## Green's Theorem From Stoke's Theorem

Stoke's Theorem
$\oint_C \mathbf{F} \cdot d \mathbf{r} = \int \int_S (\mathbf{\nabla} \times \mathbf{F}) \cdot {n} dS$
and Green's Theorem
$\oint_C Pdx +Qdy = \int \int_S (\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y})dx dy$
where
$P=P(x,y), \; Q=Q(x,y)$
(and all derivatives with respect to
$z$
are zero) are not independent theorems. In fact applying Stoke's Theorem to a vector field in the plane results in Green's Theorem.
To see this take
$\mathbf{F} = P(x,y) \mathbf{i} + Q(x,y) \mathbf{j}$

$d \mathbf{r} = dx \mathbf{i} + dy \mathbf{j}$

The left hand side of Stoke's Theorem becomes
$\oint_C \mathbf{F} \cdot d \mathbf{r} = \oint_C Pdx +Qdy$

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

Since
$\mathbf{n}=\mathbf{k}$
the right hand side of Stoke's Theorem becomes
$\int \int_S (\mathbf{\nabla} \times \mathbf{F}) \cdot {n} dS =\int \int_S (\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}) \mathbf{k} \cdot \mathbf{k} dS = \int \int_S (\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y})dx dy$