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