Green's theorem brings in a relationship between a line integral around a simple closed curve C and a double integral in the plane region D, whose boundary is C. The statement of Green's Theorem is as follows:

Suppose C is a positively oriented, piecewise smooth, simple closed curve in the plane and D is the region bounded by C. If P and Q have continuous partial derivatives on an open region containing D, then
$\int_{C}(Pdx+Qdy) = \iint_{D}(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y})dA$
The notation $\oint _{C}$ is used to indicate that the curve C satisfies the conditions stated in Green's theorem and hence we have the equivalent theorem statement
$\oint _{C}(Pdx+Qdy) = \iint_{D}(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y})dA$
If C is traversed clockwise that is in the negative direction, then -C is taken to be the positive direction.

Thus
$\oint _{C}(Pdx+Qdy) = -\oint _{-C}(Pdx + Qdy) =-\iint_{D}(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y})dA$
If we consider P and Q as components of vector field, that is when F = Pi + Qj, then the line integral in Green's theorem can be rewritten as
$\oint _{C}F.dr$ = $\oint _{C}(Pdx+Qdy)$
Green's theorem can then be written in vector form using curl of F as,
$\oint _{C}F.dr=\iint_{D}(curlF).k dA$
As F.dr = F.Tds, the above formula involves the Tangent component of F and re written as
$\oint _{C}F.Tds=\iint_{D}(curlF).k dA$           
Here the the tangential form of Green's theorem includes the curl of the vector Filed, while the normal form of the statements contains divergence of F.
$\oint _{C}F.nds=\iint_{D}divF(x,y)dA$

Green's theorem also gives a method to calculate the area of the bounded region D using a line integral.
Area of D = $\frac{1}{2}$ $\oint _{C}(xdy-ydx)$

The normal divergence vector form of Green's theorem is stated for vector fields in R2. Gauss's Divergence theorem can be considered as an extension of Green's theorem for vector fields in R3, the surface integral taking over the line integral and the double integral being replaced by a triple integral.

Divergence Theorem

Suppose E is a simple solid region and S the boundary surface of the region E with positive outward orientation. If F is a vector field whose component functions have continuous partial derivatives on an open region containing E, then
$\int \int _{S}F.ds=\int \int \int _{E}div F dV$

This is also equivalently written as
$\int \int _{S}F.ndS=\int \int \int _{E}(\bigtriangledown .F)dV$

The Flux integral of a vector field across a boundary is equal to the integral of the divergence of that vector field within the region enclosed by the boundary.

Solved Examples

Question 1: Verify Green's theorem for the line integral $\oint _{C}(xy^{2}dx-x^{2}ydy$) where C is the circle with center origin and radius 1.
Solution:
 
This means we need to prove $\oint _{C}(xy^{2}dx-x^{2}ydy)$ = $\int \int _{D}(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y})dA$

    where P = xy2 and Q = x2y  and D is the interior of the circle r = 1.

    Let us first evaluate the line integral using the parametrization x = cos θ and y = sin θ where 0 ≤ θ < 2π
    $\oint _{C}xy^{2}dx-x^{2}ydy$ = $\int_{0}^{2\pi }(-cos\theta  sin^{3}\theta -cos^{3}\theta sin\theta )d\theta $

                                               = $\int_{0}^{2\pi }-sin\theta  cos\theta (cos^{2}\theta +sin^{2}\theta )d\theta $

                                               = $-\frac{1}{2}\int_{0}^{2\pi }sin(2\theta )d\theta $

                                               = $\frac{1}{4}(cos2\theta )_{0}^{2\pi }$ = 0     -------------------(1)

   $\frac{\partial Q}{\partial x}$ = -2xy     and   $\frac{\partial P}{\partial y}$ = 2xy

   Thus we have for the double integral

   $\int \int _{D}(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y})dA$ = $\int \int _{D}4xydxdy$

                                            = $\int_{0}^{2\pi }\int_{0}^{1}-4sin\theta cos\theta r dr \theta$

                                            = $\int_{0}^{2\pi }-sin2\theta d\theta $

                                            = $\frac{1}{2}[cos2\theta ]_{0}^{2\pi }$

                                            = 0    -----------------------(2)
  Thus the evaluated value of the line integral and the double integral are both equal to 0 which verifies the Green's Theorem
 

Question 2: Use Green's theorem to evaluate $\oint _{C}F.dr$ where F(x,y) = <ex + x2y, ey - xy2> is the circle x2 + y2 = 25 and traversed in the direction.
Solution:
 
C is traversed clockwise, hence -C is the positive direction here

    Thus $\oint _{C}F.dr$ = $-\oint _{-C}(e^{x}+x^{2}y)dx_(e^{y}-xy^{2})dy$
   
P = ex + x2y      and     Q = ey - xy2.

    $\frac{\partial Q}{\partial x}$ = -y2       and   $\frac{\partial P}{\partial y}$

= x2                   Partial Derivatives

    The Region is D for the double integral is the disk enclosed by x2 + y2 ≤ 25

    $\oint _{C}F.dr$ = $-\oint _{-C}(e^{x}+x^{2}y)dx+(e^{y}-xy^{2})dy$

                           = $-\int \int _{D}(-y^{2}-x^{2})dA$

                           = $\int \int _{D}(x^{2}+y^{2})dA$

                           = $\int_{0}^{2\pi }\int_{0}^{5}(r^{2})rdrd\theta $

                           = $\int_{0}^{2\pi }\int_{0}^{5}r^{3}drd\theta $

                           = $\int_{0}^{2\pi }d\theta \int_{0}^{5}r^{3}dr$

                           = $2\pi [\frac{r^{4}}{4}]_{0}^{5}$ = 312.5π