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 u ( x , y , z ) and w ( x , y , z ) . Two statements of Green’s identities are as follows

S ( u ̲ w ) d S ̲ = V ̲ u ̲ w + u ̲ 2 w d V [1]

and

S u ̲ w v ̲ u d S ̲ = V u ̲ 2 w w ̲ 2 u d V [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 ̲ ( ϕ A ̲ ) = ( ̲ ϕ ) A ̲ + ϕ ( ̲ A ̲ ) .



Letting ϕ = u and A ̲ = ̲ w ,

̲ ( u ̲ w ) = ( ̲ u ) ( ̲ w ) + u ( ̲ ( ̲ w ) ) = ( ̲ u ) ( ̲ w ) + u ̲ 2 w

Gauss’ theorem states

S F ̲ d S ̲ = V ̲ F ̲ d V

Now, letting F ̲ = u ̲ w ,

S ( u ̲ w ) d S ̲ = V ̲ ( u ̲ w ) d V = V ( ̲ u ) ( ̲ w ) + u ̲ 2 w d V

This is Green’s identity [1]. Reversing the roles of u and w ,

S ( w ̲ u ) d S ̲ = V ( ̲ w ) ( ̲ u ) + w ̲ 2 u d V

Subtracting the last two equations yields Green’s identity [2].

Key Point 10

Green’s Identities

[1] S ( u ̲ w ) d S ̲ = V ̲ u ̲ w + u ̲ 2 w d V

[2] S u ̲ w v ̲ u d S ̲ = V u ̲ 2 w w ̲ 2 u d V

Example 37

Verify Green’s first identity for u = ( x x 2 ) y , w = x y + z 2 and the unit cube, 0 x 1 , 0 y 1 , 0 z 1 .

Solution

As w = x y + z 2 , ̲ w = y i ̲ + x j ̲ + 2 z k ̲ . Thus u ̲ w = ( x y x 2 y ) ( y i ̲ + x j ̲ + 2 z k ̲ ) and the surface integral is of this quantity (scalar product with d S ̲ ) integrated over the surface of the unit cube.

On the three faces x = 0 , x = 1 , y = 0 , the vector u ̲ w = 0 ̲ and so the contribution to the surface integral is zero.

On the face y = 1 , u ̲ w = ( x x 2 ) ( i ̲ + x j ̲ + 2 z k ̲ ) and d S ̲ = j ̲ so ( u ̲ w ) d S ̲ = x 2 x 3 and the contribution to the integral is

x = 0 1 z = 0 1 ( x 2 x 3 ) d z d x = 0 1 ( x 2 x 3 ) d x = x 3 3 x 4 4 0 1 = 1 12 .

On the face z = 0 , u ̲ w = ( x x 2 ) y ( y i ̲ + x j ̲ ) and d S ̲ = k ̲ so ( u ̲ w ) d S ̲ = 0 and the contribution to the integral is zero.

On the face z = 1 , u ̲ w = ( x x 2 ) y ( y i ̲ + x j ̲ + 2 k ̲ ) and d S ̲ = k ̲ so ( u ̲ w ) d S ̲ = 2 y ( x x 2 ) and the contribution to the integral is

x = 0 1 y = 0 1 2 y ( x x 2 ) d y d x = 0 1 ( x 2 x 3 ) d x = x = 0 1 y 2 ( x x 2 ) y = 0 1 d x = 0 1 ( x x 2 ) d x = 1 6 .

Thus, S ( u ̲ w ) d S ̲ = 0 + 0 + 0 + 1 12 + 0 + 1 6 = 1 4 .

Now evaluate V ̲ u ̲ w + u ̲ 2 w d V .

Note that ̲ u = ( 1 2 x ) y i ̲ + ( x x 2 ) j ̲ and ̲ 2 w = 2 so

̲ u ̲ w + u ̲ 2 w = ( 1 2 x ) y 2 + ( x x 2 ) x + 2 ( x x 2 ) y = x 2 x 3 + 2 x y 2 x 2 y + y 2 2 x y 2

and the integral

V ̲ u ̲ w + u ̲ 2 w d V = z = 0 1 y = 0 1 x = 0 1 ( x 2 x 3 + 2 x y 2 x 2 y + y 2 2 x y 2 ) d x d y d z = z = 0 1 y = 0 1 x 3 3 x 4 4 + x 2 y 2 3 x 3 y + x y 2 x 2 y 2 x = 0 1 d y d z = z = 0 1 y = 0 1 ( 1 12 + y 3 ) d y d z = z = 0 1 y 12 + y 2 6 y = 0 1 d z = z = 0 1 ( 1 4 ) d z = z 4 z = 0 1 = 1 4

Hence S ( u ̲ w ) d S ̲ = V ̲ u ̲ w + u ̲ 2 w d V = 1 4

6.2 Green’s theorem in the plane

This states that

C ( P d x + Q d y ) = S Q x P y d x d y

S is a 2 - D surface with perimeter C ; P ( x , y ) and Q ( x , y ) 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.

S ( ̲ × F ̲ ) d S ̲ = C F ̲ d r ̲

Now let F ̲ be the vector field P ( x , y ) i ̲ + Q ( x , y ) j ̲ i.e. there is no dependence on z and there are no components in the z direction. Now

̲ × F ̲ = i ̲ j ̲ k ̲ x y z P ( x , y ) Q ( x , y ) 0 = Q x P y k ̲

and d S ̲ = d x d y k ̲ giving ( ̲ × F ̲ ) d S ̲ = Q x P y d x d y .

Thus Stokes’ theorem becomes

S Q x P y d x d y = C F ̲ d r ̲

and Green’s theorem in the plane follows.

Key Point 11

Green’s Theorem in the Plane

C ( P d x + Q d y ) = S Q x P y d x d y
Example 38

Evaluate the line integral C ( 4 x 2 + y 3 ) d x + ( 3 x 2 + 4 y 2 2 ) d y around the rectangle 0 x 3 , 0 y 1 .

Solution

The integral could be accomplished by four line integrals but it is easier to note that ( 4 x 2 + y 3 ) d x + ( 3 x 2 + 4 y 2 2 ) d y is of the form P d x + Q d y with P = 4 x 2 + y 3 and Q = 3 x 2 + 4 y 2 2 . It is thus of a suitable form for Green’s theorem in the plane.

Note that Q x = 6 x and P y = 1 .

Green’s theorem in the plane becomes

C { ( 4 x 2 + y 3 ) d x + ( 3 x 2 + 4 y 2 2 ) d y } = y = 0 1 x = 0 3 6 x 1 d x d y = y = 0 1 3 x 2 x x = 0 3 d y = y = 0 1 24 d y = 24

The same result could be gained by evaluating four line integrals.

Example 39

Verify Green’s theorem in the plane for the integral C 4 z d y + ( y 2 2 ) d z and the triangular contour starting at the origin O = ( 0 , 0 , 0 ) and going to A = ( 0 , 2 , 0 ) and B = ( 0 , 0 , 1 ) before returning to the origin.

Solution

The whole of the contour is in the plane x = 0 and Green’s theorem in the plane becomes

C ( P d y + Q d z ) = S Q y P z d y d z

  1. Firstly evaluate C 4 z d y + ( y 2 2 ) d z .

    On O A , z = 0 and d z = 0 . As the integrand is zero, the integral will also be zero.

    On A B , z = ( 1 y 2 ) and d z = 1 2 d y . The integral is

    y = 2 0 ( 4 2 y ) d y 1 2 ( y 2 2 ) d y = 2 0 ( 5 2 y 1 2 y 2 ) d y = 5 y y 2 1 6 y 3 2 0 = 14 3

    On B O , y = 0 and d y = 0 . The integral is 1 0 ( 2 ) d z = 2 z 1 0 = 2 .

    Summing, C 4 z d y + ( y 2 2 ) d z = 8 3
  2. Secondly evaluate  S Q y P z d y d z

    In this example, P = 4 z and Q = y 2 2 . Thus P z = 4 and Q y = 2 y . Hence,

    S Q y P z d y d z = y = 0 2 z = 0 1 y 2 2 y 4 d z d y = y = 0 2 2 y z 4 z z = 0 1 y 2 d y = y = 0 2 y 2 + 4 y 4 d y = 1 3 y 3 + 2 y 2 4 y 0 2 = 8 3

    Conclusion:

    C ( P d y + Q d z ) = S Q y P z d y d z = 8 3

One very useful, special case of Green’s theorem in the plane is when Q = x and P = y . The theorem becomes

C { y d x + x d y } = S 1 ( 1 ) d x d y

The right-hand side becomes S 2 d x d y i.e. 2 A where A is the area inside the contour C . Hence

A = 1 2 C { x d y y d x }

This result is known as the area theorem .

Example 40

Verify the area theorem A = 1 2 C { x d y y d x } for the segment of the circle x 2 + y 2 = 4 lying above the line y = 1 .

Solution

Firstly, the area of the segment A D B C can be found by subtracting the area of the triangle O A D B from the area of the sector O A C B . The triangle has area 1 2 × 2 3 × 1 = 3 . The sector has area π 3 × 2 2 = 4 3 π . Thus segment A D B C has area 4 3 π 3 .

Now, evaluate the integral C { x d y y d x } around the segment. Along the line, y = 1 , d y = 0 so the integral C { x d y y d x } becomes 3 3 ( x × 0 1 × d x ) = 3 3 ( d x ) = 2 3 .

Along the arc of the circle, y = 4 x 2 = ( 4 x 2 ) 1 2 so d y = x ( 4 x 2 ) 1 2 d x . The integral C { x d y y d x } becomes

3 3 { x 2 ( 4 x 2 ) 1 2 ( 4 x 2 ) 1 2 } d x = 3 3 4 4 x 2 d x = π 3 π 3 4 1 2 cos θ 2 cos θ d θ = π 3 π 3 4 d θ = 8 3 π

So, 1 2 C { x d y y d x } = 1 2 8 3 π 2 3 = 4 3 π 3 .

Hence both sides of the theorem equal 4 3 π 3 .

Task!

Verify Green’s theorem in the plane when applied to the integral

C ( 5 x + 2 y 7 ) d x + ( 3 x 4 y + 5 ) d y

where C represents the perimeter of the trapezium with vertices at ( 0 , 0 ) , ( 3 , 0 ) , ( 6 , 1 ) and ( 1 , 1 ) .

First let P = 5 x + 2 y 7 and Q = 3 x 4 y + 5 and find Q x P y :

1

Now find Q x P y d x d y over the trapezium:

4

Now find ( P d x + Q d y ) along the four sides of the trapezium:

1.5 , 66 , 62.5 , 1   whose sum is 4.

Finally show that the two sides of the statement of Green’s theorem are equal:

Both sides are 4 .

Exercises
  1. Verify Green’s identity [1] for the functions u = x y z , w = y 2 and the unit cube 0 x 1 , 0 y 1 , 0 z 1 .
  2. Verify the area theorem for
    1. The area above y = 0 , but below y = 1 x 2 .
    2. The segment of the circle x 2 + y 2 = 1 , to the upper left of the line y = 1 x .
    1. 4 3 ,
    2. π 4 1 2 .