# Stokes in $\mathbb{R}^2$ (Manifolds)

Let $M=\mathbb{R}^2_{x,y\geq0}$ and $\omega$ a 1-form with support inside $[0,1]^2\subset M$.

What is $\int_M d\omega$? What is $\int_{\partial M}\omega$?

I would like to find them explicitly, but don't know how. I would appreciate any hint to help me in the right direction. Also I use this version of Stokes:

Stokes' Theorem: Let $M$ be a smooth, oriented $n$-manifold with boundary, and let $\omega$ be a compactly supported smooth $(n-1)$-form on $M$. Then $$\int_M d\omega = \int_{\partial M} \omega.$$

We have canonical coordinates on $M$, namely $x,y$. w.r.t these coordinates we can write $$\omega=f dx + gdy$$ for some $f,g\in C^\infty(M)$. Then $$d\omega = \left(\frac{\partial g}{\partial x}-\frac{\partial f}{\partial y}\right)dx\wedge dy$$ Hence $$\int_{[0,1]^2}d\omega=\int_0^1\int_0^1\frac{\partial g}{\partial x}-\frac{\partial f}{\partial y}dxdy$$ \begin{align} \int_{\partial[0,1]^2}\omega&=\int_{\{0\}\times[0,1]}(fdx+gdy) +\int_{\{1\}\times[0,1]}(fdx+gdy)+\int_{[0,1]\times\{0\}}(fdx+gdy)+\int_{[0,1]\times\{1\}}(fdx+gdy)\\ &=\int_{\{0\}\times[0,1]}gdy +\int_{\{1\}\times[0,1]}gdy+\int_{[0,1]\times\{0\}}fdx+\int_{[0,1]\times\{1\}}fdx \end{align} Where we use that technically Stokes' theorem says $$\int_{\partial M}i^*\omega = \int_{M}d\omega$$ where $i:\partial M\to M$ is the inclusion, and $i^*$ the pullback. Now the pullback along the inclusion of e.g. $dx$ to $\{0\}\times[0,1]$ is $0$. So in this case Stokes' theorem says that $$\int_0^1\int_0^1\frac{\partial g}{\partial x}-\frac{\partial f}{\partial y}dxdy=\int_{\{0\}\times[0,1]}gdy +\int_{\{1\}\times[0,1]}gdy+\int_{[0,1]\times\{0\}}fdx+\int_{[0,1]\times\{1\}}fdx$$
• My formulation was indeed not very clear, but I wanted to find explicit expressions for $\int_M d\omega$ and $\int_{\partial M} \omega$ and thus show that $\int_M d\omega=\int_{\partial M} \omega$ (so I should obtain the same expressions for them) – user389900 Nov 16 '16 at 21:05
• I'm not sure how to interpret something as $\int_{\{0\}\times [0,1]}gdy$ though. I assume $\int_{\{0\}\times [0,1]}gdy+\int_{\{1\}\times [0,1]}gdy=\int_0^1\int_0^1 \frac{\partial g}{\partial x} dxdy$, but why? – user389900 Nov 16 '16 at 21:11
• Do you know how integration on a manifold is defined? $\{0\}\times [0,1]$ is a 1-dimensional manifold, so just follow the definition. – user2520938 Nov 16 '16 at 21:13