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So let me write ds is equal to super small change in arc length of our path or of our curve. That's our ds. So you can imagine the area of that little rectangle right there along my curvy wall is going to be ds times the height at that point. Well, that's f of xy. And then if I take the sum, because these are infinitely narrow, these ds's have infinitely small width. If I were to take the infinite sum of all of those guys from t is equal to a to t is equal to b, from t is equal to a, I keep taking the sum of those rectangles to t is equal to b right there, that will give me my area. I'm just using the exact same logic as I did up there.
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Introduction to the line integral Multivariable Calculus Khan Academy.mp3
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Well, that's f of xy. And then if I take the sum, because these are infinitely narrow, these ds's have infinitely small width. If I were to take the infinite sum of all of those guys from t is equal to a to t is equal to b, from t is equal to a, I keep taking the sum of those rectangles to t is equal to b right there, that will give me my area. I'm just using the exact same logic as I did up there. I'm not being very mathematically rigorous, but I want to give you the intuition of what we're doing. We're really kind of just bending the base of this thing to get a curvy wall instead of a straight, direct wall like we had up here. But you're saying, Sal, this is all abstract, and how can I even calculate something like this?
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Introduction to the line integral Multivariable Calculus Khan Academy.mp3
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I'm just using the exact same logic as I did up there. I'm not being very mathematically rigorous, but I want to give you the intuition of what we're doing. We're really kind of just bending the base of this thing to get a curvy wall instead of a straight, direct wall like we had up here. But you're saying, Sal, this is all abstract, and how can I even calculate something like this? This makes no sense to me. I have an s here. I have an x and a y. I have a t. What can I do with this?
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Introduction to the line integral Multivariable Calculus Khan Academy.mp3
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But you're saying, Sal, this is all abstract, and how can I even calculate something like this? This makes no sense to me. I have an s here. I have an x and a y. I have a t. What can I do with this? Let's see if we can make some headway. And I promise you when we do it with a tangible problem, the end product of this video is going to be a little bit hairy to look at. But when we do it with an actual problem, it'll actually, I think, be very concrete, and you'll see it's not too hard to deal with.
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Introduction to the line integral Multivariable Calculus Khan Academy.mp3
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I have an x and a y. I have a t. What can I do with this? Let's see if we can make some headway. And I promise you when we do it with a tangible problem, the end product of this video is going to be a little bit hairy to look at. But when we do it with an actual problem, it'll actually, I think, be very concrete, and you'll see it's not too hard to deal with. But let's see if we can get all of this in terms of t. First of all, let's focus just on this ds. Let me re-pick up the xy-axis. If I were to re-flip the xy, let me switch colors.
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Introduction to the line integral Multivariable Calculus Khan Academy.mp3
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But when we do it with an actual problem, it'll actually, I think, be very concrete, and you'll see it's not too hard to deal with. But let's see if we can get all of this in terms of t. First of all, let's focus just on this ds. Let me re-pick up the xy-axis. If I were to re-flip the xy, let me switch colors. This is getting a little monotonous. If I were to re-flip the xy-axis like that, I'm actually only doing that same green, so you know we're dealing with the same xy-axis. That's my y-axis.
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Introduction to the line integral Multivariable Calculus Khan Academy.mp3
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If I were to re-flip the xy, let me switch colors. This is getting a little monotonous. If I were to re-flip the xy-axis like that, I'm actually only doing that same green, so you know we're dealing with the same xy-axis. That's my y-axis. That is my x-axis. And so this path right here, if I were to just draw it straight up like this, it would look something like this. It would look something like this.
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Introduction to the line integral Multivariable Calculus Khan Academy.mp3
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That's my y-axis. That is my x-axis. And so this path right here, if I were to just draw it straight up like this, it would look something like this. It would look something like this. That's my path, my arc. This is when t is equal to a. So this is t is equal to a.
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Introduction to the line integral Multivariable Calculus Khan Academy.mp3
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It would look something like this. That's my path, my arc. This is when t is equal to a. So this is t is equal to a. This is t is equal to b. Same thing, I just kind of picked it back up so we can visualize it. And we say that we have some change in arc length.
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Introduction to the line integral Multivariable Calculus Khan Academy.mp3
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So this is t is equal to a. This is t is equal to b. Same thing, I just kind of picked it back up so we can visualize it. And we say that we have some change in arc length. Let me switch colors. Let's say that this one right here. Let's say that's some small change in arc length, and we're calling that ds.
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Introduction to the line integral Multivariable Calculus Khan Academy.mp3
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And we say that we have some change in arc length. Let me switch colors. Let's say that this one right here. Let's say that's some small change in arc length, and we're calling that ds. Now, is there some way to relate ds to infinitely small changes in x or y? Well, if we think about it, and this is all a little bit hand-wavy. I'm not being mathematically rigorous, but I think it will give you the correct intuition.
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Introduction to the line integral Multivariable Calculus Khan Academy.mp3
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Let's say that's some small change in arc length, and we're calling that ds. Now, is there some way to relate ds to infinitely small changes in x or y? Well, if we think about it, and this is all a little bit hand-wavy. I'm not being mathematically rigorous, but I think it will give you the correct intuition. If you imagine, you can figure out the length of ds if you know the lengths of these super small changes in x and super small changes in y. So if this distance right here is dx, infinitesimally small change in x. This distance right here is dy, infinitesimally small change in y.
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Introduction to the line integral Multivariable Calculus Khan Academy.mp3
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I'm not being mathematically rigorous, but I think it will give you the correct intuition. If you imagine, you can figure out the length of ds if you know the lengths of these super small changes in x and super small changes in y. So if this distance right here is dx, infinitesimally small change in x. This distance right here is dy, infinitesimally small change in y. Then we can figure out ds from the Pythagorean theorem. You can say that ds is going to be, it's the hypotenuse of this triangle. It's equal to the square root of dx squared plus dy squared.
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Introduction to the line integral Multivariable Calculus Khan Academy.mp3
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This distance right here is dy, infinitesimally small change in y. Then we can figure out ds from the Pythagorean theorem. You can say that ds is going to be, it's the hypotenuse of this triangle. It's equal to the square root of dx squared plus dy squared. That seems to make things a little bit, we can get rid of the ds all of a sudden. Let's rewrite this little expression here using this sense of what ds, ds is really the square root of dx squared plus dy squared. I'm not being very rigorous, and actually it's very hard to be rigorous with differentials, but intuitively I think it makes a lot of sense.
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Introduction to the line integral Multivariable Calculus Khan Academy.mp3
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It's equal to the square root of dx squared plus dy squared. That seems to make things a little bit, we can get rid of the ds all of a sudden. Let's rewrite this little expression here using this sense of what ds, ds is really the square root of dx squared plus dy squared. I'm not being very rigorous, and actually it's very hard to be rigorous with differentials, but intuitively I think it makes a lot of sense. We can say that this integral, the area of this curvy curtain, is going to be the integral from t is equal to a to t is equal to b of f of xy. Instead of writing ds, we can write this, times the square root of dx squared plus dy squared. Now we at least got rid of this big capital S, but we still haven't solved the problem of how do you solve something, an integral, a definite integral that looks like this.
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Introduction to the line integral Multivariable Calculus Khan Academy.mp3
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I'm not being very rigorous, and actually it's very hard to be rigorous with differentials, but intuitively I think it makes a lot of sense. We can say that this integral, the area of this curvy curtain, is going to be the integral from t is equal to a to t is equal to b of f of xy. Instead of writing ds, we can write this, times the square root of dx squared plus dy squared. Now we at least got rid of this big capital S, but we still haven't solved the problem of how do you solve something, an integral, a definite integral that looks like this. We have it in terms of t here, but we only have it in terms of x's and y's here. We need to get everything in terms of t. We know x and y are both functions of t, so we can actually rewrite it like this. We can rewrite it as from t is equal to a to t is equal to b. F of xy, we can write it, f is a function of x, which is a function of t, and f is also a function of y, which is also a function of t. You give me a t, I'll be able to give you an x or a y, and once you give me an x or a y, I can figure out what f is.
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Introduction to the line integral Multivariable Calculus Khan Academy.mp3
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Now we at least got rid of this big capital S, but we still haven't solved the problem of how do you solve something, an integral, a definite integral that looks like this. We have it in terms of t here, but we only have it in terms of x's and y's here. We need to get everything in terms of t. We know x and y are both functions of t, so we can actually rewrite it like this. We can rewrite it as from t is equal to a to t is equal to b. F of xy, we can write it, f is a function of x, which is a function of t, and f is also a function of y, which is also a function of t. You give me a t, I'll be able to give you an x or a y, and once you give me an x or a y, I can figure out what f is. We have that, and then we have this part right here, I'll do it in orange, square root of dx squared plus dy squared. We still don't have things in terms of t. We need a dt someplace here in order to be able to evaluate this integral, and we'll see that in the next video when I do a concrete problem. I really want to give you a sense for the end product, the formula we're going to get at the end product of this video, where it comes from.
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Introduction to the line integral Multivariable Calculus Khan Academy.mp3
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We can rewrite it as from t is equal to a to t is equal to b. F of xy, we can write it, f is a function of x, which is a function of t, and f is also a function of y, which is also a function of t. You give me a t, I'll be able to give you an x or a y, and once you give me an x or a y, I can figure out what f is. We have that, and then we have this part right here, I'll do it in orange, square root of dx squared plus dy squared. We still don't have things in terms of t. We need a dt someplace here in order to be able to evaluate this integral, and we'll see that in the next video when I do a concrete problem. I really want to give you a sense for the end product, the formula we're going to get at the end product of this video, where it comes from. One thing we can do is if we allow ourselves to algebraically manipulate differentials, what we could do is let us multiply and divide by dt. One way to think about it, you could rewrite, let me just do this orange part right here, let's do a little side right here. If you take this orange part and write it in pink, you have dx squared, and then you have plus dy squared.
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Introduction to the line integral Multivariable Calculus Khan Academy.mp3
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I really want to give you a sense for the end product, the formula we're going to get at the end product of this video, where it comes from. One thing we can do is if we allow ourselves to algebraically manipulate differentials, what we could do is let us multiply and divide by dt. One way to think about it, you could rewrite, let me just do this orange part right here, let's do a little side right here. If you take this orange part and write it in pink, you have dx squared, and then you have plus dy squared. Let's say we just multiply it times dt over dt. That's a small change in t divided by a small change in t, that's 1, so of course you can multiply it by that. If we were to bring this part inside of the square root sign, let me rewrite this, this is the same thing as 1 over dt times the square root of dx squared plus dy squared, and then times that dt.
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Introduction to the line integral Multivariable Calculus Khan Academy.mp3
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If you take this orange part and write it in pink, you have dx squared, and then you have plus dy squared. Let's say we just multiply it times dt over dt. That's a small change in t divided by a small change in t, that's 1, so of course you can multiply it by that. If we were to bring this part inside of the square root sign, let me rewrite this, this is the same thing as 1 over dt times the square root of dx squared plus dy squared, and then times that dt. I just wanted to write it this way to show you I'm just multiplying by 1, and here I'm just taking this dt, writing it there, and leaving this over here. Now if I wanted to bring this into the square root sign, this is the same thing. This is equal to, and I'll do it very slowly just to allow you to believe that I'm not doing anything shady with the algebra, this is the same thing as the square root of 1 over dt squared, let me make the radical a little bit bigger, times dx squared plus dy squared, and then all of that times dt.
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Introduction to the line integral Multivariable Calculus Khan Academy.mp3
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If we were to bring this part inside of the square root sign, let me rewrite this, this is the same thing as 1 over dt times the square root of dx squared plus dy squared, and then times that dt. I just wanted to write it this way to show you I'm just multiplying by 1, and here I'm just taking this dt, writing it there, and leaving this over here. Now if I wanted to bring this into the square root sign, this is the same thing. This is equal to, and I'll do it very slowly just to allow you to believe that I'm not doing anything shady with the algebra, this is the same thing as the square root of 1 over dt squared, let me make the radical a little bit bigger, times dx squared plus dy squared, and then all of that times dt. I didn't do anything, you could just take the square root of this and you get 1 over dt. If I just distribute this, this is equal to the square root, and we have our dt at the end, of dx squared, or we could even write dx over dt squared, plus dy over dt squared. dx squared over dt squared is just dx over dt squared, same thing with the y's.
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Introduction to the line integral Multivariable Calculus Khan Academy.mp3
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This is equal to, and I'll do it very slowly just to allow you to believe that I'm not doing anything shady with the algebra, this is the same thing as the square root of 1 over dt squared, let me make the radical a little bit bigger, times dx squared plus dy squared, and then all of that times dt. I didn't do anything, you could just take the square root of this and you get 1 over dt. If I just distribute this, this is equal to the square root, and we have our dt at the end, of dx squared, or we could even write dx over dt squared, plus dy over dt squared. dx squared over dt squared is just dx over dt squared, same thing with the y's. Now all of a sudden this starts to look pretty interesting. Let's substitute this expression with this one, we said that these are equivalent. I'll switch colors just for the sake of it.
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Introduction to the line integral Multivariable Calculus Khan Academy.mp3
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dx squared over dt squared is just dx over dt squared, same thing with the y's. Now all of a sudden this starts to look pretty interesting. Let's substitute this expression with this one, we said that these are equivalent. I'll switch colors just for the sake of it. We have the integral from t is equal to a, let me get our drawing back, from t is equal to a to t is equal to b, of f of x of t, and y of t, they're both functions of t, and now instead of this expression, we can write the square root of, well what's dx, what's the change in x with respect to whatever this parameter is? What is dx over dt? dx over dt is the same thing, let me write this down, dx over dt is the same thing, actually I should write it here, is the same thing as g prime of t. x is a function of t, the function I wrote is g prime of t, and then this dy over dt is the same thing as h prime of t, we could say that this function of t. I just want to make that clear, we know these two functions, so we can just take their derivatives with respect to t, but I'm just going to leave it in that form.
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Introduction to the line integral Multivariable Calculus Khan Academy.mp3
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I'll switch colors just for the sake of it. We have the integral from t is equal to a, let me get our drawing back, from t is equal to a to t is equal to b, of f of x of t, and y of t, they're both functions of t, and now instead of this expression, we can write the square root of, well what's dx, what's the change in x with respect to whatever this parameter is? What is dx over dt? dx over dt is the same thing, let me write this down, dx over dt is the same thing, actually I should write it here, is the same thing as g prime of t. x is a function of t, the function I wrote is g prime of t, and then this dy over dt is the same thing as h prime of t, we could say that this function of t. I just want to make that clear, we know these two functions, so we can just take their derivatives with respect to t, but I'm just going to leave it in that form. The square root, and we take the derivative of x with respect to t squared, plus the derivative of y with respect to t squared, and then all of that times dt. This might look like some strange and convoluted formula, but this is actually something that we know how to deal with. We've now simplified this strange, this arc length problem, or this line integral, that's essentially what we're doing, we're taking an integral over a curve, or over a line, as opposed to just an interval on the x axis.
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Introduction to the line integral Multivariable Calculus Khan Academy.mp3
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dx over dt is the same thing, let me write this down, dx over dt is the same thing, actually I should write it here, is the same thing as g prime of t. x is a function of t, the function I wrote is g prime of t, and then this dy over dt is the same thing as h prime of t, we could say that this function of t. I just want to make that clear, we know these two functions, so we can just take their derivatives with respect to t, but I'm just going to leave it in that form. The square root, and we take the derivative of x with respect to t squared, plus the derivative of y with respect to t squared, and then all of that times dt. This might look like some strange and convoluted formula, but this is actually something that we know how to deal with. We've now simplified this strange, this arc length problem, or this line integral, that's essentially what we're doing, we're taking an integral over a curve, or over a line, as opposed to just an interval on the x axis. We've taken this strange line integral, that's in terms of the arc length of the line, and x's and y's, and we've put everything in terms of t, and I'm going to show you that in the next video. Everything is going to be expressed in terms of t, so this just turns into a simple, definite integral. Hopefully that didn't confuse you too much, I think you're going to see in the next video that this right here is actually a very straightforward thing to implement.
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Introduction to the line integral Multivariable Calculus Khan Academy.mp3
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We've now simplified this strange, this arc length problem, or this line integral, that's essentially what we're doing, we're taking an integral over a curve, or over a line, as opposed to just an interval on the x axis. We've taken this strange line integral, that's in terms of the arc length of the line, and x's and y's, and we've put everything in terms of t, and I'm going to show you that in the next video. Everything is going to be expressed in terms of t, so this just turns into a simple, definite integral. Hopefully that didn't confuse you too much, I think you're going to see in the next video that this right here is actually a very straightforward thing to implement. Just to remind you where it all came from, I think I got the parentheses right, this right here was just a change in our arc length, that whole thing right there was just a change in arc length, and this is just the height of our function at that point, and we're just summing it, doing an infinite sum of infinitely small lengths. This was a change in our arc length times the height, this is going to have an infinitely narrow width, and you're going to take an infinite number of these rectangles to get the area of this entire fence, or this entire curtain. That's what this definite integral will give us, and we'll actually apply it in the next video.
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Introduction to the line integral Multivariable Calculus Khan Academy.mp3
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So type two regions, type two region is a region, I'll call it R2, that's the set of all X, Ys, and Zs in three dimensions, such that, and now instead of thinking of our domain in terms of XY coordinates, we're gonna think of them in terms of YZ coordinates, such that our YZ pairs are a member of some domain, I'll call it D2, since we're talking about type two regions, and X is bounded below by some function of YZ, so I'll call it G1, G1 of YZ is less than or equal to X, which is less than or equal to some other function of YZ, G2 of YZ. And so you'll immediately see a very similar way of thinking about it, but instead of having Z vary between two functions of X and Y, as we had in a type one region, we now have X varying between two functions of Y and Z. Now let's think about some of the shapes we explored. We saw that these two right up here, the sphere and the cylinder, were type one regions, but this dumbbell, the way that I oriented it here, was not a type one region. Let's think about which of these are type two regions and what might not be a type two region. So first let's think about the sphere. So I have my axes right over here, let me scroll down a little bit, so I got my axes, and so over here our domain, we can still construct our sphere, but our domain is now going to be in the YZ plane.
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Type II regions in three dimensions Divergence theorem Multivariable Calculus Khan Academy.mp3
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We saw that these two right up here, the sphere and the cylinder, were type one regions, but this dumbbell, the way that I oriented it here, was not a type one region. Let's think about which of these are type two regions and what might not be a type two region. So first let's think about the sphere. So I have my axes right over here, let me scroll down a little bit, so I got my axes, and so over here our domain, we can still construct our sphere, but our domain is now going to be in the YZ plane. So YZ plane is this business right over here, so this will be our domain. Don't want to make it more spherical than that. So our domain is this right over here in the YZ plane, that is our D2.
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Type II regions in three dimensions Divergence theorem Multivariable Calculus Khan Academy.mp3
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So I have my axes right over here, let me scroll down a little bit, so I got my axes, and so over here our domain, we can still construct our sphere, but our domain is now going to be in the YZ plane. So YZ plane is this business right over here, so this will be our domain. Don't want to make it more spherical than that. So our domain is this right over here in the YZ plane, that is our D2. And now the lower bound, in order to construct a region, the solid region of the sphere, or the globe or whatever you want to call it, the lower bound on X would be kind of the back half of the sphere, the one that's away from us right over here. So the lower bound, so let me see how well I can wireframe it at first, the lower bound, I can do a better job than that, so it might go do something like that, and then do something like that, but this is if the domain right over here was transparent, but all we might catch, we'll just catch a glimpse of it in the back right over here. So it's the side of the sphere that's facing away from us, and then the upper bound on X would be the side of the sphere that's facing us.
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Type II regions in three dimensions Divergence theorem Multivariable Calculus Khan Academy.mp3
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So our domain is this right over here in the YZ plane, that is our D2. And now the lower bound, in order to construct a region, the solid region of the sphere, or the globe or whatever you want to call it, the lower bound on X would be kind of the back half of the sphere, the one that's away from us right over here. So the lower bound, so let me see how well I can wireframe it at first, the lower bound, I can do a better job than that, so it might go do something like that, and then do something like that, but this is if the domain right over here was transparent, but all we might catch, we'll just catch a glimpse of it in the back right over here. So it's the side of the sphere that's facing away from us, and then the upper bound on X would be the side of the sphere that's facing us. So if I were to do some contours, it might look something like this, and then look something like this, and then we would color in this entire region right over here. And X can take on all the values above that magenta surface and below this green surface, and essentially it would fill up the globe for every YZ point in our domain. So a sphere is both a type I and a type II region.
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Type II regions in three dimensions Divergence theorem Multivariable Calculus Khan Academy.mp3
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So it's the side of the sphere that's facing away from us, and then the upper bound on X would be the side of the sphere that's facing us. So if I were to do some contours, it might look something like this, and then look something like this, and then we would color in this entire region right over here. And X can take on all the values above that magenta surface and below this green surface, and essentially it would fill up the globe for every YZ point in our domain. So a sphere is both a type I and a type II region. Actually, we're going to see it's going to be a type III region as well. What about this cylinder right over here? Can we construct it or think about it in a way that it would actually be a type II region?
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Type II regions in three dimensions Divergence theorem Multivariable Calculus Khan Academy.mp3
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So a sphere is both a type I and a type II region. Actually, we're going to see it's going to be a type III region as well. What about this cylinder right over here? Can we construct it or think about it in a way that it would actually be a type II region? So let's try to do that. So let me paste it. So what if we had a domain, what if our domain was something like this?
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Type II regions in three dimensions Divergence theorem Multivariable Calculus Khan Academy.mp3
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Can we construct it or think about it in a way that it would actually be a type II region? So let's try to do that. So let me paste it. So what if we had a domain, what if our domain was something like this? It was a rectangle in the YZ plane. So it was a rectangle in the YZ plane. So this is our domain, a rectangle in the YZ plane.
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Type II regions in three dimensions Divergence theorem Multivariable Calculus Khan Academy.mp3
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So what if we had a domain, what if our domain was something like this? It was a rectangle in the YZ plane. So it was a rectangle in the YZ plane. So this is our domain, a rectangle in the YZ plane. So that would be my D2. And what if the lower bound was kind of the back side of the cylinder? So the back side of the cylinder, try to draw it as good as I can, the back side of the cylinder.
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Type II regions in three dimensions Divergence theorem Multivariable Calculus Khan Academy.mp3
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So this is our domain, a rectangle in the YZ plane. So that would be my D2. And what if the lower bound was kind of the back side of the cylinder? So the back side of the cylinder, try to draw it as good as I can, the back side of the cylinder. And so if we just saw the outside of it, it would look something like that. It's facing away from us, so we barely see it. If we could see through the cylinder or see through the little flat cut of the cylinder, it would look something like that.
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Type II regions in three dimensions Divergence theorem Multivariable Calculus Khan Academy.mp3
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So the back side of the cylinder, try to draw it as good as I can, the back side of the cylinder. And so if we just saw the outside of it, it would look something like that. It's facing away from us, so we barely see it. If we could see through the cylinder or see through the little flat cut of the cylinder, it would look something like that. So that over there would be our G1. And then our G2 would be the front side of the cylinder. The G2 could be the front side of the cylinder.
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Type II regions in three dimensions Divergence theorem Multivariable Calculus Khan Academy.mp3
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If we could see through the cylinder or see through the little flat cut of the cylinder, it would look something like that. So that over there would be our G1. And then our G2 would be the front side of the cylinder. The G2 could be the front side of the cylinder. So let me color it in as best as I can. So G2 would be the front side of the cylinder. And X can vary above G1 and below G2, and it would fill up this entire cylinder.
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Type II regions in three dimensions Divergence theorem Multivariable Calculus Khan Academy.mp3
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The G2 could be the front side of the cylinder. So let me color it in as best as I can. So G2 would be the front side of the cylinder. And X can vary above G1 and below G2, and it would fill up this entire cylinder. So we see that this same cylinder that we also saw was a type I region can also be a type II region. Now what about this hourglass thing that we saw could not be a type I region? Can this be a type II region?
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Type II regions in three dimensions Divergence theorem Multivariable Calculus Khan Academy.mp3
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And X can vary above G1 and below G2, and it would fill up this entire cylinder. So we see that this same cylinder that we also saw was a type I region can also be a type II region. Now what about this hourglass thing that we saw could not be a type I region? Can this be a type II region? Well, let's think about it. Let's think about it. I'll do it the same way.
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Type II regions in three dimensions Divergence theorem Multivariable Calculus Khan Academy.mp3
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Can this be a type II region? Well, let's think about it. Let's think about it. I'll do it the same way. We can construct a domain. So maybe our domain, it's in the Y, well, it should be in the YZ plane if we're talking about type II regions, or if we want to think of it as a type II region. So our domain could be this kind of flat hourglass shape that's in the YZ plane.
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Type II regions in three dimensions Divergence theorem Multivariable Calculus Khan Academy.mp3
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I'll do it the same way. We can construct a domain. So maybe our domain, it's in the Y, well, it should be in the YZ plane if we're talking about type II regions, or if we want to think of it as a type II region. So our domain could be this kind of flat hourglass shape that's in the YZ plane. So our domain could be a region that looks something like this in the YZ plane. So this is kind of flattened out. So this is our domain right over there.
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Type II regions in three dimensions Divergence theorem Multivariable Calculus Khan Academy.mp3
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So our domain could be this kind of flat hourglass shape that's in the YZ plane. So our domain could be a region that looks something like this in the YZ plane. So this is kind of flattened out. So this is our domain right over there. And then the lower bound on X, G1, could be a surface. It's a function of Y and Z that is kind of the backside of our hourglass. You can see, I'll try to show the contours from the underside right over there.
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Type II regions in three dimensions Divergence theorem Multivariable Calculus Khan Academy.mp3
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So this is our domain right over there. And then the lower bound on X, G1, could be a surface. It's a function of Y and Z that is kind of the backside of our hourglass. You can see, I'll try to show the contours from the underside right over there. So that could be our G1. And then our G2 could be the front side of the hourglass. So my best attempt to draw the front side of the hourglass.
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Type II regions in three dimensions Divergence theorem Multivariable Calculus Khan Academy.mp3
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You can see, I'll try to show the contours from the underside right over there. So that could be our G1. And then our G2 could be the front side of the hourglass. So my best attempt to draw the front side of the hourglass. And I could color it in. And so the way I somewhat confusingly drew it just now, you see that this hourglass oriented the way it is would actually be a type II region. Now, if we were to rotate it like this, so let me draw it like this.
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Type II regions in three dimensions Divergence theorem Multivariable Calculus Khan Academy.mp3
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So my best attempt to draw the front side of the hourglass. And I could color it in. And so the way I somewhat confusingly drew it just now, you see that this hourglass oriented the way it is would actually be a type II region. Now, if we were to rotate it like this, so let me draw it like this. So if we were to make it like this, so that the top of my hourglass is facing us, I'll try my best to draw it. So let's say the top intersects the X axis right over there. This is the bottom of my hourglass right over there.
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Type II regions in three dimensions Divergence theorem Multivariable Calculus Khan Academy.mp3
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Now, if we were to rotate it like this, so let me draw it like this. So if we were to make it like this, so that the top of my hourglass is facing us, I'll try my best to draw it. So let's say the top intersects the X axis right over there. This is the bottom of my hourglass right over there. And then it bends in and then comes back out like that. So it bends in and comes back out just like that. For the same reasons that this was not a type I region, this now would not be a type II region.
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Type II regions in three dimensions Divergence theorem Multivariable Calculus Khan Academy.mp3
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This is the bottom of my hourglass right over there. And then it bends in and then comes back out like that. So it bends in and comes back out just like that. For the same reasons that this was not a type I region, this now would not be a type II region. For any XY, you're going to see that there's multiple points. There's multiple, sorry, for any ZY, you can see there could be multiple X points that are associated with the different points of this hourglass. You can't just have a simple lower and upper bound functions right over here.
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Type II regions in three dimensions Divergence theorem Multivariable Calculus Khan Academy.mp3
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For the same reasons that this was not a type I region, this now would not be a type II region. For any XY, you're going to see that there's multiple points. There's multiple, sorry, for any ZY, you can see there could be multiple X points that are associated with the different points of this hourglass. You can't just have a simple lower and upper bound functions right over here. So this right over here is not a type II region. You could show a rationale, or this is going to be a type I region. You could create a region over here in the XY plane and have an upper and lower bound functions for Z, so you could be type I.
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Type II regions in three dimensions Divergence theorem Multivariable Calculus Khan Academy.mp3
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And what I want to do is think about the value of the line integral, let me write this down, the value of the line integral of F dot dr, where F is the vector field that I've drawn in magenta in each of these diagrams, and obviously it's different in each of these diagrams. The only part of the vector field that I've drawn is the part that's along the surface. I could have drawn the part of the vector field that's off the surface, but we're only going to be concerned with what's going on on the surface. So the vector field could be defined in this entire three-dimensional space, and here as well. And these are obviously different vector fields, and we can see that visually based on how we drew it. And the contour that we care about, remember we're going to take a line integral, so the path matters. The path that we care about is the counterclockwise boundary of our surface.
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Stokes' theorem intuition Multivariable Calculus Khan Academy.mp3
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So the vector field could be defined in this entire three-dimensional space, and here as well. And these are obviously different vector fields, and we can see that visually based on how we drew it. And the contour that we care about, remember we're going to take a line integral, so the path matters. The path that we care about is the counterclockwise boundary of our surface. So it's going to be this right over here, the counterclockwise boundary of our surface is what we're going to be taking F dot dr along. So this right over here, and let me draw the orientation, it's going to be counterclockwise. We're going to do it in every one of these situations, every one of these surfaces and every one of these Fs.
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Stokes' theorem intuition Multivariable Calculus Khan Academy.mp3
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The path that we care about is the counterclockwise boundary of our surface. So it's going to be this right over here, the counterclockwise boundary of our surface is what we're going to be taking F dot dr along. So this right over here, and let me draw the orientation, it's going to be counterclockwise. We're going to do it in every one of these situations, every one of these surfaces and every one of these Fs. And what I really want to think about is how the value of F dot dr over that contour, how it might change from example to example, and obviously the only difference between each of these is what the vector field F is doing. So first let's think about this example right over here. This part of the contour, this bottom part right over here, our vector field is going in the exact same direction as our line, as our contour.
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Stokes' theorem intuition Multivariable Calculus Khan Academy.mp3
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We're going to do it in every one of these situations, every one of these surfaces and every one of these Fs. And what I really want to think about is how the value of F dot dr over that contour, how it might change from example to example, and obviously the only difference between each of these is what the vector field F is doing. So first let's think about this example right over here. This part of the contour, this bottom part right over here, our vector field is going in the exact same direction as our line, as our contour. So it's going to be we're going to get positive values of F dot dr down here, and we're going to keep summing them up. We're taking an integral. Then as we go up the curve, or as we go kind of uphill right over here, we see that our vector field is going essentially orthogonal.
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Stokes' theorem intuition Multivariable Calculus Khan Academy.mp3
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This part of the contour, this bottom part right over here, our vector field is going in the exact same direction as our line, as our contour. So it's going to be we're going to get positive values of F dot dr down here, and we're going to keep summing them up. We're taking an integral. Then as we go up the curve, or as we go kind of uphill right over here, we see that our vector field is going essentially orthogonal. It's going perpendicular to our contour. So our F dot dr, we're not going to get any value from that. F dot dr and all of these, this part of the contour, it's all going to be 0.
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Stokes' theorem intuition Multivariable Calculus Khan Academy.mp3
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Then as we go up the curve, or as we go kind of uphill right over here, we see that our vector field is going essentially orthogonal. It's going perpendicular to our contour. So our F dot dr, we're not going to get any value from that. F dot dr and all of these, this part of the contour, it's all going to be 0. So we're not going to get anything right over there. Actually, let me just write nothing. So we get nothing right over there.
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Stokes' theorem intuition Multivariable Calculus Khan Academy.mp3
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F dot dr and all of these, this part of the contour, it's all going to be 0. So we're not going to get anything right over there. Actually, let me just write nothing. So we get nothing right over there. Or maybe I'll write 0. We'll get 0 right over there. Then up here, when we're at this part of the contour, our vector field is going in the exact opposite direction as our path.
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Stokes' theorem intuition Multivariable Calculus Khan Academy.mp3
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So we get nothing right over there. Or maybe I'll write 0. We'll get 0 right over there. Then up here, when we're at this part of the contour, our vector field is going in the exact opposite direction as our path. Up here our path is going, I guess, from the right to the left, while our vector field is going from the left to the right. So we're actually going to get negative values. We are going to get negative values up here, and they're going to sum up to a reasonable negative value.
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Stokes' theorem intuition Multivariable Calculus Khan Academy.mp3
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Then up here, when we're at this part of the contour, our vector field is going in the exact opposite direction as our path. Up here our path is going, I guess, from the right to the left, while our vector field is going from the left to the right. So we're actually going to get negative values. We are going to get negative values up here, and they're going to sum up to a reasonable negative value. If the vector field is constant, and I kind of drew it like it is, and if this length is equal to this length, then these two values are going to cancel out. When you add this positive sum to this negative sum, they're going to be 0. Then once again, when you go downhill again, the vector field is perpendicular to our actual path, so we're going to get 0.
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Stokes' theorem intuition Multivariable Calculus Khan Academy.mp3
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We are going to get negative values up here, and they're going to sum up to a reasonable negative value. If the vector field is constant, and I kind of drew it like it is, and if this length is equal to this length, then these two values are going to cancel out. When you add this positive sum to this negative sum, they're going to be 0. Then once again, when you go downhill again, the vector field is perpendicular to our actual path, so we're going to get 0. So based on the way I've described it, your line integral of F dot dr for this version of F, in this example right over here, it might all cancel out. You'll get something, if we make the assumptions that I made, this might be equal to 0. So in this example, F dot dr could be equal to 0.
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Stokes' theorem intuition Multivariable Calculus Khan Academy.mp3
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Then once again, when you go downhill again, the vector field is perpendicular to our actual path, so we're going to get 0. So based on the way I've described it, your line integral of F dot dr for this version of F, in this example right over here, it might all cancel out. You'll get something, if we make the assumptions that I made, this might be equal to 0. So in this example, F dot dr could be equal to 0. Now let's think about what's going on in this situation. In this one, just like the last one, as we go along the bottom, the vector field is going in the exact same direction as our contour, so we're going to get positive values. Once we go up the hill, the vector field is going perpendicular to our path, so it's not going to add anything really to it, so we're just going to get 0 along this part.
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Stokes' theorem intuition Multivariable Calculus Khan Academy.mp3
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So in this example, F dot dr could be equal to 0. Now let's think about what's going on in this situation. In this one, just like the last one, as we go along the bottom, the vector field is going in the exact same direction as our contour, so we're going to get positive values. Once we go up the hill, the vector field is going perpendicular to our path, so it's not going to add anything really to it, so we're just going to get 0 along this part. But then up here, our vector field has switched directions, and once again, it's going in the exact same direction as our path, so we're going to get more positive values right over there. Then as we go down here, it's not going to add anything because our vector field is perpendicular to our path, so we're going to get 0. But notice, now these two ends don't cancel out with each other.
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Stokes' theorem intuition Multivariable Calculus Khan Academy.mp3
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Once we go up the hill, the vector field is going perpendicular to our path, so it's not going to add anything really to it, so we're just going to get 0 along this part. But then up here, our vector field has switched directions, and once again, it's going in the exact same direction as our path, so we're going to get more positive values right over there. Then as we go down here, it's not going to add anything because our vector field is perpendicular to our path, so we're going to get 0. But notice, now these two ends don't cancel out with each other. We're going to get a positive value. We are going to get a positive value. What was the difference between this version of F, this vector field, and this vector field right over here?
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Stokes' theorem intuition Multivariable Calculus Khan Academy.mp3
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But notice, now these two ends don't cancel out with each other. We're going to get a positive value. We are going to get a positive value. What was the difference between this version of F, this vector field, and this vector field right over here? Well, this vector field switched directions so that the top part didn't cancel out with the bottom part. Or another way to think about it is it had some curl. There's a little bit of spinning going on.
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Stokes' theorem intuition Multivariable Calculus Khan Academy.mp3
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What was the difference between this version of F, this vector field, and this vector field right over here? Well, this vector field switched directions so that the top part didn't cancel out with the bottom part. Or another way to think about it is it had some curl. There's a little bit of spinning going on. If this was describing the velocity of a fluid, and if you were to put a stick right over there on the surface, the stick would spin. It has some spin or some curl, however you want to describe it. This right over here has no curl.
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Stokes' theorem intuition Multivariable Calculus Khan Academy.mp3
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There's a little bit of spinning going on. If this was describing the velocity of a fluid, and if you were to put a stick right over there on the surface, the stick would spin. It has some spin or some curl, however you want to describe it. This right over here has no curl. If you put a stick right over here, it would just flow with the fluid, but the stick itself would not spin. So we got a positive value for the line integral in this situation, and we also have what looks like a positive curl. Now let's think about this one.
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Stokes' theorem intuition Multivariable Calculus Khan Academy.mp3
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This right over here has no curl. If you put a stick right over here, it would just flow with the fluid, but the stick itself would not spin. So we got a positive value for the line integral in this situation, and we also have what looks like a positive curl. Now let's think about this one. In this situation, as we go along this part of our contour, our vector field F is going in the exact same direction so we're going to get positive values. Now as we go uphill, our vector field F, it's also kind of turned in that direction so we'll get more positive values. Now, as we go in that direction, our vector field F is still going in the direction of our contour, we're going to get more positive values.
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Stokes' theorem intuition Multivariable Calculus Khan Academy.mp3
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Now let's think about this one. In this situation, as we go along this part of our contour, our vector field F is going in the exact same direction so we're going to get positive values. Now as we go uphill, our vector field F, it's also kind of turned in that direction so we'll get more positive values. Now, as we go in that direction, our vector field F is still going in the direction of our contour, we're going to get more positive values. And as we go down, once again the vector field F is going in the direction of our contour So we'll get even more positive values. In this situation, the value of our line integral of F dot dr is even more positive. More, more positive.
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Stokes' theorem intuition Multivariable Calculus Khan Academy.mp3
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Now, as we go in that direction, our vector field F is still going in the direction of our contour, we're going to get more positive values. And as we go down, once again the vector field F is going in the direction of our contour So we'll get even more positive values. In this situation, the value of our line integral of F dot dr is even more positive. More, more positive. And we see that the actual vector field along the surface, and remember, the vector field might be doing all sorts of crazy things off the surface. Actually, let me draw that in that same magenta color. It might be doing all sorts of crazy things off the surface.
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Stokes' theorem intuition Multivariable Calculus Khan Academy.mp3
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More, more positive. And we see that the actual vector field along the surface, and remember, the vector field might be doing all sorts of crazy things off the surface. Actually, let me draw that in that same magenta color. It might be doing all sorts of crazy things off the surface. But what we really care about is what's happening on the surface. And because this vector field is, I guess you could say, curling, or it's spinning along the surface, it allows it to go with the boundary along all the points, and we get a very positive value for this line integral. So we have a higher curl, so more curl, more curl, it looks like, is leading to a more positive line integral.
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Stokes' theorem intuition Multivariable Calculus Khan Academy.mp3
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It might be doing all sorts of crazy things off the surface. But what we really care about is what's happening on the surface. And because this vector field is, I guess you could say, curling, or it's spinning along the surface, it allows it to go with the boundary along all the points, and we get a very positive value for this line integral. So we have a higher curl, so more curl, more curl, it looks like, is leading to a more positive line integral. Now let's think about what's happening in this situation right over here. This situation down here, our vector field is going in the same direction as our path, so we're going to get positive values. Just like the first situation, as we go up the hill like this, or up the surface like that, our vector field is perpendicular to our surface, so it's really not going to add anything to our line integral.
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Stokes' theorem intuition Multivariable Calculus Khan Academy.mp3
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So we have a higher curl, so more curl, more curl, it looks like, is leading to a more positive line integral. Now let's think about what's happening in this situation right over here. This situation down here, our vector field is going in the same direction as our path, so we're going to get positive values. Just like the first situation, as we go up the hill like this, or up the surface like that, our vector field is perpendicular to our surface, so it's really not going to add anything to our line integral. And then as we go along this top part, this first part of the top part right over here, the vector field is going against us, so it's negative right over here. We're going in the exact opposite direction of our path. And then right as we get to the end, the vector field flips direction, and we get a little bit of positive right over here, so it's a little bit of it going in the same direction.
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Stokes' theorem intuition Multivariable Calculus Khan Academy.mp3
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Just like the first situation, as we go up the hill like this, or up the surface like that, our vector field is perpendicular to our surface, so it's really not going to add anything to our line integral. And then as we go along this top part, this first part of the top part right over here, the vector field is going against us, so it's negative right over here. We're going in the exact opposite direction of our path. And then right as we get to the end, the vector field flips direction, and we get a little bit of positive right over here, so it's a little bit of it going in the same direction. And then we go back downhill. When we go back downhill, it adds nothing, because our vector field is going perpendicular to our path. So the big difference between this case and the case up here is the case up here, or actually the big difference between this, well actually I could compare between these two or these two, but the difference between this one and this one is that at least this part of the vector field has switched direction, so we get a little bit of positive value.
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Stokes' theorem intuition Multivariable Calculus Khan Academy.mp3
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And then right as we get to the end, the vector field flips direction, and we get a little bit of positive right over here, so it's a little bit of it going in the same direction. And then we go back downhill. When we go back downhill, it adds nothing, because our vector field is going perpendicular to our path. So the big difference between this case and the case up here is the case up here, or actually the big difference between this, well actually I could compare between these two or these two, but the difference between this one and this one is that at least this part of the vector field has switched direction, so we get a little bit of positive value. And one way to think about it, this one is going to be less positive than that if we take the line integral, but more positive than that. And one other way to think about it is we have a little bit of curl going on right over here, our vector field switched directions right around there, or I guess you could say that it's spinning right around there, so if you put a stick, if that was in the water, it would start spinning. But everywhere else there isn't a lot of curl, so you have some curl, but it's over a little small region of the surface.
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Stokes' theorem intuition Multivariable Calculus Khan Academy.mp3
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So the big difference between this case and the case up here is the case up here, or actually the big difference between this, well actually I could compare between these two or these two, but the difference between this one and this one is that at least this part of the vector field has switched direction, so we get a little bit of positive value. And one way to think about it, this one is going to be less positive than that if we take the line integral, but more positive than that. And one other way to think about it is we have a little bit of curl going on right over here, our vector field switched directions right around there, or I guess you could say that it's spinning right around there, so if you put a stick, if that was in the water, it would start spinning. But everywhere else there isn't a lot of curl, so you have some curl, but it's over a little small region of the surface. Over here you had curl going on over a larger portion of our surface, and so up here you had a more positive curl, more positive line integral. Here you have a curl over less of the surface, and you're going to have a less positive line integral. Now let's think about this one over here.
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Stokes' theorem intuition Multivariable Calculus Khan Academy.mp3
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But everywhere else there isn't a lot of curl, so you have some curl, but it's over a little small region of the surface. Over here you had curl going on over a larger portion of our surface, and so up here you had a more positive curl, more positive line integral. Here you have a curl over less of the surface, and you're going to have a less positive line integral. Now let's think about this one over here. This vector field along the surface, there is some curl. There's some curl going on right over there. If you put a stick in the water, if you view that as the velocity of water, the stick would spin, so you have some curl.
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Stokes' theorem intuition Multivariable Calculus Khan Academy.mp3
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Now let's think about this one over here. This vector field along the surface, there is some curl. There's some curl going on right over there. If you put a stick in the water, if you view that as the velocity of water, the stick would spin, so you have some curl. But then it switches direction again, so you have curl there as well. It's actually curl in the opposite direction. To some degree, if you were to sum all of this up, maybe it would cancel out.
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Stokes' theorem intuition Multivariable Calculus Khan Academy.mp3
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If you put a stick in the water, if you view that as the velocity of water, the stick would spin, so you have some curl. But then it switches direction again, so you have curl there as well. It's actually curl in the opposite direction. To some degree, if you were to sum all of this up, maybe it would cancel out. It makes sense that it would cancel out, because when you take the line integral around the whole thing, just like this first situation, it looks like it'll add up to zero. Because even though you have some curl, the curls cancel out each other, and so when you go to this top part of the surface, the vector field is going in the exact same direction as this bottom part of the surface. If you were to take your line integral, the same one that we care about, just like the first situation, it would be positive down here, zero as we go up the curve, and then as we go down here, the vector field has switched directions twice, so it's still going against the path of our contour, just like this first situation.
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Stokes' theorem intuition Multivariable Calculus Khan Academy.mp3
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To some degree, if you were to sum all of this up, maybe it would cancel out. It makes sense that it would cancel out, because when you take the line integral around the whole thing, just like this first situation, it looks like it'll add up to zero. Because even though you have some curl, the curls cancel out each other, and so when you go to this top part of the surface, the vector field is going in the exact same direction as this bottom part of the surface. If you were to take your line integral, the same one that we care about, just like the first situation, it would be positive down here, zero as we go up the curve, and then as we go down here, the vector field has switched directions twice, so it's still going against the path of our contour, just like this first situation. It would be negative up there, and then as we go down, it would be zero. This thing right over here also looks just like the first one, because the curls essentially cancel out. We switch direction twice.
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Stokes' theorem intuition Multivariable Calculus Khan Academy.mp3
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If you were to take your line integral, the same one that we care about, just like the first situation, it would be positive down here, zero as we go up the curve, and then as we go down here, the vector field has switched directions twice, so it's still going against the path of our contour, just like this first situation. It would be negative up there, and then as we go down, it would be zero. This thing right over here also looks just like the first one, because the curls essentially cancel out. We switch direction twice. Over here, our line integral might also be zero. The whole reason why I went through this exercise is to give you an intuition of why it might make sense that if we have more curl happening over more of this surface, why that might make the value of this line integral be larger. Hopefully, it starts to give you an intuition that maybe, just maybe, the value of this line integral, the value of F dot dr over that contour, over this contour that's going in the counterclockwise direction, we'll talk more about orientation in future videos, maybe this is equal to the sum of the curls over the surface.
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Stokes' theorem intuition Multivariable Calculus Khan Academy.mp3
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We switch direction twice. Over here, our line integral might also be zero. The whole reason why I went through this exercise is to give you an intuition of why it might make sense that if we have more curl happening over more of this surface, why that might make the value of this line integral be larger. Hopefully, it starts to give you an intuition that maybe, just maybe, the value of this line integral, the value of F dot dr over that contour, over this contour that's going in the counterclockwise direction, we'll talk more about orientation in future videos, maybe this is equal to the sum of the curls over the surface. Let's think about this. It could be a surface integral. We're going to go over the surface.
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Stokes' theorem intuition Multivariable Calculus Khan Academy.mp3
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Hopefully, it starts to give you an intuition that maybe, just maybe, the value of this line integral, the value of F dot dr over that contour, over this contour that's going in the counterclockwise direction, we'll talk more about orientation in future videos, maybe this is equal to the sum of the curls over the surface. Let's think about this. It could be a surface integral. We're going to go over the surface. What we care about is the curl of F. We don't just care about the curl of F generally, because F might be spinning in a direction in a way that's off the surface. We care about how much it's curling on the surface. What we'd want to do is we'd want to take the curl of F and dot it with the normal vector at any point of the surface, and then multiply that times the surface itself.
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Stokes' theorem intuition Multivariable Calculus Khan Academy.mp3
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We're going to go over the surface. What we care about is the curl of F. We don't just care about the curl of F generally, because F might be spinning in a direction in a way that's off the surface. We care about how much it's curling on the surface. What we'd want to do is we'd want to take the curl of F and dot it with the normal vector at any point of the surface, and then multiply that times the surface itself. Then multiply that times the surface itself. Another way of writing this, and this is just to say that the more surface where we have more curling going on, the more that the line integral, the value of the line integral might be. We saw that when we compared these three examples.
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Stokes' theorem intuition Multivariable Calculus Khan Academy.mp3
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What we'd want to do is we'd want to take the curl of F and dot it with the normal vector at any point of the surface, and then multiply that times the surface itself. Then multiply that times the surface itself. Another way of writing this, and this is just to say that the more surface where we have more curling going on, the more that the line integral, the value of the line integral might be. We saw that when we compared these three examples. Another way of writing all of this is the surface integral. Let me write the surface in that same brown color. The surface integral of the curl of F, which would just be another vector that tells us how much we are spinning generally, but we want to care how much we're spinning along the surface.
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Stokes' theorem intuition Multivariable Calculus Khan Academy.mp3
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We saw that when we compared these three examples. Another way of writing all of this is the surface integral. Let me write the surface in that same brown color. The surface integral of the curl of F, which would just be another vector that tells us how much we are spinning generally, but we want to care how much we're spinning along the surface. We're dotting it with the normal vector. Another way to write this whole thing is to say dot ds. If you take essentially the sum across the entire surface of how much we're curling, how much we're spinning along that surface, then maybe, just maybe, this will be equal to the value of the line integral as we go around the boundary of the surface.
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Stokes' theorem intuition Multivariable Calculus Khan Academy.mp3
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So after introducing the idea of fluid rotation in a vector field like this, let's start tightening up our grasp on this intuition to get something that we can actually apply formulas to. So a vector field like the one that I had there that's two dimensional is given by a function that has a two dimensional input and a two dimensional output. And it's common to write the components of that output as the functions p and q. So each one of those p and q takes in two different variables as its input, p and q. And what I wanna do here is talk about this idea of curl. And you might write it down as just curl, curl of v, the vector field, which takes in the same inputs that the vector field does. And because this is the two dimensional example, I might write just to distinguish it from three dimensional curl, which is something we'll get later on, 2D curl of v. So you're kind of thinking of this as a differential thing in the same way that you have a derivative, d dx, it's gonna take in some kind of function.
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2d curl formula.mp3
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So each one of those p and q takes in two different variables as its input, p and q. And what I wanna do here is talk about this idea of curl. And you might write it down as just curl, curl of v, the vector field, which takes in the same inputs that the vector field does. And because this is the two dimensional example, I might write just to distinguish it from three dimensional curl, which is something we'll get later on, 2D curl of v. So you're kind of thinking of this as a differential thing in the same way that you have a derivative, d dx, it's gonna take in some kind of function. And you give it a function and it gives you a new function, the derivative. Here, you think of this 2D curl as like an operator. You give it a function, a vector field function, and it gives you another function, which in this case will be scalar valued.
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2d curl formula.mp3
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And because this is the two dimensional example, I might write just to distinguish it from three dimensional curl, which is something we'll get later on, 2D curl of v. So you're kind of thinking of this as a differential thing in the same way that you have a derivative, d dx, it's gonna take in some kind of function. And you give it a function and it gives you a new function, the derivative. Here, you think of this 2D curl as like an operator. You give it a function, a vector field function, and it gives you another function, which in this case will be scalar valued. And the reason it's scalar valued is because at every given point, you want it to give you a number. So if I look back at the vector field that I have here, we want that at a point like this where there's a lot of counterclockwise rotation happening around it, for the curl function to return a positive number. But at a point like this where there's some counter, where there's clockwise rotation happening around it, we want the curl to return a negative number.
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2d curl formula.mp3
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You give it a function, a vector field function, and it gives you another function, which in this case will be scalar valued. And the reason it's scalar valued is because at every given point, you want it to give you a number. So if I look back at the vector field that I have here, we want that at a point like this where there's a lot of counterclockwise rotation happening around it, for the curl function to return a positive number. But at a point like this where there's some counter, where there's clockwise rotation happening around it, we want the curl to return a negative number. So let's start thinking about what that should mean. And a good way to understand this two-dimensional curl function and start to get a feel for it is to imagine the quintessential 2D curl scenario. Well, let's say you have a point, and this here is gonna be our point x, y, sitting off somewhere in space.
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2d curl formula.mp3
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But at a point like this where there's some counter, where there's clockwise rotation happening around it, we want the curl to return a negative number. So let's start thinking about what that should mean. And a good way to understand this two-dimensional curl function and start to get a feel for it is to imagine the quintessential 2D curl scenario. Well, let's say you have a point, and this here is gonna be our point x, y, sitting off somewhere in space. And let's say there's no vector attached to it, as in the values p and q at x and y are zero. And then let's say that to the right of it, you have a vector pointing straight up. Above it, in the vector field, you have a vector pointing straight to the left.
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2d curl formula.mp3
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Well, let's say you have a point, and this here is gonna be our point x, y, sitting off somewhere in space. And let's say there's no vector attached to it, as in the values p and q at x and y are zero. And then let's say that to the right of it, you have a vector pointing straight up. Above it, in the vector field, you have a vector pointing straight to the left. To its left, you have one pointing straight down. And below it, you have one pointing straight to the right. So in terms of the functions, what that means is this vector to its right, whatever point it's evaluated at, that's gonna be q is greater than zero.
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2d curl formula.mp3
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Above it, in the vector field, you have a vector pointing straight to the left. To its left, you have one pointing straight down. And below it, you have one pointing straight to the right. So in terms of the functions, what that means is this vector to its right, whatever point it's evaluated at, that's gonna be q is greater than zero. So this function q that corresponds to the y component, the up and down component of each vector, when you evaluate it at this point to the right of our x, y point, q is gonna be greater than zero. Whereas if you evaluate it to the left over here, q would be less than zero, less than zero, in our kind of perfect curl will be positive example. And then these bottom guys, if you start thinking about what this means for, you'd have a rightward vector below and a leftward vector above, the one below it, whatever point you're evaluating that at, p, which gives us the kind of left-right component of these vectors, since it's the first component of the output, would have to be positive.
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2d curl formula.mp3
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So in terms of the functions, what that means is this vector to its right, whatever point it's evaluated at, that's gonna be q is greater than zero. So this function q that corresponds to the y component, the up and down component of each vector, when you evaluate it at this point to the right of our x, y point, q is gonna be greater than zero. Whereas if you evaluate it to the left over here, q would be less than zero, less than zero, in our kind of perfect curl will be positive example. And then these bottom guys, if you start thinking about what this means for, you'd have a rightward vector below and a leftward vector above, the one below it, whatever point you're evaluating that at, p, which gives us the kind of left-right component of these vectors, since it's the first component of the output, would have to be positive. And then above it, above it here, when you evaluate p at that point, it would have to be negative. Whereas p, if you did it on the left and right points, would be equal to zero because there's kind of no x component. And similarly, q, if you did it on the top and bottom points since there's no up and down component of those vectors, would also be zero.
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2d curl formula.mp3
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And then these bottom guys, if you start thinking about what this means for, you'd have a rightward vector below and a leftward vector above, the one below it, whatever point you're evaluating that at, p, which gives us the kind of left-right component of these vectors, since it's the first component of the output, would have to be positive. And then above it, above it here, when you evaluate p at that point, it would have to be negative. Whereas p, if you did it on the left and right points, would be equal to zero because there's kind of no x component. And similarly, q, if you did it on the top and bottom points since there's no up and down component of those vectors, would also be zero. So this is just the very specific, almost contrived scenario that I'm looking at. And I wanna say, hey, if this should have positive curl, maybe if we look at the information, the partial derivative information to be specific, about p and q in a scenario like this, it'll give us a way to quantify the idea of curl. And first let's look at p. So p starts positive, and then as y increases, as the y value of our input increases, it goes from being positive to zero to negative.
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2d curl formula.mp3
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And similarly, q, if you did it on the top and bottom points since there's no up and down component of those vectors, would also be zero. So this is just the very specific, almost contrived scenario that I'm looking at. And I wanna say, hey, if this should have positive curl, maybe if we look at the information, the partial derivative information to be specific, about p and q in a scenario like this, it'll give us a way to quantify the idea of curl. And first let's look at p. So p starts positive, and then as y increases, as the y value of our input increases, it goes from being positive to zero to negative. So we would expect that the partial derivative of p with respect to y, so as we change that y component moving up in the plane and look at the x component of the vectors, that should be negative. That should be negative in circumstances where we want positive curl. So all of this we're looking at cases, you know, the quintessential case where curl is positive.
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2d curl formula.mp3
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And first let's look at p. So p starts positive, and then as y increases, as the y value of our input increases, it goes from being positive to zero to negative. So we would expect that the partial derivative of p with respect to y, so as we change that y component moving up in the plane and look at the x component of the vectors, that should be negative. That should be negative in circumstances where we want positive curl. So all of this we're looking at cases, you know, the quintessential case where curl is positive. So evidently this is a fact that corresponds to positive curl. Whereas q, let's take a look at q. It starts negative when you're at the left, and then becomes zero, then it becomes positive.
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2d curl formula.mp3
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So all of this we're looking at cases, you know, the quintessential case where curl is positive. So evidently this is a fact that corresponds to positive curl. Whereas q, let's take a look at q. It starts negative when you're at the left, and then becomes zero, then it becomes positive. So here as x increases, q increases. So we're expecting that a partial derivative of q with respect to x should be positive. Or at the very least, that situations where the partial derivative of q with respect to x is positive corresponds to positive two-dimensional curl.
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2d curl formula.mp3
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It starts negative when you're at the left, and then becomes zero, then it becomes positive. So here as x increases, q increases. So we're expecting that a partial derivative of q with respect to x should be positive. Or at the very least, that situations where the partial derivative of q with respect to x is positive corresponds to positive two-dimensional curl. And in fact, it turns out, these guys tell us all you need to know. We can say as a formula that the 2D curl, 2D curl of our vector field v as a function of x and y is equal to the partial derivative of q with respect to x, partial derivative of q with respect to x, and then I'm gonna subtract off the partial of p with respect to y, because I want when this is negative for that to correspond with more positive 2D curl. So I'm gonna subtract off partial of p with respect to y.
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2d curl formula.mp3
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Or at the very least, that situations where the partial derivative of q with respect to x is positive corresponds to positive two-dimensional curl. And in fact, it turns out, these guys tell us all you need to know. We can say as a formula that the 2D curl, 2D curl of our vector field v as a function of x and y is equal to the partial derivative of q with respect to x, partial derivative of q with respect to x, and then I'm gonna subtract off the partial of p with respect to y, because I want when this is negative for that to correspond with more positive 2D curl. So I'm gonna subtract off partial of p with respect to y. And this right here is the formula for two-dimensional curl. Which basically, you can think of it as a measure at any given point, you're asking how much does the surrounding information to that point look like this setup, like this perfect counterclockwise rotation setup. And the more it looks like this setup, the more this value will be positive.
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2d curl formula.mp3
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One of the most fundamental ideas in all of physics is the idea of work. And when you first learn work, you just say, oh, that's just force times distance. But then later on, when you learn a little bit about vectors, you realize that the force isn't always going in the same direction as your displacement. So you learn that work is really the magnitude of the force in the direction, or the component of the force in the direction, in direction of displacement. Displacement is just distance with some direction. In direction of displacement times the magnitude of the displacement. Or you could just kind of say times the distance displaced.
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Line integrals and vector fields Multivariable Calculus Khan Academy.mp3
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So you learn that work is really the magnitude of the force in the direction, or the component of the force in the direction, in direction of displacement. Displacement is just distance with some direction. In direction of displacement times the magnitude of the displacement. Or you could just kind of say times the distance displaced. I'm not being too particular about it. And the classic example, maybe you have an ice cube or some type of block. I just say ice so that it's not a lot of friction.
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Line integrals and vector fields Multivariable Calculus Khan Academy.mp3
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Or you could just kind of say times the distance displaced. I'm not being too particular about it. And the classic example, maybe you have an ice cube or some type of block. I just say ice so that it's not a lot of friction. Maybe it's standing on a bigger sheet of lake or ice or something. And maybe you're pulling on that ice cube at an angle. Let's say you're pulling at an angle like that.
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Line integrals and vector fields Multivariable Calculus Khan Academy.mp3
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