greens theorem proof

7/28/2019 Greens Theorem Proof

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Theory Supplement Section L 63

L PROOFS OF LINE INTEGRAL THEOREMS

In this section we give proofs of the Fundamental Theorem of Calculus for Line Integrals and

Greens Theorem.

Proof of the Fundamental Theorem of Calculus for Line Integrals

Suppose that C is a curve from point P to point Q. Then the Fundamental Theorem of Calculus for

Line Integrals says C

grad f dr = f(Q) f(P).

The Fundamental Theorem for Line Integrals can be derived from the Fundamental Theorem for

ordinary definite integrals.

Suppose that (x(t), y(t)), for a t b, is a parameterization of C, with endpoints P =(x(a), y(a)) and Q = (x(b), y(b)). Thus, the values of f along C are given by the single variablefunction h(t) = f(x(t), y(t)).

Using this parameterization ofC we have

C

grad f dr =

ba

(fx(x(t), y(t))i + fy(x(t), y(t))j ) (x(t)i + y(t)j )dt

=

ba

(fx(x(t), y(t))x(t) + fy(x(t), y(t))y

(t))dt.

By the chain rule

dh

dt

=f

x

dx

dt

+f

y

dy

dt

= fxx(t) + fyy

(t).

By the Fundamental Theorem of Calculus, this gives us

C

grad f dr =

ba

h(t)dt

= h(b) h(a) = f(Q) f(P).

Proof of Greens Theorem

In this section we will give a proof of Greens Theorem based on the change of variables formula

for double integrals. Suppose that C is a simple closed curve surrounding a region R in the planeand oriented so that the region is on the left as we move around the curve. Assume the vector fieldF , defined at every point ofR, is given in components by

F (x, y) = F1(x, y)i + F2(x, y)j ,

where F1 and F2 are continuously differentiable. We wish to show that

C

F dr =

R

F2

x

F1

y

dxdy.


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64 Theory Supplement Section L

a b

c

d

C1

C4 C2

C3

R

x

y

Figure L.46: A rectangular region R with boundary C brokeninto C1, C2, C3, and C4

Proof for Rectangles

We prove Greens Theorem first when R is a rectangular region, as shown in Figure L.46. The

line integral in Greens theorem can be written as

C

F dr = C1

F dr + C2

F dr + C3

F dr + C4

F dr

=

ba

F1(x, c) dx +

dc

F2(b, y) dy

ba

F1(x, d) dx

dc

F2(a, y) dy

=

dc

(F2(b, y) F2(a, y)) dy +

ba

(F1(x, d) + F1(x, c)) dx.

On the other hand, the double integral in Greens theorem can be written as an iterated integral.

We evaluate the inside integral using the Fundamental Theorem of Calculus.R

F2

x

F1

y

dxdy =

R

F2

xdxdy +

R

F1

ydxdy

=dc

ba

F2

x dxdy +ba

dc

F1

y dydx

=

dc

(F2(b, y) F2(a, y)) dy +

ba

(F1(x, d) + F1(x, c)) dx.

Since the line integral and the double integral are equal, we have proved Greens theorem for rect-

angles.

Proof for Regions Parameterized by Rectangles

a b

c

d

D1

D4 D2

D3

T

s

t

C4

C3

C2

C1

R

x

y

Figure L.47: A curved region R in the xy-plane corresponding to a rectangular region T in the st-plane

Now we prove Greens Theorem for a region R which can be transformed into a rectangular

region. Suppose we have a smooth change of coordinates

x = x(s, t), y = y(s, t).


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Consider a curved region R in the xy-plane corresponding to a rectangular region T in the st-

plane, as in Figure L.47. We suppose that the change of coordinates is one-to-one on the interior of

T.

We prove Greens theorem for R using Greens theorem for T and the change of variables

formula for double integrals given on page 823. First we express the line integral around CC

F dr ,

as a line integral in the st-plane around the rectangle D = D1 + D2 + D3 + D4. In vector notation,the change of coordinates is

r = r (s, t) = x(s, t)i + y(s, t)j

and so

F dr = F (r (s, t)) r

sds + F (r (s, t))

r

tdt.

We define a vector fieldG on the st-plane with components

G1 = F r

sand G2 = F

r

t.

Then, ifu is the position vector of a point in the st-plane, we have F dr = G1 ds+G2 dt = G du .Problem 5 at the end of this section asks you to show that the formula for line integrals along

parameterized paths leads to the following result:

C

F dr =

D

G du .

In addition, using the product rule and chain rule we can show that

G2

s

G1

t=

F2

x

F1

y

xs ysxt

yt

.

(See Problem 6 at the end of this section.) Hence, by the change of variables formula for double

integrals on page 823,

R

F2

x

F1

y

dxdy =

T

F2

x

F1

y

xs

ys

xt

yt

dsdt =T

G2

s

G1

t

ds dt.

Thus we have shown that C

F dr =

D

G du

and that R

F2

x

F1

y

dxdy =

T

G2

s

G1

t

ds dt.

The integrals on the right are equal, by Greens Theorem for rectangles; hence the integrals on the

left are equal as well, which is Greens Theorem for the region R.

Pasting Regions Together

Lastly we show that Greens Theorem holds for a region formed by pasting together regions

which can be transformed into rectangles. Figure L.48 shows two regions R1 and R2 that fit together


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66 Theory Supplement Section L

C1R

C2

C

CR1

R2

x

y

Figure L.48: Two regions R1 and R2pasted together to form a region R

to form a region R. We break the boundary of R into C1, the part shared with R1, and C2, the part

shared with R2. We let C be the part of the boundary of R1 which it shares with R2. So

Boundary ofR = C1 + C2, Boundary ofR1 = C1 + C, Boundary ofR2 = C2 + (C).

Note that when the curve C is considered as part of the boundary of R2, it receives the opposite

orientation from the one it receives as the boundary of R1. Thus

Boundary ofR1

F dr +

Boundary ofR2

F dr =

C1+C

F dr +

C2+(C)

F dr

=

C1

F dr +

C

F dr +

C2

F dr

C

F dr

=

C1

F dr +

C2

F dr

=

Boundary ofR

F dr .

So, applying Greens Theorem for R1 and R2, we get

R

F2

x

F1

y

dxdy =

R1

F2

x

F1

y

dxdy +

R2

F2

x

F1

y

dxdy

=

Boundary ofR1

F dr +

Boundary ofR2

F dr

=

Boundary ofR

F dr ,

which is Greens Theorem for R. Thus, we have proved Greens Theorem for any region formed by

pasting together regions that are smoothly parameterized by rectangles.

Example1 Let R be the annulus (ring) centered at the origin with inner radius 1 and outer radius 2. Using polar

coordinates, show that the proof of Greens Theorem applies to R. See Figure L.49.

Solution In polar coordinates, x = r cos t and y = r sin t, the annulus corresponds to the rectangle in thert-plane 1 r 2, 0 t 2. The sides t = 0 and t = 2 are pasted together in the xy-planealong the x-axis; the other two sides become the inner and outer circles of the annulus. Thus R is

formed by pasting the ends of a rectangle together.


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1 2

2

r

t

1 2x

y

t = 0t = 2

r = 2

r=1

Figure L.49: The annulus R in the xy-plane and the corresponding rectangle 1 r 2, 0 t 2in the rt-plane

Problems for Section L

1. Let R be the annulus centered at (1, 2) with inner ra-dius 2 and outer radius 3. Show that R can be parameter-ized by a rectangle.

2. Let R be the region under the first arc of the graph of

the sine function. Show that R can be parameterized by

a rectangle.

3. Let f(x) and g(x) be two smooth functions, and supposethat f(x) g(x) for a x b. Let R be the regionf(x) y g(x), a x b.

(a) Sketch an example of such a region.(b) For a constant x0, parameterize the vertical line seg-

ment in R where x = x0. Choose your parameter-ization so that the parameter starts at 0 and ends at

1.

(c) By putting together the parameterizations in part (b)

for different values of x0, show that R can be pa-

rameterized by a rectangle.

4. Let f(y) and g(y) be two smooth functions, and supposethat f(y) g(y) for c y d. Let R be the regionf(y) x g(y), c y d.

(a) Sketch an example of such a region.

(b) For a constant y0, parameterize the horizontal line

segment in R where y = y0. Choose your parame-terization so that the parameter starts at 0 and ends

at 1.

(c) By putting together the parameterizations in part (b)

for different values ofy0, show that R can be param-

eterized by a rectangle.

5. Use the formula for calculating line integrals by parame-

terization to prove the statement on page 65:C

F dr =

D

G du .

6. Use the product rule and the chain rule to prove the for-

mula on page 65:

G2

s

G1

t=

F2

x

F1

y

xs

y

sxt

y

t

.

greens theorem proof

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