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**16.4** Green’s Theorem

### Green’s Theorem

Green’s Theorem gives the relationship between a line
*integral around a simple closed curve C and a double *
*integral over the plane region D bounded by C. (See *

*Figure 1. We assume that D consists of all points inside C *
*as well as all points on C.)*

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### Green’s Theorem

In stating Green’s Theorem we use the convention that the
**positive orientation of a simple closed curve C refers to a ***single counterclockwise traversal of C. Thus if C is given by *
**the vector function r(t), a***≤ t ≤ b, then the region D is *

**always on the left as the point r(t) traverses C.**

(See Figure 2.)

**Figure 2**

(a) Positive orientation (b) Negative orientation

### Green’s Theorem

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### Example 1

Evaluate

### ∫

_{C}*x*

^{4}

*dx + xy dy, where C is the triangular curve*consisting of the line segments from (0, 0) to (1, 0), from (1, 0) to (0, 1), and from (0, 1) to (0, 0).

Solution:

Although the given line integral could be evaluated as usual by the methods, that would involve setting up three

separate integrals along the three sides of the triangle, so let’s use Green’s Theorem instead.

*Notice that the region D enclosed *
*by C is simple and C has positive *
orientation (see Figure 4).

**Figure 4**

*Example 1 – Solution*

*If we let P(x, y) = x*^{4} *and Q(x, y) = xy, then we have*

cont’d

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### Green’s Theorem

In Example 1 we found that the double integral was easier to evaluate than the line integral.

But sometimes it’s easier to evaluate the line integral, and Green’s Theorem is used in the reverse direction.

*For instance, if it is known that P(x, y) = Q(x, y) = 0 on the *
*curve C, then Green’s Theorem gives*

*no matter what values P and Q assume in the region D.*

### Green’s Theorem

Another application of the reverse direction of Green’s
*Theorem is in computing areas. Since the area of D is *

### ∫∫

_{D}*1 dA, we wish to choose P and Q so that*

There are several possibilities:

*P(x, y) = 0 * *P(x, y) = –y* *P(x, y) = y*
*Q(x, y) = x* *Q(x, y) = 0* *Q(x, y) = x*

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### Green’s Theorem

Then Green’s Theorem gives the following formulas for the
*area of D:*

### Green’s Theorem

Formula 5 can be used to explain how planimeters work.

**A planimeter is a mechanical instrument used for **

measuring the area of a region by tracing its boundary curve.

These devices are useful in all the sciences: in biology for measuring the area of leaves or wings, in medicine for

measuring the size of cross-sections of organs or tumors, in forestry for estimating the size of forested regions from photographs.

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### Green’s Theorem

Figure 5 shows the operation of a polar planimeter: the pole is fixed and, as the tracer is moved along the

boundary curve of the region, the wheel partly slides and partly rolls perpendicular to the tracer arm.

The planimeter measures the distance that the wheel rolls and this is proportional to the area of the enclosed region.

A Keuffel and Esser polar planimeter

**Figure 5**

### Extended Versions of

### Green’s Theorem

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### Extended Versions of Green’s Theorem

Although we have proved Green’s Theorem only for the
*case where D is simple, we can now extend it to the case *
*where D is a finite union of simple regions. *

*For example, if D is the region shown in Figure 6, then we *
*can write D = D*_{1} *U D*_{2}*, where D*_{1} *and D*_{2} are both simple.

**Figure 6**

### Extended Versions of Green’s Theorem

*The boundary of D*_{1} *is C*_{1} *U C*_{3} *and the boundary of D*_{2} is
*C*_{2} *U (–C*_{3}*) so, applying Green’s Theorem to D*_{1} *and D*_{2}
separately, we get

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### Extended Versions of Green’s Theorem

If we add these two equations, the line integrals along
*C*_{3} *and –C*_{3} cancel, so we get

*which is Green’s Theorem for D = D*_{1} *U D*_{2}, since its
*boundary is C = C*_{1} *U C*_{2}.

The same sort of argument allows us to establish Green’s Theorem for any finite union of nonoverlapping simple regions (see Figure 7).

**Figure 7**

### Example 4

Evaluate *, where C is the boundary of the *
*semiannular region D in the upper half-plane between the *
*circles x*^{2} *+ y*^{2} *= 1 and x*^{2} *+ y*^{2} = 4.

Solution:

*Notice that although D is not simple, the y-axis divides it *
into two simple regions (see Figure 8).

In polar coordinates we can write
*D = {(r, *θ) | 1 ≤ r ≤ 2, 0 ≤ θ ≤ π}

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*Example 4 – Solution*

Therefore Green’s Theorem gives

cont’d

### Extended Versions of Green’s Theorem

Green’s Theorem can be extended to apply to regions with holes, that is, regions that are not simply-connected.

*Observe that the boundary C of the region D in Figure 9 *
*consists of two simple closed curves C*_{1} *and C*_{2}.

We assume that these boundary curves are oriented so
*that the region D is always on the left as the curve C is *
traversed.

Thus the positive direction is counterclockwise for the outer

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### Extended Versions of Green’s Theorem

*If we divide D into two regions *
*D′ and D′′ by means of the lines *
shown in Figure 10 and then apply
*Green’s Theorem to each of D*′ and
*D*′′, we get

**Figure 10**

### Extended Versions of Green’s Theorem

Since the line integrals along the common boundary lines are in opposite directions, they cancel and we get

*which is Green’s Theorem for the region D.*