# How do I show that the following is a definition of an alternating form?

Suppose $$E$$ is an open set in $$\mathbb{R}^n$$. A differential form of order $$k$$ in $$E$$ (briefly, a k-form in E) is a function $$\omega$$, which assigns to each $$k$$ surface $$\Phi$$ in $$E$$ a number $$\omega(\Phi) = \int_{\Phi} \omega$$ $$\int_{\Phi}\omega = \int_{D} \sum a_{i_1 \cdots i_k}(\Phi(u)) \frac{\partial(x_{i_1},\cdots,x_{i_k})}{\partial(u_{i_1},\cdots,u_{i_k})}du \cdots \cdots \cdots (1)$$ where $$\omega$$ is symbollically representated as $$\omega = \sum a_{i_1 \cdots i_k}(x) dx_{i_1} \wedge \cdots \wedge dx_{i_k}$$

The definition of differential forms that was given to us in class is that $$\omega(x)$$ is an alternating $$k$$ tesnor. How are these two definitions related?How do I show that $$(1)$$ is an alternating $$k$$ tensor?

What I understand by differential form is the function your professor gave and define the integration of the differential form over a $$k$$ surface $$\Phi$$ as in $$(1)$$. The two definitions have different natures, but the main purpose is to formalize integration on surfaces and curves in $$\mathbb{R}^{n}$$.
It is clear that when a differenial form $$\alpha = \sum_{i_1< \cdots > i_k} \alpha_{i_1\cdots i_k} dx^{i_1} \wedge \cdots \wedge dx^{i_k}$$ is given, one uses it to define integration for any $$k$$ surface $$\Phi$$ using (1). One can show that any continuous differential form $$\alpha$$ can be recovered (by continuous I mean that $$\alpha_{i_1\cdots i_k}$$ are continuous functions), if we know $$\int_\Phi \alpha$$ for all $$k$$-surface $$\Phi$$.
To see this, let $$x = (x^1, \cdots, x^n) \in E$$ and any $$i_1 < \cdots < i_k$$. Let $$R>0$$ so that $$B^n_R(x) \subset E$$. Let $$\Phi$$ be the parametrized surface given by
$$\Phi : B^k_R(0) \to E, \ \ \Phi(u^1, \cdots, u^k) = (x^1 (u) , \cdots, x^n(u)),$$ where $$x^j (u) = \begin{cases} x^{i_l} + u^{i_l}, & \text{ if } j = i_l, l =1, \cdots, k, \\ x^j & \text{ otherwise}. \end{cases}$$
For any $$0, let $$\Phi_r = \Phi|_{B^k_r(0)}$$. Then $$\int_{\Phi_r} \alpha = \int_{B^k_r(0)} \alpha_{i_1\cdots i_k} (\Phi (u)) du$$ and by continuity of $$\alpha_{i_1\cdots i_k}$$,
$$\alpha_{i_1\cdots i_k} (x) = \lim_{r\to 0} \frac{1}{|B^k_r(0)|} \int_{\Phi_r} \alpha.$$