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

$\int \; <!--nolimits\; -->{\int \; <!--nolimits\; -->}_S\left(u\underset{̲}{\nabla }w\right)\cdot \underset{̲}{dS}=\int \; <!--nolimits\; -->\int \; <!--nolimits\; -->{\int \; <!--nolimits\; -->}_V\left\{\underset{̲}{\nabla }u\cdot \underset{̲}{\nabla }w+u{\underset{̲}{\nabla }}^{2}w\right\}dV$ [1]

and

$\int \; <!--nolimits\; -->{\int \; <!--nolimits\; -->}_S\left\{u\underset{̲}{\nabla }w-v\underset{̲}{\nabla }u\right\}\cdot \underset{̲}{dS}=\int \; <!--nolimits\; -->\int \; <!--nolimits\; -->{\int \; <!--nolimits\; -->}_V\left\{u{\underset{̲}{\nabla }}^{2}w-w{\underset{̲}{\nabla }}^{2}u\right\}dV$ [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 $\underset{̲}{\nabla }\cdot \left(\varphi \underset{̲}{A}\right)=\left(\underset{̲}{\nabla }\varphi \right)\cdot \underset{̲}{A}+\varphi \left(\underset{̲}{\nabla }\cdot \underset{̲}{A}\right)$ .



Letting $\varphi =u$ and $\underset{̲}{A}=\underset{̲}{\nabla }w$ ,

$\underset{̲}{\nabla }\cdot \left(u\underset{̲}{\nabla }w\right)=\left(\underset{̲}{\nabla }u\right)\cdot \left(\underset{̲}{\nabla }w\right)+u\left(\underset{̲}{\nabla }\cdot \left(\underset{̲}{\nabla }w\right)\right)=\left(\underset{̲}{\nabla }u\right)\cdot \left(\underset{̲}{\nabla }w\right)+u{\underset{̲}{\nabla }}^{2}w$

Gauss’ theorem states

$\int \; <!--nolimits\; -->{\int \; <!--nolimits\; -->}_S\underset{̲}{F}\cdot \underset{̲}{dS}=\int \; <!--nolimits\; -->\int \; <!--nolimits\; -->{\int \; <!--nolimits\; -->}_V\underset{̲}{\nabla }\cdot \underset{̲}{F}dV$

Now, letting $\underset{̲}{F}=u\underset{̲}{\nabla }w$ ,

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

$\int \; <!--nolimits\; -->{\int \; <!--nolimits\; -->}_S\left(w\underset{̲}{\nabla }u\right)\cdot \underset{̲}{dS}=\int \; <!--nolimits\; -->\int \; <!--nolimits\; -->{\int \; <!--nolimits\; -->}_V\left\{\left(\underset{̲}{\nabla }w\right)\cdot \left(\underset{̲}{\nabla }u\right)+w{\underset{̲}{\nabla }}^{2}u\right\}dV$

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

Example 37

Verify Green’s first identity for $u=\left(x-x^2\right)y$ , $w=xy+z^2$ and the unit cube, $0\le x\le 1$ , $0\le y\le 1$ , $0\le z\le 1$ .

Solution

As $w=xy+z^2$ , $\underset{̲}{\nabla }w=y\underset{̲}{i}+x\underset{̲}{j}+2z\underset{̲}{k}$ . Thus $u\underset{̲}{\nabla }w=\left(xy-x^{2}y\right)\left(y\underset{̲}{i}+x\underset{̲}{j}+2z\underset{̲}{k}\right)$ and the surface integral is of this quantity (scalar product with $\underset{̲}{dS}$ ) integrated over the surface of the unit cube.

On the three faces $x=0$ , $x=1$ , $y=0$ , the vector $u\underset{̲}{\nabla }w=\underset{̲}{0}$ and so the contribution to the surface integral is zero.

On the face $y=1$ , $u\underset{̲}{\nabla }w=\left(x-x^2\right)\left(\underset{̲}{i}+x\underset{̲}{j}+2z\underset{̲}{k}\right)$ and $\underset{̲}{dS}=\underset{̲}{j}$ so $\left(u\underset{̲}{\nabla }w\right)\cdot \underset{̲}{dS}=x^2-x^3$ and the contribution to the integral is

${\int \; <!--nolimits\; -->}_{x=0}^1{\int \; <!--nolimits\; -->}_{z=0}^1\left(x^2-x^3\right)dzdx={\int \; <!--nolimits\; -->}_0^1\left(x^2-x^3\right)dx={\left[\frac{x^3}{3}-\frac{x^4}{4}\right]}_0^1=\frac{1}{12}$ .

On the face $z=0$ , $u\underset{̲}{\nabla }w=\left(x-x^2\right)y\left(y\underset{̲}{i}+x\underset{̲}{j}\right)$ and $\underset{̲}{dS}=-\underset{̲}{k}$ so $\left(u\underset{̲}{\nabla }w\right)\cdot \underset{̲}{dS}=0$ and the contribution to the integral is zero.

On the face $z=1$ , $u\underset{̲}{\nabla }w=\left(x-x^2\right)y\left(y\underset{̲}{i}+x\underset{̲}{j}+2\underset{̲}{k}\right)$ and $\underset{̲}{dS}=\underset{̲}{k}$ so $\left(u\underset{̲}{\nabla }w\right)\cdot \underset{̲}{dS}=2y\left(x-x^2\right)$ and the contribution to the integral is

${\int \; <!--nolimits\; -->}_{x=0}^1{\int \; <!--nolimits\; -->}_{y=0}^{1}2y\left(x-x^2\right)dydx={\int \; <!--nolimits\; -->}_0^1\left(x^2-x^3\right)dx={\int \; <!--nolimits\; -->}_{x=0}^1{\left[y^2\left(x-x^2\right)\right]}_{y=0}^{1}dx={\int \; <!--nolimits\; -->}_0^1\left(x-x^2\right)dx=\frac{1}{6}$ .

Thus, $\int \; <!--nolimits\; -->{\int \; <!--nolimits\; -->}_S\left(u\underset{̲}{\nabla }w\right)\cdot \underset{̲}{dS}=0+0+0+\frac{1}{12}+0+\frac{1}{6}=\frac{1}{4}$ .

Now evaluate $\int \; <!--nolimits\; -->\int \; <!--nolimits\; -->{\int \; <!--nolimits\; -->}_V\left\{\underset{̲}{\nabla }u\cdot \underset{̲}{\nabla }w+u{\underset{̲}{\nabla }}^{2}w\right\}dV$ .

Note that $\underset{̲}{\nabla }u=\left(1-2x\right)y\underset{̲}{i}+\left(x-x^2\right)\underset{̲}{j}$ and ${\underset{̲}{\nabla }}^{2}w=2$ so

$\underset{̲}{\nabla }u\cdot \underset{̲}{\nabla }w+u{\underset{̲}{\nabla }}^{2}w=\left(1-2x\right)y^2+\left(x-x^2\right)x+2\left(x-x^2\right)y=x^2-x^3+2xy-2x^{2}y+y^2-2x{y}^2$

and the integral

Hence $\int \; <!--nolimits\; -->{\int \; <!--nolimits\; -->}_S\left(u\underset{̲}{\nabla }w\right)\cdot \underset{̲}{dS}=\int \; <!--nolimits\; -->\int \; <!--nolimits\; -->{\int \; <!--nolimits\; -->}_V\left[\underset{̲}{\nabla }u\cdot \underset{̲}{\nabla }w+u{\underset{̲}{\nabla }}^{2}w\right]dV=\frac{1}{4}$

6.2 Green’s theorem in the plane

This states that

${\oint }_C\left(Pdx+Qdy\right)=\int \; <!--nolimits\; -->{\int \; <!--nolimits\; -->}_S\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)dxdy$

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

$\int \; <!--nolimits\; -->{\int \; <!--nolimits\; -->}_S\left(\underset{̲}{\nabla }\times \underset{̲}{F}\right)\cdot \underset{̲}{dS}={\oint }_C\underset{̲}{F}\cdot \underset{̲}{dr}$

Now let $\underset{̲}{F}$ be the vector field $P\left(x,y\right)\underset{̲}{i}+Q\left(x,y\right)\underset{̲}{j}$ i.e. there is no dependence on $z$ and there are no components in the $z-$ direction. Now

$\underset{̲}{\nabla }\times \underset{̲}{F}=\left|\begin{array}{ccc}\hfill \underset{̲}{i}\hfill & \hfill \underset{̲}{j}\hfill & \hfill \underset{̲}{k}\hfill \\ \hfill \hfill \\ \hfill \frac{\partial }{\partial x}\hfill & \hfill \frac{\partial }{\partial y}\hfill & \hfill \frac{\partial }{\partial z}\hfill \\ \hfill \hfill \\ \hfill P\left(x,y\right)\hfill & \hfill Q\left(x,y\right)\hfill & \hfill 0\hfill \end{array}\right|=\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)\underset{̲}{k}$

and $\underset{̲}{dS}=dxdy\underset{̲}{k}$ giving $\left(\underset{̲}{\nabla }\times \underset{̲}{F}\right)\cdot \underset{̲}{dS}=\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)dxdy$ .

Thus Stokes’ theorem becomes

$\int \; <!--nolimits\; -->{\int \; <!--nolimits\; -->}_S\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)dxdy={\oint }_C\underset{̲}{F}\cdot \underset{̲}{dr}$

and Green’s theorem in the plane follows.

Example 38

Evaluate the line integral ${\oint }_C\left[\left(4x^2+y-3\right)dx+\left(3x^2+4y^2-2\right)dy\right]$ around the rectangle $0\le x\le 3$ , $0\le y\le 1$ .

Solution

The integral could be accomplished by four line integrals but it is easier to note that $\left[\left(4x^2+y-3\right)dx+\left(3x^2+4y^2-2\right)dy\right]$ is of the form $Pdx+Qdy$ with $P=4x^2+y-3$ and $Q=3x^2+4y^2-2$ . It is thus of a suitable form for Green’s theorem in the plane.

Note that $\frac{\partial Q}{\partial x}=6x$ and $\frac{\partial P}{\partial y}=1$ .

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 ${\oint }_C\left[4zdy+\left(y^2-2\right)dz\right]$ and the triangular contour starting at the origin $O=\left(0,0,0\right)$ and going to $A=\left(0,2,0\right)$ and $B=\left(0,0,1\right)$ 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

${\oint }_C\left(Pdy+Qdz\right)=\int \; <!--nolimits\; -->{\int \; <!--nolimits\; -->}_S\left(\frac{\partial Q}{\partial y}-\frac{\partial P}{\partial z}\right)dydz$

1. Firstly evaluate ${\oint }_C\left\{4zdy+\left(y^2-2\right)dz\right\}$ .
   
   On $OA$ , $z=0$ and $dz=0$ . As the integrand is zero, the integral will also be zero.
   
   On $AB$ , $z=\left(1-\frac{y}{2}\right)$ and $dz=-\frac{1}{2}dy$ . The integral is
   
   ${\int \; <!--nolimits\; -->}_{y=2}^0\left(\left(4-2y\right)dy-\frac{1}{2}\left(y^2-2\right)dy\right)={\int \; <!--nolimits\; -->}_2^0\left(5-2y-\frac{1}{2}{y}^2\right)dy={\left[5y-y^2-\frac{1}{6}{y}^3\right]}_2^0=-\frac{14}{3}$
   
   On $BO$ , $y=0$ and $dy=0$ . The integral is ${\int \; <!--nolimits\; -->}_1^0\left(-2\right)dz={\left[-2z\right]}_1^0=2$ .
   
   Summing, ${\oint }_C\left(4zdy+\left(y^2-2\right)dz\right)=-\frac{8}{3}$
2. Secondly evaluate  $\int \; <!--nolimits\; -->{\int \; <!--nolimits\; -->}_S\left(\frac{\partial Q}{\partial y}-\frac{\partial P}{\partial z}\right)dydz$

   In this example, $P=4z$ and $Q=y^2-2$ . Thus $\frac{\partial P}{\partial z}=4$ and $\frac{\partial Q}{\partial y}=2y$ . Hence,
   
   

   $$\begin{array}{rcll}\int \; <!--nolimits\; -->{\int \; <!--nolimits\; -->}_S\left(\frac{\partial Q}{\partial y}-\frac{\partial P}{\partial z}\right)dydz& =& {\int \; <!--nolimits\; -->}_{y=0}^2{\int \; <!--nolimits\; -->}_{z=0}^{1-y∕2}\left(2y-4\right)dzdy& \\ & =& {\int \; <!--nolimits\; -->}_{y=0}^2{\left[2yz-4z\right]}_{z=0}^{1-y∕2}dy={\int \; <!--nolimits\; -->}_{y=0}^2\left(-y^2+4y-4\right)dy& \\ & =& {\left[-\frac{1}{3}{y}^3+2y^2-4y\right]}_0^2=-\frac{8}{3}& \end{array}$$

   Conclusion:

   ${\oint }_C\left(Pdy+Qdz\right)=\int \; <!--nolimits\; -->{\int \; <!--nolimits\; -->}_S\left(\frac{\partial Q}{\partial y}-\frac{\partial P}{\partial z}\right)dydz=-\frac{8}{3}$

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

${\oint }_C\left\{-ydx+xdy\right\}=\int \; <!--nolimits\; -->{\int \; <!--nolimits\; -->}_S\left(1-\left(-1\right)\right)dxdy$

The right-hand side becomes $\int \; <!--nolimits\; -->{\int \; <!--nolimits\; -->}_{S}2dxdy$ i.e. $2A$ where $A$ is the area inside the contour $C$ . Hence

$A=\frac{1}{2}{\oint }_C\left\{xdy-ydx\right\}$

This result is known as the area theorem .

Example 40

Verify the area theorem $A=\frac{1}{2}{\oint }_C\left\{xdy-ydx\right\}$ for the segment of the circle $x^2+y^2=4$ lying above the line $y=1$ .

Solution

Firstly, the area of the segment $ADBC$ can be found by subtracting the area of the triangle $OADB$ from the area of the sector $OACB$ . The triangle has area $\frac{1}{2}\times 2\sqrt{3}\times 1=\sqrt{3}$ . The sector has area $\frac{\pi }{3}\times 2^2=\frac{4}{3}\pi$ . Thus segment $ADBC$ has area $\frac{4}{3}\pi -\sqrt{3}$ .

Now, evaluate the integral ${\oint }_C\left\{xdy-ydx\right\}$ around the segment. Along the line, $y=1$ , $dy=0$ so the integral ${\int \; <!--nolimits\; -->}_C\left\{xdy-ydx\right\}$ becomes ${\int \; <!--nolimits\; -->}_{-\sqrt{3}}^{\sqrt{3}}\left(x\times 0-1\times dx\right)={\int \; <!--nolimits\; -->}_{-\sqrt{3}}^{\sqrt{3}}\left(-dx\right)=-2\sqrt{3}$ .

Along the arc of the circle, $y=\sqrt{4-x^2}={\left(4-x^2\right)}^{1∕2}$ so $dy=-x{\left(4-x^2\right)}^{-1∕2}dx$ . The integral ${\int \; <!--nolimits\; -->}_C\left\{xdy-ydx\right\}$ becomes

So, $\frac{1}{2}{\oint }_C\left\{xdy-ydx\right\}=\frac{1}{2}\left[\frac{8}{3}\pi -2\sqrt{3}\right]=\frac{4}{3}\pi -\sqrt{3}.$

Hence both sides of the theorem equal $\frac{4}{3}\pi -\sqrt{3}.$

Task!

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

${\oint }_C\left\{\left(5x+2y-7\right)dx+\left(3x-4y+5\right)dy\right\}$

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

First let $P=5x+2y-7$ and $Q=3x-4y+5$ and find $\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}$ :

Answer

Now find $\int \; <!--nolimits\; -->\int \; <!--nolimits\; -->\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)dxdy$ over the trapezium:

Answer

Now find $\int \; <!--nolimits\; -->\left(Pdx+Qdy\right)$ along the four sides of the trapezium:

Answer

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

Answer

Exercises

1. Verify Green’s identity [1] for the functions $u=xyz$ , $w=y^2$ and the unit cube $0\le x\le 1$ , $0\le y\le 1$ , $0\le z\le 1$ .
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 $y=1-x$ .

Answer