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## Multivariable calculus

# 2D divergence theorem

Using Green's Theorem to establish a two dimensional version of the Divergence Theorem. Created by Sal Khan.

## Want to join the conversation?

- Should it not be the double integral over the region R of (-dQ/dy - dP/dx) * dA? Since the vector field is f = P(x,y) i + Q(x,y) j ?(5 votes)
- Excellent question!

The short answer: no

The long answer:

Although the vector field F = P(x,y) i + Q(x,y) j, the*unit normal vector***n**= (**dy i - dx j**)/**ds**. (You divide by ds because**n**is a*unit*vector)

Thus, when you take F ⋅**n**, you get (P(x,y) * dy - Q(x,y) * dx)/ds

So, ∳(F ⋅**n**) ds = ∲[(P(x,y) * dy - Q(x,y) * dx)/ds] * ds = ∳(P(x,y) * dy - Q(x,y) * dx

Remember Green's Theorem? ∳P(x,y) * dx + Q(x,y) * dy = ∫∫ (∂Q/∂X - ∂P/∂Y) dA

Notice that the dx term is*added*to the dy term ^^. We can take our F ⋅**n**line integral and rearrange it to match Green's Theorem (right now, our F ⋅**n**line integral has the dy term*subtracted*by the dx term):

∳(P(x,y) * dy - Q(x,y) * dx = ∳(-Q(x,y)) * dx + P(x,y) * dy

Now that our line integral is in the correct dx + dy form, we can finally apply Green's Theorem properly:

∳(-Q(x,y)) * dx + P(x,y) * dy = ∫∫∂P/∂X - (-∂Q/∂Y) dA = ∫∫(∂P/∂X + ∂Q/∂Y) * dA

I hope this clarified things :)(3 votes)

- What is Flux that Sal mentions?(1 vote)
- The formula that this video is about int(F*n ds) is the formula for flux, which has common applications in Physics and Engineering. His discussion relating to divergence gives an intuitive meaning behind the term, rather than simply using the term.(2 votes)

- Does it matter if "F" is capital?(1 vote)
- As a general rule, when a function is referred to with a capital F it is an integral, as opposed to f being used for a function that isn't an integral. This isn't a hard and fast rule but a matter of convenient notation.(2 votes)

- At5:42, sal was trying to say "how many", but he didn't and I thought he will say "how many" in the beginning. I want to know, is F-(vector field) dot n-hat represent an acceleration in somehow? and how would be possible to know how many? Should I know the total, for example, gases bounded in that contour first, then, minus the escaped-time multiplied by "how fast", to get how many? What does how fast mean if I am talking about bees exiting from a hole in bee house? Is it the divergence?(2 votes)
- But isn't that what we're doing? Since we're equating the integral on the left to an integral of the divergence, on the right?(1 vote)

- It appears to me that if Sal had picked a point (say, below), where the slope of the tangent was in the positive direction, then the normal would work out with a POSITIVE j component and there would be no negative to cancel the other negative so the partials would have been dP/dx - dQ/dy and the divergence would have been incorrect. I must be missing something. help!(2 votes)
- yi + (-x)j is always 90 degree clockwise so it does work at all points.(1 vote)

- Does anyone know if there is a place to discuss the Divergence Theorem with R^3??(1 vote)
- You can find videos and discussion on that at http://www.khanacademy.org/math/calculus/divergence_theorem_topic/divergence_theorem/v/3-d-divergence-theorem-intuition.(1 vote)

- I am taking from the question above--what is the physical analog to the curve--diffusion through a membrane is fairly intuitive (3d flux), but why would evaluating diffusion through a curve be interesting--maybe as an exercise or maybe some 3d phenomena reduces in dimension.(1 vote)
- As I understand it, there aren't really that many direct applications (Although Luke's Frog Race answer to the question above would qualify), but a lot of the time when dealing with 3d phenomena, it can be easier to reduce it to two dimensions and work from there.(1 vote)

- The transcript is completely broken. None of the timecodes are right. Can fix it?(1 vote)
- Since we're using the unit normal vector, does that mean that the specific parametrization of the curve doesn't matter? (ie, y=sin(t), x=cos(t) would produce the same result as y=sin(2t), x=cos(2t))(1 vote)
- Look at the result, I would say that the 'specific' parametrization doesn't matter, provided your parametrization maps the same curve (t or 2t, you will end up with the same region R).(1 vote)

- Why is that normal unit vector ñ perpendicular to do curve, not the vector field F, since ñ is multiplied by F, i don get it(1 vote)
- If Piper Fowler-Wright gave you the answer you were seeking then disregard this. But if your question was why a point on the vector field "doted" with a unit-normal vector will represent how much the vector-field vector is going in the normals direction then here is an answer:

Let's say that the unit normal vector was going purely in the x-direction, then the dot-product will only scale the vector-filed vectors x-value and disregard the rest. With this intuition, you can imagine that the same is true for all unit vectors. I recommend testing "doting" some arbitrary unit vectors with some random vector and it will become clear to you.(1 vote)

## Video transcript

Sada kada znamo malo o tome kako napraviti vektor na bilo kojoj tacki na grafiku -- to smo radili u prosloj lekciji-- Zelim poceti sa istrazivanjem zanimljivih izraza Izraz glasi : Integral je oko zatvorene petlje i mi cemo ici u pozitivnom skalaru. Orijentacija okretanja kazaljke na satu vektora oznacenim sa "F" sa normalnim vektorom na bilo kojoj tacki petlje ds (Napisacu ds u ljubicastoj boji) ds. Prvo cemo da damo koncept onoga o cemu se prica i onda cemo to obradjivati da vidimo da li mozemo da se nosimo s time a interesantno je to da cemo ipak to obradjivati Koristicemo greenovu teoremu i zapravo cemo radit isa 2 dimenzije verzija teoreme divergencije sto zvuci veoma komplikovano ali na svu srecu cemo mi to uprostiti za one koji zapravi shvataju bar malo gradiva Prvo da razmislimo o ovome. Nacrtajmo pravougaonik uradicemo to belom bojojm tacno je ovde gde je nasa y osa a ovde je nasa x osa sada mi dozvolite da nam nacrtam liniju koja ce izgledati ovako nekako -- Nacrtacu je plavom bojom Tako da ce moja kriva izgldati ovako i ona ide u pozitivnom smeru kretanja kazaljki na satu bas ovako sada kada imamo nas vekttor, i samo da potsetim videli smo ovo vise puta Ovom vektoroskom polju moze da bude dodeljen bilo koji vektor u delu x-y i bice -- moze biti definisana kao funkcija od x i y, koju cu ja nazvati sa "P" neke funkcije od x i y ... i onda j komponenta, ili ono sa cime smo pomnozili komponenty j ili vertikalna komponenta za bilo koju tacku x y tako da neke funkcije x y plus neke ostale skalarne funckije x i y j tako da ako mi date bilo koju tacku za nju ce postojati vektor za svaku tacku je pripisan vektor koji joj pripada tako mi definisemo ovu funkciju ali izraz je zapravo sto mi pricamo o integralu linije mi brinime posebno za tacke na ovoj krivoj oko ove konture tacno ovde tako da najbolje je da razmislimo sta je zapravo ovaj deo i sta nam on govori pre nego sto saberemo sve ove delove tako da ako uzmemo u obzir samo tacku f razmislimo samo o tacki na ovoj krivoj tako da tacka na ovoj krivoj moze da bude bas ovde pripojena sa tom tackom postoji vektor to je sta polje vektora radi tako da f moze da izgleda ovako ... i mi cemo da ga obelezimo sa tackom normalnog vektora
u toj tacki tako da normalan vektor treba da izgleda tako to bi trebalo da bude n tacka ovo je vektor u toj tacki