A.1. EXERCISES 457

= limn→∞

∑R∈Iδn

∫∂R

f ·dγ = limn→∞

∫Iδn

±(Qx−Py)dm2 =∫

Ui

±(Qx−Py)dm2

where the ± adjusts for whether the interior rectangles are all oriented positively (counterclockwise) or all oriented negatively (clockwise). It was assumed these rectangles areoriented counter clockwise and so the + sign would be used. ■

This has proved the general form of Green’s theorem which is stated in the followingtheorem.

Theorem A.0.4 Let J be a rectifiable simple closed curve in R2 having inside Uiand outside Uo. Let P,Q be functions with the property that Qx,Py ∈ L1 (Ui) and P,Q areC1 on Ui. Assume also P,Q are continuous on J∪Ui. Then there exists an orientation for J(Remember there are only two.) such that for

f(x,y) = (P(x,y) ,Q(x,y)) ,∫

Jf ·dγ =

∫Ui

(Qx−Py)dm2.

Proof: In the construction of the regions, an orientation was imparted to J. The abovecomputation shows

∫J f ·dγ =

∫Ui(Qx−Py)dm2 ■

A.1 Exercises1. Consider the following diagram.

(x2,y2)

(x1,y1)

z1

z2

αβ

UiJrJl

In this diagram there is a polygonal curve from z1 to z2 in the inside componentof JC,Ui in which there are no horizontal segments. Here z1,z2 are very close tothe top and bottom points respectively. This intersects a horizontal line shown infinitely many points and for one of the segments, it crosses the horizontal line an oddnumber of times. Pick the part of J the simple closed curve which corresponds tothat segment. Explain why this divides J into two simple closed curves and that ifthe horizontal line is y = c, one Jordan curve has height at least |y1− c| and the otherhaving height at least |c− y2|. For the existence of the curves from z1 to (x1,y1)and from z2 to (x2,y2) shown in the picture, consult Problem 30 on Page 222. Thisapproach to splitting up the simple closed curve into smaller regions is in Apostol[2].