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?
 A: 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 $\mathbb{R}^{n}$.
A: 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 $0<r<R$, 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 $\alpha_{i_1\cdots i_k}$,
$$ \alpha_{i_1\cdots i_k} (x) = \lim_{r\to 0} \frac{1}{|B^k_r(0)|} \int_{\Phi_r} \alpha.$$
