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Let's revisit the line integral F.n ds

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right over here because I want to make sure we have the proper conception

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and I was little "loosey goosey" with it in the last video

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and in this video I want to get a little bit more exacting and actually use units

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so that we really understand what's going on here

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So I've drawn our path "C"

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and we're traversing it in the positive counterclockwise direction

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and then I've taken a few sample points for F

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at any point in the x-y plane

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that associates a 2-dimensional vector

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maybe at that point the 2-dimensional vector looks like that,

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maybe at that point the 2-dimensional vector looks like that

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and then n is of course the unit normal vector at any point on our curve

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the outward pointing unit vector at any point on our curve

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Now in the last video, I talked about F as being some type of a velocity function

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that at any point it gives you the velocity of the particles there

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and that wasn't exactly right

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in order to really understand what's happening here

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in order to really conceptualize this as kind of flux through the boundary

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the rate of mass exiting this boundary here

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we actually have to introduce a density aspect to F

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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

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so the scalar part right over here Rho of x,y

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Rho is a Greek letter often used to represent density of some kind

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in this case its mass density

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so at any given x,y point this tells us what the mass density is

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mass density will be some mass in a 2-dimensional world

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so it's mass per area

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and if we want particular units for our example-

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once again, this isn't the only way that this can be conceived of

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there's other applications, but this is the easiest way for my brain to process it

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we can imagine this as kilogram per square meter

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and this right over here is the velocity vector

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it tells us what is the velocity of the particles of that point

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so this is kind of saying, "How much particles do you have at a kind of a point?

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How dense are they?" and this is "How fast are they going and in what direction?"

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and this whole thing is a vector, it's a velocity vector

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but the components right over here

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M of x,y is just a number and you multiply that times a vector

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so M of x,y right over here is going to be a scalar function

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when you multiply by i it becomes a vector

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that's going to give you a speed

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and then N of x,y is also going to give you a speed

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and it tells you a speed in a j direction

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so it becomes a vector and a speed in the i direction becomes a vector as well

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but these speeds, the units of speed (let me write this over here)

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so now we're talking about in particular M of x,y and N of x,y

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that would be in units of distance per time and so maybe for this example we'll say

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the units are meters per second

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So let's think about the units will be for this function

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if we distribute the Rho, because really at any given x,y, it really is just a number

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so if we do that, we're going to get F- I'm not going to keep writing F of x,y

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we'll just understand that F, Rho, M and N are functions of x,y

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F is going to be equal to Rho times M times the unit vector i

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plus Rho times N times the unit vector j

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now what are the units here? what's Rho times M- what units are we going to get there?

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and we're gonna get the same units when we do Rho times N

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we'll we're gonna have, if we pick these particular units, we're going to have

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kilograms per meter squared times meters per second

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so a little bit of dimensional analysis here

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this meter in the numerator will cancel out with one of the meters in the denominator

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and we are left with something kind of strange

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kilograms per meter second

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which is essentially what the- if you view this vector has a magnitude in some direction

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the magnitude component is going to have these units right over here

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and then we're going to take this and we're dotting it with N

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N just only gives us a direction

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it is a unit-less vector- it's only specifying a direction at any point in the curve

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so when I take a dot product with this, it's going to give us essentially what is the magnitude of F

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going in the direction of N. So this right over here, when you take the dot, it's essentially

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a part of the magnitude of F going in N's direction

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and it's going to have the same exact units as F

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so the units of this part, you're going to have kilograms per meter second

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and let me make this very clear-

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So let's say we're focusing on this point over here

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F looks like that, its magnitude, the length of that vector is going to be in kilograms per meter second

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then we have a normal vector right over there

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and when you take the dot product, you're essentially saying

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"What's the magnitude that's going in the normal direction?"

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so essentially, what's the magnitude of that vector right over there

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it's going to be in kilograms per meter second

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and we're multiplying it times ds

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we're multiplying it times this infinitesimally small segment of the curve

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we're going to multiply that times ds

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well, what are the units of ds? it's going to be unit of length

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we'll just go with meters

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so this right over here is going to be meters

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there's this whole integral, you're going to have kilogram per meter second times meters

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so if you have kilograms per meter second and you were to multiply that times meters, what do you get?

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well this meters is going to cancel out that meters and then you get something that kinda starts to make sense

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you have kilogram per second

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and so this hopefully this makes it clear what's going on here

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this is telling us how much mass is crossing that little ds

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that little section of the curve per second

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and if you were to add up- and that's what integrals are all about,

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adding up an infinite number of these infinitesimally small ds's

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if you add all of that up

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you're going to get- the value of this entire integral is going to be in kilograms

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and kilograms per second, and it's essentially going to say,

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"How much mass is exiting this this curve at any given point? Or at any given time?"

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So this whole integral (let me rewrite it) of F.n ds tells us

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the mass exiting the curve per second

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and this should also be consistent in the last video

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we saw that this is equivalent to- and this is where we kinda view it as a 2-dimensional divergence theorem

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in the last video, we saw that this is equivalent to the double integral over the area of the divergence of f

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which is essentially just- well, I could write it 2 ways

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the divergence of f and this right over here, that's just the partial of the i component with respect to x

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(let me write it over here, I don't want to do this too fast and loose)

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so this right over here is going to be the partial of Rho M (let me write it like this), Rho M

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with respect to x plus the partial of the y component

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Rho N with respect to y times each little chunk of area

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Well, what are the units of this going to be right over here?

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we know what Rho M is- Rho M gives us kilogram per meter second

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but if we take the derivative with respect to meters again,

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the units for either of these characters

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are going to be kilograms per meter second per second

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because we're taking the derivative with respect - sorry per METER, we're taking the derivative

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with respect to another unit of distance so you're going to take per meter

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so you're going to have another meter right in the denominator

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that's going to be the units here

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and then you're multiplying it times an area

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so that would be meters squared, this right over here is square meters

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they cancel it out, and once again

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this whole part here that you're summing up

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gives us kilograms per second, so you're having a bunch of kilograms per second

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and you're just adding them up throughout the entire area right over here

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So hopefully this makes a little more sense, about how kinda how to conceptualize this vector function F

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if it confuses you, try your best to ignore it I guess

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for me at least, this helped me out having a stronger conception of what vector F could kind of represent