# Stokes theorem for manifolds with corners?

Maybe this is an elementary question, but I'm unable to find the appropriate reference for it. The Stokes theorem tells us that, for a $n+1$-dimensional manifold $M$ with boundary $\partial M$ and any differentiable $n$-form $\omega$ on $M$, we have $\int_{\partial M} \omega = \int_M d\omega$.

But Stokes theorem is also true, say, for a cone $M = \{(x,y,z) \in \mathbb{R}^3 \ \vert\ \ x^2 + y^2 = z^2, 0\leq z \leq 1 \}$, or a square in the plane, $M =\{(x,y) \in \mathbb{R}^2 \ \vert\ 0 \leq x, y\leq 1 \}$ which are not manifolds. So my questions are:

1. Are these cone and square examples of what I think are called "manifold with corners"?
2. If this is so, where can I find a reference for a version of Stokes' theorem for manifolds with corners?
3. If "manifold with corners" is not, which is the appropriate setting (and a reference) for a Stokes' theorem that includes those examples?

Any hints will be appreciated.

EDIT: Since thanking individually everyone would be too long, let me edit my question to acknowledge all of your answers. Thank you very much: I've found what I was looking for and more.

• You may find usefull the article of Joyce "On manifolds with corners" arxiv.org/abs/0910.3518. Square is considered as a manifold with corners, but the cone usually not, it seems. – Dmitri Panov Jan 25 '10 at 10:13
• A cone is not a manifold with corners. A cone on the other hand is a stratified space, and the proof of Stokes' that Orbicular mentions works for them, and even more general objects. – Ryan Budney Jan 25 '10 at 14:04
• You may want to take a look at Brian Conrad's notes on differential geometry: math.stanford.edu/~conrad/diffgeomPage/handouts.html – Johnson Jia Jan 25 '10 at 22:39

Triangulate your manifold $M$ so that $\partial M$ is triangulated as well. Then prove Stokes' theorem for diffeomorphic images of a standard simplex, as in de Rham's "Variétées différentiables".