- Stokes' theorem intuition
- Green's and Stokes' theorem relationship
- Orienting boundary with surface
- Orientation and stokes
- Orientations and boundaries
- Conditions for stokes theorem
- Stokes example part 1
- Stokes example part 2
- Stokes example part 3
- Stokes example part 4
- Stokes' theorem
- Evaluating line integral directly - part 1
- Evaluating line integral directly - part 2
Parameterizing the surface. Created by Sal Khan.
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- i dont understand why is sal using the unit circle, isnt the intersection of the shape of an ellipse?(5 votes)
- Because all the X and Y points will lie within the Unit Circle. He then completed the vector expressing the Z points in terms of the X and Y points. No Z he was concerned with "lived" outside the projection of the Ellipse on the X,Y planes: the shadow of the ellipse on the XY planes is a unit circle(13 votes)
- So when paramterizing for stokes theorem, you have to considered every point inside the region, but for just the line integral, its just the edges and not inside the region for example? like its a disk, not a circle, but for fevaluating line integral directly, its a cirlce, not a disk. Can someone please helpme out on this i have a final in 6 days on this.(4 votes)
- Before evaluating any surface integral, one needs to take into account every point at which you are computing the integral by parameterizing the surface using two independent variables. Before evaluating a line integral, one must take every point covered along the path using a one variable parameterization. Creating a vector representation of either a surface or closed path is often the most difficult part of applying Stokes Theorem. From my experience, it takes practice to develop that skill to a second nature degree. Usually, once you have conceived of a suitable parameterization, the problem involves calculating a single, double, or triple integral. I hope this helps.(2 votes)
- Why should we split the surface into three surfaces and separately apply stokes law like in the previous video where he calculated the surface integral... how can we parametrize taking it as one surface , wont taking a variable radius r and a variable angle trace out a solid and not a surface.? And is this surface open at the top or the base?(4 votes)
- The vector-valued function that is created in this video does not define the surface S but rather the region bounded by the curve c. This occurs because z is defined explicitly as a function of y and therefore can only take on values sitting on the plane y+z=2. This is also the reason that the function does not define a solid, i.e. z cannot take on values in the range: [0,y+z=2).(1 vote)
- I'm still struggling to understand when to use when to use i-j+k vs. i+j+k. Are there any videos that explain when it's appropriate to use one verses the other? Thanks!(1 vote)
- If we were to use this parameterization, could we build a 3-d Knot? This seems like a really good way to describe a basic set of them. like a clove hitch, which I feel can be done like a torrus.(0 votes)
Now that we've set up our surface integral, we can attempt to parametrise the surface. And one way to think about is we want our x and y values to take on all of the values inside of the unit circle, what I'm shading in right over here. And that our z values can be a function of the y values. We can express this equation right here, z is equal to 2 minus y. And then we could figure out how high above to go to get our z value. And by doing that, we'll be able to essentially get to every point that sits on our surface. And so first let's think about how we can get every x in y value inside of the unit circle. So let's just focus on the xy plane. We're kind of rotated around a little bit, so it looks a little bit more traditional. So this is my x-axis and then my y-axis would look something like that. Let me draw it a little bit different. This is my y-axis. And then if I were to draw the unit circle, some kind of the base of this thing, or at least where it intersects the xy plane-- actually this thing would keep going down, if I wanted to draw the x squared plus y squared equals 1. But if I draw where it intersects the xy plane, we get the unit circle. So let me just draw it. That's my best attempt at drawing a unit circle. We get the unit circle and we need to think of using parameters so that we can get every x and y-coordinate that's inside of the unit circle. And to think about that, I'll introduce one parameter that's essentially the angle with the x-axis. And I'll call that parameter theta. So theta is the angle with the x-axis. And so theta will essentially sweep things all the way around. So theta can go between 0 and 2 pi. So theta will take on values between 0 and 2 pi. And if we just fix the radius at some point, say radius 1, that would only give us all of the points on the unit circle. But we want all the points inside of it too. So we need to vary the radius as well. So let's introduce another parameter, let's call it r, that is the radius. So for any given r, if we keep changing theta, we would essentially sweep out a circle of that radius. And if you change radius a little bit more, you'll sweep out another circle. And if you vary radius between 0 and 1, you'll get all of the circles that will fill out this entire area. So the radius is going to go between 0 and 1. Another way of thinking about it is for any given theta, if you keep varying the radius, you'll sweep out all of the points on this line. And then as you change theta, it'll sweep out the entire circle. So either way you think about it. So with that, let's actually define x and y in those terms. So we could say that x is equal to-- so the x value whatever r is, the x value is going to be r cosine theta. It's going to be that component, it's going to be r cosine theta. And then the y component-- this is just basic trigonometry-- is going to be r sine theta. And then the z component, we already said z can be expressed as a function of y. Right over here we can rewrite this as z is equal to 2 minus y. That'll tell us how high to go so we end up on that plane. So if z is equal to 2 minus y and if y is r sine theta, we can rewrite z as being equal to 2 minus r sine theta. So there, we're done. That's our parametrization, if we wanted to write this as a position vector with two parameters-- I'll call it lowercase s, it's already used r. Lowercase s, this is our surface, and it's going to be parametrized with r and theta. We can write it as r cosine theta i plus r sine of theta j plus r plus 2 minus r sine of theta k.