When discussing the " >idea behind Green’s theorem, I claimed that the “microscopic” circulation of a two-dimensional vector field F was

“microscopic circulation” = -. |

-. |

does represent “circulation per unit area,” which is probably a better term for it than “microscopic circulation.”

Remeber that the circulation of F around the closed curve C is

∫
_{C}F ⋅ ds. |

. |

→ -, |

assuming C is oriented counterclockwise.

We will sketch a proof of this for a rectangular curve C, oriented counterclockwise.
Let the lower-left point of C be (a,b), its width be Δx, and its height be Δy.
Label the edges of the rectangle by C_{1}, C_{2}, C_{3}, and C_{4}.

.

We assume box small enough to approximate F as constant along each edge.

Along the curve C_{1}, the value of y is constantly b, but the value of x changes from
a to a + Δx. We assume we can ignore that change in x (since the box is
small), and simply approximate F(x,y) as F(a,b) all along the segment
C_{1}.

Along the curve C_{2}, the value of x is constantly a + Δx, but y changes from b to
b + Δy. Since we assume we can ignore the change in y, we approximate F(x,y) as
F(a + Δx,b) all along segment C_{2}.

Similiarly, along C_{3}, y = b + Δy, and we approximate x as a (even though it
ranges between a and a + Δx). We approximate F(x,y) as F(a,b + Δy) all along
segment C_{3}.

Along C_{4}, x = a, and we approximage y as b + Δy (even though it ranges
between b and b + Δy). We approximate F(x,y) as F(a,b) all along segment
C_{4}.

We summarize these approximations in the following figure.

.

Next, to compute the integral ∫
_{C}F ⋅ds, we remember that it is the same thing as
integrating the scalar-valued function F ⋅ T, where T is the unit tangent vector
along C:

∫
_{C}F ⋅ ds = ∫
_{C}F ⋅ Tds. |

We need to compute F ⋅ T along each segment of C. The tangent vector T is constant along each segment, and we are approximating F as constant along each segment, so the dot product F ⋅ T will be constant along each segment of C.

Along C_{1}, the path is directed in the positive x direction, so the unit tangent
vector is T = (1, 0). The dot product F ⋅ T is simply the first component of
F(a,b):

F ⋅ T = F_{1}(a,b). |

Along C_{2}, the path is directed in the positive y direction, so T = (0, 1), and F ⋅ T
is simply the second component of F(a + Δx,b):

F ⋅ T = F_{2}(a + Δx,b). |

Along C_{3}, the path is directed in the negative x direction, so the unit tangent
vector is T = (-1, 0). The dot product F ⋅ T is minus the first component of
F(a,b + Δy):

F ⋅ T = -F_{1}(a,b + Δy). |

Along C_{4}, the path is directed in the negative y direction, so T = (0,-1), and
F ⋅ T is minus the second component of F(a,b):

F ⋅ T = -F_{2}(a,b). |

We summarize these findings in the following figure.

.

The integrals are now easy to compute. Along C_{1}, F ⋅ T is constant, so
the integral is simply its value F_{1}(a,b) times the length of the segment
Δx:

∫
_{C1}F ⋅ ds = ∫
_{C1}F ⋅ Tds = ∫
_{C1}F_{1}(a,b)ds = F_{1}(a,b)Δx. |

∫
_{C2}F ⋅ ds = F_{2}(a + Δx,b)Δy. |

∫
_{C3}F ⋅ ds = -F_{1}(a,b + Δy)Δx, |

∫
_{C4}F ⋅ ds = -F_{2}(a,b)Δy. |

The integral around all of C is just the sum along the four segments

∫
_{C}F ⋅ ds | = ∫
_{C1}F ⋅ ds + ∫
_{C2}F ⋅ ds + ∫
_{C3}F ⋅ ds + ∫
_{C4}F ⋅ ds | ||

= F_{1}(a,b)Δx + F_{2}(a + Δx,b)Δy - F_{1}(a,b + Δy)Δx - F_{2}(a,b)Δy. |

The circulation per unit area is the integral divided by the area of the rectangle, which is ΔxΔy

= , |

where I simply rearranged the terms in the numerator.

Half of the numerator is multiplied by Δy and half is multiplied byΔx. If we separate these into two fractions, we can cancel the Δy in the first fraction with the Δy in the demoninator

= . |

In the second fraction, we can cancel the Δx,

= . |

= -. |

Now, we let the curve C shrink down to a point. This means that Δx → 0 and Δy → 0. In this limit, the two fractions become something familar: partial derivatives of F.

lim _{Δx,Δy→0} | = lim _{Δx→0} - lim _{Δy→0} | ||

= (a,b) -(a,b). |

We have shown the the circulation per unit area around the point (x,y) = (a,b) is

(a,b) -(a,b). |

∫
∫
_{D}dA = ∫
_{∂D}F ⋅ ds. |

where C = ∂D is the path going counterclockwise around D.