# Conceptual clarification for 2D divergence theorem | Multivariable Calculus | Khan Academy

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