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# Scalar field line integral independent of path direction

## Video transcript

in the last video we saw that if we had some curve in the XY plane and we just parameterize it in a very general sense like this we could generate another parameterization that essentially is the same curve but goes in the opposite direction it starts here and it goes here as T goes from A to B as opposed to the first parameterization we started with T equal a over here and it went up like that and the question I want to answer in this in this video is how a line integral of a scalar field over this curve so this is my scalar field it's a function of x and y how a line integral over a scalar field over this curve relates to that's a line integral of that same scalar field over the reverse curve over the curve going in the other direction so the question is does it even matter whether we we move in this direction or that direction when we're taking the line integral of a scalar field and in the next video we'll talk about whether it matters on a vector field and let's see if we can get a little intuition to our answer before we even prove our answer so let me draw let me draw a little diagram here let me do it a little bit lower because I think I'm going to need a little bit more real estate so let me draw the y-axis that is the x-axis let me draw the vertical axis just like that that is Z let me draw a scalar field here so I'll just draw it at some surface I'll draw a part of it that is my scalar field that is f of XY right there for any point on the XY plane we can associate a height that defines this this this surface this scalar field and let me put a curve down there so let's say that this is the curve C this is the curve C just like that and the way we define it first we start over here and we move in that direction that was our curve C and we know from several videos ago that the way to visualize what this line integral means is we're essentially trying to figure out the area of a curtain that has this curve as its base and then its ceiling is defined by this surface by the scalar field so we're literally just trying to find the area of this curvy you got a piece of paper or wall or whatever you want to view it that's what this thing is now if we take the same integral but we take the reverse curve instead of going in that direction we're now going in the opposite direction we're not taking a curve where we're going from the top to the bottom but the idea is still the same you know I don't know which one see which one's - see you know I could have defined this one as I could have defined this path going from that way is see and then the minus C path would have started here and gone back up so it seems in either case no matter what I'm doing I'm going to try to figure out the area of this curved piece of paper so my intuition tells me that either of these are going to give me the answer the the area of this curved piece of paper so maybe they should be equal to each other I haven't proved anything very rigorously yet but it seems that they should be equal to each other right in this case let's say I'm taking a let me just make it very clear I'm taking I'm taking a D s a little change in distance let me do it in a different color a little change in distance and I'm multiplying it by the height and I'm multiplying it by the height to find kind of a differential of the area and then I'm going to add a bunch of these together to get the whole area here I'm doing the same thing I'm taking a little D s and remember the d s is always going to be positive the way we parameterize dit it's the hardest word in the English language for me to say so here - we're taking a D s we're going to multiply it by the height so once again we should take take the area and I want actually differentiate that relative to when you take a normal integral and you take a normal integral from A to B of let's say f of X DX we know that when we switch the boundaries of the integration that it makes the integral negative that equals the negative of the integral from B to a of f of X DX and the reason why this is the case is if you imagine if you imagine this is a this is B that is my f of X when you do this way your D X's are always going to be positive when you go in that direction your D X's are always going to be positive right the each increment the right boundary is going to be higher than the left boundary so your D X's are positive in this situation your DX's are negative the heights are always going to be the same they're always going to be f of X but here your change in X is a negative change in X when you go from B to a and that's why you get a negative integral in either case here our path changes but our DS's are going to be positive and the way I've drawn the surface it's above the XY plane this sort the f of X Y is also going to be positive so that also kind of gives the same intuition that this should be the exact same area but let's prove it let's prove it to ourselves so let's start off with our first parameter is parameterization just like we did in the last video we have X is equal to X of T y is equal to Y of T and we're dealing with this from T goes from A to B and we know we're going to need the derivatives of these so let me do write that right down right now we can write DX DT is equal to X prime of T and dy DT we write that a little bit neater dy DT is equal to Y prime Y prime of T this is nothing nothing I'm nothing groundbreaking I've done so far but we know the integral we know the integral over C of f of X Y F is a scalar field not a vector field D s is equal to the integral from t is equal to a to T is equal to B of f of X of T y of T times the square root of DX DT squared which is the same thing as X prime of T squared X prime of T squared plus dy DT squared which the same thing as Y prime of T squared plus y prime of T squared all that on the radical times DT that's this integral is exactly that given this parameter ization parameterization have so much trouble saying that now let's do the minus C version I'll do that in we all do that in this orange color so - in my the - C version actually let me do the - C version down here the - C version we have X is equal to you remember this actually just from up here this was from the last video X is equal to a plus B minus T X is equal to a plus B minus T Y is sorry X doesn't equal a plus B minus DX is equal to X of a plus B minus T got ahead of myself X is equal to X of a plus B minus T Y is equal to Y of a plus B minus T and then T goes from A to B a go T goes from A to B and this is just exactly what we did in that last video X is equal to X of a plus B minus T Y is equal to Y of a plus B minus T same curve just going in a different direction as T increases from A to B but let's get the derivatives so I'll do it in the derivative color maybe so DX DT for this path it's going to be a little different when to do the chain rule now derivative of the inside with respect to T well these are constants minus T derivative of minus T with respect to T is minus 1 so it's minus 1 times the derivative of the outside with respect to the inside well that's just X prime of a plus B minus T or we could rewrite this as this is just minus X prime of a plus B minus T dy DT dy DT same logic derivative of the inside is minus 1 with respect to T right derivative of minus T is just minus 1 times the derivative of the outside with respect to the inside so Y prime of a plus B minus T same thing as minus y prime a plus B minus T so what's so given all of that given all of that what is this integral going to be equal to the integral of minus C of the scalar field F of XY d s what is this going to be equal to well it's going to be the integral from get almost pattern match it T is equal to a 2 T is equal to B of f of X but now X is no longer X of T X now equals X of a plus B minus D X of a plus B minus T it's a little bit hairy but I don't think anything here is groundbreaking hopefully it's not too confusing and once again Y is no longer y of T y is y of a plus B minus T y of a plus B minus T and then times the square root I'll just switch colors times the square root of DX DT squared what's DX DT squared DX DT squared is just this thing squared or this thing squared and this thing if I have - anything squared that's the same thing as the anything squared right this is equal to minus X prime of a plus B minus T squared which is the same thing as just X prime of a plus B minus T squared right you lose that minus information when you square it so that's going to be equal to X prime of a plus B minus T squared the whole what result function the squared plus dy DT squared by the same logic that's going to be you lose the negative when you square it Y prime of a plus B minus T squared extend the radical and then all of that DT all of that DT so that that's the the surface integral over the regular curve see this is the not surface the line integral thought we're not doing surface integrals yeah this is a line integral over the curve C this is the line integral over the curve minus C they don't look equal just yet this looks a lot more convoluted than that one does so let's see if we can simplify it a little bit and we can simplify it perhaps by making a substitution let's let let me make get a nice substitution color let's let u equal two a plus B minus T so first we're going to figure out the boundaries of our interpret well actually let's just figure out what's D u so D u DT D u DT the derivative of U with respect to T is just going to be equal to minus 1 or we could say that D u if we multiply both sides by the differential DT is equal to minus DT and let's figure out our boundaries of integration when T is equal to a what is U is U U is equal to a plus B minus a which is equal to B and then when T is equal to B U is equal to a plus B minus B which is equal to a so if we do the substitution on this crazy hairy-looking integral let's should simplify a little bit and it changes our so this integral is going to be the same thing as the integral from u when T is a U is B when T is B U is a and f of X of this thing right here is just u X of U so let's simplify it a good bit and Y of this thing right here is just u Y of U times the square root times the squared in the same color so times the square root of x prime X prime of U squared plus y prime of U squared Y prime of u squared instead of a DT we have to write a or we can write if we multiply both sides of this by - we have DT is equal to minus D U so instead of a DT we have to put a minus D U here so this is x minus D you or just so we don't think this is a subtraction let's just put that negative sign out here in the front just like that so going from B to a of this thing right like that and just to make just to make the boundaries of integration make a little bit more sense because we know that a is less than B let's swap them and I said at the beginning of this video if you swap thee for just a standard regular run-of-the-mill integral if you swap if you if you have something going from B to a of f of X DX or D you and I maybe I should write it this way f of u D u this is equal to the minus of the integral from A to B of f of f of u D U and we did that by the logic that I graphed up here that here when you switch the order your d use will become the negatives of each other when you actually visualize that we're actually finding the area under the curve so let's do that let's swap the boundaries of integration right here and if we do that that'll negate this negative or make it a positive so this is going to be equal to the integral from A to B I'm dropping the negative sign because I swap these two things so I'm going to take the negative of a negative which is a positive of f of X of U Y of U times the square root times the square root of x prime of U squared X prime of u squared plus y prime of U squared D U now remember this all everything we just did with the substitution this was all this is equal to just to remember what we're doing this was the integral of the minus curve of our scalar field F of XY DS now how does this compare to when we take the regular curve how does this compare to that let me copy and paste it to see no I'm using the wrong tool let me copy and paste it to see how they compare let me copy and then let me paste it down here edit paste so how do these two things compare let's take a close look well they actually look pretty similar right over here in this and for the minus curve we have a bunch of views over here for the positive curve we have a bunch of T's but then in the exact same places these integrals are the exact are the exact same integrals if you make a u substitution here if you just make the substitution you is equal to tea this thing is going to be the integral from A to B of what's going to be this exact same thing of f of X of u x of u Y of U times the square root of another parenthesis square root of x prime of u squared plus y prime of u squared D u these two things are identical so we did all this substitution everything but we got the exact same integrals so hopefully that satisfies you that it doesn't matter what direction we go on the curve as long as the shape of the curve is the same doesn't matter if we go forward or backward on the curve we're going to get the same answer and I think that meets our intuition because in either case in either case we're finding the area of this curtain