Definition 12.1.1Here and elsewhere, an open connected set will be called a region unless anotheruse for this term is specified.
Green’s theorem is an important theorem which relates line integrals to integrals over a surface
in the plane. It can be used to establish Stoke’s theorem but is interesting for it’s own sake.
Historically, it was important in the development of complex analysis. I will first establish
Green’s theorem for regions of a particular sort and then show that the theorem holds for many
other regions also. Suppose a region is of the form indicated in the following picture in which
U = {(x,y) : x ∈ (a,b) and y ∈ (b(x),t(x))}
= {(x,y) : y ∈ (c,d) and x ∈ (l(y),r(y))}.
PICT
I will refer to such a region as being convex in both the x and y directions. For sufficiently simple
regions like those just described, it is easy to see what is meant by counter clockwise motion over the
curve.
It will always be assumed in this section that the bounding curves are piecewise C^{1} meaning that there
is a parametrization t → R
(t)
≡
(x(t),y(y))
for t ∈
[a,b]
and a partition of
[a,b]
,
{x0,⋅⋅⋅,xn}
such that
R has derivatives which are continuous on each
[xk−1,xk]
and R is continuous on
[a,b]
. Thus these curves
are of bounded variation thanks to Lemma 4.2.2. One can then compute the line integrals by adding
together the integrals over the sub-intervals thanks to Lemma 4.2.9. In particular, one writes for one of
these integrals
∫ xk ∫
F (R (t))⋅R′(t)dt = F ⋅dR
xk−1 R([xk−1,xk])
Orientation is assumed to come from increasing t.
Lemma 12.1.2Let F
(x,y)
≡
(P (x,y) ,Q (x,y))
be a C^{1}vector field defined near U where U is a regionof the sort indicated in the above picture which is convex in both the x and y directions. Suppose also thatthe functions, r,l,t, and b in the above picture are all C^{1}functions and denote by ∂U the boundary of Uoriented such that the direction of motion is counter clockwise. (As you walk around U on ∂U, the points ofU are on your left.) Then
∫ ∫ ∫ ∫
d r(y) ∂Q- b t(x)∂P-
= c l(y) ∂xdxdy − a b(x) ∂y dydx
∫ d ∫ b
= (Q (r(y),y) − Q(l(y),y))dy + (P (x,b(x))) − P (x,t(x))dx. (12.2)
c a
Now consider the left side of 12.1. Denote by V the vertical parts of ∂U and by H the horizontal
parts.
∫
F⋅dR =
∂U
∫
= ∂U ((0,Q )+ (P,0)) ⋅dR
∫ d ∫
= (0,Q (r(s),s))⋅(r′(s),1)ds + (0,Q (r (s),s))⋅(±1,0)ds
c∫ H∫
− d(0,Q (l(s),s))⋅(l′(s),1)ds + b(P (s,b(s)),0)⋅(1,b′(s))ds
c a
∫ ∫ b
+ (P (s,b(s)) ,0)⋅(0,±1)ds − (P (s,t(s)),0)⋅(1,t′(s))ds
∫ dV ∫ d a ∫ b ∫ b
= Q(r(s),s)ds− Q (l(s),s)ds+ P (s,b(s))ds− P (s,t(s))ds
c c a a
Corollary 12.1.3Let everything be the same as in Lemma 12.1.2but only assume the functionsr,l,t, and b are continuous and piecewise C^{1}functions. Then the conclusion this lemma is still valid.
Proof:The details are left for you. All you have to do is to break up the various line integrals into the
sum of integrals over sub intervals on which the function of interest is C^{1}.
From this corollary, it follows 12.1 is valid for any triangle for example.
,U_{m} and the open sets, U_{k} have the property that
no two have nonempty intersection and their boundaries intersect only in a finite number of
piecewise smooth curves. Then 12.1 must hold for U ≡∪_{i=1}^{m}U_{i}, the union of these sets. This is
because
∫ ( )
∂Q-− ∂P- dm =
U ∂x ∂y 2
m ∫ ( )
= ∑ ∂Q-− ∂P- dm
k=1 Uk ∂x ∂y 2
m∑ ∫ ∫
= F ⋅dR = F ⋅dR
k=1 ∂Uk ∂U
because if Γ = ∂U_{k}∩ ∂U_{j}, then its orientation as a part of ∂U_{k} is opposite to its orientation as a part of
∂U_{j} and consequently the line integrals over Γ will cancel, points of Γ also not being in ∂U. It is obvious
from the definition of Lebesgue measure given earlier that the intersection of two of these in
a smooth curve has measure zero. Thus adding an integral with respect to m_{2} over such a
curve yields 0. I am not trying to be completely general here. I am just noting that when you
paste together simple shapes like triangles and rectangles, this kind of cancelation will take
place. As part of the development of a general Green’s theorem given later, it is shown that
whenever you have a curve of bounded variation, it will have two dimensional Lebesgue measure
zero.
As an illustration, consider the following picture for two such U_{k}.
PICT
You can see that you could paste together many such simple regions composed of triangles or rectangles
and obtain a region on which Green’s theorem will hold even though it is not convex in each direction. This
approach is developed much more in the book by Spivak [115] who pastes together boxes as part of his
treatment of a general Stokes theorem.
Similarly, if U ⊆ V and if also ∂U ⊆ V and both U and V are open sets for which 12.1 holds,
then the open set, V ∖
(U ∪∂U )
consisting of what is left in V after deleting U along with
its boundary also satisfies 12.1. Roughly speaking, you can drill holes in a region for which
12.1 holds and get another region for which this continues to hold provided 12.1 holds for the
holes. To see why this is so, consider the following picture which typifies the situation just
described.
PICT
Then
∫ ∫ (∂Q ∂P )
F⋅dR = --- − --- dA
∂V V ∂x ∂y
∫ (∂Q ∂P) ∫ (∂Q ∂P)
= ∂x-− ∂y- dA + ∂x-− ∂y- dA
∫U ∫ ( V∖U )
= F ⋅dR + ∂Q-− ∂P- dA
∂U V ∖U ∂x ∂y
and so
∫ ( ) ∫ ∫
∂Q-− ∂P- dA = F ⋅dR − F ⋅dR
V∖U ∂x ∂y ∂V ∂U
which equals
∫
F ⋅dR
∂(V∖U)
where ∂V is oriented as shown in the picture. (If you walk around the region, V ∖ U with the area on the
left, you get the indicated orientation for this curve.)