9.12 Green’s Theorem

INTRODUCTION

One of the most important theorems in vector integral calculus relates a line integral around a piecewise-smooth simple closed curve C to a double integral over the region R bounded by the curve.

Line Integrals Along Simple Closed Curves

We say the positive direction around a simple closed curve C is that direction a point on the curve must move, or the direction a person must walk on C, in order to keep the region R bounded by C to the left. See FIGURE 9.12.1(a). Roughly, as shown in Figure 9.12.1(b) and 9.12.1(c), the positive and negative directions correspond to the counterclockwise and clockwise directions, respectively. Line integrals on simple closed curves are written

and so on. The symbols and refer, in turn, to integrations in the positive and negative directions.

3 visual representations. Visual representation a. Caption. Positive direction. Figure. The region inside an elliptical curve labeled C is shaded and labeled R. A person walks along the curve C while keeping the shaded region to the left. Visual representation b. Caption. Positive direction. Figure. The region inside a closed curve is shaded and labeled R. Several arrows placed along the curve C point to a counter clockwise direction. Visual representation c. Caption. Negative direction. Figure. The region inside a closed curve is shaded and labeled R. Several arrows placed along the curve C point to a clockwise direction.

FIGURE 9.12.1 Directions around curve C

THEOREM 9.12.1 Green’s Theorem in the Plane

Suppose that C is a piecewise-smooth simple closed curve bounding a simply connected region R. If P, Q, ∂P/∂y, and ∂Q/∂x are continuous on R, then

(1)

PARTIAL PROOF:

We shall prove (1) only for a region R that is simultaneously of Type I and Type II:

R: g1(x) ≤ y ≤ g2(x), a ≤ x ≤ b

R: h1(y) ≤ x ≤ h2(y), c ≤ y ≤ d.

Using FIGURE 9.12.2(a), we have

(2)

Similarly, from Figure 9.12.2(b),

(3)

2 graphs. Graph a. Caption. R as a Type 1 region. Graph. A closed curve consisting of 2 smooth curves, is graphed on an x y coordinate plane. The first smooth curve, labeled y = g subscript 1(x), begins at a marked point at x = a, goes down to the right, reaches a low point, goes up to the right, and ends a second marked point at x = b. The second curve, labeled y = g subscript 2(x), begins at the second marked point at x = b, goes up to the left, reaches a high point, goes down to the left, and ends at the first marked point at x = a. The region inside the closed curve is shaded and labeled R. Graph b. Caption. R as a Type 2 region. Graph. A closed curve consisting of 2 smooth curves, is graphed on an x y coordinate plane. The first smooth curve, labeled x = h subscript 2(y), begins at a marked point at y = c, goes up to the right, changes direction and goes up to the left, reaches a high point, and ends a second marked point at y = d. The second curve, labeled x = h subscript 1(y), begins at the second marked point at y = d, goes down to the left, changes direction and goes down to the right, reaches a low point, and ends at the first marked point at y = c. The region inside the closed curve is shaded and labeled R.

FIGURE 9.12.2 Region R in Theorem 9.12.1

Adding the results in (2) and (3) yields (1). ≡

The result given in Theorem 9.12.1 is named after George Green, the same English mathematical physicist who developed the Green’s functions discussed in Section 3.10. The words in the plane suggest that Theorem 9.12.1 generalizes to 3-space. It does—read on.

Although the foregoing proof is not valid, the theorem is applicable to more complicated regions, such as those shown in FIGURE 9.12.3. The proof consists of decomposing R into a finite number of subregions to which (1) can be applied and then adding the results.

A graph. A closed curve is graphed in an x y coordinate plane. The curve begins at a point in the first quadrant, goes up and to the right, reaches a high point, goes down to the right, reaches a low point, goes up to the right, changes direction and goes to the left, reaches a high point, goes down to the left, reaches a low point vertically above the first low point, goes up to the left, reaches a high point vertically above the first high point, goes down to the left, changes direction and goes down to the right, and ends at the starting point. The region inside the closed curve is shaded and labeled R. A dashed vertical line connects the first and third high points. A dashed horizontal line connects the first high point and the second low point. A second dashed vertical line connects the first and second low points. The region inside the closed curve, to the left of the first dashed vertical line, is labeled R subscript 1. The region inside the closed curve, to the right of the first dashed vertical line and above the dashed horizontal line, is labeled R subscript 2. The region inside the closed curve, under the dashed horizontal line and to the left of the second dashed vertical line, is labeled R subscript 3. The region inside the closed curve, to the right of the second dashed vertical line, is labeled R subscript 4.

FIGURE 9.12.3 Subregions of R

EXAMPLE 1 Using Green’s Theorem

Evaluate (x2 – y2) dx + (2y – x) dy, where C consists of the boundary of the region in the first quadrant that is bounded by the graphs of y = x2 and y = x3.

SOLUTION

If P(x, y) = x2 – y2 and Q(x, y) = 2y – x, then ∂P/∂y = –2y and ∂Q/∂x = –1. From (1) and FIGURE 9.12.4, we have

≡

A graph. A closed curve, consisting of 2 smooth curves, is graphed on an x y coordinate plane. The first smooth curve, labeled y = x^3, begins at the marked origin, goes up to the right with increasing steepness, and ends at the marked point (1, 1). The second smooth curve, labeled y = x^2, begins at the marked point (1, 1), goes down to the left with decreasing steepness, to the left of the first curve, and ends at the marked origin. The region bound by the 2 smooth curves, is shaded and labeled R.

FIGURE 9.12.4 Curve C in Example 1

We note that the line integral in Example 1 could have been evaluated in a straightforward manner using the variable x as a parameter. However, in the next example, ponder the problem of evaluating the given line integral in the usual manner.

EXAMPLE 2 Using Green’s Theorem

Evaluate (x5 + 3y) dx + (2x – ) dy, where C is the circle (x – 1)2 + (y – 5)2 = 4.

SOLUTION

Identifying P(x, y) = x5 + 3y and Q(x, y) = 2x – , we have ∂P/∂y = 3 and ∂Q/∂x = 2. Hence, (1) gives

Now the double integral dA gives the area of the region R bounded by the circle of radius 2 shown in FIGURE 9.12.5. Since the area of the circle is π22 = 4π, it follows that

= –4π. ≡

A graph. A circle, labeled (x minus 1)^2 + (y minus 5)^2 = 4, is graphed on an x y coordinate plane. The center of the circle is in the first quadrant. The area of the shaded region inside the circle is 4 pi. 2 arrows on the circle are oriented counter clockwise.

FIGURE 9.12.5 Curve C in Example 2

EXAMPLE 3 Work Done by a Force

Find the work done by the force F = (–16y + sin x2)i + (4ey + 3x2)j acting along the simple closed curve C shown in FIGURE 9.12.6.

A graph. A closed curve, consisting of 2 line segments and a circular smooth curve, is graphed on an x y coordinate plane. The first line segment, labeled C subscript 1: y = x, begins at the origin, goes up and to the right, and ends at the point in the first quadrant. The circular smooth curve, labeled C subscript 2: x^2 + y^2 = 1, begins at the point the first line segment ends, goes up to the left, reaches a high point on the y axis, and ends at a point in the second quadrant that is symmetric to the point in the first quadrant. The second line segment, labeled C subscript 3: y = negative x, begins at the point the circular smooth curve ends, goes down to the right, and ends at the origin. The region inside the closed curve is shaded and labeled R.

FIGURE 9.12.6 Curve C in Example 3

SOLUTION

From (12) of Section 9.8 the work done by F is given by

W = F · dr = (–16y + sin x2) dx + (4ey + 3x2) dy

and so by Green’s theorem W = (6x + 16) dA. In view of the region R, the last integral is best handled in polar coordinates. Since R is defined by 0 ≤ r ≤ 1, π/4 ≤ θ ≤ 3π/4,

≡

EXAMPLE 4 Green’s Theorem Not Applicable

Let C be the closed curve consisting of the four straight line segments C1, C2, C3, C4 shown in FIGURE 9.12.7. Green’s theorem is not applicable to the line integral

since P, Q, ∂P/∂y, and ∂Q/∂x are not continuous at the origin. ≡

A graph. A closed curve, consisting of 4 line segments, is graphed on an x y coordinate plane. The first line segment, labeled C subscript 1: y = negative 2, begins at a point in the third quadrant, goes horizontally to the right, and ends at a point in the fourth quadrant that is symmetric to the point in the third quadrant. The second line segment, labeled C subscript 2: x = 2, begins at the point the first line segment ends, goes vertically up, and ends at a point in the first quadrant that is symmetric to the point in the fourth quadrant. The third line segment, labeled C subscript 3: y = 2, begins at the point the second line segment ends, goes horizontally to the left, and ends at a point in the second quadrant that is symmetric to the point in the first quadrant. The fourth line segment begins at the point the third line segment ends, goes vertically down, and ends at the point in the third quadrant. The region inside the closed curve is shaded and labeled R.

FIGURE 9.12.7 Curve C in Example 4

Region with Holes

Green’s theorem can also be extended to a region R with “holes,” that is, a region bounded between two or more piecewise-smooth simple closed curves. In FIGURE 9.12.8(a) we have shown a region R bounded by a curve C that consists of two simple closed curves C1 and C2; that is, C = C1 ∪ C2. The curve C is positively oriented, since if we traverse C1 in a counterclockwise direction and C2 in a clockwise direction, the region R is always to the left. If we now introduce crosscuts as shown in Figure 9.12.8(b), the region R is divided into two subregions, R1 and R2. By applying Green’s theorem to R1 and R2, we obtain

(4)

2 visual representations. Visual representation a. A closed curve labeled C subscript 2 is placed inside a closed curve labeled C subscript 1. 3 arrows placed on the curve labeled C subscript 1 are oriented counter clockwise. 2 arrows placed on the curve labeled C subscript 2 are orientred clockwise. The region inside the closed curve C subscript 1, excluding the region inside the closed curve C subscript 2, is shaded and labeled R. Visual representation b. A closed curve labeled C subscript 2 is placed inside a closed curve labeled C subscript 1. 3 arrows placed on the curve labeled C subscript 1 follow a counter clockwise direction. 2 arrows placed on the curve labeled C subscript 2 are oriented clockwise. 2 horizontal lines placed on either sides of the curve C subscript 2, connect the curve C subscript 2 to the curve C subscript 1. The region at the bottom bound by the curves C subscript 1 and C subscript 2, and the 2 horizontal lines, is shaded and labeled R subscript 1. The region at the top bound by the curves C subscript 1 and C subscript 2, and the 2 horizontal lines, is shaded in another color and labeled R subscript 2. 3 arrows placed along the boundary of the region R subscript 1, are oriented counter clockwise. 4 arrows placed along the boundary of the region R subscript 2, are oriented clockwise.

FIGURE 9.12.8 Boundary of R is C = C1 ∪ C2

The last result follows from the fact that the line integrals on the crosscuts (paths with opposite orientations) cancel each other. See (8) of Section 9.8.

EXAMPLE 5 Region with a Hole

Evaluate where C = C1 ∪ C2 is the boundary of the shaded region R shown in FIGURE 9.12.9.

A graph. A circle, labeled C subscript 2, x^2 + y^2 = 1, is graphed inside a square labeled C subscript 1. The centers of the square and the circle are both at the origin. 4 arrows placed on the closed curve labeled C subscript 1 are oriented counter clockwise. 2 arrows placed on the curve labeled C subscript 2 are oriented clockwise. The region inside the closed curve C subscript 1, excluding the region inside the closed curve C subscript 2, is shaded and labeled R.

FIGURE 9.12.9 Boundary C in Example 5

SOLUTION

Because P(x, y) = , Q(x, y) = , and the partial derivatives

are continuous on the region R bounded by C, it follows from (4) that

≡

As a consequence of the discussion preceding Example 5 we can establish a result for line integrals that enables us, under certain circumstances, to replace a complicated closed path with a path that is simpler. Suppose, as shown in FIGURE 9.12.10, that C1 and C2 are two nonintersecting piecewise-smooth simple closed paths that have the same counterclockwise orientation. Suppose further that P and Q have continuous first partial derivatives such that

in the region R bounded between C1 and C2. Then from (4) above and (8) of Section 9.8 we have

P dx + Q dy + P dx + Q dy = 0

or (5)

A closed curve labeled C subscript 2 is placed inside a closed curve labeled C subscript 1. 3 arrows placed on the curve labeled C subscript 1 are oriented counter clockwise. 3 arrows placed on the curve labeled C subscript 2 are oriented counter clockwise. The region inside the closed curve C subscript 1, is shaded and labeled R.

FIGURE 9.12.10 Curves C1 and C2 in (5)

EXAMPLE 6 Example 4 Revisited

Evaluate the line integral in Example 4.

SOLUTION

One method of evaluating the line integral is to write

and then evaluate the four integrals on the line segments C1, C2, C3, and C4. Alternatively, if we note that the circle C′: x2 + y2 = 1 lies entirely within C (see FIGURE 9.12.11), then from Example 5 it is apparent that P = –y/(x2 + y2) and Q = x/(x2 + y2) have continuous first partial derivatives in the region R bounded between C and C′. Moreover,

in R. Hence, it follows from (5) that

Using the parameterization x = cos t, y = sin t, 0 ≤ t ≤ 2π for C′ we obtain

(6)

≡

A graph. A circle, labeled C prime, is graphed inside a square labeled C. The centers of the square and the circle are both at the origin. 4 arrows placed on the closed curve labeled C are oriented counter clockwise. 2 arrows placed on the curve labeled C prime are oriented counter clockwise. The region inside the closed curve C is shaded and labeled R.

FIGURE 9.12.11 Curves C and C′ in Example 6

It is interesting to note that the result in (6),

is true for every piecewise-smooth simple closed curve C with the origin in its interior. We need only choose C′ to be x2 + y2 = a2, where a is small enough so that the circle lies entirely within C.

9.12 Exercises Answers to selected odd-numbered problems begin on page ANS-25.

In Problems 1–4, verify Green’s theorem by evaluating both integrals.

  1. (x – y) dx + xy dy = (y + 1) dA, where C is the triangle with vertices (0, 0), (1, 0), (1, 3)
  2. 3x2y dx + (x2 – 5y) dy = (2x – 3x2) dA, where C is the rectangle with vertices (–1, 0), (1, 0), (1, 1), (–1, 1)
  3. –y2 dx + x2 dy = R (2x + 2y) dA, where C is the circle x = 3 cos t, y = 3 sin t, 0 ≤ t ≤ 2π
  4. –2y2 dx + 4xy dy = 8y dA, where C is the boundary of the region in the first quadrant determined by the graphs of y = 0, y = , y = –x + 2

In Problems 5–14, use Green’s theorem to evaluate the given line integral.

  1. 2y dx + 5x dy, where C is the circle (x – 1)2 + (y + 3)2 = 25
  2. (x + y2) dx + (2x2 – y) dy, where C is the boundary of the region determined by the graphs of y = x2, y = 4
  3. (x4 – 2y3) dx + (2x3 – y4) dy, where C is the circle x2 + y2 = 4
  4. (x – 3y) dx + (4x + y) dy, where C is the rectangle with vertices (–2, 0), (3, 0), (3, 2), (–2, 2)
  5. 2xy dx + 3xy2 dy, where C is the triangle with vertices (1, 2), (2, 2), (2, 4)
  6. e2x sin 2y dx + e2x cos 2y dy, where C is the ellipse 9(x – 1)2 + 4(y – 3)2 = 36
  7. xy dx + x2 dy, where C is the boundary of the region determined by the graphs of x = 0, x2 + y2 = 1, x ≥ 0
  8. dx + 2 tan−1 x dy, where C is the triangle with vertices (0, 0), (0, 1), (–1, 1)
  9. y3 dx + (xy + xy2) dy, where C is the boundary of the region in the first quadrant determined by the graphs of y = 0, x = y2, x = 1 – y2
  10. xy2 dx + 3 cos y dy, where C is the boundary of the region in the first quadrant determined by the graphs of y = x2, y = x3

In Problems 15 and 16, evaluate the given integral on any piecewise-smooth simple closed curve C.

  1. ay dx + bx dy
  2. P(x) dx + Q(y) dy

In Problems 17 and 18, let R be the region bounded by a piecewise-smooth simple closed curve C. Prove the given result.

  1. x dy = – y dx = area of R
  2. –y dx + x dy = area of R

In Problems 19 and 20, use the results of Problems 17 and 18 to find the area of the region bounded by the given closed curve.

  1. The hypocycloid x = a cos3t, y = a sin3t, a > 0, 0 ≤ t ≤ 2π
  2. The ellipse x = a cos t, y = b sin t, a > 0, b > 0, 0 ≤ t ≤ 2π
    1. Show that
      –y dx + x dy = x1y2 – x2y1,
      where C is the line segment from the point (x1, y1) to (x2, y2).
    2. Use part (a) and Problem 18 to show that the area A of a polygon with vertices (x1, y1), (x2, y2), ..., (xn, yn), labeled counterclockwise, is
  4. Use part (b) of Problem 21 to find the area of the quadrilateral with vertices (–1, 3), (1, 1), (4, 2), and (3, 5).

In Problems 23 and 24, evaluate the given line integral where C = C1 ∪ C2 is the boundary of the shaded region R.

  1. (4x2 – y3) dx + (x3 + y2) dy
    A graph. A circle, labeled C subscript 2, x^2 + y^2 = 1, is graphed inside a circle labeled C subscript 1: x^2 + y^2 = 4. The centers of the circles are both at the origin. 4 arrows placed on the closed curve labeled C subscript 1 are oriented counter clockwise. 2 arrows placed on the closed curve labeled C subscript 2 are oriented clockwise. The region inside the closed curve C subscript 1, excluding the region inside the closed curve C subscript 2, is shaded and labeled R.

    FIGURE 9.12.12 Boundary C for Problem 23

  2. (cos x2 – y) dx +
    A graph. A vertical ellipse, labeled C subscript 2, 4 x^2 + y^2 = 16, is graphed inside a diamond labeled C subscript 1. The centers of the diamond and the ellipse are both at the origin. The vertices of the diamond are at the following points: (negative, 0), (0, negative 6), (6, 0), and (0, 6). 4 arrows placed on the closed curve labeled C subscript 1 follow a counter clockwise direction. The ellipse passes through the following points: (negative 2, 0), (0, 4), (2, 0), and (0, negative 2). 2 arrows placed on the closed curve labeled C subscript 2 are oriented clockwise. The region inside the closed curve C subscript 1, excluding the region inside the closed curve C subscript 2, is shaded and labeled R.

    FIGURE 9.12.13 Boundary C for Problem 24

In Problems 25 and 26, proceed as in Example 6 to evaluate the given line integral.

  1. , where C is the ellipse x2 + 4y2 = 4
  2. where C is the circle x2 + y2 = 16

In Problems 27 and 28, use Green’s theorem to evaluate the given double integral by means of a line integral. [Hint: Find appropriate functions P and Q.]

  1. x2 dA; R is the region bounded by the ellipse x2/9 + y2/4 = 1
  2. [1 – 2 (y – 1)] dA; R is the region in the first quadrant bounded by the circle x2 + (y – 1)2 = 1 and x = 0

In Problems 29 and 30, use Green’s theorem to find the work done by the given force F around the closed curve in FIGURE 9.12.14.

A graph. A closed curve, consisting of 2 smooth curves and 2 line segments, and a shaded region are graphed on an x y coordinate plane. The first smooth curve, labeled x^2 + y^2 = 1, begins on the y axis, goes down and to the right following a circular path, and ends on the x axis. The first line segment begins at the point the first smooth curve ends, goes horizontally to the right, and ends at a point on the x axis. The second smooth curve, labeled x^2 + y^2 = 4, begins at the point the first line segment ends, goes up to the left, following a circular path, and ends at a point on the y axis. The second line segment begins at the point the second smooth curve ends, goes vertically down, and ends at the point the first smooth curve begins. 4 arrows placed on the closed curve follow a counter clockwise direction. The region inside the closed curve is shaded.

FIGURE 9.12.14 Curve for Problems 29 and 30

  1. F = (x – y)i + (x + y)j
  2. F = –xy2i + x2yj
  3. Let P and Q be continuous and have continuous first partial derivatives in a simply connected region of the xy-plane. If P dx + Q dy is independent of the path, show that P dx + Q dy = 0 on every piecewise-smooth simple closed curve C in the region.
  4. Let R be a region bounded by a piecewise-smooth simple closed curve C. Show that the coordinates of the centroid of the region are given by
  5. Find the work done by the force F = –yi + xj acting along the cardioid r = 1 + cos θ.