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In this video I will attempt to prove or -actually in the next several

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videos attempt to prove a special case version

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of stokes' theorem or essentially stokes'

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theorem for a special case and I'm doing this

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because that the proof will be a little bit

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simpler but at the same time it's pretty convincing

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and the special case we're going to assume is

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that the surface we're dealing with is function

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of x and y so if you give me any particular x and y

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it only determines one point on that surface

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so a surface like this would be the case

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so it's kind of a mapping of this region of the x y

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plane into three demensions so for any x y we can

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figure out the height so essentially z is going to be a function of

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x and y and we can get a point on the surface

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so this proof would not apply to a surface

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that's like a sphere or something like that where any

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point on the x y plane could actually determine two

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points on our surface but this is a pretty good start

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the other thing that we are going to assume, we are going to assume

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that z which is essentially a function of x and y

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and that this function of x and y has continuous

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second order derivatives so continuous

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second derivatives and the reason why I'm

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going to make that assumption

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is it's going to help us in our proof later on

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it's going to allow us to say " that the partial of our surface

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or the partial of z with respect to x and then taking the derivative of that

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with respect to y is going to be the same as

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the partial of z with respect to y and taking the n

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derivative of that with respect to x

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and in order to be able to make this statement we have to

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assume that z or or z right over here z is a function of

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x and y has continuous second order derivatives

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and over here we've just written our vector field f

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that we're going to deal with when we're trying

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to play with stokes' theorem

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and we'll assume that it has continuous first-order

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derivatives now with that out of the way

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let's think about what stokes' theorem tells us,

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and then we'll think about for this particular case

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how we can write it out and hopefully

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we will see the two things are equal!

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so let me write it out

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so stoke's theorem tells us that

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if x dot dr over some path

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and the path that we care about is essentially

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this path right over here I'll do it in blue

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it's this path right over here

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this is the boundary this is the boundary

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of our surface

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so this is c right over here

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stokes' theorem tells us that this should be the same

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thing this should be equivalent to the surface

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integral over our surface

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--over our surface of curl of f--- curl of f DOT ds

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dot- dotted with the surface itself

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and so in this video I want to focus

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or probably this and the next video

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I want to focus on the second half

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I want to focus this I want to see how we

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can express this given the assumptions

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we've made and then after that we're going to

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see how we can express this given the same assumptions

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and then hopefully we'll find that we get them to

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be equal to each other!

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so let's just start figuring out what the curl

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of f is equal to

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so the curl of f

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the curl of f

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is equal to

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you can view it as the dell operator crossed

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with our vector field f which is equal to

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can write our components so i- let's do it in different colors-

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i, j and k components

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i j and k components and then I need to

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write my dell operator or my partial

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operator or my partial operator as I guess I could call them

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so the partial with respect to x

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the partial with respect to y

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the partial with respect to z

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and then I have to write i, j and k

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components of my vector field f

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and I will do that in gre- well, I'll do that in blue

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and so we have p which is a function of x y and z

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q which is a function of x y and z

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and r which is a function of x y and z

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and so this is going to evaluate as

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it's going to be i* so blank out that column that row

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it's going to be the partial of r with respect to y

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minus the partial of q with respect to z

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and then checkerboard pattern

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minus j and then times the partial of r