is we get this. Now we're going in the reverse direction, but it's all the same difference. All of that times dy. Just like we saw in the last video, this expression right here. Actually, let me draw that dy a little further out so it doesn't get all congested. Let me do that dy out here.
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Green's theorem proof (part 2) Multivariable Calculus Khan Academy.mp3
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All of that times dy. Just like we saw in the last video, this expression right here. Actually, let me draw that dy a little further out so it doesn't get all congested. Let me do that dy out here. This expression, this entire expression, is the same exact thing. That entire expression is the exact same thing as the integral from x is equal to I can just write it here. Let me write it in the same colors.
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Green's theorem proof (part 2) Multivariable Calculus Khan Academy.mp3
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Let me do that dy out here. This expression, this entire expression, is the same exact thing. That entire expression is the exact same thing as the integral from x is equal to I can just write it here. Let me write it in the same colors. x2 of y to x1 of y of the partial. I'll do it in the orange color. Of the partial of q with respect to x dx.
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Green's theorem proof (part 2) Multivariable Calculus Khan Academy.mp3
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Let me write it in the same colors. x2 of y to x1 of y of the partial. I'll do it in the orange color. Of the partial of q with respect to x dx. This, I want to make it very clear, this is, at least in my mind, the first part. When I first saw it, it was a little confusing. If you just saw an integral like this, if this is the inside of a double integral, and it is.
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Green's theorem proof (part 2) Multivariable Calculus Khan Academy.mp3
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Of the partial of q with respect to x dx. This, I want to make it very clear, this is, at least in my mind, the first part. When I first saw it, it was a little confusing. If you just saw an integral like this, if this is the inside of a double integral, and it is. The outside is what we saw there. It's the integral from a to b dy. If you just saw this in a double integral, what you would do is you would take the anti-derivative of this.
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Green's theorem proof (part 2) Multivariable Calculus Khan Academy.mp3
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If you just saw an integral like this, if this is the inside of a double integral, and it is. The outside is what we saw there. It's the integral from a to b dy. If you just saw this in a double integral, what you would do is you would take the anti-derivative of this. The anti-derivative of this with respect to x. The anti-derivative of the partial of q with respect to x. With respect to x is going to be just q of xy.
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Green's theorem proof (part 2) Multivariable Calculus Khan Academy.mp3
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If you just saw this in a double integral, what you would do is you would take the anti-derivative of this. The anti-derivative of this with respect to x. The anti-derivative of the partial of q with respect to x. With respect to x is going to be just q of xy. Since it's a definite integral, you would evaluate it at x1 of y. Then subtract from that this function evaluated at x2 of y, which is exactly what we did. Hopefully, you appreciate that.
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Green's theorem proof (part 2) Multivariable Calculus Khan Academy.mp3
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With respect to x is going to be just q of xy. Since it's a definite integral, you would evaluate it at x1 of y. Then subtract from that this function evaluated at x2 of y, which is exactly what we did. Hopefully, you appreciate that. Then we got our result, which is very similar to the last result. What does this double integral represent? What does it represent?
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Green's theorem proof (part 2) Multivariable Calculus Khan Academy.mp3
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Hopefully, you appreciate that. Then we got our result, which is very similar to the last result. What does this double integral represent? What does it represent? It represents anything. If you have any double integral that goes from, if you imagine this is some function. Let me draw it in three dimensions.
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Green's theorem proof (part 2) Multivariable Calculus Khan Academy.mp3
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What does it represent? It represents anything. If you have any double integral that goes from, if you imagine this is some function. Let me draw it in three dimensions. This is really almost a review of what we did in the last video. That's the y-axis. That's our x-axis.
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Green's theorem proof (part 2) Multivariable Calculus Khan Academy.mp3
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Let me draw it in three dimensions. This is really almost a review of what we did in the last video. That's the y-axis. That's our x-axis. That's our z-axis. This is some function of x and y. It's some surface you could imagine on the xy-plane.
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Green's theorem proof (part 2) Multivariable Calculus Khan Academy.mp3
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That's our x-axis. That's our z-axis. This is some function of x and y. It's some surface you could imagine on the xy-plane. It's some surface. We could call that the partial of q with respect to x. What this double integral is, this is essentially defining a region.
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Green's theorem proof (part 2) Multivariable Calculus Khan Academy.mp3
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It's some surface you could imagine on the xy-plane. It's some surface. We could call that the partial of q with respect to x. What this double integral is, this is essentially defining a region. You could view this dx times dy as a small differential of area. The region under question, the boundary points are from y going from, at the bottom, it goes from x2 of y, which we saw was a curve that looks something like this. That's the lower y.
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Green's theorem proof (part 2) Multivariable Calculus Khan Academy.mp3
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What this double integral is, this is essentially defining a region. You could view this dx times dy as a small differential of area. The region under question, the boundary points are from y going from, at the bottom, it goes from x2 of y, which we saw was a curve that looks something like this. That's the lower y. Over here, if we draw it in two dimensions, this was the lower y curve. The upper y curve is x1 of y. The upper y curve looks something like that.
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Green's theorem proof (part 2) Multivariable Calculus Khan Academy.mp3
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That's the lower y. Over here, if we draw it in two dimensions, this was the lower y curve. The upper y curve is x1 of y. The upper y curve looks something like that. The upper y curve goes something like that. x varies from the lower y curve to the upper y curve. That's what we're doing right here.
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Green's theorem proof (part 2) Multivariable Calculus Khan Academy.mp3
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The upper y curve looks something like that. The upper y curve goes something like that. x varies from the lower y curve to the upper y curve. That's what we're doing right here. Then y varies from a to b. This is essentially saying, let's take the double integral over this region, over this region right here, of this function. It's essentially the volume, this is the ceiling, and this boundary is essentially the wall.
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Green's theorem proof (part 2) Multivariable Calculus Khan Academy.mp3
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That's what we're doing right here. Then y varies from a to b. This is essentially saying, let's take the double integral over this region, over this region right here, of this function. It's essentially the volume, this is the ceiling, and this boundary is essentially the wall. It's the volume of that room. I don't know what it would look like when it comes up here, but you can imagine something like that. It would be the volume of that.
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Green's theorem proof (part 2) Multivariable Calculus Khan Academy.mp3
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It's essentially the volume, this is the ceiling, and this boundary is essentially the wall. It's the volume of that room. I don't know what it would look like when it comes up here, but you can imagine something like that. It would be the volume of that. That's what we're taking. This is the identical result we got in the last video. This is a pretty neat thing.
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Green's theorem proof (part 2) Multivariable Calculus Khan Academy.mp3
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It would be the volume of that. That's what we're taking. This is the identical result we got in the last video. This is a pretty neat thing. All of a sudden, this vector, and q of xy, I didn't draw it out like I did the last time. This q of xy only has things in the j direction. If I were to draw it, it's a vector field.
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Green's theorem proof (part 2) Multivariable Calculus Khan Academy.mp3
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