Green’s theorem in the plane 379

theorem which expresses a double integral over a plane region R as a line integral taken
along a closed curve forming the boundary of R. This theorem is usually referred to as
Green’s theorem.? It can be stated in several ways; the most common is in the form of the
identity :

(11.17)

The curve C which appears on the right is the boundary of the region R, and the integra-
tion symbol indicates that the curve is to be traversed in the counterclockwise direction,
as suggested by the example shown in Figure 11.13.

FIGURE 11.13 The curve C is the boundary of R, traversed in a counterclockwise

direction.

Two types of assumptions are required for the validity of this identity. First, conditions
are imposed on the functions P and to ensure the existence of the integrals. The usual
assumptions are that P and are continuously differentiable on an open set containing
the region R. This implies continuity of P and Q on C as well as continuity of and
on R, although the theorem is also valid under less stringent hypotheses. Second,
there are conditions of a geometric nature that are imposed on the region R and its boundary
curve C. The curve C may be any simple closed curve. The term “rectifiable”
means, of course, that C has a finite arc length. To explain what is meant by a simple
closed curve, we refer to the vector-valued function which describes the curve.
Suppose C is described by a continuous vector-valued function a defined on an interval
[a, b]. If a(a) = a(b), the curve is closed. A closed curve such that for every
pair of values in the half-open interval (a, is called a simple closed curve. This
means that, except for the end points of the interval [a, b], distinct values of lead to
distinct points on the curve. A circle is the prototype of a simple closed curve.
Simple closed curves that lie in a plane are usually called Jordan curves in honor of
Camille Jordan a famous French mathematician who did much of the pioneer-
ing work on such concepts as simple closed curves and arc length. Every Jordan curve C

In honor of George Green an English mathematician who wrote on the applications of

mathematics to electricity and magnetism, fluid flow, and the reflection and refraction of light and sound.
The theorem which bears Green’s name appeared earlier in the researches of Gauss and Lagrange.
380 Multiple integrals

decomposes the plane into two disjoint open connected sets having the curve C as their
common boundary. One of these regions is bounded and is called the interior (or inner
region) of C. (An example is the shaded region in Figure 11.13.) The other is unbounded
and is called the exterior (or outer region) of C. For some familiar Jordan curves such as
circles, ellipses, or elementary polygons, it is intuitively evident that the curve divides
the plane into an inner and an outer region, but to prove that this is true for an arbitrary
Jordan curve is not easy. Jordan was the first to point out that this statement requires
proof; the result is now known as the Jordan curve theorem. Toward the end of the 19th
century Jordan and others published incomplete proofs. In 1905 the American mathe-
matician Oswald Veblen (1880-1960) gave the first complete proof of this theorem.
Green’s theorem is valid whenever C is a rectifiable Jordan curve, and the region R is the
union of C and its interior.? Since we have not defined line integrals along arbitrary
rectifiable curves, we restrict our discussion here to piecewise smooth curves.
There is another technical difficulty associated with the formulation of Green’s theorem.
We have already remarked that, for the validity of the identity in the curve C
must be traversed in the counterclockwise direction. Intuitively, this means that a man
walking along the curve in this direction always has the region R to his left. Again, for
some familiar Jordan curves, such as those mentioned earlier, the meaning of the expression
“traversing a curve in the counterclockwise direction” is intuitively evident. However, in a
strictly rigorous treatment of Green’s theorem one would have to define this expression in
completely analytic terms, that is, in terms of the vector-valued function a that describes
the curve. One possible definition is outlined in Section 11.24.
Having pointed out some of the difficulties associated with the formulation of Green’s
theorem, we shall state the theorem in a rather general form and then indicate briefly why
it is true for certain special regions. In this discussion the meaning of “counterclockwise”
will be intuitive, so the treatment is not completely rigorous.

THEOREM GREEN’S THEOREM FOR PLANE REGIONS BOUNDED BY PIECEWISE SMOOTH

JORDAN CURVES. Let P and Q be scalar fields that are continuously differentiable on an
open set S in the xy-plane. Let C be apiecewise smooth Jordan curve, and let R denote the
union of C and its interior. Assume R is subset of S. Then we have the identity

(11.18)
R

where the line integral is taken around C in the counterclockwise direction.

Note: The identity in (11.18) is equivalent to the two formulas

(11.19) =
R

A proof of Green’s theorem for regions of this generality can be found in Chapter 10 of the author’s
Mathematical Analysis.
 Green’s theorem in the 381

and
(11.20)

In fact, if both of these are true, (11.18) follows by addition. Conversely, if (I 1.18) is true
we may obtain (11.19) and (11.20) as special cases by taking = 0 and Q = respectively.

Proof for special regions. We shall prove (1 I for a region of Type I. Such a region
has the form

where and g are continuous on [a, b] with The boundary C of R consists of four
parts, a lower arc (the graph an upper arc (the graph ofg), and two vertical line
segments, traversed in the directions indicated in Figure 11.14.

Y
, =

FIGURE 11.14 Proof of Green’s theorem for a FIGURE 11.15 Proof of Green’s theorem for
special region. a more general region.

First we evaluate the double integral dx by iterated integration. Integrat-

ing first with respect to we have

On the other hand, the line integral P dx can be written as follows:

since the line integral along each vertical segment is zero. To evaluate the integral along
382 Multiple integrals

we use the vector representation a(t) = ti + and obtain

To evaluate the integral along we use the representation a(t) = ti + and take into
account the reversal in direction to obtain

dx = g(t)] dt.
Therefore we have
dx = dt -- g(t)] dt.

Comparing this equation with the formula in (11.21) we obtain (11.20).

A similar argument can be used to prove (11.19) for regions of Type II. In this way a
proof of Green’s theorem is obtained for regions that are of both Type I and Type II.
Once this is done, the theorem can be proved for those regions R that can be decomposed
into a finite number of regions that are of both types. “Crosscuts” are introduced as shown
in Figure 11.15, the theorem is applied to each subregion, and the results are added
together. The line integrals along the crosscuts cancel in pairs, as suggested in the figure,
and the sum of the line integrals along the boundaries of the subregions is equal to the line
integral along the boundary of R.

11.20 Some applications of Green’s theorem

The following examples illustrate some applications of Green’s theorem.

1. Use Green’s theorem to compute the work done by the force field
E X A MP L E
(y + 3x)i in moving a particle once around the ellipse + = 4
in the counterclockwise direction.

Solution.
C is the ellipse. Since = -2, Green’s theorem gives us

P dx + Q = (-2) =
R

where a(R) is the area of the region enclosed by the ellipse. Since this ellipse has
a = and = 2, its area is = and the value of the line integral is

EXAMPLE 2. Evaluate the line integral (5 dx dy , where C is

the square with vertices (0, 0), (1, 0), (1, (0, traversed counterclockwise.

Solution. Here P = 5 Q = and 3x. Hence,

by Green’s theorem, we have

R
 A necessary and condition for a vector to be a gradient

where is the x-coordinate of the centroid of the square. Since is obviously the value
of the line integral is

EXAMPLE 3. Area expressed as a line integral. The double integral for the area a(R) of a
region R can be expressed in the form

R R

where P and are such that = 1. For example, we can take y) =

and y) = . If R is the region enclosed by a Jordan curve C we can apply Green’s
theorem to express a(R) as a line integral,

a(R) = xdy.
c

If the boundary curve C is described by parametric equations, say

x = X(t), y =
the line integral for area becomes

a(R) + dt
a

11.21 A necessary and sufficient condition for a two-dimensional vector field to be a

gradient
Let y) = be a vector field that is continuously differentiable on
an open set S in the plane. Iff is a gradient on S we have

(11.22)

everywhere on S. In other words, the condition (11.22) is necessary for f to be a gradient.

As we have already noted, this condition is not sufficient. For example, the vector field

= +
+

satisfies (11.22) everywhere on the set S = O)}, but f is not a gradient on S.

In Theorem 10.9 we proved that condition (11.22) is both necessary and sufficient for
to be a gradient on if the set S is convex. With the help of Green’s theorem we can extend
this result to a more general class of plane sets known as connected sets. They are
defined as follows.
384 Multiple integrals

DEFINITION OF A SIMPLY CONNECTED PLANE SET. Let S be an open connected set in the
plane. Then S is called simply connected for every Jordan curve C which lies in the inner
region of C is also a subset of S.

An (the set of points lying between two concentric circles) is not simply connected
because the inner region of a circle concentric with the bounding circles and of radius
between theirs is not a subset of the Intuitively speaking, we say that S is simply
connected when it has no “holes.” Another way to describe simple connectedness is to say
that a curve in S connecting any two points may be continuously deformed into any
other curve in S joining these two points, with all intermediate curves during the
deformation lying completely in S. An alternative definition, which can be shown to be
equivalent to the one given here, states that an open connected set S is simply connected if
its complement (relative to the whole plane) is connected. For example, an is not
simply connected because its complement is disconnected. An open connected set that is
not simply connected is called multiply connected.

THEOREM y) = + be a vectorjeld that is continuously

differentiable on an open simply connected set S in the plane. Then is a gradient on S if and
only if we have

(11.23) everywhere on S .

Proof. As already noted, condition (11.23) is necessary for f to be a gradient. We shall

prove now that it is also sufficient.
It can be shown that in any open connected plane set S, every pair of points a and
can be joined by a simple step-polygon, that is, by a polygon whose edges are parallel to
the coordinate axes and which has no self-intersections. If the line integral a to
has the same value for every simple step-polygon in S joining a to then exactly the same
argument used to prove Theorem 10.4 shows that f is a gradient on S. Therefore, we need

I
I

a - +
11.16 Independence of the path in a simply connected region.
 Exercises

only verify that the line integral off from to has the same value for every simple
polygon in S joining a to x.
Let and be two simple step-polygons in S joining a to x. Portions of these polygons
may coincide along certain line segments. The remaining portions will intersect at most a
finite number of times, and will form the boundaries of a finite number of polygonal
regions, say . . . , R,. Since S is assumed to be simply connected, each of the regions
is a subset of S. An example is shown in Figure 11.16. The solid line represents ,
the dotted line represents and the shaded regions represent . . . , . (These two
particular polygons coincide along the segment
We observe next that the line integral off from a to along plus the integral from
back to a along is zero because the integral along the closed path is a sum of integrals
taken over the line segments common to and plus a sum of integrals taken around
the boundaries of the regions R,. The integrals over the common segments cancel in pairs,
since each common segment is traversed twice, in opposite directions, and their sum is
zero. The integral over the boundary of each region is also zero because, by Green’s
theorem we may write

and the integrand of the double integral is zero because of the hypothesis =
It follows that the integral from a to along is equal to that along As we have
already noted, this implies that f is a gradient in S.

Use Green’s theorem to evaluate the line integral +x when

(a) C is the square with vertices (0, 0),
(b) C is the square with vertices ( 1, 1).
(c) C is the square with vertices (0, f2).
(d) C is the circle of radius 2 and center at the origin.
C has the vector equation a(t) = 2 i 2 0
If y) = and y) = + + evaluate the line integral P d x +
Q dy around the boundary of the square of side determined by the inequalities a and
a.
Let C be a simple closed curve in the xy-plane and let denote the moment of inertia (about
the z-axis) of the region enclosed by C. Show that an integer exists such that

Given two scalar fields and v that are continuously differentiable on an open set containing
the circular disk R whose boundary is the circle + = 1 . Define two vector fieldsfand
as follows:

Find the value of the double integral dx dy if it is known that on the boundary of R we
have = 1 and y) = .