5.2 Green’s Theorem (Classical Version)

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Theorem 5.2.1. Green’s Theorem Let U _ R2 be connected and simply

connected (has no holes in it), and has boundary a simple closed curve, that

is a loop which does not intersect itself, say `.

Let V be a smooth vector field

V

_

x

y

_

=

_

P(x, y)

Q(x, y)

_

defined on a region which is open and contains U and its boundary loop.

58 CHAPTER 5. GREEN’S THEOREM

Figure 5.8: Adding paths around four sub-squares

Then Z

`

V =

ZZ

U

_

@Q

@x

−

@P

@y

_

where the loop ` is traversed in the positive (anticlockwise) sense.

“Proof” I shall prove it for the particular case where U is a square, figure 5.8.

If ABCD is a square, Q is the midpoint of AB, M of BC, N of CD, and P

of DA, and if E is the centre of the square, then

Z

ABCD

V =

Z

AQEP

V +

Z

QBME

V +

Z

MCNE

V +

Z

NDPE

V

This is trivial, since we get the integral around each subsquare by adding up

the integral around each edge; the inner lines are traversed twice in opposite

directions and so cancel out.

We can continue subdividing the squares as finely as we like, and the sum

of the path integral around all the little squares is still going to be the path

integral around the big one.

But the path integral around a very small square can be approximated by

_

@Q

@x

−

@P

@y

_

5.2. GREEN’S THEOREM (CLASSICAL VERSION) 59

evaluated at the centre of the square and multiplied by its area, as we saw in

the last section. And the limit of this sum is precisely the definition of the

Riemann integral of

@Q

@x

−

@P

@y

over the region enclosed by the square. _

Remark 5.2.1. To do it for more general regions we might hope the boundary

is reasonable and fill it with squares. This is not terribly convincing, but

we can reason that other regions also have path integral over the boundary

approximated by

_

@Q

@x

−

@P

@y

_

× (area enclosed by shape)

Remark 5.2.2. The result for a larger collection of shapes will be proved

later.

Exercise 5.2.1. Try to prove it for a triangular region, say a right-angled

triangle, by chopping the triangle up into smaller triangles.

Example 5.2.1. of Green’s Theorem in use: Evaluate

R

c sin(x3)dx +

xy+6 dy where c is the triangular path starting at i, going by a straight line

to j, then down to the origin, then back to i.

Solution: It is not too hard to do this the long way, but Green’s Theorem

tells us that the result is the same as

ZZ

U

_

@Q

@x

−

@P

@y

_

dx dy

where U is the inside of the triangle, P(x, y) = sin(x3) and Q(x, y) = xy + 6

This gives Z 1

0

Z 1−x

0

(y − 0) dy dx

which I leave you to verify is 1/6.

Remark 5.2.3. While we are talking about integrals around simple loops,

there is some old fashioned notation for such integrals, they often used to

write I

c

f(t) dt

The little circle in the integral told you c was supposed to be a simple loop.

Sometimes they had an arrow on the circle.

60 CHAPTER 5. GREEN’S THEOREM

The only known use for these signs is to impress first year students with how

clever you are, and it doesn’t work too well.

Example 5.2.2. Find:

I

S1

(loge(x6 + 152) + 17y) dx + (

p

1 + y58 + x) dy

You must admit this looks horrible regarded as a path integral. It is easily

seen however to be Z

D2

(1 − 17)

this is −16 times the area of the unit disc which is of course _ So we get

−16_

Remark 5.2.4. So Green’s Theorem can be used to scare the pants off

people who have just learnt to do line integrals. This is obviously extremely

useful.

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