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Current time:0:00Total duration:7:30

I think we're now ready to get into the meat of the proof and the last video we said if we can just prove that each of these parts are equal to each other we've essentially have proven that that is equal to that because here in yellow is another way of writing the flux across the surface and here in green is another way of writing the triple integral over our region of the divergence of F and what I want to do in this video and probably the next video is prove that these two are equivalent e to each other I'm going to and I'm going to prove it using the fact that our original region is a simple solid region in particular is a type is a type 1 region and then that's essentially going to be it because you can use the exact same argument and the fact that it is a type 2 region to prove this type 2 region the exact same argument to prove this and the exact same argument and the fact that it is a type 3 region type 3 region to prove that so I'm going to assume it's a type 1 which I can it's a type 1 n 2 it's a type 1 2 & 3 region so given the fact that's type 1 I'm now going to prove this relationship right over here and then I'll leave it up to you to do the exact same argument with a type 2 and a type 3 region so let's get going so type 1 region just to remind ourselves a type 1 region is a so type type 1 region type 1 region is a region that is equal to the set of all X YS and Z's where or I should say such that the XY pairs are a member of a domain and the xy-plane and Z is bounded by two functions Z's lower bound is f1 of X and Y and that's going to be less than or equal to Z less than or equal to Z and Z's upper bound we can call it f2 f2 of f2 of X Y and then let me close the set notation right over here and let me just draw a general version of a type 1 a type 1 region so let me draw my x y&z axes so this is my z axis this is my x axis and there is my y axis and so we might have a region D we might have a region D here so our region I'll draw it as a little circle a little circle right over here this is our region D and for any XY in our region D you can take you can evaluate the function f you can figure out an f1 I should say so this might define an f1 which we can kind of imagine as a surface or a little thing that's at the bottom of a cylinder if you want so this every XY there when you evaluate or when you figure out what the corresponding f1 of those points in this domain would be you might get a surface that looks something like this it doesn't have to be flat but hopefully this gives the idea it doesn't have to be completely flat it can be curved or whatever else but this just shows that every XY when you evaluate it right over here to so it gets associated with a point this lower bound surface right over here and I'll draw a dotted line to show that's only for the XY z-- in this domain and then we have an upper bound surface an upper bound surface that might be up here give me any XY when I evaluate f2 I get this surface up here I get this surface up here and once again they don't have to look the same this could be like a dome or it could be slanted or who knows what it might be but this will give you the general idea and then Z fills up the region remember the region isn't just the surface of the figure it's the entire volume inside of it so when Z varies between that surface and that surface for any given X Y in our domain we fill up we fill up the entire region so we fill up we can fill up this entire region and so we and this is the way I drew it looks like a cylinder but it doesn't have to be a cylinder like this and this these two surfaces might touch each other in which case there would be no side of the cylinder that they could be smooth they could be lumpier than this they might be inclined in some way but hopefully this gives this is a good generalization of a type 1 region now a type 1 region you can kind of think of it it can be broken up into three parts it can be broken up into surface or the surfaces of a type 1 region I should say can be broken up into three parts it can be broken up into it can be broken up into let's call that surface one surface one let's call this right over here surface to the top of the cylinder or whatever kind of lumpy top it might be and let's call the side if these two surfaces don't touch each other let's call that surface three there might not necessarily even be a surface three if these two touch each others in the case of the sphere but let's just assume that there actually is a surface three so if we're evaluating the surface integral we'll think about this double integral in this or this triple integral in a second but let's think about how we can rewrite this surface integral right over here instead of this entire surface is s1 plus s2 plus s3 so we can essentially break this up into three separate surface integrals so let's do that so remember we're just focusing on this part right over here so the surface integral the surface integral of R times K dot n the dot product of K and n DS can be rewritten as the let me write it this way can be rewritten can be re-written as the surface integral over s 2 of R times K dot n DS plus the surface integral plus the surface I'm just breaking up the surface here plus the surface integral over s one of our times K K dot n DS plus the surface integral over surface three of the same thing R times K dot n-d-s-u have drawn it this is actually the case s three is if those surfaces don't touch each other and for a type 1 situation right over here the normal vector at any given point on this kind of the side of the cylinder for this type 1 region if there is this in-between region there are always isn't in the sphere there wouldn't be the surface in which case this would be zero but if there is this surface in a type 1 region the one that essentially connects the boundaries of the top and the bottom then the normal vector will never have a K component the normal vector will always be pointing flow I doubt it will only have an I and J component so if you take the this normal vector right over here does not have a K component and you are dotting it with a K vector then the dot product of two things that orthogonal the K vector goes like that you're going to get zero so this thing is going to be this thing is going to be 0 because K dot n is going to be 0 in this situation for the surface k dot n is going to be equal to 0 so this part right over here this part right over here simplifies to this right over here now in the next video what we can do is essentially re express these in terms of surface integrals but in terms of double integrals over this domain right over here we'll kind of evaluate these surface integrals