Stokes' theorem proof part 3

We've now laid the groundwork so we can express this surface integral, which is the right-hand side of the way we've written Stokes' theorem. We can now express this as a double integral over the domain of the parameters that we care about. And we're going to do that in this video. And then in the next series of videos, we'll do the same thing for this expression. We're actually going to do that using Green's theorem. What we're going to do is we're going to see we're going to get the same expressions, which will show us that Stokes' theorem is true, at least for this special class of surfaces that we are studying right here. But they're pretty general. Now let's now try to do that. So our surface integral-- I'm just going to rewrite it down here. It's the surface integral, so over our surface of the curl of f. Actually, let me go a little bit lower. So we have our surface integral of the curl of F dot ds. Well, we've already figured out what our curl of F is here two videos ago. And we've almost figured out what ds is. ds is the cross product of these two vectors times dA. The cross product of these two vectors is this right over here. So we could just write that ds is going to be equal to this thing times dA. This is the cross product of the partial of r with respect to x and the partial of r with respect to y. And then we have to multiply that times dA right over there. So this expression is just going to be the dot product of the curl of F, which is this stuff up here dotted with this stuff down here. And essentially, we're going to take the dot product of this vector and that vector and then multiply it times this-- we can actually consider this to be a scalar value. So let's do that. So this is going to be equal to-- and when we do this, all of these we are now going to start operating in the domain of our parameters. So it's going to turn from a surface integral into a double integral over that domain, over that region that we care about. This is the domain of our parameters, this region R. And this is how we've manipulated any of the surface integrals that we've come across so far. We've turned them into double integrals over the domain of the parameters. So this is going to turn into a double integral over the domain of our parameters, which is the region R. It's the region R in the xy plane right up here. And now we can take the dot product of the curl of F with dotted ds, which is all of this business right over here. And let me see if I can show both of them on the screen at the same time. There you go. So first let's think about the x components. So you have that right over there. And then you have this right over here. You multiply the two. The negative we can swap the order right over there. You have the partial of z with respect to x times-- and we're going to swap this order right over here. The partial of Q with respect to z minus the partial of R with respect to y. Now let's think about the j component. You have negative z sub y times all of this up here, at least the stuff that's multiplied times the j component. This negative can cancel out with that negative, so you have plus z sub y, the partial of z with respect to y, times the partial of R with respect to x minus the partial of P with respect to z. Let me make that clear, that is an R right over here. And then we have the k component. And the k component is actually the easiest because it's just 1 here. So it's just going to be 1 times-- and I'll just do it in that same color-- 1 times the partial of Q with respect to x minus the partial of P with respect to y. And then finally, we just have this dA over here. And this dA is multiplied by everything. So we'll put some parentheses, and we'll write dA. So we're done. We've expressed our surface integral as a double integral over the domain of our parameters. And what we're going to do in the next few videos is do the same thing with this using Green's theorem. And we're going to see that we get the exact same value.