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4.4: Green's Theorem

( \newcommand{\kernel}{\mathrm{null}\,}\)

Green’s theorem is an example from a family of theorems which connect line integrals (and their higher-dimensional analogues) with the definite integrals we studied in Section 3.6. We will first look at Green’s theorem for rectangles, and then generalize to more complex curves and regions in R2.

Green’s theorem for rectangles

Suppose F:R2R2 is C1 on an open set containing the closed rectangle

D=[a,b]×[c,d],

and let F1 and F2 be the coordinate functions of F. If C denotes the boundary of D, oriented in the clockwise direction, then we may decompose C into the four curves C1, C2, C3, and C4 shown in Figure 4.4.1.

Screen Shot 2021-08-23 at 10.18.32.png
Figure 4.4.1: The boundary of a rectangle decomposed into four smooth curves

Then

α(t)=(t,c),

atb, is a smooth parametrization of C1,

β(t)=(b,t),

ctd, is a smooth parametrization of C2,

γ(t)=(t,d),

atb, is a smooth parametrization of C3, and

δ(t)=(a,t),

ctd, is a smooth parametrization of C4. Now

CFds=C1Fds+C2Fds+C3Fds+C4Fds=C1Fds+C2FdsC3FdsC4Fds

and

C1Fds=ba((F1(t,c),F2(t,c))(1,0)dt=baF1(t,c)dt,,

C2Fds=dc((F1(b,t),F2(b,t))(0,1)dt=ccF2(b,t)dt,

C3Fds=ba((F1(t,d),F2(t,d))(1,0)dt=baF1(t,d)dt,

and

C4Fds=dc((F1(a,t),F2(a,t))(0,1)dt=ccF2(a,t)dt,

Hence, inserting (???) through (???) into (4.4.6),

CFds=baF1(t,c)dt+dcF2(b,t)dtbaF1(t,d)dtdcF2(a,t)dt=dc(F2(b,t)F2(a,t))dtba(F1(t,d)F1(t,c))dt.

Now, by the Fundamental Theorem of Calculus, for a fixed value of t,

baxF2(x,t)dx=F2(b,t)F2(a,t)

and

dcyF1(t,y)dy=F1(t,d)F1(t,c).

Thus, combining (???) and (???) with (4.4.12), we have

CFds=dcbaxF2(x,t)dxdtbadcyF1(t,y)dydt=dcbaxF2(x,y)dxdybadcyF1(x,y)dydx=dcba(xF2(x,y)yF1(x,y))dxdy

If we let p=F1(x,y),q=F2(x,y), and D=C (a common notation for the boundary of D), then we may rewrite (4.4.16) as

Dpdx+qdy=D(qxpy)dxdy.

This is Green’s theorem for a rectangle.

Example 4.4.1

If D=[1,3]×[2,5], then

Dxydx+xdy=D(xxyxy)dxdy=3152(1x)dydx=313(1x)dx=3x|3132x2|31=6.

Clearly, this is simpler than evaluating the line integral directly.

Green’s theorem for regions of Type III

Green’s theorem holds for more general regions than rectangles. We will confine ourselves here to discussing regions known as regions of Type III, but it is not hard to generalize to regions which may be subdivided into regions of this type (for an example, see Problem 12). Recall from Section 3.6 that we say a region D in R2 is of Type I if there exist real numbers a<b and continuous functions α:RR and β:RR such that

D={(x,y):axb,α(x)yβ(x)}.

We say a region D in R2 is of Type II if there exist real numbers c and d and continuous functions γ:RR and δ:RR such that

D={(x,y):cyd,γ(y)xδ(y)}.

Definition 4.4.1

We call a region D in R2 which is both of Type I and of Type II a region of Type III.

Example 4.4.2

In Section 3.6, we saw that the triangle T with vertices at (0,0), (1,0), and (1,1) and the closed disk

D=ˉB2((0,0),1)={(x,y):x2+y21}

are of both Type I and Type II. Thus T and D are regions of Type III. We also saw that the region E beneath the graph of y=x2 and above the interval [−1,1] is of Type I, but not of Type II. Hence E is not of Type III.

Example 4.4.3

Any closed rectangle in R2 is a region of Type III, as is any closed region bounded by an ellipse.

Now suppose D is a region of Type III and D is the boundary of D, that is, the curve enclosing D, oriented counterclockwise. Let F:R2R2 be a C1 vector field, with coordinate functions p=F1(x,y) and q=F2(x,y). We will first prove that

Dpdx=Dpydxdy.

Since D is, in particular, a region of Type I, there exist continuous functions α and β such that

D={(x,y):axb,α(x)yβ(x)}.

In addition, we will assume that α and β are both differentiable (without this assumption the line integral of F along D would not be defined). As with the rectangle in the previous proof, we may decompose D into four curves, C1, C2, C3, and C4, as shown in Figure 4.4.2.

Screen Shot 2021-08-23 at 10.37.30.png
Figure 4.4.2: Decomposing the boundary of a region of Type I

Then

φ1(t)=(t,α(t)),

atb, is a smooth parametrization of C1,

φ2(t)=(b,t),

α(b)tβ(b), is a smooth parametrization of C2,

φ3(t)=(t,β(t)),

atb, is a smooth parametrization of C3, and

φ4(t)=(a,t),

α(a)tβ(a), is a smooth parametrization of C4. Now

Dpdx=C1pdx+C2pdxC3pdxC4pdx,

where

C1pdx=ba(F1(t,α(t)),0)(1,α(t))dt=baF1(t,α(t))dt,

C2pdx=β(b)α(b)(F1(b,t),0)(0,1)dt=β(b)α(b)0dt=0,

C3pdx=ba(F1(t,β(t)),0)(1,β(t))dt=baF1(t,β(t))dt,

and

C4pdx=β(a)α(a)(F1(a,t),0)(0,1)dt=β(a)α(a)0dt=0.

Hence

Dpdx=baF1(t,α(t))dtbaF1(t,β(t))dt=ba(F1(t,β(t))F1(t,α(t)))dt

Now, by the Fundamental Theorem of Calculus,

β(t)α(t)yF1(t,y)dy=F1(t,β(t))F1(t,α(t)),

and so

Dpdx=baβ(t)α(t)yF1(t,y)dydt=baβ(x)α(x)yF1(x,y)dydx=Dpydxdy.

A similar calculation, treating D as a region of Type II, shows that

Dqdy=Dqxdxdy.

(You are asked to verify this in Exercise 7.) Putting (4.4.37) and (???) together, we have

DFds=Dpdx+qdy=Dpydxdy+Dqxdxdy=D(qxpy)dxdy

Green's Theorem 4.4.1

Suppose D is a region of Type III, D is the boundary of D with counterclockwise orientation, and the curves describing D are differentiable. Let F:R2R2 be a C1 vector field, with coordinate functions p=F1(x,y) and q=F2(x,y). Then

Dpdx+qdy=D(qxpy)dxdy.

Example 4.4.4

Let D be the region bounded by the triangle with vertices at (0,0), (2,0), and (0,3), as shown in Figure 4.4.3.

Screen Shot 2021-08-24 at 08.11.35.png
Figure 4.4.3: A triangle with counterclockwise orientation

If we orient D in the counterclockwise direction, then

D(3x2+y)dx+5xdy=D(x(5x)y(3x2+y))dxdy=D(51)dxdy=4Ddxdy=(4)(3)=12,

where we have used the fact that the area of D is 3 to evaluate the double integral.

The line integral in the previous example reduced to finding the area of the region D. This can be exploited in the reverse direction to compute the area of a region. For example, given a region D with area A and boundary D, it follows from Green’s theorem that

A=Ddxdy=Dpdx+qdy

for any choice of p and q which have the property that

qxpy=1.

For example, letting p=0 and q=x, we have

A=Dxdy

and, letting p=y and q=0, we have

A=Dydx.

The next example illustrates using the average of (???) and (???) to find A:

A=12(DxdyDydx)=12Dxdyydx.

Example 4.4.5

Let A be the area of the region D bounded by the ellipse with equation

x2a2+y2b2=1,

where a>0 and b>0, as shown in Figure 4.4.4.

Screen Shot 2021-08-24 at 08.12.37.png
Figure 4.4.4 The ellipse x2a2+y2b2=1 with counterclockwise orientation

Since we may parametrize D, with counterclockwise orientation, by

φ(t)=(acos(t),bsin(t)),

0t2π, we have

A=12Dxdyydx=122π0(bsin(t),acos(t))(asin(t),bcos(t)dt=122π0(absin2(t)+abcos2(t))dt=ab22π0dt=(ab2)(2π)=πab.


This page titled 4.4: Green's Theorem is shared under a CC BY-NC-SA 1.0 license and was authored, remixed, and/or curated by Dan Sloughter via source content that was edited to the style and standards of the LibreTexts platform.

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