[proofplan] We recognise the integrand $P\,dx + Q\,dy$ as a smooth $1$-form $\omega \in \Omega^1(V)$ and compute its [exterior derivative](/theorems/1525) in the global coordinates $(x,y)$, obtaining $d\omega = \bigl(\partial_x Q - \partial_y P\bigr)\,dx \wedge dy$. The [Generalised Stokes' Theorem](/theorems/3555) applied to the compact oriented manifold-with-boundary $M$ then yields $\int_M d\omega = \int_{\partial M} \omega$. We close the argument by identifying these form-integrals with the classical objects in the statement: the bulk integral with the [Lebesgue integral](/page/Lebesgue%20Integral) of $\partial_x Q - \partial_y P$ over $M$, and the boundary integral with the line integral of $P\,dx + Q\,dy$ over $\partial M$. [/proofplan]

[step:Introduce the $1$-form $\omega = P\,dx + Q\,dy$ on $V$]Let $(x_1, x_2) = (x, y)$ denote the standard global coordinates on $\mathbb{R}^2$, and write $dx, dy \in \Omega^1(\mathbb{R}^2)$ for the associated coordinate $1$-forms. Define the smooth $1$-form \begin{align*} \omega: V &\to \Lambda^1 T^*\mathbb{R}^2 \\ p &\mapsto P(p)\, dx_p + Q(p)\, dy_p. \end{align*} Since $P, Q \in C^\infty(V; \mathbb{R})$ by hypothesis and $dx, dy$ are smooth $1$-forms on $\mathbb{R}^2$, we have $\omega \in \Omega^1(V)$.[/step]

[guided]The strategy of the proof is to reformulate Green's identity in the language of differential forms, where it becomes a special case of the [Generalised Stokes' Theorem](/theorems/3555). The first step is to package the integrand on the left-hand side of Green's formula — the expression $P\,dx + Q\,dy$ that appears under the line integral — as a single geometric object, namely a smooth $1$-form $\omega$ on $V$. We use the standard global coordinates $(x_1, x_2) = (x, y)$ on $\mathbb{R}^2$. Associated to these coordinates are the coordinate $1$-forms $dx, dy \in \Omega^1(\mathbb{R}^2)$, defined by $dx_p(\partial_x|_p) = 1$, $dx_p(\partial_y|_p) = 0$, and analogously for $dy$. Define \begin{align*} \omega: V &\to \Lambda^1 T^*\mathbb{R}^2 \\ p &\mapsto P(p)\, dx_p + Q(p)\, dy_p. \end{align*} Why is $\omega$ smooth? A $1$-form on an [open set](/page/Open%20Set) of $\mathbb{R}^2$ is smooth iff its coefficient functions in the global coordinate basis $(dx, dy)$ are smooth. Here the coefficients are precisely $P$ and $Q$, which are smooth on $V$ by hypothesis. Thus $\omega \in \Omega^1(V)$.[/guided]

[step:Compute $d\omega = (\partial_x Q - \partial_y P)\,dx \wedge dy$]We apply the [Coordinate Formula for the Exterior Derivative](/theorems/3564) to $\omega$ on the chart $V \subseteq \mathbb{R}^2$ with global coordinates $(x, y)$. Since $\omega = P\,dx + Q\,dy$ has coefficient functions $\omega_1 = P$ and $\omega_2 = Q$, the formula gives \begin{align*} d\omega &= dP \wedge dx + dQ \wedge dy \\ &= \left(\frac{\partial P}{\partial x}\,dx + \frac{\partial P}{\partial y}\,dy\right) \wedge dx + \left(\frac{\partial Q}{\partial x}\,dx + \frac{\partial Q}{\partial y}\,dy\right) \wedge dy. \end{align*} By [graded commutativity of the wedge product](/theorems/3560), $dx \wedge dx = 0$, $dy \wedge dy = 0$, and $dy \wedge dx = -\,dx \wedge dy$. Substituting and collecting, \begin{align*} d\omega &= \frac{\partial P}{\partial y}\,dy \wedge dx + \frac{\partial Q}{\partial x}\,dx \wedge dy = \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) dx \wedge dy. \end{align*}[/step]

[guided]To apply [Stokes' theorem](/theorems/1530) we need to know the [exterior derivative](/theorems/1525) of $\omega$. We use the [Coordinate Formula for the Exterior Derivative](/theorems/3564): for a $1$-form $\eta = \sum_i \eta_i\,dx_i$ written in a coordinate chart, \begin{align*} d\eta = \sum_{i} d\eta_i \wedge dx_i. \end{align*} The hypotheses of that theorem require $\eta$ to be a smooth $1$-form on an open subset of $\mathbb{R}^n$, which is exactly our setting: $\omega \in \Omega^1(V)$ with $V$ open in $\mathbb{R}^2$. Applying the formula with $\omega_1 = P$, $\omega_2 = Q$, \begin{align*} d\omega = dP \wedge dx + dQ \wedge dy. \end{align*} For a smooth scalar function $f$ on an [open set](/page/Open%20Set) of $\mathbb{R}^2$, the same coordinate formula (applied to the $0$-form $f$) gives $df = \partial_x f\,dx + \partial_y f\,dy$. Therefore \begin{align*} d\omega = \left(\frac{\partial P}{\partial x}\,dx + \frac{\partial P}{\partial y}\,dy\right) \wedge dx + \left(\frac{\partial Q}{\partial x}\,dx + \frac{\partial Q}{\partial y}\,dy\right) \wedge dy. \end{align*} Now we simplify using [Graded Commutativity of the Wedge Product](/theorems/3560): for $1$-forms $\alpha, \beta$ on a manifold, $\alpha \wedge \beta = -\,\beta \wedge \alpha$; in particular, $\alpha \wedge \alpha = 0$. Thus $dx \wedge dx = 0$ and $dy \wedge dy = 0$, killing two of the four terms, while $dy \wedge dx = -\,dx \wedge dy$. The surviving terms are \begin{align*} d\omega = \frac{\partial P}{\partial y}\,dy \wedge dx + \frac{\partial Q}{\partial x}\,dx \wedge dy = \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) dx \wedge dy. \end{align*} This is exactly the expression appearing on the right-hand side of Green's identity, paired with the standard $\mathbb{R}^2$ volume form $dx \wedge dy$.[/guided]

[step:Apply the Generalised Stokes' Theorem to $\omega$ on $M$]The manifold $M \subseteq V$ is, by hypothesis, a compact, oriented smooth $2$-dimensional submanifold with boundary, with $\partial M$ carrying the induced boundary orientation. The form $\omega \in \Omega^1(V)$ is smooth on an open neighbourhood of $M$; in particular, its restriction to $M$ is a smooth, compactly supported $1$-form on the compact oriented manifold $M$ (compactness of $M$ forces every smooth form on $M$ to have compact support). The [Generalised Stokes' Theorem](/theorems/3555) therefore applies and gives \begin{align*} \int_M d\omega = \int_{\partial M} \omega. \end{align*}[/step]

[guided]We now apply the [Generalised Stokes' Theorem](/theorems/3555), which asserts: if $N$ is a compact oriented smooth $n$-manifold with boundary and $\eta$ is a smooth $(n-1)$-form on $N$, then $\int_N d\eta = \int_{\partial N} \iota^*\eta$, where $\iota: \partial N \hookrightarrow N$ is the inclusion and $\partial N$ carries the induced Stokes orientation. We verify the hypotheses for $N = M$ and $\eta = \omega|_M$: (i) *$M$ is a compact oriented smooth $n$-manifold with boundary with $n = 2$.* This is exactly the hypothesis of Green's theorem. (ii) *$\omega|_M$ is a smooth $(n-1) = 1$-form on $M$.* Since $\omega \in \Omega^1(V)$ and $M \subseteq V$, the restriction $\omega|_M$ is smooth on $M$. Compactness of $M$ ensures the support condition is automatic. (iii) *Orientation conventions match.* The orientation of $M$ is the standard one of $\mathbb{R}^2$ (i.e. $dx \wedge dy > 0$), and $\partial M$ carries the induced boundary orientation by hypothesis. The conclusion of [Stokes' Theorem](/theorems/1530) gives \begin{align*} \int_M d\omega = \int_{\partial M} \omega, \end{align*} where on the right we are implicitly pulling $\omega$ back along the inclusion $\partial M \hookrightarrow V$ before integrating; we suppress this pullback in the notation, as is standard.[/guided]

[step:Identify $\int_M d\omega$ with the Lebesgue integral of $\partial_x Q - \partial_y P$]Since $M$ is open in $\mathbb{R}^2$ as a $2$-manifold (its interior is a non-empty open subset of $\mathbb{R}^2$, and the boundary $\partial M$ has $\mathcal{L}^2$-measure zero), and the orientation of $M$ is the standard one, the integral of any continuous top-degree form $f\,dx \wedge dy$ over $M$ coincides with the [Lebesgue integral](/page/Lebesgue%20Integral) of its coefficient function: \begin{align*} \int_M f\, dx \wedge dy = \int_M f\, d\mathcal{L}^2. \end{align*} Applying this with $f = \partial_x Q - \partial_y P$ (continuous on $V$ since $P, Q \in C^\infty(V; \mathbb{R})$) and using the formula for $d\omega$ from the previous step, \begin{align*} \int_M d\omega = \int_M \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) dx \wedge dy = \int_M \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) d\mathcal{L}^2. \end{align*}[/step]

[guided]We need to translate the form-integral $\int_M d\omega$ into the classical area integral that appears in Green's formula. The key fact is that integrals of top-degree forms over open subsets of $\mathbb{R}^n$ — and more generally over full-dimensional submanifolds with boundary in $\mathbb{R}^n$ — coincide with Lebesgue integrals, provided the orientation chosen is the standard one. Concretely: by definition, the integral of a compactly supported top form $f\,dx_1 \wedge \dots \wedge dx_n$ over an oriented chart domain $(U, \varphi)$ with $\varphi$ orientation-preserving is $\int_{\varphi(U)} (f \circ \varphi^{-1})(y)\,d\mathcal{L}^n(y)$. For $M \subseteq \mathbb{R}^2$ a full-dimensional oriented submanifold with boundary, the identity chart already realises this, and we obtain \begin{align*} \int_M f\, dx \wedge dy = \int_M f\, d\mathcal{L}^2 \end{align*} for any continuous (in fact $\mathcal{L}^2$-integrable) function $f$ on $M$. The contribution from the boundary $\partial M$ is null because $\partial M$ is a $1$-manifold and therefore has $\mathcal{L}^2$-measure zero. Here $f = \partial_x Q - \partial_y P$, which is continuous on $V$ because $P, Q \in C^\infty(V;\mathbb{R})$; restricted to the compact set $M$, it is bounded, hence $\mathcal{L}^2$-integrable on $M$. Substituting the formula for $d\omega$ from the previous step yields \begin{align*} \int_M d\omega = \int_M \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) dx \wedge dy = \int_M \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) d\mathcal{L}^2, \end{align*} which is exactly the right-hand side of Green's identity.[/guided]

[step:Identify $\int_{\partial M} \omega$ with the line integral $\oint_{\partial M} P\,dx + Q\,dy$]The integral of the $1$-form $\omega = P\,dx + Q\,dy$ along the positively oriented closed $1$-manifold $\partial M$ is, by the very definition of the line integral, the classical line integral $\oint_{\partial M}(P\,dx + Q\,dy)$. Concretely, if $\gamma: [a,b] \to \partial M$, $t \mapsto (\gamma_1(t), \gamma_2(t))$, is any orientation-preserving smooth parametrisation of a connected component of $\partial M$, then by the change-of-variables formula for integrals of $1$-forms along curves, \begin{align*} \int_{\gamma([a,b])} \omega = \int_a^b \left[P(\gamma(t))\,\gamma_1'(t) + Q(\gamma(t))\,\gamma_2'(t)\right] d\mathcal{L}^1(t), \end{align*} which is the standard definition of $\int_\gamma P\,dx + Q\,dy$. Summing over the (finitely many) connected components of $\partial M$ yields \begin{align*} \int_{\partial M} \omega = \oint_{\partial M}\left(P\,dx + Q\,dy\right). \end{align*}[/step]

[guided]The remaining identification translates the form-integral $\int_{\partial M}\omega$ on the right of Stokes' identity into the classical line integral on the left of Green's identity. This is essentially tautological: classical line-integral notation $\oint_\gamma P\,dx + Q\,dy$ is *defined* as the integral of the $1$-form $\omega = P\,dx + Q\,dy$ along $\gamma$. Let us spell this out. The boundary $\partial M$ is a compact oriented smooth $1$-manifold, hence a finite disjoint union of oriented circles $\gamma^{(1)}, \dots, \gamma^{(k)}$. Pick any orientation-preserving smooth parametrisation $\gamma: [a,b] \to \partial M$ of one such component $\gamma^{(j)}$, with $\gamma(a) = \gamma(b)$, and write $\gamma(t) = (\gamma_1(t), \gamma_2(t))$. Then by the definition of the integral of a $1$-form over an oriented $1$-manifold via pullback, \begin{align*} \int_{\gamma^{(j)}} \omega = \int_{[a,b]} \gamma^*\omega. \end{align*} Now $\gamma^* dx = \gamma_1'(t)\,d\mathcal{L}^1(t)$ and $\gamma^* dy = \gamma_2'(t)\,d\mathcal{L}^1(t)$ — this is the elementary calculation of the pullback of a coordinate $1$-form by a smooth curve. Hence \begin{align*} \gamma^*\omega = \bigl[P(\gamma(t))\,\gamma_1'(t) + Q(\gamma(t))\,\gamma_2'(t)\bigr]\, d\mathcal{L}^1(t), \end{align*} and therefore \begin{align*} \int_{\gamma^{(j)}}\omega = \int_a^b \bigl[P(\gamma(t))\,\gamma_1'(t) + Q(\gamma(t))\,\gamma_2'(t)\bigr]\,d\mathcal{L}^1(t). \end{align*} The right-hand side is, by definition, the classical line integral $\int_{\gamma^{(j)}} P\,dx + Q\,dy$. Summing over the components, \begin{align*} \int_{\partial M}\omega = \sum_{j=1}^k \int_{\gamma^{(j)}}\omega = \sum_{j=1}^k \oint_{\gamma^{(j)}}(P\,dx + Q\,dy) =: \oint_{\partial M}(P\,dx + Q\,dy). \end{align*}[/guided]

[step:Combine the identifications to obtain Green's identity] Chaining the equalities established in the previous three steps, \begin{align*} \oint_{\partial M}(P\,dx + Q\,dy) = \int_{\partial M}\omega \stackrel{\text{Stokes}}{=} \int_M d\omega = \int_M\left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) d\mathcal{L}^2, \end{align*} which is Green's identity. $\blacksquare$ [/step]