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Studying for a test? Prepare with these 11 lessons on Green's, Stokes', and the divergence theorems.

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# Stokes' theorem proof part 1

Video transcript

In this video I will attempt to prove or -actually in the next several videos attempt to prove a special case version of stokes' theorem or essentially stokes' theorem for a special case and I'm doing this because that the proof will be a little bit simpler but at the same time it's pretty convincing and the special case we're going to assume is that the surface we're dealing with is function of x and y so if you give me any particular x and y it only determines one point on that surface so a surface like this would be the case so it's kind of a mapping of this region of the x y plane into three demensions so for any x y we can figure out the height so essentially z is going to be a function of x and y and we can get a point on the surface so this proof would not apply to a surface that's like a sphere or something like that where any point on the x y plane could actually determine two points on our surface but this is a pretty good start the other thing that we are going to assume, we are going to assume that z which is essentially a function of x and y and that this function of x and y has continuous second order derivatives so continuous second derivatives and the reason why I'm going to make that assumption is it's going to help us in our proof later on it's going to allow us to say " that the partial of our surface or the partial of z with respect to x and then taking the derivative of that with respect to y is going to be the same as the partial of z with respect to y and taking the n derivative of that with respect to x and in order to be able to make this statement we have to assume that z or or z right over here z is a function of x and y has continuous second order derivatives and over here we've just written our vector field f that we're going to deal with when we're trying to play with stokes' theorem and we'll assume that it has continuous first-order derivatives now with that out of the way let's think about what stokes' theorem tells us, and then we'll think about for this particular case how we can write it out and hopefully we will see the two things are equal! so let me write it out so stoke's theorem tells us that if x dot dr over some path and the path that we care about is essentially this path right over here I'll do it in blue it's this path right over here this is the boundary this is the boundary of our surface so this is c right over here stokes' theorem tells us that this should be the same thing this should be equivalent to the surface integral over our surface --over our surface of curl of f--- curl of f DOT ds dot- dotted with the surface itself and so in this video I want to focus or probably this and the next video I want to focus on the second half I want to focus this I want to see how we can express this given the assumptions we've made and then after that we're going to see how we can express this given the same assumptions and then hopefully we'll find that we get them to be equal to each other! so let's just start figuring out what the curl of f is equal to so the curl of f the curl of f is equal to you can view it as the dell operator crossed with our vector field f which is equal to can write our components so i- let's do it in different colors- i, j and k components i j and k components and then I need to write my dell operator or my partial operator or my partial operator as I guess I could call them so the partial with respect to x the partial with respect to y the partial with respect to z and then I have to write i, j and k components of my vector field f and I will do that in gre- well, I'll do that in blue and so we have p which is a function of x y and z q which is a function of x y and z and r which is a function of x y and z and so this is going to evaluate as it's going to be i* so blank out that column that row it's going to be the partial of r with respect to y minus the partial of q with respect to z and then checkerboard pattern minus j and then times the partial of r