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So plus x times y squared. And then there could be some function of y here. So plus some, I don't know, I'll call it g of y. Because there could have been some function of y here. If it's a pure function of y, when you take the derivative or the partial with respect to x, this would have disappeared. So it would reappear when we take the anti-derivative. So let me, just to be clear, let me make it clear that f is going to be a function of x and y.
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Example of closed line integral of conservative field Multivariable Calculus Khan Academy.mp3
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Because there could have been some function of y here. If it's a pure function of y, when you take the derivative or the partial with respect to x, this would have disappeared. So it would reappear when we take the anti-derivative. So let me, just to be clear, let me make it clear that f is going to be a function of x and y. So this, we just took the, I guess you could say the anti-derivative with respect to x. Let's see if we take the anti-derivative with respect to y and then we can reconcile the two. So based on this, f of xy is going to have to look like, so let's take the anti-derivative with respect to y here.
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Example of closed line integral of conservative field Multivariable Calculus Khan Academy.mp3
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So let me, just to be clear, let me make it clear that f is going to be a function of x and y. So this, we just took the, I guess you could say the anti-derivative with respect to x. Let's see if we take the anti-derivative with respect to y and then we can reconcile the two. So based on this, f of xy is going to have to look like, so let's take the anti-derivative with respect to y here. So remember, you just treat x like it's just some number. It could be a k, it could be an m, it could be a 5. It's just some number.
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Example of closed line integral of conservative field Multivariable Calculus Khan Academy.mp3
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So based on this, f of xy is going to have to look like, so let's take the anti-derivative with respect to y here. So remember, you just treat x like it's just some number. It could be a k, it could be an m, it could be a 5. It's just some number. So if x is just some, the anti-derivative of 2y is y squared. And if x is just a number there, the anti-derivative of this with respect to y is just going to be xy squared. Don't believe me?
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Example of closed line integral of conservative field Multivariable Calculus Khan Academy.mp3
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It's just some number. So if x is just some, the anti-derivative of 2y is y squared. And if x is just a number there, the anti-derivative of this with respect to y is just going to be xy squared. Don't believe me? Take the partial of this with respect to y. Treat x like a constant, you'll get 2 times xy. And then you, with no exponent there.
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Example of closed line integral of conservative field Multivariable Calculus Khan Academy.mp3
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Don't believe me? Take the partial of this with respect to y. Treat x like a constant, you'll get 2 times xy. And then you, with no exponent there. And of course, since we took the anti-derivative with respect to x, there might be some function of x here. We were just basing it off of that information. Now, given that, this information says f of xy is going to have to look something like this.
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Example of closed line integral of conservative field Multivariable Calculus Khan Academy.mp3
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And then you, with no exponent there. And of course, since we took the anti-derivative with respect to x, there might be some function of x here. We were just basing it off of that information. Now, given that, this information says f of xy is going to have to look something like this. This information tells us f of xy is going to have to look something like that. Let's see if there is an f of xy that looks like both of them, essentially. So let's see.
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Example of closed line integral of conservative field Multivariable Calculus Khan Academy.mp3
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Now, given that, this information says f of xy is going to have to look something like this. This information tells us f of xy is going to have to look something like that. Let's see if there is an f of xy that looks like both of them, essentially. So let's see. We have, on this one, we have xy squared here. We have an xy squared there. So good.
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Example of closed line integral of conservative field Multivariable Calculus Khan Academy.mp3
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So let's see. We have, on this one, we have xy squared here. We have an xy squared there. So good. That looks good. And then over here, we have an f of x. We have something that's a pure function of x.
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Example of closed line integral of conservative field Multivariable Calculus Khan Academy.mp3
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So good. That looks good. And then over here, we have an f of x. We have something that's a pure function of x. And here we have something that is a pure function of x. So these two things could be the same thing. And then, here we have a pure function of y that might be there.
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Example of closed line integral of conservative field Multivariable Calculus Khan Academy.mp3
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We have something that's a pure function of x. And here we have something that is a pure function of x. So these two things could be the same thing. And then, here we have a pure function of y that might be there. But it didn't really show up anywhere over here. So we could just say, hey, that's going to be 0. 0 is a pure function of y.
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Example of closed line integral of conservative field Multivariable Calculus Khan Academy.mp3
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And then, here we have a pure function of y that might be there. But it didn't really show up anywhere over here. So we could just say, hey, that's going to be 0. 0 is a pure function of y. You could have something called g of y is equal to 0. And then we get that f of xy is equal to x to the third over 3 plus xy squared. And the gradient of this is going to be equal to f. And we've already established that.
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Example of closed line integral of conservative field Multivariable Calculus Khan Academy.mp3
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0 is a pure function of y. You could have something called g of y is equal to 0. And then we get that f of xy is equal to x to the third over 3 plus xy squared. And the gradient of this is going to be equal to f. And we've already established that. But just to hit the point home, let's take the gradient of it. Just so you don't believe this little stuff that I did right there. Let's take the gradient.
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Example of closed line integral of conservative field Multivariable Calculus Khan Academy.mp3
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And the gradient of this is going to be equal to f. And we've already established that. But just to hit the point home, let's take the gradient of it. Just so you don't believe this little stuff that I did right there. Let's take the gradient. The gradient of f is equal to, and sometimes people put a little vector there because you're getting a vector out of it. You could put a little vector on top of that gradient sign. The gradient of f is going to be what?
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Example of closed line integral of conservative field Multivariable Calculus Khan Academy.mp3
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Let's take the gradient. The gradient of f is equal to, and sometimes people put a little vector there because you're getting a vector out of it. You could put a little vector on top of that gradient sign. The gradient of f is going to be what? The partial of this with respect to x times i. So the partial of this with respect to x, the derivative here is 3 divided by 3 is 1. So just x squared plus the derivative of this with respect to x is y squared times i plus the partial with respect to y.
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Example of closed line integral of conservative field Multivariable Calculus Khan Academy.mp3
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The gradient of f is going to be what? The partial of this with respect to x times i. So the partial of this with respect to x, the derivative here is 3 divided by 3 is 1. So just x squared plus the derivative of this with respect to x is y squared times i plus the partial with respect to y. Well, the partial with respect to y of this is 0. The partial with respect to y of this is 2xy to the first. So it's 2xy times j.
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Example of closed line integral of conservative field Multivariable Calculus Khan Academy.mp3
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So just x squared plus the derivative of this with respect to x is y squared times i plus the partial with respect to y. Well, the partial with respect to y of this is 0. The partial with respect to y of this is 2xy to the first. So it's 2xy times j. And this is exactly equal to f, our f that we wrote up there. So we've established that f can definitely be written, f is definitely the gradient of some potential scalar function there. So f is conservative.
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Example of closed line integral of conservative field Multivariable Calculus Khan Academy.mp3
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So it's 2xy times j. And this is exactly equal to f, our f that we wrote up there. So we've established that f can definitely be written, f is definitely the gradient of some potential scalar function there. So f is conservative. And that tells us that this closed loop line integral of f is going to be equal to 0. And we are done. We've explored the actual parameterization of the path.
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Example of closed line integral of conservative field Multivariable Calculus Khan Academy.mp3
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So this over here is the input space, it's just a copy of the xy-plane. And the output space is also two-dimensional, so the output space in this case, also the two-dimensional plane. And what I'm gonna do, I'm just gonna first play an example of one of these transformations and then go through the details of the underlying function and how you can understand the transformation as a result. So here's what it looks like, here's what we're gonna be going towards. Very complicated, a lot of points moving, lots of different things happening here. And what's common with this sort of thing when you're thinking about moving from two dimensions to two dimensions, given that it's really the same space, the xy-plane, you often just think about the input and output space all at once and instead just watch a copy of that plane move on to itself. And by the way, when I say watch, I don't mean that you'll always have an animation like this just sort of sitting in front of you.
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Transformations, part 2 Multivariable calculus Khan Academy.mp3
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So here's what it looks like, here's what we're gonna be going towards. Very complicated, a lot of points moving, lots of different things happening here. And what's common with this sort of thing when you're thinking about moving from two dimensions to two dimensions, given that it's really the same space, the xy-plane, you often just think about the input and output space all at once and instead just watch a copy of that plane move on to itself. And by the way, when I say watch, I don't mean that you'll always have an animation like this just sort of sitting in front of you. When I think about transformations, it's usually a very vague thought in the back of my mind somewhere, but it helps to understand what's really going on with the function. I'll talk about that more at the end, but first let's just go into what this function is. So the one that I told the computer to animate here is f of xy, as the input, is equal to x squared plus y squared as the x component of the output, and x squared minus y squared as the y component of the output.
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Transformations, part 2 Multivariable calculus Khan Academy.mp3
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And by the way, when I say watch, I don't mean that you'll always have an animation like this just sort of sitting in front of you. When I think about transformations, it's usually a very vague thought in the back of my mind somewhere, but it helps to understand what's really going on with the function. I'll talk about that more at the end, but first let's just go into what this function is. So the one that I told the computer to animate here is f of xy, as the input, is equal to x squared plus y squared as the x component of the output, and x squared minus y squared as the y component of the output. So just to help start understanding this, let's take a relatively simple point like the origin. So here the origin, which is zero, zero, and let's think about what happens to that. f of zero, zero.
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Transformations, part 2 Multivariable calculus Khan Academy.mp3
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So the one that I told the computer to animate here is f of xy, as the input, is equal to x squared plus y squared as the x component of the output, and x squared minus y squared as the y component of the output. So just to help start understanding this, let's take a relatively simple point like the origin. So here the origin, which is zero, zero, and let's think about what happens to that. f of zero, zero. Well, x and y are both zero, so that top is zero. And same with the bottom. That bottom also equals zero, which means it's taking the point zero, zero to itself.
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Transformations, part 2 Multivariable calculus Khan Academy.mp3
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f of zero, zero. Well, x and y are both zero, so that top is zero. And same with the bottom. That bottom also equals zero, which means it's taking the point zero, zero to itself. And if you watch the transformation, what this means is that the point zero stays fixed. It's like you can hold your thumb down on it and nothing really happens to it. And in fact, we call this a fixed point of the function as a whole.
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Transformations, part 2 Multivariable calculus Khan Academy.mp3
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That bottom also equals zero, which means it's taking the point zero, zero to itself. And if you watch the transformation, what this means is that the point zero stays fixed. It's like you can hold your thumb down on it and nothing really happens to it. And in fact, we call this a fixed point of the function as a whole. And that kind of terminology doesn't really make sense unless you're thinking of the function as a transformation. So let's look at another example here. Let's take a point like one, one.
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Transformations, part 2 Multivariable calculus Khan Academy.mp3
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And in fact, we call this a fixed point of the function as a whole. And that kind of terminology doesn't really make sense unless you're thinking of the function as a transformation. So let's look at another example here. Let's take a point like one, one. f of one, one. So in the input space, let's just kind of start this thing over so we're only looking at the input. In the input space, one, one is sitting right here.
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Transformations, part 2 Multivariable calculus Khan Academy.mp3
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Let's take a point like one, one. f of one, one. So in the input space, let's just kind of start this thing over so we're only looking at the input. In the input space, one, one is sitting right here. And we're wondering where that's gonna move. So when we plug it in, x squared plus y squared, it's gonna be one squared plus one squared. And then the bottom, x squared minus y squared, one squared minus y squared.
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Transformations, part 2 Multivariable calculus Khan Academy.mp3
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In the input space, one, one is sitting right here. And we're wondering where that's gonna move. So when we plug it in, x squared plus y squared, it's gonna be one squared plus one squared. And then the bottom, x squared minus y squared, one squared minus y squared. Whoop, minus one squared. I'm plugging things in here. So that's two, zero.
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Transformations, part 2 Multivariable calculus Khan Academy.mp3
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And then the bottom, x squared minus y squared, one squared minus y squared. Whoop, minus one squared. I'm plugging things in here. So that's two, zero. Two, zero. Which means we expect this point to move over to two, zero in some way. So if we watch the transformation, we expect to watch that point move over to here.
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Transformations, part 2 Multivariable calculus Khan Academy.mp3
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So that's two, zero. Two, zero. Which means we expect this point to move over to two, zero in some way. So if we watch the transformation, we expect to watch that point move over to here. And again, it can be hard to follow because there's a lot of moving parts. But if you're careful as you watch it, the point will actually land right there. And you can, in principle, do this for any given point and understand how it moves from one to another.
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Transformations, part 2 Multivariable calculus Khan Academy.mp3
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So if we watch the transformation, we expect to watch that point move over to here. And again, it can be hard to follow because there's a lot of moving parts. But if you're careful as you watch it, the point will actually land right there. And you can, in principle, do this for any given point and understand how it moves from one to another. But you might ask, hey Grant, what is the point of all of this? We have other ways of visualizing functions that are more precise and kind of less confusing, to be honest. Like vector fields are a great way for functions like this.
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Transformations, part 2 Multivariable calculus Khan Academy.mp3
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And you can, in principle, do this for any given point and understand how it moves from one to another. But you might ask, hey Grant, what is the point of all of this? We have other ways of visualizing functions that are more precise and kind of less confusing, to be honest. Like vector fields are a great way for functions like this. Graphs were a great way for functions with one input and one output. Why think in terms of transformations? And the main reason is conceptual.
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Transformations, part 2 Multivariable calculus Khan Academy.mp3
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Like vector fields are a great way for functions like this. Graphs were a great way for functions with one input and one output. Why think in terms of transformations? And the main reason is conceptual. It's not like you'll have an animation sitting in front of you. And it's not like you're gonna, by hand, evaluate a bunch of points and think of how they move. But there's a lot of different concepts in math and with functions where when you understand it in terms of a transformation, it gives you a more nuanced understanding.
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Transformations, part 2 Multivariable calculus Khan Academy.mp3
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And the main reason is conceptual. It's not like you'll have an animation sitting in front of you. And it's not like you're gonna, by hand, evaluate a bunch of points and think of how they move. But there's a lot of different concepts in math and with functions where when you understand it in terms of a transformation, it gives you a more nuanced understanding. Things like derivatives or the variations of the derivative that you're gonna learn with multivariable calculus. There's different ways of understanding it in terms of like stretching or squishing space and things like this that doesn't really have a good analog in terms of graphs or vector fields. So it adds a new color to your understanding.
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Transformations, part 2 Multivariable calculus Khan Academy.mp3
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But there's a lot of different concepts in math and with functions where when you understand it in terms of a transformation, it gives you a more nuanced understanding. Things like derivatives or the variations of the derivative that you're gonna learn with multivariable calculus. There's different ways of understanding it in terms of like stretching or squishing space and things like this that doesn't really have a good analog in terms of graphs or vector fields. So it adds a new color to your understanding. Also, transformations are a super important part of linear algebra. There will come a point when you start learning the connection between linear algebra and multivariable calculus. And if you have a strong conception of transformations both in the context of linear algebra and in the context of multivariable calculus, you'll be in a much better position to understand the connection between those two fields.
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Transformations, part 2 Multivariable calculus Khan Academy.mp3
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And in order to evaluate a surface integral, we had to take the parameterization, take its partial with respect to s and with respect to t. We did that in the first video. Then we had to take its cross product. We did that in the second video. Now we're ready to take the magnitude of the cross product. We have to take the magnitude of the cross product. And then we can evaluate it inside of a double integral. And we will have solved, or we would have computed an actual surface integral, something you see very few times in your education career.
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Example of calculating a surface integral part 3 Multivariable Calculus Khan Academy.mp3
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Now we're ready to take the magnitude of the cross product. We have to take the magnitude of the cross product. And then we can evaluate it inside of a double integral. And we will have solved, or we would have computed an actual surface integral, something you see very few times in your education career. So this is kind of exciting. So this was the cross product right here. Now let's take the magnitude of this thing.
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Example of calculating a surface integral part 3 Multivariable Calculus Khan Academy.mp3
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And we will have solved, or we would have computed an actual surface integral, something you see very few times in your education career. So this is kind of exciting. So this was the cross product right here. Now let's take the magnitude of this thing. And you might remember, the magnitude of any vector is kind of a Pythagorean theorem. And in this case, it's going to be kind of the distance formula, the Pythagorean theorem in three dimensions. So the magnitude, this right here is equal to, just as a reminder, is equal to this right here.
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Example of calculating a surface integral part 3 Multivariable Calculus Khan Academy.mp3
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Now let's take the magnitude of this thing. And you might remember, the magnitude of any vector is kind of a Pythagorean theorem. And in this case, it's going to be kind of the distance formula, the Pythagorean theorem in three dimensions. So the magnitude, this right here is equal to, just as a reminder, is equal to this right here. It's equal to the partial of r with respect to s cross with this partial of r with respect to t. Let me copy it and paste it. That is equal to that right there, put an equal sign. These two quantities are equal.
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Example of calculating a surface integral part 3 Multivariable Calculus Khan Academy.mp3
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So the magnitude, this right here is equal to, just as a reminder, is equal to this right here. It's equal to the partial of r with respect to s cross with this partial of r with respect to t. Let me copy it and paste it. That is equal to that right there, put an equal sign. These two quantities are equal. Now we want to figure out the magnitude. So if we want to take the magnitude of this thing, so if we are interested in taking the magnitude of that thing, that's going to be equal to, well, this is just a scalar that's multiplying everything. So let's just write the scalar out there.
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Example of calculating a surface integral part 3 Multivariable Calculus Khan Academy.mp3
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These two quantities are equal. Now we want to figure out the magnitude. So if we want to take the magnitude of this thing, so if we are interested in taking the magnitude of that thing, that's going to be equal to, well, this is just a scalar that's multiplying everything. So let's just write the scalar out there. So b plus a cosine of s times the magnitude of this thing right here. And the magnitude of this thing right here is going to be the sum of, you can imagine it's the square root of this thing dotted with itself. Or you could say it's the sum of the squares of each of these terms to the 1 half power.
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Example of calculating a surface integral part 3 Multivariable Calculus Khan Academy.mp3
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So let's just write the scalar out there. So b plus a cosine of s times the magnitude of this thing right here. And the magnitude of this thing right here is going to be the sum of, you can imagine it's the square root of this thing dotted with itself. Or you could say it's the sum of the squares of each of these terms to the 1 half power. So let me write it like that. So it's equal to, let me write the sum of the squares. So if you square this, you get a squared cosine squared of s sine squared of t. That's that term.
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Example of calculating a surface integral part 3 Multivariable Calculus Khan Academy.mp3
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Or you could say it's the sum of the squares of each of these terms to the 1 half power. So let me write it like that. So it's equal to, let me write the sum of the squares. So if you square this, you get a squared cosine squared of s sine squared of t. That's that term. Plus, let me color code it. That's that term. Plus, I'll do the magenta, plus that term squared, plus a squared cosine squared of s cosine squared of t. That's that term.
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Example of calculating a surface integral part 3 Multivariable Calculus Khan Academy.mp3
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So if you square this, you get a squared cosine squared of s sine squared of t. That's that term. Plus, let me color code it. That's that term. Plus, I'll do the magenta, plus that term squared, plus a squared cosine squared of s cosine squared of t. That's that term. And then finally, I'll do another color, this term squared. So plus a squared sine squared of s. That's going to be all of this business to the 1 half power. This right here is the same thing as the magnitude of this right here.
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Example of calculating a surface integral part 3 Multivariable Calculus Khan Academy.mp3
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Plus, I'll do the magenta, plus that term squared, plus a squared cosine squared of s cosine squared of t. That's that term. And then finally, I'll do another color, this term squared. So plus a squared sine squared of s. That's going to be all of this business to the 1 half power. This right here is the same thing as the magnitude of this right here. And this is just a scalar that's multiplying by both of these terms. So let's see if we can do anything interesting here, if this can be simplified in any way. We have a squared cosine squared of s. We have an a squared cosine squared of s here.
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Example of calculating a surface integral part 3 Multivariable Calculus Khan Academy.mp3
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This right here is the same thing as the magnitude of this right here. And this is just a scalar that's multiplying by both of these terms. So let's see if we can do anything interesting here, if this can be simplified in any way. We have a squared cosine squared of s. We have an a squared cosine squared of s here. So let's factor that out from both of these terms and see what happens. So I'm just going to rewrite the second part. So this is going to be a squared cosine squared of s times sine squared of t plus cosine squared of t. Sine squared of t plus cosine squared of t. And then you're going to have this plus a squared sine squared of s. And of course, all of that is to the 1 half power.
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Example of calculating a surface integral part 3 Multivariable Calculus Khan Academy.mp3
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We have a squared cosine squared of s. We have an a squared cosine squared of s here. So let's factor that out from both of these terms and see what happens. So I'm just going to rewrite the second part. So this is going to be a squared cosine squared of s times sine squared of t plus cosine squared of t. Sine squared of t plus cosine squared of t. And then you're going to have this plus a squared sine squared of s. And of course, all of that is to the 1 half power. Now what is this? Well, we have sine squared of t plus cosine squared of t. That's nice. That's equal to 1, the most basic of trig identities.
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Example of calculating a surface integral part 3 Multivariable Calculus Khan Academy.mp3
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So this is going to be a squared cosine squared of s times sine squared of t plus cosine squared of t. Sine squared of t plus cosine squared of t. And then you're going to have this plus a squared sine squared of s. And of course, all of that is to the 1 half power. Now what is this? Well, we have sine squared of t plus cosine squared of t. That's nice. That's equal to 1, the most basic of trig identities. So this expression right here simplifies to a squared cosine squared of s plus this over here. a squared sine squared of s. And then all of that to the 1 half power. You might immediately recognize, you can factor out an a squared.
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Example of calculating a surface integral part 3 Multivariable Calculus Khan Academy.mp3
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That's equal to 1, the most basic of trig identities. So this expression right here simplifies to a squared cosine squared of s plus this over here. a squared sine squared of s. And then all of that to the 1 half power. You might immediately recognize, you can factor out an a squared. This is equal to a squared times cosine squared of s plus sine squared of s. And all of that to the 1 half power. I'm just focusing on this term right here. I'll write this in a second.
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Example of calculating a surface integral part 3 Multivariable Calculus Khan Academy.mp3
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You might immediately recognize, you can factor out an a squared. This is equal to a squared times cosine squared of s plus sine squared of s. And all of that to the 1 half power. I'm just focusing on this term right here. I'll write this in a second. But once again, cosine squared plus sine squared of anything is going to be equal to 1. As long as it's the same anything, it's equal to 1. So this term is a squared to the 1 half power, or the square root of a squared, which is just going to be equal to a.
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Example of calculating a surface integral part 3 Multivariable Calculus Khan Academy.mp3
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I'll write this in a second. But once again, cosine squared plus sine squared of anything is going to be equal to 1. As long as it's the same anything, it's equal to 1. So this term is a squared to the 1 half power, or the square root of a squared, which is just going to be equal to a. So all of this, all of that crazy business right here, just simplifies to a. So this cross product here simplifies to this times a, which is a pretty neat and clean simplification. So let me rewrite this.
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Example of calculating a surface integral part 3 Multivariable Calculus Khan Academy.mp3
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So this term is a squared to the 1 half power, or the square root of a squared, which is just going to be equal to a. So all of this, all of that crazy business right here, just simplifies to a. So this cross product here simplifies to this times a, which is a pretty neat and clean simplification. So let me rewrite this. So this, that simplifies to a times that. And what's that? a times b, so it's ab plus a squared cosine of s. Plus a squared cosine of s. So already we've gotten pretty far.
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Example of calculating a surface integral part 3 Multivariable Calculus Khan Academy.mp3
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So let me rewrite this. So this, that simplifies to a times that. And what's that? a times b, so it's ab plus a squared cosine of s. Plus a squared cosine of s. So already we've gotten pretty far. It's nice when you do something so beastly. And eventually it gets to something reasonably simple. And just to review what we had to do, what our mission was several videos ago, is we want to evaluate what this thing is over the region over which the surface is defined.
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Example of calculating a surface integral part 3 Multivariable Calculus Khan Academy.mp3
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a times b, so it's ab plus a squared cosine of s. Plus a squared cosine of s. So already we've gotten pretty far. It's nice when you do something so beastly. And eventually it gets to something reasonably simple. And just to review what we had to do, what our mission was several videos ago, is we want to evaluate what this thing is over the region over which the surface is defined. So s going from 0 to 2 pi, and t going from 0 to 2 pi over this region. So we want to integrate this over that region. So that region, we're going to vary s from 0 to 2 pi, so ds.
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Example of calculating a surface integral part 3 Multivariable Calculus Khan Academy.mp3
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And just to review what we had to do, what our mission was several videos ago, is we want to evaluate what this thing is over the region over which the surface is defined. So s going from 0 to 2 pi, and t going from 0 to 2 pi over this region. So we want to integrate this over that region. So that region, we're going to vary s from 0 to 2 pi, so ds. And then we're going to vary t from 0 to 2 pi, dt. And this is what we're evaluating. We're evaluating the magnitude of the cross product of these two partial derivatives of our original parameterization.
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Example of calculating a surface integral part 3 Multivariable Calculus Khan Academy.mp3
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So that region, we're going to vary s from 0 to 2 pi, so ds. And then we're going to vary t from 0 to 2 pi, dt. And this is what we're evaluating. We're evaluating the magnitude of the cross product of these two partial derivatives of our original parameterization. So this is what we can put in there. Things are getting simple all of a sudden. Or simpler, ab plus a squared cosine of s. And what is this equal to?
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Example of calculating a surface integral part 3 Multivariable Calculus Khan Academy.mp3
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We're evaluating the magnitude of the cross product of these two partial derivatives of our original parameterization. So this is what we can put in there. Things are getting simple all of a sudden. Or simpler, ab plus a squared cosine of s. And what is this equal to? So this is going to be equal to, well, we just take the antiderivative of the inside with respect to s. So the antiderivative, so let me do the outside of our integral. So we're still going to have to deal with the 0 to 2 pi. And our dt right here.
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Example of calculating a surface integral part 3 Multivariable Calculus Khan Academy.mp3
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Or simpler, ab plus a squared cosine of s. And what is this equal to? So this is going to be equal to, well, we just take the antiderivative of the inside with respect to s. So the antiderivative, so let me do the outside of our integral. So we're still going to have to deal with the 0 to 2 pi. And our dt right here. But the antiderivative with respect to s right here is going to be, ab is just a constant, so it's going to be abs plus, what's the antiderivative of cosine of s? It's sine of s. So plus a squared sine of s. And we're going to evaluate it from 0 to 2 pi. And what is this going to be equal to?
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Example of calculating a surface integral part 3 Multivariable Calculus Khan Academy.mp3
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And our dt right here. But the antiderivative with respect to s right here is going to be, ab is just a constant, so it's going to be abs plus, what's the antiderivative of cosine of s? It's sine of s. So plus a squared sine of s. And we're going to evaluate it from 0 to 2 pi. And what is this going to be equal to? Let's put our boundaries out again. Or the t integral that we're going to have to do in a second. 0 to 2 pi dt.
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Example of calculating a surface integral part 3 Multivariable Calculus Khan Academy.mp3
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And what is this going to be equal to? Let's put our boundaries out again. Or the t integral that we're going to have to do in a second. 0 to 2 pi dt. When you put 2 pi here, you're going to get ab times 2 pi, or 2 pi ab. So you're going to have 2 pi ab plus a squared sine of 2 pi. Sine of 2 pi is 0, so there's not going to be any term there.
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Example of calculating a surface integral part 3 Multivariable Calculus Khan Academy.mp3
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0 to 2 pi dt. When you put 2 pi here, you're going to get ab times 2 pi, or 2 pi ab. So you're going to have 2 pi ab plus a squared sine of 2 pi. Sine of 2 pi is 0, so there's not going to be any term there. And then minus 0 times ab, which is 0. And then you're going to have minus a squared sine of 0, which is also 0. So all of the other terms are all 0.
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Example of calculating a surface integral part 3 Multivariable Calculus Khan Academy.mp3
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Sine of 2 pi is 0, so there's not going to be any term there. And then minus 0 times ab, which is 0. And then you're going to have minus a squared sine of 0, which is also 0. So all of the other terms are all 0. So that's what we're left with. It's simplified nicely. So now we just have to take the antiderivative of this with respect to t. And this is a constant in t. So this is going to be equal to, take the antiderivative with respect to t, 2 pi abt.
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Example of calculating a surface integral part 3 Multivariable Calculus Khan Academy.mp3
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So all of the other terms are all 0. So that's what we're left with. It's simplified nicely. So now we just have to take the antiderivative of this with respect to t. And this is a constant in t. So this is going to be equal to, take the antiderivative with respect to t, 2 pi abt. And we need to evaluate that from 0 to 2 pi, which is equal to, so we put 2 pi in there. You have a 2 pi for t. It'll be a 2 pi times 2 pi ab, or we should say 2 pi squared times ab minus 0 times this thing. Well, that's just going to be 0, so we're going to have to write it down.
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Example of calculating a surface integral part 3 Multivariable Calculus Khan Academy.mp3
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So now we just have to take the antiderivative of this with respect to t. And this is a constant in t. So this is going to be equal to, take the antiderivative with respect to t, 2 pi abt. And we need to evaluate that from 0 to 2 pi, which is equal to, so we put 2 pi in there. You have a 2 pi for t. It'll be a 2 pi times 2 pi ab, or we should say 2 pi squared times ab minus 0 times this thing. Well, that's just going to be 0, so we're going to have to write it down. So we're done. This is the surface area of the torus. So this is equal to, this is exciting.
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Example of calculating a surface integral part 3 Multivariable Calculus Khan Academy.mp3
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Well, that's just going to be 0, so we're going to have to write it down. So we're done. This is the surface area of the torus. So this is equal to, this is exciting. It just kind of snuck up on us. This is equal to 4 pi squared ab, which is kind of a neat formula. Because it's a very neat and clean, it has the 2 pi, which is kind of the diameter of a circle.
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Example of calculating a surface integral part 3 Multivariable Calculus Khan Academy.mp3
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So this is equal to, this is exciting. It just kind of snuck up on us. This is equal to 4 pi squared ab, which is kind of a neat formula. Because it's a very neat and clean, it has the 2 pi, which is kind of the diameter of a circle. We're squaring it, which kind of makes sense, because we're taking the product of, you can kind of imagine the product of these two circles. I'm speaking in very abstract, general terms, but that kind of feels good. And then we're taking the product of those two radiuses.
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Example of calculating a surface integral part 3 Multivariable Calculus Khan Academy.mp3
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Because it's a very neat and clean, it has the 2 pi, which is kind of the diameter of a circle. We're squaring it, which kind of makes sense, because we're taking the product of, you can kind of imagine the product of these two circles. I'm speaking in very abstract, general terms, but that kind of feels good. And then we're taking the product of those two radiuses. Remember, let me just copy this thing down here. Let me, actually, let me copy this thing, because this is our exciting result. Let me copy this.
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Example of calculating a surface integral part 3 Multivariable Calculus Khan Academy.mp3
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And then we're taking the product of those two radiuses. Remember, let me just copy this thing down here. Let me, actually, let me copy this thing, because this is our exciting result. Let me copy this. So copy. So all of this work that we did simplified to this, which is exciting. We now know that if you have a torus where the radius of the cross section is a, and the radius from the center of the torus to the middle of the cross sections is b, that the surface area of that torus is going to be 4 pi squared times a times b, which I think is a pretty, pretty neat outcome.
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Example of calculating a surface integral part 3 Multivariable Calculus Khan Academy.mp3
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Instead of writing the surface as a capital sigma, I've written it as a capital S. Instead of writing d lowercase sigma, I wrote d uppercase S. But this is still a surface integral of the function y. And the surface we care about is x plus y squared minus z is equal to zero. X between zero and one, y between zero and two. Now this one might be a little bit more straightforward than the last one we did, or at least I hope it's a little bit more straightforward. Because we can explicitly define z in terms of x and y. And actually we could even explicitly define x in terms of y and z. But I'll do it the other way.
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Surface integral ex2 part 1 Parameterizing the surface Multivariable Calculus Khan Academy.mp3
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Now this one might be a little bit more straightforward than the last one we did, or at least I hope it's a little bit more straightforward. Because we can explicitly define z in terms of x and y. And actually we could even explicitly define x in terms of y and z. But I'll do it the other way. It's a little bit easier for me to visualize. So if you add z to both sides of this equation right over here, you get x plus y squared is equal to z. Or z is equal to x plus y squared.
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Surface integral ex2 part 1 Parameterizing the surface Multivariable Calculus Khan Academy.mp3
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But I'll do it the other way. It's a little bit easier for me to visualize. So if you add z to both sides of this equation right over here, you get x plus y squared is equal to z. Or z is equal to x plus y squared. And this is actually pretty straightforward. This surface is pretty easy to visualize, or we can give our best attempt at visualizing it. So if that is our z-axis, and that is our x-axis, and that is the y-axis, we care about the region x between zero and one.
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Surface integral ex2 part 1 Parameterizing the surface Multivariable Calculus Khan Academy.mp3
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Or z is equal to x plus y squared. And this is actually pretty straightforward. This surface is pretty easy to visualize, or we can give our best attempt at visualizing it. So if that is our z-axis, and that is our x-axis, and that is the y-axis, we care about the region x between zero and one. So maybe this is x equals one, and y between zero and two. So let's say this is one, this is two in the y area. So this is, we essentially care about the surface over this, over this region of the xy plane.
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Surface integral ex2 part 1 Parameterizing the surface Multivariable Calculus Khan Academy.mp3
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So if that is our z-axis, and that is our x-axis, and that is the y-axis, we care about the region x between zero and one. So maybe this is x equals one, and y between zero and two. So let's say this is one, this is two in the y area. So this is, we essentially care about the surface over this, over this region of the xy plane. And then we can think about what the surface actually looks like. This isn't the surface, this is just kind of the range of x and y's that we actually care about. And so let's think about the surface.
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Surface integral ex2 part 1 Parameterizing the surface Multivariable Calculus Khan Academy.mp3
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So this is, we essentially care about the surface over this, over this region of the xy plane. And then we can think about what the surface actually looks like. This isn't the surface, this is just kind of the range of x and y's that we actually care about. And so let's think about the surface. When x and y are zero, z is zero. So we're gonna be sitting, let me do this in a, I'm doing it in green. Z is going to be right over there.
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Surface integral ex2 part 1 Parameterizing the surface Multivariable Calculus Khan Academy.mp3
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And so let's think about the surface. When x and y are zero, z is zero. So we're gonna be sitting, let me do this in a, I'm doing it in green. Z is going to be right over there. And now as y increases, or if we, when x is equal to zero, if we're just talking about the zy plane, z is going to be equal to y squared. So z is going to be equal to y squared. So this might be z is equal to four, this is z is equal to two, one, three.
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Surface integral ex2 part 1 Parameterizing the surface Multivariable Calculus Khan Academy.mp3
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Z is going to be right over there. And now as y increases, or if we, when x is equal to zero, if we're just talking about the zy plane, z is going to be equal to y squared. So z is going to be equal to y squared. So this might be z is equal to four, this is z is equal to two, one, three. So z is going to do something like this. It's going to be a parabola in the zy plane. It's gonna look something like that.
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Surface integral ex2 part 1 Parameterizing the surface Multivariable Calculus Khan Academy.mp3
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So this might be z is equal to four, this is z is equal to two, one, three. So z is going to do something like this. It's going to be a parabola in the zy plane. It's gonna look something like that. Now when y is equal to zero, z is just equal to x. So as x goes to one, z will also go to one. So z will go like this.
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Surface integral ex2 part 1 Parameterizing the surface Multivariable Calculus Khan Academy.mp3
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It's gonna look something like that. Now when y is equal to zero, z is just equal to x. So as x goes to one, z will also go to one. So z will go like this. The scales of the axes aren't, they're not drawn to scale. The z is a little bit more compressed than the x or y, the way I've drawn them. And then from this point right over here, you add the y squared, and so you get something that looks, something that looks, so this is, this is this point there.
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Surface integral ex2 part 1 Parameterizing the surface Multivariable Calculus Khan Academy.mp3
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So z will go like this. The scales of the axes aren't, they're not drawn to scale. The z is a little bit more compressed than the x or y, the way I've drawn them. And then from this point right over here, you add the y squared, and so you get something that looks, something that looks, so this is, this is this point there. And then this point, when y is equal to two, and x is equal to one, you have z is equal to five. It's gonna look something like this. Something like this.
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Surface integral ex2 part 1 Parameterizing the surface Multivariable Calculus Khan Academy.mp3
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And then from this point right over here, you add the y squared, and so you get something that looks, something that looks, so this is, this is this point there. And then this point, when y is equal to two, and x is equal to one, you have z is equal to five. It's gonna look something like this. Something like this. And then you're going to have the straight line like that. And this point is right over there. And this surface is the surface that we are going to take, or the surface over which we're going to, we're going to, we're going to evaluate the surface integral of the function y.
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Surface integral ex2 part 1 Parameterizing the surface Multivariable Calculus Khan Academy.mp3
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Something like this. And then you're going to have the straight line like that. And this point is right over there. And this surface is the surface that we are going to take, or the surface over which we're going to, we're going to, we're going to evaluate the surface integral of the function y. And so one way you could think about it, y could be maybe the mass density of this surface. And so when you multiply y times each ds, you're essentially figuring out the mass of that little chunk, and then you're figuring out the mass of this entire surface. And so one way you can imagine as we go more and more in that direction, as y is increasing, this thing is getting more and more dense.
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Surface integral ex2 part 1 Parameterizing the surface Multivariable Calculus Khan Academy.mp3
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And this surface is the surface that we are going to take, or the surface over which we're going to, we're going to, we're going to evaluate the surface integral of the function y. And so one way you could think about it, y could be maybe the mass density of this surface. And so when you multiply y times each ds, you're essentially figuring out the mass of that little chunk, and then you're figuring out the mass of this entire surface. And so one way you can imagine as we go more and more in that direction, as y is increasing, this thing is getting more and more dense. So this part, this part of the surface is more dense, is more dense than as y becomes lower and lower. And then that would actually give us the mass. But with that out of the way, let's actually evaluate it.
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Surface integral ex2 part 1 Parameterizing the surface Multivariable Calculus Khan Academy.mp3
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And so one way you can imagine as we go more and more in that direction, as y is increasing, this thing is getting more and more dense. So this part, this part of the surface is more dense, is more dense than as y becomes lower and lower. And then that would actually give us the mass. But with that out of the way, let's actually evaluate it. And so as you know, the first step is to figure out a parameterization. And it should be pretty straightforward because we can write z explicitly in terms of x and y. And so we can actually use x and y as the actual parameters, or if we wanna just substitute it with different parameters, we could.
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Surface integral ex2 part 1 Parameterizing the surface Multivariable Calculus Khan Academy.mp3
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But with that out of the way, let's actually evaluate it. And so as you know, the first step is to figure out a parameterization. And it should be pretty straightforward because we can write z explicitly in terms of x and y. And so we can actually use x and y as the actual parameters, or if we wanna just substitute it with different parameters, we could. But let me, so let's just write, let me do that. So let me just write x is equal to, x is equal to, and in the spirit of using different notation, instead of using s and t, I'll use u and v. x is equal to u. Let's say y is equal to v. And then z is going to be equal to u plus v squared.
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Surface integral ex2 part 1 Parameterizing the surface Multivariable Calculus Khan Academy.mp3
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And so we can actually use x and y as the actual parameters, or if we wanna just substitute it with different parameters, we could. But let me, so let's just write, let me do that. So let me just write x is equal to, x is equal to, and in the spirit of using different notation, instead of using s and t, I'll use u and v. x is equal to u. Let's say y is equal to v. And then z is going to be equal to u plus v squared. And so our surface written as a vector position function or position vector function, our surface, we can write it as r, which is going to be a function of u and v. And it's going to be equal to ui plus v plus vj plus u plus v squared, u plus v squared k. And then u is going to be between zero and one because x is just equal to u or u is equal to x. So u is going to be equal to, is going to be between zero and one. And then v is going to be, v is going to be between zero and two.
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Surface integral ex2 part 1 Parameterizing the surface Multivariable Calculus Khan Academy.mp3
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In the last video, we saw that if we had some curve in the xy plane, and we just parameterized it in a very general sense like this, we could generate another parameterization that essentially is the same curve but goes in the opposite direction. It starts here and it goes here as t goes from a to b, as opposed to the first parameterization. We started with t equal a over here, and it went up like that. And the question I want to answer in this video is how a line integral of a scalar field over this curve, so this is my scalar field, it's a function of x and y, how a line integral over a scalar field over this curve relates to a line integral of that same scalar field over the reverse curve, over the curve going in the other direction. So the question is, does it even matter whether we move in this direction or that direction when we're taking the line integral of a scalar field? And in the next video, we'll talk about whether it matters on a vector field. Let's see if we can get a little intuition to our answer before we even prove our answer.
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Scalar field line integral independent of path direction Multivariable Calculus Khan Academy.mp3
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And the question I want to answer in this video is how a line integral of a scalar field over this curve, so this is my scalar field, it's a function of x and y, how a line integral over a scalar field over this curve relates to a line integral of that same scalar field over the reverse curve, over the curve going in the other direction. So the question is, does it even matter whether we move in this direction or that direction when we're taking the line integral of a scalar field? And in the next video, we'll talk about whether it matters on a vector field. Let's see if we can get a little intuition to our answer before we even prove our answer. So let me draw a little diagram here. Actually, let me do it a little bit lower because I think I'm going to need a little bit more real estate. So let me draw the y-axis.
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Scalar field line integral independent of path direction Multivariable Calculus Khan Academy.mp3
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Let's see if we can get a little intuition to our answer before we even prove our answer. So let me draw a little diagram here. Actually, let me do it a little bit lower because I think I'm going to need a little bit more real estate. So let me draw the y-axis. That is the x-axis. Let me draw the vertical axis just like that. That is z.
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Scalar field line integral independent of path direction Multivariable Calculus Khan Academy.mp3
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So let me draw the y-axis. That is the x-axis. Let me draw the vertical axis just like that. That is z. Let me draw a scalar field here. So I'll just draw it as some surface. I'll draw part of it.
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Scalar field line integral independent of path direction Multivariable Calculus Khan Academy.mp3
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That is z. Let me draw a scalar field here. So I'll just draw it as some surface. I'll draw part of it. That is my scalar field. That is f of x, y right there. For any point on the xy-plane, we can associate a height that defines this surface, this scalar field.
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Scalar field line integral independent of path direction Multivariable Calculus Khan Academy.mp3
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I'll draw part of it. That is my scalar field. That is f of x, y right there. For any point on the xy-plane, we can associate a height that defines this surface, this scalar field. And let me put a curve down there. So let's say that this is the curve C, just like that. And the way we define it first, we start over here and we move in that direction.
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Scalar field line integral independent of path direction Multivariable Calculus Khan Academy.mp3
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For any point on the xy-plane, we can associate a height that defines this surface, this scalar field. And let me put a curve down there. So let's say that this is the curve C, just like that. And the way we define it first, we start over here and we move in that direction. That was our curve C. And we know from several videos ago that the way to visualize what this line integral means is we're essentially trying to figure out the area of a curtain that has this curve as its base, and then its ceiling is defined by this surface, by the scalar field. So we're literally just trying to find the area of this curvy piece of paper or wall or whatever you want to view it. That's what this thing is.
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Scalar field line integral independent of path direction Multivariable Calculus Khan Academy.mp3
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And the way we define it first, we start over here and we move in that direction. That was our curve C. And we know from several videos ago that the way to visualize what this line integral means is we're essentially trying to figure out the area of a curtain that has this curve as its base, and then its ceiling is defined by this surface, by the scalar field. So we're literally just trying to find the area of this curvy piece of paper or wall or whatever you want to view it. That's what this thing is. Now, if we take the same integral but we take the reverse curve, instead of going in that direction, we're now going in the opposite direction. We're not taking a curve where we're going from the top to the bottom. But the idea is still the same.
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Scalar field line integral independent of path direction Multivariable Calculus Khan Academy.mp3
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That's what this thing is. Now, if we take the same integral but we take the reverse curve, instead of going in that direction, we're now going in the opposite direction. We're not taking a curve where we're going from the top to the bottom. But the idea is still the same. I don't know which one's C, which one's minus C. I could have defined this path going from that way as C, and then the minus C path would have started here and gone back up. So it seems in either case, no matter what I'm doing, I'm going to try to figure out the area of this curved piece of paper. So my intuition tells me that either of these are going to give me the area of this curved piece of paper.
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Scalar field line integral independent of path direction Multivariable Calculus Khan Academy.mp3
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But the idea is still the same. I don't know which one's C, which one's minus C. I could have defined this path going from that way as C, and then the minus C path would have started here and gone back up. So it seems in either case, no matter what I'm doing, I'm going to try to figure out the area of this curved piece of paper. So my intuition tells me that either of these are going to give me the area of this curved piece of paper. So maybe they should be equal to each other. I haven't proved anything very rigorously yet, but it seems that they should be equal to each other. In this case, let me just make it very clear.
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Scalar field line integral independent of path direction Multivariable Calculus Khan Academy.mp3
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So my intuition tells me that either of these are going to give me the area of this curved piece of paper. So maybe they should be equal to each other. I haven't proved anything very rigorously yet, but it seems that they should be equal to each other. In this case, let me just make it very clear. I'm taking a ds, a little change in distance, and I'm multiplying it by the height to find kind of a differential of the area. And then I'm going to add a bunch of these together to get the whole area. Here I'm doing the same thing.
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Scalar field line integral independent of path direction Multivariable Calculus Khan Academy.mp3
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In this case, let me just make it very clear. I'm taking a ds, a little change in distance, and I'm multiplying it by the height to find kind of a differential of the area. And then I'm going to add a bunch of these together to get the whole area. Here I'm doing the same thing. I'm taking a little ds. And remember, the ds is always going to be positive the way we've parameterized it. It's the hardest word in the English language for me to say.
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Scalar field line integral independent of path direction Multivariable Calculus Khan Academy.mp3
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Here I'm doing the same thing. I'm taking a little ds. And remember, the ds is always going to be positive the way we've parameterized it. It's the hardest word in the English language for me to say. So here too, we're taking a ds, and we're going to multiply it by the height. So once again, we should take the area. And I want to actually differentiate that relative to when you take a normal integral from a to b of, let's say, f of x dx, we know that when we switch the boundaries of the integration, that it makes the integral negative.
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Scalar field line integral independent of path direction Multivariable Calculus Khan Academy.mp3
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It's the hardest word in the English language for me to say. So here too, we're taking a ds, and we're going to multiply it by the height. So once again, we should take the area. And I want to actually differentiate that relative to when you take a normal integral from a to b of, let's say, f of x dx, we know that when we switch the boundaries of the integration, that it makes the integral negative. That equals the negative of the integral from b to a of f of x dx. And the reason why this is the case is if you imagine this is a, this is b, that is my f of x. When you do it this way, your dx's are always going to be positive.
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Scalar field line integral independent of path direction Multivariable Calculus Khan Academy.mp3
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And I want to actually differentiate that relative to when you take a normal integral from a to b of, let's say, f of x dx, we know that when we switch the boundaries of the integration, that it makes the integral negative. That equals the negative of the integral from b to a of f of x dx. And the reason why this is the case is if you imagine this is a, this is b, that is my f of x. When you do it this way, your dx's are always going to be positive. When you go in that direction, your dx's are always going to be positive, right? The each increment, the right boundary is going to be higher than the left boundary. So your dx's are positive.
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Scalar field line integral independent of path direction Multivariable Calculus Khan Academy.mp3
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When you do it this way, your dx's are always going to be positive. When you go in that direction, your dx's are always going to be positive, right? The each increment, the right boundary is going to be higher than the left boundary. So your dx's are positive. In this situation, your dx's are negative. The heights are always going to be the same. They're always going to be f of x, but here your change in x is a negative change in x when you go from b to a, and that's why you get a negative integral.
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Scalar field line integral independent of path direction Multivariable Calculus Khan Academy.mp3
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