Value of exterior derivative as of limit of integral

Question

Let $w$ be a $(n-1)$-form in $\mathbb{R}^n$, and $e_1,…,e_n$ be a basis for the tangent space at $x\in\mathbb{R}^n$. Let $B(x,r)$ be the ball centered at $x$ with radius $r$ and $S(x,r)$ it's boundary. Then:
$$dw_x(e_1,…,e_n)=\lim_{r\rightarrow0}\frac{1}{vol.B(x,r)}\int_{S(x,r)}w$$
Where $vol. B(x,r)$ is the volume of the ball, given by:
$\int_{B(x,r)}dx_1\wedge…\wedge{x_n}$.
I'm aware of this approach: Prove that $f(x) = \lim_{\epsilon\rightarrow 0}\frac{1}{Vol(B_\epsilon(x))}\int_{B_\epsilon(x)}f(y)dV$
However, I was wondering if it would be possible to solve this using Stoke's Theorem, that is, my idea was to use the volume formula:
$vol.B(x,r) = 1/n\int_{S(x,r)}\sum_{i=1}^n(-1)^{i+1}x_idx_1\wedge…\wedge\hat{x_i}\wedge …\wedge x_n$
. And suppose $w= \sum_{i=1}^n(-1)^{i+1} f_idx_1\wedge…\wedge\hat{x_i}\wedge …\wedge x_n$ so that $dw = \sum \frac{\partial f_i}{\partial x_i}dx_1\wedge…\wedge{x_n}$.
Then the equation becomes:
$$\sum \frac{\partial f_i}{\partial x_i}(x)=\frac{1}{n}\lim_{r\rightarrow0}\int_{S(x,r)}\frac{\sum_{i=1}^n(-1)^{i+1} f_i(x)dx_1\wedge…\wedge\hat{x_i}\wedge …\wedge x_n}{\sum_{i=1}^n(-1)^{i+1}x_idx_1\wedge…\wedge\hat{x_i}\wedge …\wedge x_n}$$
But I'm not sure how to group things together then…

Answer 1

The correct proof uses Stokes's Theorem together with the continuity result you've linked. What you've written makes no sense, as you can neither divide differential forms nor integrate a function (as opposed to an $(n-1)$-form) over the sphere. But we have \begin{align*} d\omega(x)(e_1,\dots,e_n) &= \sum \frac{\partial f}{\partial x_i}(x) = \lim_{r\to 0^+} \frac 1{\text{Vol}(B(x,r))} \int_{B(x,r)}d\omega \\ &= \lim_{r\to 0^+} \frac 1{\text{Vol}(B(x,r))} \int_{S(x,r)}\omega, \end{align*} as desired.