GREEN’S THEOREM IN THE PLANE

9. GREEN’S THEOREM IN THE PLANE

The fundamental theorem of calculus says that the integral of the derivative of a function is the function, or more precisely:

f (t) dt = f (b) − f(a).

a dt

We are going to consider some useful generalizations of this theorem to two and three dimensions. The divergence theorem and Stokes’ theorem (Sections 10 and

11) are very important in electrodynamics and other applications; in this section we will find two-dimensional forms of these theorems. First we develop an underlying useful theorem relating an area integral to the line integral around its boundary (see applications in examples and problems and also Chapter 14, Section 3).

Figure 9.1

Recall that we know how to evaluate line integrals (Section 8), and that we learned in Chapter 5 to evaluate double integrals over areas in the (x, y) plane. We are going to consider areas (such as those in Figure 9.1 or in Chapter 5, Figure 2.7) for which we can evaluate the double integral over the area either with respect to x first or with respect to y first. Look at Figure 9.1. We want to find a relation between a double integral over the area A and a line integral around the curve C, for simple closed curves C. (A simple curve does not cross itself; for example, it is not

a figure 8.) Now in Figure 9.1, the upper part of C between points 1 and 2 is given by an equation y = y u (x) and the lower part by an equation y = y √ l (x). (Think √ of solving the equation of a circle for y u (x) = 1−x 2 and y l (x) = − 1−x 2 .) Similarly in Figure 9.1, we can find x l (y) and x r (y) for the left and right parts of

C between points 3 and 4. Let P (x, y) and Q(x, y) be continuous functions with continuous first derivatives.

We are going to show that the double integral of ∂P (x, y)/∂y over the area A is equal to the line integral of P around C. We write the double integral using Figure 9.1 to integrate first with respect to y, and do the y integration by equation (9.1) with t = y to get:

∂P (x, y)

∂P (x, y)

(9.2) dy dx =

[P (x, y u ) − P (x, y l )] dx ∂y

dx

dy =

∂y

P (x, y l ) dx −

P (x, y u ) dx.

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Now we have our answer—we just have to recognize it! Think how you would evaluate the line integral of P (x, y)dx along the lower part of C in Figure 9.1 from point 1 to point 2. You would substitute y = y l (x) into P (x, y) and integrate from x = a to b (see Section 8).

b (9.3)

P (x, y l ) dx = line integral of P dx

a along lower part of C from point 1 to point 2.

This is one of the terms in (9.2). Similarly, to find the line integral of P (x, y) dx along the upper part of C from point 2 to point 1, we substitute y = y u (x) and integrate from b to a.

a (9.4)

P (x, y u ) dx = line integral of P dx

b along upper part of C from point 2 to point 1.

Combining (9.3) and (9.4) gives us the line integral all the way around C in the counterclockwise direction, that is, so that A is always on our left as we go around

C. (The symbol means an integral around a closed curve back to the starting point.) Then, from (9.2), we have

Repeating the calculation but integrating first with respect to x, we find

∂Q d ∂Q (9.6)

[Q(x r , y) − Q(x l , y)]dy ∂x

Adding (9.5) and (9.6) and using the notation ∂A to mean the boundary of A (that is, C) we have

Green’s theorem in the plane: ∂P

The line integral is counterclockwise around the boundary of area A. Using Green’s theorem we can evaluate either a line integral around a closed path

or a double integral over the area inclosed, whichever is easier to do. If the area is not of the simple type we have assumed, it may be possible to cut it into pieces (see Figure 9.2) so that our proof applies to each piece. Then the line integrals along the dotted cuts in Figure 9.2 are in opposite directions for adjacent pieces, and so

Section 9 Green’s Theorem in the Plane 311

cancel. Thus the theorem is valid for this more general area and its inclosing curve. In fact, w can even close up Figure 9.2 creating an area with a hole in the middle. e

Figure 9.2

Green’s theorem still holds, but now the line integral consists of a counterclockwise integral around the outside plus a clockwise integral around the hole as you can see in Figure 9.2. We say that this area is not ”simply connected”—see further discussion of this in Section 11.

Example 1. In Example 1, Section 8, we found the line integral (8.2) along several paths (Figure 8.2). Sup- pose we want the line integral in Figure 8.2 around the closed loop (Figure 9.3) from (0, 0) to (2, 1) and back as shown. From Section 8, Example 1, this is the work done along path 2 minus the work done along path 3 (since we are now going in the opposite

Figure 9.3 direction); we find W 2 −W 3 = 2 3 5 − 3 = −1. Let us evaluate this using Green’s

theorem. From (8.2) and (9.7) we have

(xy) dx dy

as before. Example 2. In Section 8, we discussed conservative forces for which work done is inde-

pendent of the path. By Green’s theorem (9.7), the work done by a force F around

a closed path in the (x, y) plane is

If (∂F y /∂x) − (∂F x /∂y) = 0 (note that this is the z component of curl F = 0) then W around any closed path is zero, which means that the work from one point to another is independent of the path (also see Section 11).

The functions P (x, y) and Q(x, y) in (9.7) are arbitrary; we may choose them to suit our purposes. Note that a two-dimensional vector function iV x (x, y) + jV y (x, y)

312 Vector Analysis Chapter 6

contains two functions, V x and V y . In the next two examples, we are going to define P and Q in terms of V x and V y in order to obtain two useful results.

Example 3. We define: (9.8)

Q=V x , P = −V y , where V = iV x + jV y . Then

by (7.2) with V z = 0. Along the curve bounding an area A (Figure 9.4) the vector

Figure 9.4

dr = i dx + j dy (tangent)

is a tangent vector, and the vector

(9.11) n ds = i dy − j dx (outward normal), where n is a unit vector and ds =

2 + dy 2 ,

is a normal vector (perpendicular to the tangent) pointing out of area A. Using (9.11) and (9.8), we can write

(9.12) P dx + Q dy = −V y dx + V x dy = (iV x + jV y ) · (i dy − j x)

= V · n ds.

Then substitute (9.9) and (9.12) into (9.7) to get

This is the divergence theorem in two dimensions. It can be extended to three dimensions (also see Section 10). Let τ represent a volume; then ∂τ (read boundary of τ ) means the closed surface area of τ . Let dτ mean a volume element and let dσ

Section 9 Green’s Theorem in the Plane 313

mean an element of surface area. At each point of the surface, let n be a unit vector perpendicular to the surface and pointing outward. Then the divergence theorem in three dimensions says (also see Section 10)

div V dτ =

V · n dσ. Divergence theorem

Example 4. To see another application of (9.7) to vector functions, we let (9.15)

Q=V y , P=V x , where V = iV x + jV y . Then

= (curl V) · k

by (7.3) with V z = 0. Equations (9.10) and (9.15) give (9.17)

P dx + Q dy = (iV x + jV y ) · (i dx + j dy) = V · dr. Substituting (9.16) and (9.17) into (9.7), we get

(curl V) · k dx dy =

V · dr.

∂A A

This is Stokes’ theorem in two dimensions. It can be extended to three dimensions (Section 11). Let σ be an open surface (for example, a hemisphere); then ∂σ means the curve bounding the surface (Figure 9.5). Let n be a unit vector normal to the surface. Then Stokes’ theorem in three dimensions is (also see Section 11)

Figure 9.5

(curl V) · n dσ =

V · dr. Stokes’ theorem.

The direction of integration for the line integral is as shown in Figure 9.5 (see also Section 11).

PROBLEMS, SECTION 9

1. Write out the equations corresponding to (9.3) and (9.4) for R Q dy between points 3 and 4 in Figure 9.2, and add them to get (9.6).

In Problems 2 to 5 use Green’s theorem [formula (9.7)] to evaluate the given integrals. 2. H 2x dy − 3y dx around the square with vertices (0, 2), (2, 0), (−2, 0), and (0, −2).

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3. H C xy dx + x 2 dy, where C is as sketched.

4. R e x

C cos y dx −e sin y dy, where C is the broken line from A = (ln 2, 0) to D = (0, 1) and then from D to B =

(− ln 2, 0). Hint: Apply Green’s theorem to the integral around the closed curve ADBA.

5. R C (ye x −1) dx+e x dy, where C is the semicircle through (0, −10), (10, 0), and (0, 10). (Compare Problem 4.)

6. For a simple closed curve C in the plane show by Green’s theorem that the area inclosed is

A=

2 C (x dy − y dx).

7. Use Problem 6 to show that the area inside the ellipse x = a cos θ, y = b sin θ, 0 ≤ θ ≤ 2π, is A = πab.

8. Use Problem 6 to find the area inside the curve x 2/3 +y 2/3 = 4. 9. Apply Green’s theorem with P = 0, Q = 1 x 2 2 to the triangle with vertices (0, 0),

(0, 3), (3, 0). You will then have RR x dx dy over the triangle expressed as a very simple line integral. Use this to locate the centroid of the triangle. (Compare Chapter 5, Section 3.)

Evaluate each of the following integrals in the easiest way you can. 10. H (2y dx − 3x dy) around the square bounded by x = 3, x = 5, y = 1 and y = 3.

C (x sin x − y) dx + (x − y ) dy, where C is the triangle in the (x, y) plane with vertices (0, 0), (1, 1), and (2, 0).

11. R 2

√ −x ) dx + (2xy + 3) dy along the x axis from (0, 0) to ( √

5, 0) and then along a circular arc from (

12. R (y 2 2

5, 0) to (1, 2).