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Stokes' theorem proof part 6

More manipulating the integrals... Created by Sal Khan.

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  • blobby green style avatar for user Sarah-Laura Narcisse
    I had an hard time understanding the anology between N and M and (p+RZx) and (Q+RZy) . I don't understand how we can replace those terms by M and N?
    (7 votes)
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    • leaf green style avatar for user Flynn Clubbaire
      M and N are functions of x and y -- (p+RZx) and (Q+RZy) are also functions of x and y. This is only possible because Z can be written as a function of x and y, which is exactly the assumption that Sal made at the very beginning of this video. When sal found the original line integral involving M and N, he did so with M and N being arbitrary -- meaning that we can actually pretend that M and N are (p+RZx) and (Q+RZy) respectively. Now, here's the key -- Were you to plug in (p+RZx) and (Q+RZy) as M and N into the line integral involving C_1, you would get an identical value to the line integral involving C. This is because each line integral actually traverses a different field. the C_1 integral traverses through the field defined by (p+RZx), and (Q+RZy), whereas the C integral traversed the field defined by P, Q, R. Sal managed to simplify the P, Q, R traversal into a scalar field defined in terms of (p+RZx) and (Q+RZy), and, this scalar field being a function only of x and y, is identical for all paths that look identical along the x and y axes -- C_1 and C are identical along said axes, they differ only in their Z components. This means that you could change the C in the original (p+RZx) and (Q+RZy) line integral to a C_1, and get the exact same result. What Sal noticed is that pretending (p+RZx) = M and (Q+RZy) = N, then substituting those into the C_1 integral is exactly the same as swapping the C with a C_1, since (p+RZx) and (Q+RZy) appear everywhere you would expect to see an M and an N. Hence why he is able to pretend they are the same, and interchange them freely! While it appears he is replacing M's and N's everywhere, what he's really doing is swapping a C for a C_1, but going through the algebra required to prove that that is possible, as opposed to making an intuitive argument.
      (6 votes)
  • leaf green style avatar for user Blake Jones
    At ish, I am a little confused about how having the result of F dot dr in the same form as the line integral over c1 (i.e. some M times dx + some N times dy) leads to said product of F dot dr necessarily being equal to the line integral over c1 specifically. I see how it represents the same form as the general line integral, but I am not sure why that necessarily makes it equal to the line integral specifically over c1. I hope that question makes sense; any clarification would be appreciated.
    (5 votes)
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    • aqualine seed style avatar for user Bobak McCann
      Those things, the (P + Rzx)dx/dt + (Q..... are how much the curve C changes in the x-direction and y-direction. For a tiny change in t, dt, the curve C's x and y directions change by (P + Rzx)dx/dt + (Q..... Now, for the curve C1, its x and y components change by the SAME AMOUNT as the curve C for a tiny change in t. C1 does not have a change in z, but whatever the change in x and y for the curve C is, C1 will have that same change, and if C's change for those is (P + Rzx)dx/dt + (Q..... then so will C1's change for x and y. Hope that helps, I wish that there was a way on Khan Academy to input math text and symbols
      (5 votes)
  • blobby green style avatar for user P
    1. Could Sal have just integrated F.dr over C or was it necessary to integrate over C1.
    2. Could Greens theorem have applied to the integral F.dr over C?; and
    3. Whats the difference between a surface and a region?

    My thanks in advance.
    (1 vote)
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Video transcript

Where we had left off, we had expressed our line integral over the boundary of our surface in terms of dt, in the dt domain. We expressed what f dot dr is going to be equal to. What I'm going to do in this video is do a little bit of algebraic manipulation, and then we will actually apply Green's theorem. And whether or not we have time in this video, we'll then manipulate that to show that it's the exact same thing that we saw in the earlier part, where we evaluated the surface integral. And so let's do that. So this integral right over here is the same thing as the integral from a to b. And then what I'm going to do is I'm going to group the things that are being multiplied by dx dt. So if I group them and then factor out a dx dt, so if I take this part right here and this part right over here, I'm essentially going to distribute the R. Let me make it clear. I'm going to distribute the R. And so I'm going to group that and that right over there, I will be left with-- and I'll factor out the dx dt. I will be left with P plus R times the partial of z with respect to x times dx dt. And I'll do the same thing for the dy dt's. So that's that part right over there. And then the R is going to get distributed in this thing right over here. So it's going to be plus Q plus R-- the R is going to get distributed-- times the partial of z with respect to y. And I'm going to factor out a dy dt. And then we can't forget all of that is going to be multiplied by dt. Now, this right over here is interesting, because it's starting to look very similar to what we had up here, where we just had our theoretical vector field. In fact, let me copy and paste it. So actually, I don't know if I'm on the right a layer of my work right over here. So let me copy and paste. No, that didn't work. Let me try it one more time. So if I try to copy, I'll go a layer down. I'm using an art program for this. Copy-- and then I think this will work-- and paste. There we go. So this is a result that we had before. This is kind of a template to look at. But what is going on over here if we just look at this template? We see that we're in the t domain. We're integrating over t right over here. But then we have these things. We have some function that's a function of x and y times dx dt, and then some function, the function of x and y times dy dt, and we're integrating with respect to dt. Well, that's exactly what we're doing right over here. We can distribute the dt, and we have something that looks exactly like this right over here, where M is analogous to this piece right over here. Let me make it clear. M, this piece right over here, looks a lot like M in our example right over here. It's being multiplied by dx dt, and then this dt, which you can distribute, and this piece right over here, looks a lot like N. And so, we can say that, well, we have something that looks like this. We can rewrite it like this and go back into the-- kind of deparametrize it. So this thing is going to be equal to now the line integral of C1. We are in the xy plane. We started with the curve C, but now we're going to go to C1. It's completely analogous. These are only functions of x's and y's. Everything here is. So now, this is going to be the line integral over C1-- and I could even draw it as that, if I like-- of M dx, and that makes sense. Because if you multiply dt times dx dt, the dt's cancel out, and you're just left with dx. So M times dx, so let me write it that way. So it's going to be P plus R times the partial of z with respect to x dx plus n. Let me scroll to the right a little bit. Plus n, which is Q plus R times the partial of z with respect to y dy. And then this is really interesting, because this path that we are now concerned with, it's completely analogous. I hope you don't think I'm doing some voodoo here. This statement is completely analogous to this statement, where M could be this and N could be this. And so, we can revert it back to now the path C1 that sits in the xy plane. Not our original boundary C, but now we're just dealing with a boundary in the xy plane. So it reverts to this. But what's powerful about getting it to this point, is we can now apply Green's theorem to this to essentially turn this into a double integral over the region that this path surrounds, the original region over this region R. And when we change how we manipulate that, when we play around with it, we're going to see that we get the exact same result that we got in earlier videos. And I'll leave you there, and I'll see if I can do that in the next video.