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 the seemingly more general Stoke’s
theorem but is interesting for it’s own sake. Historically, theorems like it were 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.
Lemma 28.1.1Let F
(x,y)
≡
(P (x,y),Q (x,y))
be a C^{1}vector field defined near Uwhere U is a region of the sort indicated in the above picture which is convex in both thex and y directions. Suppose also that the functions r,l,t, and b in the above picture areall C^{1}functions and denote by ∂U the boundary of U oriented such that the direction ofmotion is counter clockwise. (As you walk around U on ∂U, the points of U are on yourleft.) Then
∫ ∫ ∫ ( ∂Q ∂P )
P dx+ Qdy ≡ F⋅dR = ∂x-− -∂y dA. (28.1)
∂U ∂U U
∫ ( ) ∫ d∫ r(y) ∫ b∫ t(x)
∂Q-− ∂P- dA = ∂Q-dxdy − ∂P-dydx
U ∂x ∂y c l(y) ∂x a b(x) ∂y
∫
d
= c (Q (r(y),y)− Q(l(y),y)) dy
∫ b
+ (P (x,b(x)))− P (x,t(x))dx. (28.2)
a
Now consider the left side of 28.1. Denote by V the vertical parts of ∂U and by H the
horizontal parts.
∫ ∫
F ⋅dR = ((0,Q) +(P,0))⋅dR
∂U ∂U
∫ d ′ ∫
= c (0,Q (r (s),s))⋅(r(s),1)ds+ H (0,Q (r(s),s)) ⋅(±1, 0) ds
∫ d ∫ b
− (0,Q(l(s),s))⋅(l′(s),1)ds+ (P (s,b(s)),0)⋅(1,b′(s))ds
∫c ∫ a
b ′
+ V (P (s,b(s)),0)⋅(0,±1)ds− a (P (s,t(s)),0)⋅(1,t (s))ds
∫ d ∫ d ∫ 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 28.1.2Let everything be the same as in Lemma 28.1.1but onlyassume the functions r,l,t, and b are continuous and piecewise C^{1}functions. Thenthe 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 28.1 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 28.1 must hold for U ≡∪_{i=1}^{m}U_{i}, the union of
these sets. This is because
∫ ( )
∂Q-− ∂P- dA =
U ∂x ∂y
∫ ( )
∑m ∂Q- ∂P-
= Uk ∂x − ∂y dA
k=m1∫ ∫
= ∑ 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. As an illustration, consider the following picture for two such
U_{k}.
PICT
Similarly, if U ⊆ V and if also ∂U ⊆ V and both U and V are open sets for which
28.1 holds, then the open set V ∖
(U ∪ ∂U)
consisting of what is left in V after deleting U
along with its boundary also satisfies 28.1. Roughly speaking, you can drill holes in a
region for which 28.1 holds and get another region for which this continues to hold
provided 28.1 holds for the holes. To see why this is so, consider the following picture
which typifies the situation just described.
PICT
Then
∫ ∫ ( )
F ⋅dR = ∂Q- − ∂P- dA
∂V V ∂x ∂y
∫ ( ) ∫ ( )
= ∂Q- − ∂P- dA + ∂Q- − ∂P- dA
U ∂x ∂y V∖U ∂x ∂y
∫ ∫ ( ∂Q ∂P )
= F⋅dR + ∂x-− ∂y- dA
∂U V ∖U
and so
∫ ( ∂Q- ∂P-) ∫ ∫
V∖U ∂x − ∂y dA = ∂V F ⋅dR − ∂U F ⋅dR
which equals
∫
∂(V∖U)F ⋅dR
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.)
You can see that 28.1 is valid quite generally. This verifies the following
theorem.
Theorem 28.1.3(Green’s Theorem) Let U be an openset in the plane and let ∂U bepiecewise smooth and let F
(x,y)
=
(P (x,y),Q (x,y))
be a C^{1}vector field defined near U. Thenit is often^{1}the case that
∫ ∫ ( ∂Q ∂P )
F ⋅dR = ∂x-(x,y)− ∂y-(x,y) dA.
∂U U
Here is an alternate proof of Green’s theorem from the divergence theorem.
Theorem 28.1.4(Green’s Theorem) Let U be an openset in the plane and let∂U be piecewise smooth and let F
(x,y)
=
(P (x,y),Q (x,y))
be a C^{1}vector field definednear U. Then it is often the case that
∫ ∫ ( )
F ⋅dR = ∂Q-(x,y)− ∂P-(x,y) dA.
∂U U ∂x ∂y
Proof: Suppose the divergence theorem holds for U. Consider the following
picture.
PICT
Since it is assumed that motion around U is counter clockwise, the tangent vector
′ ′
(x ,y)
is as shown. The unit exterior normal is a multiple of
(x′,y′,0)× (0,0,1) = (y′,− x′,0).
Use your right hand and the geometric description of the cross product to verify
this. This would be the case at all the points where the unit exterior normal
exists.
Now let F
(x,y)
=
(Q (x,y),− P (x,y))
. Also note the area (length) element on the
bounding curve ∂U is
∘ -----------
(x ′)2 + (y′)2
dt. Suppose the boundary of U consists of m smooth
curves, the i^{th} of which is parameterized by
(x ,y)
i i
with the parameter t ∈
[a ,b]
i i
. Then
by the divergence theorem,
∫ ∫ ∫
U (Qx − Py )dA = U div(F)dA = ∂U F ⋅ndS
∑m ∫ bi
= (Q (xi(t),yi(t)),− P (xi(t),yi(t)))
i=1 ai
◜∘-----d◞S◟-----◝
------1------ ′ ′ ′2 ′ 2
⋅∘ --′2----′-2 (yi,− xi) (xi) + (yi) dt
(xi) +(yi)
∑m ∫ bi ′ ′
= a (Q (xi(t),yi(t)),− P (xi(t),yi(t)))⋅(yi,− xi)dt
i=m1 ∫i ∫
= ∑ biQ (x (t),y (t))y′(t) + P (x (t),y (t)) x′(t)dt ≡ Pdx + Qdy
i=1 ai i i i i i i ∂U
This proves Green’s theorem from the divergence theorem. ■
Proposition 28.1.5Let U be an open set in ℝ^{2}for which Green’s theorem holds.Then
∫
Area of U = F ⋅dR
∂U
where F
(x,y)
=
1
2
(− y,x)
,
(0,x)
, or
(− y,0)
.
Proof: This follows immediately from Green’s theorem. ■
Example 28.1.6Use Proposition 28.1.5to find the area of the ellipse
x2 y2
a2 + b2 ≤ 1.
You can parameterize the boundary of this ellipse as
One way to do this is to parameterize the boundary of U and then compute the line
integral directly. It is easier to use Green’s theorem. The desired line integral
equals