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Current time:0:00Total duration:11:49

where we left off in the last video we did expressed our line integral around the boundary of our surface so this is f dot d R where our path is this boundary right over here this path see we had expressed it in terms of a line integral around this boundary around path c1 which is the boundary of region R and the reason why this is going to be valuable to us is now we can directly apply greens theorem to this to turn this into a double integral over this region right over here the region the region that it is actually bounding so let's actually do that we're just applying greens theorem here so greens greens theorem greens theorem actually let me box this off right over here this was just something that we wrote to remind ourselves in the last video so let me just box it off right over here but when you apply greens theorem you get that this is going to be the exact same thing as the double integral over the region that c1 bounds that region R that's in our XY plane of the partial of this with respect to X so the partial with respect to X partial let me write that a little bit partial with respect to X of this business so I'll do that in that same green color Q plus R times the partial of Z with respect to Y - minus the partial with respect to Y of P plus R times the partial of Z with respect to X P plus R times the partial of Z with respect to X and then da a little differential of our region D a D a and so let's actually calculate this right now let's actually you take the partials of each of these expressions and we'll see that if we expand them and then simplify we'll get something very similar or actually hopefully identical to this right over here and that'll show that this line integral for the special case is the same thing as this surface integral which will prove Stokes theorem for this special case so let's do it so let's just apply this the partial derivative operators so first we want to take the partial derivative with respect to Q and we need to remind ourselves and we did this way up here where we first thought about it we saw that Q well P Q and R they're each functions of XY and Z so we're assuming that's how they're represented and if Z was not a function of X then if we were to take the partial of Q with respect to X we would just write it as a partial of Q with respect to X but we know that we've assumed that Z is itself a function of x and y so if we're taking the partial with respect to X over here we have to think first well how can we how can Q directly change with respect to X and then how can it change due to something else changing due to X and so that other thing that could change due to X is Z Y is independent of X but Z is a function of X so let's keep that in mind so we're going to do the multivariable chain rule so when we try to take the derivative of this part we're taking essentially of this whole function with respect to X we have to think about how will Q change directly with respect to X and to that we have to add how Q could change do the changes and other variables due to changes in X and the other only variable that Q is a function of that could change due to a change in X is Z so Q could also change to Z because G Z has changed due to X so that operated here is the partial with respect to X plus the partial of Q with respect to Z times the partial of Z with respect to X if Q was purely if we if we rewrote Q so that Z was substituted with X's and Y's because it is a function of X and Y's then we would just have to write this first term right over here but we're assuming this is expressed as a function of XY and Z and Z itself is a function of X and so that's why we had to use the multivariable chain rule now let's move to the next part and both of these both of these might have some X's in them so we have to use the product rule right over here so first we can take the derivative first we have we could take the derivative of R with respect to X and then multiply that times Z sub y and then we have to take the derivative of Z sub y with respect to X and multiply that times R so it's going to be this is going to be plus so the partial of if we take the derivative of this with respect to X same exact logic are can change directly due to X and it can change due to Y and we want to see and multiply that times how Z could change due to X once again that you could view this as the multivariable chain rule in action here but of course we take the derivative of the first term times the second term do the second term in magenta times the second term so the partial of Z with respect to Y plus the derivative of the second term which is the partial of Z with respect to Y and then taking the partial of that with respect to X which we could just write as that times the first term times our x our our so that's the partial with respect to X of all of this business right over here and then we need to subtract the partial of this with respect to Y and we're going to use the exact same logic so then we're going to subtract and I'll put it in parenthesis like that so P could change directly due to Y I will circle P let me do it with a color that I haven't used yet P could change directly due to Y so this is the so we could say the the partial of P with respect to Y but it can also change due to Z changing because of Y so plus the partial with respect to Z times the partial of Z with respect to Y plus plus and I'll do our maybe in that same color plus the derivative of our well we already figured that but actually now it's with respect to X not with respect to Y I have to be careful so it's going to be the partial of R with respect to Y plus the partial of R with respect to Z times the partial of Z with respect to Y times z sub x times Z sub x times Z sub X plus now we take the derivative the second term times the first term the derivative of the partial of Z with respect to X then with respect to Y is going to be Z so the partial of Z with respect to X and we're gonna take the partial of that with respect to Y and they're going to multiply that times R then we're going to multiply that times are and now let's see if we can if we can expand this out and hopefully think simplify and just reminder I'm just working on the inside of this double integral I'll rewrite the double the the outside the the double integral and the DA once I get this all cleaned up a little bit so let me rewrite it a little bit so this is equal to this is equal to the partial of Q I'll try to color code it the same way and this is really just algebra at this point the partial of Q with respect to X plus the partial of Q with respect to Z times the partial of Z with respect to X plus plus the partial of R with respect to x times the partial of Z with respect to Y and then plus the partial of R with respect to Z times the partial of Z with respect to x times the partial of Z with respect to Y and then we have this term right over here which I'll just do in purple plus the purple plus the partial of Z with respect to Y and then with respect to x times R and now we're going to subtract all of this business right over here so minus - doing the blue minus the partial of P with respect to Y minus the partial of P with respect to Z times the partial of Z with respect to Y and then we're going to subtract from that minus the partial of R with respect to Y minus tribute this times the partial of Z with respect to X minus the partial of R this gets a little bit tedious but hopefully it'll get us to where we need to go the partial of R with respect to Z times the partial of Z with respect to Y times the partial of Z with respect to X and then finally this term right over here minus this because we have that negative out there minus the partial of Z with respect to Y then with respect to X and then with respect to Y R and let's see if we can simplify things so the first thing this and this look to be the same we just can commute the order in which we actually multiply but these are the exact same term so that is going to cancel out with that and because we assumed because we assumed that the second because we assumed way up here that we have continuous second derivatives of the function Z Z is a function of x and y that that is equal to that we can now say we can now say that these two right over here going to be the negatives of each other or that they are going to cancel out and so this simplify things a good bit and let's see if I can write it in a way let me group terms in a way that might start to make sense and actually I'm going to try to see if I can make them similar to this so I have all the terms that have a Z sub X and the Z sub y and then the rest of them so Z sub X I'll do in blue so you have the terms that have a Z sub X in it so you have this term right over here and this term right over here we can factor out the Z sub X and we get the partial of Z with respect to x times the partial of Q with respect to Z minus the partial of R with respect to Y and then let's do the and I want to get the colors the same way to I did yellow next plus I have all the terms with the partial of Z with respect to Y so it's that term and that term right over here so it becomes plus the partial of Z with respect to Y times the partial of R with respect to X minus the partial of P with respect to Z and then we have these last two terms and I use the color green up there so I'll use the color green again so for these two terms I'll just write plus the partial of Q with respect to X minus the partial of P with respect to Y so our double our double integrals has kind of I can't really say simplified but we can rewrite it like this and we don't want to forget this was all because this was all a simplification of our double integral over the region da this is what we have been able to use in Green's theorem and the multivariable chain rule and whatever else we be able to say that that line integral around the boundary of our surface is the same thing as this and now we can compare that to what our surface integral was so let's see if I have space so copy and then let me see if I have space up here to paste it well it doesn't look like I actually have much space to paste it although I'll try anyway so if I paste it you see that they are identical they are identical our line integral is identical to this we get the exact same thing so our line integral F dot d are around around this Patsey simplified to this and our surface integral simplified to this so using the assumptions we had they both simplify to the same thing so now we know for this special case our line integral is equal to our surface integral and we are done