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Current time:0:00Total duration:3:12

now we can work on the triple integral part of our of our problem or our proof this right over here and so I can rewrite that so this is the triple integral the triple integral over our region which we're assuming is a type 1 region of the partial of R with respect to Z which we could write like this doesn't matter partial of R with respect to Z and I'll D V and we can rewrite this as we can assume we're going to integrate with respect to Z first so I'm going to integrate with respect to Z we do that in another color I'm going to integrate with respect to Z first the lower bound on Z in our type 1 region the lower bound is f1 the upper bound is f2 so we're going to integrate from f1 f1 of X Y to f2 of X Y f2 of X Y and I'm going to integrate the partial of R with respect to Z so let me do that in that same yellow color partial of R with respect to Z and then I have DZ and then I'll have to integrate with respect to Y and X or with respect to x and y so it's DX dy or dy DX I can just write that as da da so what you could think of it we can evaluate the yellow part and then we're just going to take the double integral over over the XY domain so this is just going to be over the X Y domain so let me put some brackets here or some just to make it clear what we're going to do so all we're doing is we're integrating with respect to Z first and we have the bounds there well this is pretty straight for this is all going to be equal to I'll write the outside first the double integral over the domain and I have the DA the DA right over here so let me give myself some real estate da well what's the antiderivative of this this is just R and this is just R or R of X Y Z evaluated when Z is f1 or when Z is f2 and from that we evaluate when Z is f1 so this is just going to be R of XY and Z and we evaluate when Z is equal to that and from that we subtract when Z is equal to that so that's going to be equal to so our our of X Y Z evaluated when Z is equal to that is R of X Y f2 of X Y and from that we need to subtract R when Z is this minus R of X Y f1 of X Y and then make sure that we got our parentheses we got our parentheses now this is exactly what we saw in the last video it is exactly that which shows that this is exactly this so when we assumed it was a type 1 region we got that this is exactly equal to this you do the exact same argument with the type 2 region to show that this is equal to this type 3 region show this is equal to that and you have your divergence theorem proved and we can consider ourselves consider ourselves done