> [!NOTE] Theorem (Green's Theorem) > Let $C\subset \mathbb{R}^{2}$ be a [[Parametrized Curve|parametrised curve]] that is traversed anti-clockwise (positively oriented), simple and closed. Let $D$ be the region bounded by $C.$ > > For any two-variable functions $P,Q$ that have continuous partial derivatives on $D,$ we have $\oint_{C} (P\,dx+Q\,dy) =\int \int_{D} \left( \frac{ \partial Q }{ \partial x } - \frac{ \partial P }{ \partial y } \right) \, dx \, dy $where $\oint_{C} (P\,dx+Q\,dy)$ equals the [[Line Integral of Vector Field on Subset of Real n-Space|line integral]] $\oint \underline{F} \cdot d\underline{r}$ where $\underline{F}=(P,Q)$ and $\underline{r}=(x,y).$ **Proof** If it can be shown that $\oint_{C} P\,dx =\int \int_{D} \left(- \frac{ \partial P }{ \partial y } \right) \, dx \, dy $and $\oint_{C} Q\,dy =\int \int_{D} \left( \frac{ \partial Q }{ \partial x } \right) \, dx \, dy $are true, then the result follows immediately. **Proof by Stoke's theorem**: Let $\underline{F}(x,y,z)=(P(x,y),Q(x,y),0)$ be a vector field. Then its curl is given by $\nabla \times \underline{F} = \begin{pmatrix} 0 \\ 0 \\ Q_{x}-P_{y} \end{pmatrix}$ Let $D$ denote the region bounded by $C.$ Clearly $\hat{\underline{n}}=(0,0,1)$ is the outward pointing unit normal to $D$ since it lies in the $x$-$y$ plane. Thus by [[Stokes' Theorem|Stoke's theorem]], $\oint_{C} \underline{F} \cdot d\underline{r} = \oint_{C} (P\,dx+Q\,dy) = \int \int_{D} \nabla \times \underline{F} \cdot \underline{\hat{n}} \, dx \, dy = \int \int_{D} (Q_{x}-P_{y}) \, dx \, dy $ # Applications See [[Planimeter]].