General Stokes agrees with both planar Green formulas

The general Stokes theorem: Assume ${\mathrm{A}\mathrm{C}}_{\omega }$. Let $M$ be an oriented smooth $n$-manifold with boundary, $n\ge 1$, and let $\eta \in {\mathrm{\Omega }}_c^{n-1}(M)$. With $j:\mathrm{\partial }M\hookrightarrow M$ and the outward-normal-first orientation, ${\int }_{M}d\eta ={\int }_{\mathrm{\partial }M}{j}^{\ast }\eta \mathrm{.}$ An empty boundary contributes zero; in dimension one its integral is a finite signed sum of point values.

Green's theorem is the curl statement for a planar field lifted to ${\mathbb{R}}^3$: Let $D$ be a finite elementary Green region with its supplied decomposition, positively oriented, and let $P,Q$ be $C^1$ on an open $U\subseteq {\mathbb{R}}^2$ containing $D$. Define the lift $\tilde{F}(x,y,z):=(P(x,y),Q(x,y),0)((x,y,z)\in U\times \mathbb{R}),$ a field on the open set $U\times \mathbb{R}\subseteq {\mathbb{R}}^3$. Then $\tilde{F}$ is $C^1$, its curl has first and second coordinates identically $0$ and third coordinate ${\mathrm{\partial }}_{x}Q-{\mathrm{\partial }}_{y}P$ at every point, independent of $z$, and the circulation of the planar field around the positive boundary chain equals the integral of the third coordinate of the curl of the lift: ${\int }_{\mathrm{\partial }D}(P,Q)\cdot d\mathbf{r}={\iint }_D(\mathrm{curl}\tilde{F}{)}_z(x,y,0)dA\mathrm{.}$

The planar divergence theorem: the flux form of Green's theorem: Let $D$ be a finite elementary Green region with its supplied decomposition, positively oriented, and let $F=(F_x,F_y)$ be $C^1$ on an open $U\subseteq {\mathbb{R}}^2$ containing $D$. Then ${\int }_{\mathrm{\partial }D}(-F_y)dx+F_{x}dy={\iint }_D({\mathrm{\partial }}_x{F}_x+{\mathrm{\partial }}_y{F}_y)dA,$ the right-hand integrand being the divergence of $F$ as a field on an open subset of ${\mathbb{R}}^2$. Moreover, if $\sigma :[\alpha ,\beta ]\to {\mathbb{R}}^2$ is one of the arcs of the positive boundary chain and its derivative is nowhere zero on a piece with continuous derivative extension $v$, then on that piece ${\int }_{\sigma }(-F_y)dx+F_{x}dy={\int }_{\sigma }⟨F,\nu ⟩ds,\nu :=\frac{(v_2,-v_1)}{\parallel v{\parallel }_2},$ where $\nu$ is the unit vector obtained from the tangent $v$ by a quarter turn clockwise.

Computing form integrals by finite parametrizations: Let $n\ge 1$, let $M^n$ be oriented, and let $\omega \in {\mathrm{\Omega }}_c^n(M)$. For $1\le i\le m$ let $D_i\subset {\mathbb{R}}^n$ be bounded open Jordan domains and $F_i:{\bar{D}}_i\to M$ continuous and smooth up to the boundary in target coordinates: near each parameter point, a target coordinate representative extends smoothly to a Euclidean neighborhood. Suppose $F_i{\mathrm{\mid }}_{D_i}$ is an orientation-preserving diffeomorphism onto an open $W_i\subset M$, the $W_i$ are pairwise disjoint, and $\mathrm{supp}\omega \subset {\bigcup }_i{\bar{W}}_i$. Then ${\int }_M\omega ={\sum }_{i=1}^m{\int }_{D_i}{F}_i^{\ast }\omega \mathrm{.}$ An empty family is allowed when the support is empty. No nonsingularity of $D{F}_i$ on $\mathrm{\partial }{D}_i$, and no $M$-valued extension across a genuine target boundary, is assumed.