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
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.
