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## Multivariable calculus

### Course: Multivariable calculus>Unit 5

Lesson 4: 2D divergence theorem

# Conceptual clarification for 2D divergence theorem

Understanding the line integral as flux through a boundary. Created by Sal Khan.

## Video transcript

Let's revisit the line integral F.n ds right over here because I want to make sure we have the proper conception and I was little "loosey goosey" with it in the last video and in this video I want to get a little bit more exacting and actually use units so that we really understand what's going on here So I've drawn our path "C" and we're traversing it in the positive counterclockwise direction and then I've taken a few sample points for F at any point in the x-y plane that associates a 2-dimensional vector maybe at that point the 2-dimensional vector looks like that, maybe at that point the 2-dimensional vector looks like that and then n is of course the unit normal vector at any point on our curve the outward pointing unit vector at any point on our curve Now in the last video, I talked about F as being some type of a velocity function that at any point it gives you the velocity of the particles there and that wasn't exactly right in order to really understand what's happening here in order to really conceptualize this as kind of flux through the boundary the rate of mass exiting this boundary here we actually have to introduce a density aspect to F So right over here, I've rewritten F, and I've rewritten it as of a product of a scalar function and a vector function so the scalar part right over here Rho of x,y Rho is a Greek letter often used to represent density of some kind in this case its mass density so at any given x,y point this tells us what the mass density is mass density will be some mass in a 2-dimensional world so it's mass per area and if we want particular units for our example- once again, this isn't the only way that this can be conceived of there's other applications, but this is the easiest way for my brain to process it we can imagine this as kilogram per square meter and this right over here is the velocity vector it tells us what is the velocity of the particles of that point so this is kind of saying, "How much particles do you have at a kind of a point? How dense are they?" and this is "How fast are they going and in what direction?" and this whole thing is a vector, it's a velocity vector but the components right over here M of x,y is just a number and you multiply that times a vector so M of x,y right over here is going to be a scalar function when you multiply by i it becomes a vector that's going to give you a speed and then N of x,y is also going to give you a speed and it tells you a speed in a j direction so it becomes a vector and a speed in the i direction becomes a vector as well but these speeds, the units of speed (let me write this over here) so now we're talking about in particular M of x,y and N of x,y that would be in units of distance per time and so maybe for this example we'll say the units are meters per second So let's think about the units will be for this function if we distribute the Rho, because really at any given x,y, it really is just a number so if we do that, we're going to get F- I'm not going to keep writing F of x,y we'll just understand that F, Rho, M and N are functions of x,y F is going to be equal to Rho times M times the unit vector i plus Rho times N times the unit vector j now what are the units here? what's Rho times M- what units are we going to get there? and we're gonna get the same units when we do Rho times N we'll we're gonna have, if we pick these particular units, we're going to have kilograms per meter squared times meters per second so a little bit of dimensional analysis here this meter in the numerator will cancel out with one of the meters in the denominator and we are left with something kind of strange kilograms per meter second which is essentially what the- if you view this vector has a magnitude in some direction the magnitude component is going to have these units right over here and then we're going to take this and we're dotting it with N N just only gives us a direction it is a unit-less vector- it's only specifying a direction at any point in the curve so when I take a dot product with this, it's going to give us essentially what is the magnitude of F going in the direction of N. So this right over here, when you take the dot, it's essentially a part of the magnitude of F going in N's direction and it's going to have the same exact units as F so the units of this part, you're going to have kilograms per meter second and let me make this very clear- So let's say we're focusing on this point over here F looks like that, its magnitude, the length of that vector is going to be in kilograms per meter second then we have a normal vector right over there and when you take the dot product, you're essentially saying "What's the magnitude that's going in the normal direction?" so essentially, what's the magnitude of that vector right over there it's going to be in kilograms per meter second and we're multiplying it times ds we're multiplying it times this infinitesimally small segment of the curve we're going to multiply that times ds well, what are the units of ds? it's going to be unit of length we'll just go with meters so this right over here is going to be meters there's this whole integral, you're going to have kilogram per meter second times meters so if you have kilograms per meter second and you were to multiply that times meters, what do you get? well this meters is going to cancel out that meters and then you get something that kinda starts to make sense you have kilogram per second and so this hopefully this makes it clear what's going on here this is telling us how much mass is crossing that little ds that little section of the curve per second and if you were to add up- and that's what integrals are all about, adding up an infinite number of these infinitesimally small ds's if you add all of that up you're going to get- the value of this entire integral is going to be in kilograms and kilograms per second, and it's essentially going to say, "How much mass is exiting this this curve at any given point? Or at any given time?" So this whole integral (let me rewrite it) of F.n ds tells us the mass exiting the curve per second and this should also be consistent in the last video we saw that this is equivalent to- and this is where we kinda view it as a 2-dimensional divergence theorem in the last video, we saw that this is equivalent to the double integral over the area of the divergence of f which is essentially just- well, I could write it 2 ways the divergence of f and this right over here, that's just the partial of the i component with respect to x (let me write it over here, I don't want to do this too fast and loose) so this right over here is going to be the partial of Rho M (let me write it like this), Rho M with respect to x plus the partial of the y component Rho N with respect to y times each little chunk of area Well, what are the units of this going to be right over here? we know what Rho M is- Rho M gives us kilogram per meter second but if we take the derivative with respect to meters again, the units for either of these characters are going to be kilograms per meter second per second because we're taking the derivative with respect - sorry per METER, we're taking the derivative with respect to another unit of distance so you're going to take per meter so you're going to have another meter right in the denominator that's going to be the units here and then you're multiplying it times an area so that would be meters squared, this right over here is square meters they cancel it out, and once again this whole part here that you're summing up gives us kilograms per second, so you're having a bunch of kilograms per second and you're just adding them up throughout the entire area right over here So hopefully this makes a little more sense, about how kinda how to conceptualize this vector function F if it confuses you, try your best to ignore it I guess for me at least, this helped me out having a stronger conception of what vector F could kind of represent