Stokes' theorem proof part 2

Let's now parameterize our surface. And then, we can figure out what ds would actually look like. And so I will define my position vector function for our surface as r. And I'm going to make it a function of two parameters because we're going to have to define a surface right over here. And I can actually use x and y as my parameters because the surface can be defined as a function of x and y. So I'll say that my parameterization is going to be a function of x and y. And in my i direction, it's going to be x times i. In my j direction, it's going to be y times j. And then, in the k direction, well, that's just going to be z. And z is a function of x and y. And whenever you do a parameterization of a surface, you have to think about, well, what are the constraints on the domain for your parameters? And so the constraints of the domain for my parameters, I'm going to say that every pair of xy-- every xy coordinate-- it has to be a member-- this is the symbol for member-- it has to be a member of this little region right over here. We could call this the domain of the parameters. It has to be a member of r. Actually, we assume that up here. And actually, I should have written this as the coordinates of xy-- the xy pairs that are a member of r-- that's going to help define our surface. An xy that is not a member of r, then we're not going to consider that the z of that xy as part of the surface. Only the z of xy's where the xy's are part of this region. So now that we have a parameterization for our surface, we're ready to start thinking about what ds might be. And we need to think about this a little bit carefully. So first, I'll just write something. And then, we can confirm whether this really will be the case. ds-- and we've seen this before about why this is the case-- is going to be the cross product of the partial derivative of this with respect to each of the parameters times the little chunk of area in that domain. You could view it as the partial of r of this parameterization with respect to x crossed with the partial of r with respect to y. And then, that whole thing-- and we actually want this to be a vector, not just the absolute value or the magnitude of this vector. We actually want this to be a vector. That thing times-- we could put in some order dx dy or we could write dy dx. And if we want to be general just to say that it's a little differential of our region right over here, we can just write-- instead of writing dx dy or dy dx-- we will write da. And the reason why I said we need to be careful is we need to make sure, based on how we've parameterized this position vector function, based on how we parameterized it, whether this cross product really points in the right direction, the direction that we need to be oriented in. Because, remember, if we're traversing a boundary like this, we want to make sure that the surface is oriented the right way. And the way we think about-- if we were to twist a cap like this, the cap would move up. Or if you were to walk around this boundary in this direction with the surface to your left, your head would point up. And so we need to make sure that this vector, which really defines the orientation of the surface, is definitely going to be pointed up or above the surface as opposed to going below the surface. And so let's think about that a little bit. The partial derivative with respect to x-- well, as x gets bigger, it's going to go in that direction along the surface. And the partial with respect to y, as y gets bigger, it's going to go in that direction along the surface. If I take r cross y-- and we could use the right-hand rule here. We put our index finger in the direction of the first thing we're taking the cross product of, our middle finger in the direction of the second thing. So just like that. We bend our middle finger. We don't care what the other fingers are doing. So I'll just draw them right there. Then, the thumb will go in the direction of the cross product. So in this case, the thumb is going to point up. And that's exactly what we wanted to happen. So this actually is the right ordering. The partial with respect to y cross the partial with respect to x actually would not have been right. That would have given us the other orientation. We would have done that if this boundary actually went the other way around. But this is the right orientation given the way that we are going to traverse the boundary. Now, with that out of the way, let's actually calculate this cross product. So let me just rewrite it. So this is going to be equal to-- well, I'll just focus on the cross product right now. The partial of r with respect to x crossed with the partial of r with respect to y is going to be equal to-- and we've done this many times already. We'll just do it in more general terms now. It's going to be equal to the determinant of this matrix-- i, j, and k. Let me do those in different colors. I think that's a helpful way of thinking about it. So i, j, and k. Actually, I'll use this magenta color. i, j, and k. Let me write this line right over here. We want to write the different components of the partial with respect to x. And so the partial of the i component with respect to x is just going to be 1. The partial of the j component with respect to x is going to be 0. And the partial of this with respect to x, well, we can just write that as a partial derivative of the function z with respect to x. And so r sub x-- or the partial with respect to x-- is the vector 1, i plus 0, j plus z sub x, k. And we'll do the same thing for this piece right over here. The partial with respect to y-- it's i component, well, this is going to be 0. It's j component is going to be 1. The partial of this with respect to y is 1. And the partial of this with respect to y is the partial of z with respect to y. And actually, I forgot to write k right over here in our parameterization. So now with all of this set up, we are ready to figure out what the cross product is. The cross product is going to be equal to-- so our i component is going to be 0 times the partial with respect to y minus 1 times the partial of z with respect to x. So we get negative partial of z with respect to x. And then, checkerboard pattern. We'll have minus j times-- so ignore that column, that row-- 1 times the partial with respect to y. So that's z sub y-- the partial of z with respect to y-- minus 0 times this. So we're just left with that right over there. And then, finally, we're going to have plus k. And here we have 1 times 1 minus 0 times 0. So it's going to be k times 1. So we can just write k over there. And so we can write, the cross product really is just equal to negative z sub x times i minus z sub y times j plus k.