# Value of exterior derivative as of limit of integral

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...

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.