# Green’s Theorem

Let R be a region in the xy-plane that is bounded by a closed, piecewise smooth curve C, and let

$\mathbf{F} = P\left( {x,y} \right)\mathbf{i} + Q\left( {x,y} \right)\mathbf{j}$

be a continuous vector function with continuous first partial derivatives $${\frac{{\partial P}}{{\partial y}}}, {\frac{{\partial Q}}{{\partial x}}}$$ in a some domain containing R. Then Green's theorem states that

$\iint\limits_R {\left( {\frac{{\partial Q}}{{\partial x}} - \frac{{\partial P}}{{\partial y}}} \right)dxdy} = \oint\limits_C {Pdx + Qdy} ,$

where the symbol $$\oint\limits_C {}$$ indicates that the curve (contour) $$C$$ is closed and integration is performed counterclockwise around this curve.

If $$Q = x,$$ $$P = -y,$$ Green's formula yields:

$S = \iint\limits_R {dxdy} = \frac{1}{2}\oint\limits_C {xdy - ydx} ,$

where $$S$$ is the area of the region $$R$$ bounded by the contour $$C.$$

We can also write Green's Theorem in vector form. For this we introduce the so-called curl of a vector field. Let

$\mathbf{F} = P\left( {x,y,z} \right)\mathbf{i} + Q\left( {x,y,z} \right)\mathbf{j} + R\left( {x,y,z} \right)\mathbf{k}$

be a vector field. Then the curl of the vector field $$\mathbf{F}$$ is called the vector denoted by $$\text{rot}\,\mathbf{F}$$ or $$\nabla \times \mathbf{F}$$, which is equal to

$\text{rot}\,\mathbf{F} = \nabla \times \mathbf{F} = \left| {\begin{array}{*{20}{c}} \mathbf{i} & \mathbf{j} & \mathbf{k}\\ {\frac{\partial }{{\partial x}}}&{\frac{\partial }{{\partial y}}}&{\frac{\partial }{{\partial z}}}\\ P&Q&R \end{array}} \right| = \left( {\frac{{\partial R}}{{\partial y}} - \frac{{\partial Q}}{{\partial z}}} \right)\mathbf{i} + \left( {\frac{{\partial P}}{{\partial z}} - \frac{{\partial R}}{{\partial x}}} \right)\mathbf{j} + \left( {\frac{{\partial Q}}{{\partial x}} - \frac{{\partial P}}{{\partial y}}} \right)\mathbf{k}.$

In terms of curl, Green's Theorem can be written as

$\iint\limits_R {\left( \text{rot}\,\mathbf{F} \right) \cdot \mathbf{k}\,dxdy} = \oint\limits_C {\mathbf{F} \cdot d\mathbf{r}} .$

Note that Green's Theorem is simply Stoke's Theorem applied to a $$2$$-dimensional plane.

## Solved Problems

Click or tap a problem to see the solution.

### Example 1

Using Green's theorem, evaluate the line integral $\oint\limits_C {xydx + \left( {x + y} \right)dy} ,$ where $$C$$ is the curve bounding the unit disk $$R.$$

### Example 2

Using Green's formula, evaluate the line integral $\oint\limits_C {\left( {x - y} \right)dx + \left( {x + y} \right)dy},$ where $$C$$ is the circle ${x^2} + {y^2} = {a^2}.$

### Example 1.

Using Green's theorem, evaluate the line integral $\oint\limits_C {xydx + \left( {x + y} \right)dy} ,$ where $$C$$ is the curve bounding the unit disk $$R.$$

Solution.

The components of the vector field are

$P\left( {x,y} \right) = xy,\;\; Q\left( {x,y} \right) = x + y.$

Using the Green's formula

$\iint\limits_R {\left( {\frac{{\partial Q}}{{\partial x}} - \frac{{\partial P}}{{\partial y}}} \right)dxdy} = \oint\limits_C {Pdx + Qdy} ,$

we transform the line integral into the double integral:

$I = \oint\limits_C {xydx + \left( {x + y} \right)dy} = \iint\limits_R {\left( {\frac{{\partial \left( {x + y} \right)}}{{\partial x}} - \frac{{\partial \left( {xy} \right)}}{{\partial y}}} \right)dxdy} = \iint\limits_R {\left( {1 - x} \right)dxdy} .$

Converting the double integral into polar coordinates, we have

$I = \int\limits_R {\left( {1 - x} \right)dxdy} = \int\limits_0^{2\pi } {\int\limits_0^1 {\left( {1 - r\cos \theta } \right)rdrd\theta } } = \int\limits_0^{2\pi } {\left[ {\int\limits_0^1 {\left( {r - {r^2}\cos \theta } \right)dr} } \right]d\theta } = \int\limits_0^{2\pi } {\left[ {\left. {\left( {\frac{{{r^2}}}{2} - \frac{{{r^3}}}{3}\cos \theta } \right)} \right|_{r = 0}^1} \right]d\theta } = \int\limits_0^{2\pi } {\left( {\frac{1}{2} - \frac{{\cos \theta }}{3}} \right)d\theta } = \left. {\left( {\frac{\theta }{2} - \frac{{\sin \theta }}{3}} \right)} \right|_0^{2\pi } = \pi .$

### Example 2.

Using Green's formula, evaluate the line integral $\oint\limits_C {\left( {x - y} \right)dx + \left( {x + y} \right)dy},$ where $$C$$ is the circle ${x^2} + {y^2} = {a^2}.$

Solution.

First we identify the components of the vector field:

$P = x - y,\;\;\; Q = x + y$

and find the partial derivatives:

$\frac{{\partial Q}}{{\partial x}} = \frac{{\partial \left( {x + y} \right)}}{{\partial x}} = 1,\;\; \frac{{\partial P}}{{\partial y}} = \frac{{\partial \left( {x - y} \right)}}{{\partial x}} = - 1.$

Hence, the line integral can be written in the form

$I = \oint\limits_C {\left( {x - y} \right)dx + \left( {x + y} \right)dy} = \iint\limits_R {\left( {1 - \left( { - 1} \right)} \right)dxdy} = 2\iint\limits_R {dxdy} .$

In the last expression the double integral $$\iint\limits_R {dxdy}$$ is equal numerically to the area of the disk $${x^2} + {y^2} = {a^2},$$ which is $$\pi {a^2}.$$ Then the integral is

$I = 2\iint\limits_R {dxdy} = 2\pi {a^2}.$

See more problems on Page 2.