General Stokes' Theorem

Theorem
If $\omega$ is any smooth $(n-1)$-form with compact support on a smooth $n$-dimensional manifold $X^n$, with oriented boundary $\partial X$ then


 * $\displaystyle \int_{\partial X} \omega = \int_X \mathrm d \omega$

where $\mathrm d \omega$ is the exterior derivative of $\omega$.

Proof
Choose a finite family of coordinate systems $V_1,\dots,V_k$ on $X$ and a partition of unity $f_1,\dots,f_k$ with $f_1+\dots+f_k=1$ on $\mathrm{supp}\,\omega$ with $\mathrm{supp}\,f_i\subset V_i$ and $\mathrm{supp}\,f_i$ compact.

Put $\omega_i = f_i \omega$, then we have $\omega = \omega_1+\dots+\omega_k$ and also $\mathrm{supp}\,\omega_i\subset V_i$ and $\mathrm{supp}\,\omega_i$ compact. If Stokes' theorem holds, then we have


 * $\int_X d\omega=\sum^k_{i=1}\int_xd\omega_i=\sum^k_{i=1}\int_{\partial X}\omega_i=\int_{\partial X}\omega$

Therefore, we are reduced to showing that the theorem holds for specific patches of $X$.

Suppose $V$ is a boundary type domain, and let $\mathrm{supp}\,\omega\subset V$ with $\mathrm{supp}\,\omega$ compact, and let


 * $\omega=f(y^1,\dots,y^n)\,dy^2\wedge\dots\wedge dy^n$

and then we have


 * $d\omega=\frac{\partial f}{\partial y^1}(y^1,\dots,y^n)\,dy^1\wedge dy^2\wedge\dots\wedge dy^n$

so that


 * $\int_{X}d\omega=\int_{y(V)}\frac{\partial f}{\partial x^1}dx^1\dots dx^n=0=\int_{\partial X}\omega$

because $\int^1_{-1}\frac{\partial f}{\partial x^1}dx^1=0-0=0$ and the pullback of $\omega$ to the boundary is zero.

Suppose $V$ is a interior type domain,