Is Stokes' Theorem natural in the sense of category theory?

Here's an attempt to make the "dinaturality" observation (more) precise. I will be leaving out many details that I haven't worked out, so I may go wrong somewhere; I hope the general outline makes sense though.

• Let $\mathbb N$ be the poset of natural numbers in the usual ordering (or $\mathbb N$ could be the universal chain complex, i.e. category with the same objects given by natural numbers, $\mathrm{Hom}(n,m) = \mathbb{R}$ if $m \in \{n,n+1\}$, $0$ else, and all composites with nonidentity maps equal to zero. Then all the functors here are $\mathbb R$-linear).
• Let $\mathcal V$ be the category of topological vector spaces or some suitable similar category.
• Fix a manifold $X$ (or some other sort of smooth space).

Then we have functors

• $C: \mathbb N^\mathrm{op} \to \mathcal V$ where $C_n$ is the vector space freely generated by smooth maps $Y \to X$ where $Y$ is a compact, $n$-dimensional, oriented manifold with boundary, and the induced map $\partial: C_{n+1} \to C_n$ is the boundary map.
• $\Omega: \mathbb N \to \mathcal V$ is the de Rham complex; $\Omega_n = \Omega_n(X)$ is the space of $n$-forms on $X$ and the induced map $\mathrm d: \Omega_n \to \Omega_{n+1}$ is the exterior derivative.

Assuming that $\mathcal V$ has a suitable tensor product defined, we obtain a functor

• $C \otimes \Omega: \mathbb N ^\mathrm{op} \times \mathbb N \to \mathcal V$.

while there is also the constant functor

• $\mathbb R: \mathbb N ^\mathrm{op} \times \mathbb N \to \mathcal V$

Then Stokes' theorem says that we have an extranatural transformation

• $\int : C \otimes \Omega \to \mathbb R$ which, given a map $Y \to X$ and a form $\omega$ on $X$, pulls the form back to $Y$ and integrates it (returning 0 if it's the wrong dimension).

Interestingly, this means that integration should descend to a map out of the coend $\int : \int^{n \in \mathbb N} C_n \otimes \Omega_n \to \mathbb R$ (that first integral means integration of differential forms while the second means a coend). I'm not sure what the value of this coend is or how much it depends on the details I've left ambiguous. I suppose it probably has something to do with the de Rham cohomology of $X$?

There is some discussion here:

http://ncatlab.org/nlab/show/Stokes+theorem

and a reformulation. I must confess I didn't spend a lot of time on it, as I don't know what are $(\infty, 1)$-categories, etc.

To me, the greatest thing about Stokes' theorem is that it paves the way for de Rham's theorem. Indeed we can't even state the latter without the former. The de Rham theorem is very classical and important in geometry.