10.8. GREEN’S THEOREM 261

Theorem 10.8.5 Let U be a region in the uv plane2 for which Green’s theoremholds such that ∂U is oriented counter clockwise around this curve such that Green’s the-orem holds. Also let r(u,v) = (x,y)T be any C2

(U)

map with V = r(U) and supposedet(Dr(u,v))≥ 0. Then Green’s theorem holds for V also.

Proof: Let (x,y)→ P(x,y) ,Q(x,y) be C1(V)

functions. Then by the change of vari-ables formula, ∫

V(Qx (x,y)−Py (x,y))dm2 (x,y) =∫

U(Qx (x(u,v) ,y(u,v))−Py (x(u,v) ,y(u,v)))

∣∣∣∣ xu xvyu yv

∣∣∣∣dm2 (u,v) (10.15)

Now consider the integrand. It is

(Qx−Py)(xuyv− xvyu) = Qxxuyv−Qxxvyu−Pyxuyv +Pyxvyu (10.16)

Let F(x,y) = (P(x,y) ,Q(x,y)) .

(xv,yv) · (F◦ r)u− (xu,yu) · (F◦ r)v = xv (Pxxu +Pyyu)+ yv (Qxxu +Qyyu)

− [xu (Pxxv +Pyyv)+ yu (Qxxv +Qyyv)] = Qxxuyv +Pyxvyu− (Pyyvxu +Qxxvyu)

This is the same thing as 10.16. Thus 10.15 reduces to∫U(rv ·Fu− ru ·Fv)dm2 (u,v) (10.17)

where F = F◦ r to save notation. This integrand is of the form

(rv ·F)u− rvu ·F−((ru ·F)v− ruv ·F) = (rv ·F)u− (ru ·F)v

by equality of mixed partial derivatives. Thus 10.17 equals∫U(rv ·F)u− (ru ·F)v dm2 (u,v) =

∫∂U

ru ·Fdu+ rv ·Fdv

=∫

∂U(F◦ r) ·

(drdt

)dt =

∫∂V

F·dr =∫

∂VP(x,y)dx+

∫Q(x,y)dy

By Green’s theorem applied to (ru ·F,rv ·F) = (ru ·F◦ r,rv ·F◦ r) . Recall motion around∂U is counter clockwise with the u,v axes oriented as shown above. Now the curve ∂Uis piecewise smooth and a typical smooth piece is t → (u(t) ,v(t)) . Then on ∂V we havet→ r(u(t) ,v(t)) = (x,y) and dr

dt = ruu′+ rvv′ which is the explanation of the last line. ■The above is a reasonably good theorem and is enough for most applications to complex

analysis but it requires a map which takes U to V and the best version of this theorem onlyrequires a map from S1 to ∂U . It is in an appendix. This will also include the informationthat the Green’s theorem specifies an orientation over ∂V . Anyway, the main message ofthe above theorem is that Green’s theorem holds for very general situations. In applications,one can usually see that the theorem will hold based on the considerations discussed above.

2For a general version see the advanced calculus book by Apostol. This is presented in the appendix also. Thegeneral versions involve the concept of a rectifiable Jordan curve. You need to be able to take the area integraland to take the line integral around the boundary.