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 18.1.1Let 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
∫ ∫ ∫ ( )
P dx+ Qdy ≡ F⋅dR = ∂Q-− ∂P- dA. (18.1)
∂U ∂U U ∂x ∂y
∫ (∂Q ∂P ) ∫ d∫ r(y) ∂Q ∫ b∫ t(x)∂P
U -∂x − ∂y- dA = c l(y) ∂x-dxdy− a b(x) ∂y-dydx
∫ d
= (Q (r(y) ,y) − Q (l(y),y))dy
c∫ b
+ (P (x,b(x)))− P (x,t(x))dx. (18.2)
a
Now consider the left side of (18.1). Denote by V the vertical parts of ∂U and by H the horizontal
parts.
∫ ∫
F⋅dR = ((0,Q)+ (P,0))⋅dR
∂U ∂U
∫ d ′ ∫
= (0,Q (r(s),s))⋅(r (s),1)ds + (0,Q (r (s),s))⋅(±1,0)ds
c∫ d H∫ 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 18.1.2Let everything be the same as in Lemma 18.1.1but 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 (18.1) 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 (18.1) 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 )
= ---− --- dA
k=1 Uk ∂x ∂y
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. 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 (18.1) holds,
then the open set V ∖
(U ∪ ∂U)
consisting of what is left in V after deleting U along with its
boundary also satisfies (18.1). Roughly speaking, you can drill holes in a region for which (18.1)
holds and get another region for which this continues to hold provided (18.1) 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) ∫ (∂Q ∂P)
= ---− --- dA + ---− --- dA
∫U ∂x ∂∫y ( V∖U ∂)x ∂y
= 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.)
You can see that (18.1) is valid quite generally. This verifies the following theorem.
Theorem 18.1.3(Green’s Theorem) Let U be an openset in the plane and let ∂U be piecewisesmooth and let F
(x,y)
=
(P (x,y),Q (x,y))
be a C^{1}vector field defined near U. Then it isoften^{1}the case that
∫ ∫ ( )
F⋅dR = ∂Q-(x,y)− ∂P-(x,y) dA.
∂U U ∂x ∂y
Here is an alternate proof of Green’s theorem from the divergence theorem.
Theorem 18.1.4(Green’s Theorem) Let U be an openset in the plane and let ∂U be piecewise smoothand let F
(x,y)
=
(P (x,y),Q (x,y))
be a C^{1}vector field defined near U. Then it is often the casethat
∫ ∫ ( )
∂Q- ∂P-
∂U F⋅dR = U ∂x (x,y)− ∂y (x,y) dA.
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
∘----2-----2
(x′) + (y′)
dt. Suppose the boundary of U consists of m smooth curves, the i^{th} of
which is parameterized by
(xi,yi)
with the parameter t ∈
[ai,bi]
. Then by the divergence
theorem,
∫ ∫ ∫
(Qx − Py) dA = div(F)dA = F ⋅ndS
U U ∂U
∫
∑m bi
= ai (Q (xi(t),yi(t)),− P (xi(t),yi(t)))
i=1
◜∘-----d◞S◟-----◝
⋅∘-----1------(y′,− x ′) (x′)2 + (y′)2dt
(x′)2 + (y′)2 i i i i
∫ i i
∑m bi ′ ′
= ai (Q (xi(t),yi(t)),− P (xi(t),yi(t)))⋅(yi,− xi)dt
i=m1∫ b ∫
= ∑ iQ (xi(t),yi(t))y′(t) +P (xi(t),yi(t))x ′(t)dt ≡ Pdx + Qdy
i=1 ai i i ∂U
This proves Green’s theorem from the divergence theorem. ■
Proposition 18.1.5Let 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 18.1.6Use Proposition 18.1.5to find the area of the ellipse
x2 y2
-2 + 2-≤ 1.
a b
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