6 Green’s Identities (3D)
Like Gauss’ theorem, Green’s identities equate surface integrals to volume integrals. However, Green’s identities are concerned with two scalar fields and . Two statements of Green’s identities are as follows
[1]
and
[2]
6.1 Proof of Green’s identities
Green’s identities can be derived from Gauss’ theorem and a vector derivative identity.
Vector identity (1) from subsection 6 of 28.2 states that
.
Letting
and
,
Gauss’ theorem states
Now, letting ,
This is Green’s identity [1]. Reversing the roles of and ,
Subtracting the last two equations yields Green’s identity [2].
Example 37
Verify Green’s first identity for , and the unit cube, , , .
Solution
As , . Thus and the surface integral is of this quantity (scalar product with ) integrated over the surface of the unit cube.
On the three faces , , , the vector and so the contribution to the surface integral is zero.
On the face , and so and the contribution to the integral is
.
On the face , and so and the contribution to the integral is zero.
On the face , and so and the contribution to the integral is
.
Thus, .
Now evaluate .
Note that and so
and the integral
Hence
6.2 Green’s theorem in the plane
This states that
is a - surface with perimeter ; and are scalar functions.
This should not be confused with Green’s identities.
6.3 Justification of Green’s theorem in the plane
Green’s theorem in the plane can be derived from Stokes’ theorem.
Now let
be the vector field
i.e. there is no dependence on
and there are no components in the
direction. Now
and
giving
.
Thus Stokes’ theorem becomes
and Green’s theorem in the plane follows.
Example 38
Evaluate the line integral around the rectangle , .
Solution
The integral could be accomplished by four line integrals but it is easier to note that
is of the form
with
and
. It is thus of a suitable form for Green’s theorem in the plane.
Note that
and
.
Green’s theorem in the plane becomes
The same result could be gained by evaluating four line integrals.
Example 39
Verify Green’s theorem in the plane for the integral and the triangular contour starting at the origin and going to and before returning to the origin.
Solution
The whole of the contour is in the plane and Green’s theorem in the plane becomes
-
Firstly evaluate
.
On , and . As the integrand is zero, the integral will also be zero.
On , and . The integral is
On , and . The integral is .
Summing, -
Secondly evaluate
In this example, and . Thus and . Hence,
Conclusion:
One very useful, special case of Green’s theorem in the plane is when and . The theorem becomes
The right-hand side becomes i.e. where is the area inside the contour . Hence
This result is known as the area theorem .
Example 40
Verify the area theorem for the segment of the circle lying above the line .
Solution
Firstly, the area of the segment
can be found by subtracting the area of the triangle
from the area of the sector
. The triangle has area
. The sector has area
. Thus segment
has area
.
Now, evaluate the integral
around the segment. Along the line,
,
so the integral
becomes
.
Along the arc of the circle,
so
. The integral
becomes
So,
Hence both sides of the theorem equal
Task!
Verify Green’s theorem in the plane when applied to the integral
where represents the perimeter of the trapezium with vertices at , , and .
First let and and find :
Now find over the trapezium:
Now find along the four sides of the trapezium:
whose sum is 4.
Finally show that the two sides of the statement of Green’s theorem are equal:
Both sides are .
Exercises
- Verify Green’s identity [1] for the functions , and the unit cube , , .
-
Verify the area theorem for
- The area above but below .
- The segment of the circle , to the upper left of the line .
-
- ,
- .