2-D Divergence theorem
Conceptual clarification for 2-D Divergence Theorem Understanding the line integral as flux through a boundary
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- 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
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