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let's now prove the divergence theorem which tells us that the flux across the surface of a vector field and our vector field we're going to think about is f so the flux across that surface and I could call that f dot n where n is the normal vector of the surface and I can multiply that times DS so this is equal to the triple integral summing up summing up throughout the volume of that region of that region summing up that volume of the of the divergence of F of the divergence of F and we've done several videos explaining the intuition here but now we are actually going to prove it and of course we're times each little cube are each little differential cube of volume and we're going to make an assumption here we're gonna assume that we're dealing with a simple solid region simple simple solid solid region and what this means more formally is that the region we're thinking about can be a type 1 type 2 and type 3 region or I should say it is all of the three so it is types types one two and three and there are videos that go into what each of these regions are but a lot of the basic shapes fall into this simple solid region like a sphere or a cylinder of some kind they can be types one type two and type three and for a lot of situations that aren't simple solid regions you can break them up into simple solid regions but let's just prove it for this case for this case right over here so let's just assume that our vector field F our vector field F can be written as P which is a function of XY and Z times I plus Q which is a function of XY and Z times J plus plus R which is a function of XY and Z x times K so let's think about what each of these sides of the equation would come out to be well first of all what is going to be f dot n so let's think about that a little bit F dot n f dot n f dot n is going to be equal to this this component right over here x ends I component plus this component right over here x ends J component Plus this component here x ends K component so we could write it as this is could it could be written as P times P times R I'll just write P open parenthesis the dot product of I and n I I dotted n i dot dotted and and let me make sure I write I as a vector as a unit vector I want to be clear what's going to happen right over here if you take the dot product of I and n you're just going to get the I component the the the the scaling factor of the I component of the N normal vector and we're just gonna multiply that times P so it's exactly so that's essentially the product of the X component so I guess you could say the magnitude of the X components and then to that we are going to add Q times J J dotted with N and once again when you dot J with N you get the magnitude of the J component of the of the normal vector right over there and then times or plus I should say plus R times K dotted with in k dotted with then this isn't how we normally see it but I think it's reasonable to say that this is actually true this is right over here is going to be equal to P times the magnitude of the I component of ends normal vector which is exactly what we wanted a dot product this is the same thing for the J component this is the same thing for the K component and you could try it out to find n is equal to I don't know M times I plus n times J plus o times K or something like that and you'll see that this actually does work out fine so what would how can we simplify this expression right up here well we can rewrite this as we can rewrite the left-hand side as so the surface integral of F of F and now let me write multiple ways FDS which is equal to the surface integral f the surface integral of F dot n times the scaler D s is equal to the double integral of the surface of all of this business right over here is equal to the double integral over the surface of let me just copy and paste that of all of that business so let me copy and then let me paste so all of that all of that business right over there and I just noticed that I forgot to put the little unit vector symbol little caret right over there put some parentheses put some parentheses and then we are left with our D s and then this all of this can be rewritten as this can be written as the surface integral of P times this business and I'll just do it in the same color of P times the dot product of I and n DS plus the surface integral of Q times the dot product of J and n DS plus the surface integral of r times the dot product of K and n I forgot to carry K&N D s so I just broke it up but we were taking the integral of this sum and so I just took it as I just rewrote it as the sum of the integrals so that's the left-hand side right over here now let's think about the right-hand side what is the divergence of F and actually I'm going to take some space up here what is the divergence of F well the divergence of F based on this expression of F is just going to be let me just write it over here real small the divergence of F is going to be the partial of P with respect to let me just in a new color because I'm using the yellow too much the divergence of F the divergence of F is going to be the partial of P with respect to X plus the partial of Q with respect to Y plus the partial of R plus the partial of R with respect to Z so we can this second or this triple integral right over here can be written it as the triple integral of the partial of P with respect to X plus the partial of Q with respect to Y plus the partial of R with respect to Z well this thing once again instead of writing it as a triple integral of this Psalm we could write it as a sum of triple integrals so this thing right over here can be re-written as the triple integral triple integral over our 3-dimensional region if you let me copy and paste that so I don't have to keep rewriting it so let me copy it so it's going to be equal to the triple integral of the partial of P with respect to X DV plus let me paste that again it's a triple integral of the partial of Q with respect to Y DV plus once again triple integral triple integral of the partial the partial of R with respect to Z partial of R with respect to Z D V so we've essentially restated restated our our divergence theorem this is our surface integral and the divergence set theorem says that this needs to be equal to this business right over here we've just written in a different way and so what I'm going to do in order to prove it is just show that each of these corresponding terms are equal to each other that these are equal to each other that these these are equal to each other and that let me do it enough and these are equal to each other and in particular we're going to focus the proof on this and we're going to use the fact that our region is a type 1 region it's a type 1 type 2 and type 3 but we're going to use it the fact that it's a type 1 region to prove that these two things are equivalent and then you can use the fact that it's also a type 2 and type 3 region to make the exact same argument as to why 2 this is equal to this and why this is equal to that