In physics and mathematics, Green's theorem gives the relationship between a line integral around a simple closed curve C and a double integral over the plane region D bounded by C. It is the twodimensional special case of the more general Stokes' theorem, and is named after British mathematician George Green.
Let C be a positively oriented, piecewise smooth, simple closed curve in the plane R^{2}, and let D be the region bounded by C. If L and M are functions of (x, y) defined on an open region containing D and have continuous partial derivatives there, then
For positive orientation, an arrow pointing in the counterclockwise direction may be drawn in the small circle in the integral symbol.
In physics, Green's theorem is mostly used to solve twodimensional flow integrals, stating that the sum of fluid outflows at any point inside a volume is equal to the total outflow summed about an enclosing area.
Contents 
The following is a proof of the theorem for the simplified area D, a type I region where C_{2} and C_{4} are vertical lines. A similar proof exists for when D is a type II region where C_{1} and C_{3} are straight lines. The general case can be deduced from this special case by approximating the domain D by a union of simple domains.
If it can be shown that
and
are true, then Green's theorem is proven in the first case.
Define the type I region D as pictured on the right by:
where g_{1} and g_{2} are continuous functions on [a, b]. Compute the double integral in (1):
Now compute the line integral in (1). C can be rewritten as the union of four curves: C_{1}, C_{2}, C_{3}, C_{4}.
With C_{1}, use the parametric equations: x = x, y = g_{1}(x), a ≤ x ≤ b. Then
With C_{3}, use the parametric equations: x = x, y = g_{2}(x), a ≤ x ≤ b. Then
The integral over C_{3} is negated because it goes in the negative direction from b to a, as C is oriented positively (counterclockwise). On C_{2} and C_{4}, x remains constant, meaning
Therefore,
Combining (3) with (4), we get (1). Similar computations give (2).
Green's theorem is a special case of Stokes' theorem, when applied to a region on the xyplane:
We can augment the twodimensional field into a threedimensional field with a zcomponent that is always 0: .
Start with the left side of Green's theorem:
Then by Stokes' Theorem:
The surface S is just the region in the plane D, with the unit normals pointing up (in +z direction) to match the "positive orientation" definitions for both theorems.
The expression inside the integral becomes
Thus we get the right side of Green's theorem
Considering only twodimensional vector fields, Green's theorem is equivalent to the following twodimensional analogue of the divergence theorem:
where is the outwardpointing unit normal vector on the boundary.
To see this, consider the unit normal in the right side of the equation. Since is a vector pointing tangential along a curve, and the curve C is the positivelyoriented (i.e. counterclockwise) curve along the boundary, an outward normal would be a vector which points 90° to the right, which would be (dy, − dx). The length of this vector is . So
Now let the components of . Then the right hand side becomes
which by Green's theorem becomes
Topics in Calculus  

Fundamental theorem Limits of functions Continuity Mean value theorem

In mathematics, Green's theorem gives the relationship between a line integral around a simple closed curve C and a double integral over the plane region D bounded by C. It is the twodimensional special case of the more general Stokes' theorem, and is named after British mathematician George Green.
Let C be a positively oriented, piecewise smooth, simple closed curve in the plane $\backslash mathbb\{R\}$^{2}, and let D be the region bounded by C. If L and M are functions of (x, y) defined on an open region containing D and have continuous partial derivatives there, then
For positive orientation, an arrow pointing in the counterclockwise direction may be drawn in the small circle in the integral symbol.
In physics, Green's theorem is mostly used to solve twodimensional flow integrals, stating that the sum of fluid outflows at any point inside a volume is equal to the total outflow summed about an enclosing area. In plane geometry, and in particular, area surveying, Green's theorem can be used to determine the area and centroid of plane figures solely by integrating over the perimeter.
Contents 
The following is a proof of the theorem for the simplified area D, a type I region where C_{2} and C_{4} are vertical lines. A similar proof exists for when D is a type II region where C_{1} and C_{3} are straight lines. The general case can be deduced from this special case by approximating the domain D by a union of simple domains.
If it can be shown that
and
are true, then Green's theorem is proven in the first case.
Define the type I region D as pictured on the right by:
where g_{1} and g_{2} are continuous functions on [a, b]. Compute the double integral in (1):
$\backslash iint\_D\; \backslash frac\{\backslash partial\; L\}\{\backslash partial\; y\}\backslash ,\; dA$  $=\backslash int\_a^b\backslash ,\backslash int\_\{g\_1(x)\}^\{g\_2(x)\}\; \backslash frac\{\backslash partial\; L\}\{\backslash partial\; y\}\; (x,y)\backslash ,dy\backslash ,dx$ 
$=\; \backslash int\_a^b\; \backslash Big\backslash \{L(x,g\_2(x))\; \; L(x,g\_1(x))\; \backslash Big\backslash \}\; \backslash ,\; dx\backslash qquad\backslash mathrm\{(3)\}$ 
Now compute the line integral in (1). C can be rewritten as the union of four curves: C_{1}, C_{2}, C_{3}, C_{4}.
With C_{1}, use the parametric equations: x = x, y = g_{1}(x), a ≤ x ≤ b. Then
With C_{3}, use the parametric equations: x = x, y = g_{2}(x), a ≤ x ≤ b. Then
The integral over C_{3} is negated because it goes in the negative direction from b to a, as C is oriented positively (counterclockwise). On C_{2} and C_{4}, x remains constant, meaning
Therefore,
$\backslash int\_\{C\}\; L\backslash ,\; dx$  $=\; \backslash int\_\{C\_1\}\; L(x,y)\backslash ,\; dx\; +\; \backslash int\_\{C\_2\}\; L(x,y)\backslash ,\; dx\; +\; \backslash int\_\{C\_3\}\; L(x,y)\backslash ,\; dx\; +\; \backslash int\_\{C\_4\}\; L(x,y)\backslash ,\; dx$ 
$=\; \backslash int\_a^b\; L(x,g\_2(x))\backslash ,\; dx\; +\; \backslash int\_a^b\; L(x,g\_1(x))\backslash ,\; dx\backslash qquad\backslash mathrm\{(4)\}$ 
Combining (3) with (4), we get (1). Similar computations give (2).
Green's theorem is a special case of Stokes' theorem, when applied to a region on the xyplane:
We can augment the twodimensional field into a threedimensional field with a zcomponent that is always 0: $\backslash mathbf\{F\}\; =\; (L,\; M,\; 0)$.
Start with the left side of Green's theorem:
Then by Stokes' Theorem:
The surface $S$ is just the region in the plane $D$, with the unit normals pointing up (in +z direction) to match the "positive orientation" definitions for both theorems.
The expression inside the integral becomes
Thus we get the right side of Green's theorem
Considering only twodimensional vector fields, Green's theorem is equivalent to the following twodimensional version of the divergence theorem:
where $\backslash mathbf\{\backslash hat\; n\}$ is the outwardpointing unit normal vector on the boundary.
To see this, consider the unit normal in the right side of the equation. Since $d\backslash mathbf\{r\}\; =\; (\; dx,\; dy)$ is a vector pointing tangential along a curve, and the curve C is the positivelyoriented (i.e. counterclockwise) curve along the boundary, an outward normal would be a vector which points 90° to the right, which would be $(\; dy,\; dx)$. The length of this vector is $\backslash sqrt\{dx^2\; +\; dy^2\}\; =\; ds$. So $\backslash mathbf\{\backslash hat\; n\}\backslash ,ds\; =\; (\; dy,\; dx).$
Now let the components of $\backslash mathbf\{F\}\; =\; (\; P,\; Q)$. Then the right hand side becomes
which by Green's theorem becomes
Green's theorem can be used to compute area by line integral.^{[1]} The area of D is given by:
Provided we choose L and M such that:
Then the area is given by:
Possible formulas for the area of D include^{[1]}:
