Silly doubt on differential forms notation

Question

A little fact is confusing me concerning Differential forms. Besides all kinds of motivation, a differential $p-$form is just the following tensor:

$$\textbf{F} = F_{\mu_{1}...\mu_{p}}(x^{\nu})\textbf{dx} ^{\mu_{1}}\wedge...\wedge \textbf{dx}^{\mu_{p}}\tag{1}$$

Well, it is clear that $F_{\mu_{1}...\mu_{p}}(x^{\nu})$ are components of a covariant tensor field.

On the other hand, a volume integral can be calculated, formally as:

$$ \int_{V}\rho(x^{\nu})\textbf{dx} ^{\alpha}\wedge \textbf{dx}^{\beta} \wedge \textbf{dx}^{\gamma} =: \int_{V}\omega \tag{2}$$

The thing is, since we integrating a $3-$form I would expect a $(0,3)$-tensor field components, instead we have an scalar field $\rho(x^{\nu})$. I mean, I would expect something like:

$$ \int_{V}\rho_{\alpha \beta \gamma}(x^{\nu})\textbf{dx} ^{\alpha}\wedge \textbf{dx}^{\beta} \wedge \textbf{dx}^{\gamma} =: \int_{V}\omega \tag{3}$$

So, my doubt is: what suppose to mean then the object $\omega$ in $(2)$?

Answer 1

3-forms in $\mathbb{R}^3$ form a one-dimensional space. Any three-form has the structure $\rho(x)dx_1\wedge dx_2\wedge dx_3$.

Answer 2

Integrating the volume form means that you are integrating the form of top degree, that is the degree of the form coincides with the dimension of the space.

I'd guess that you are working in the 3-space, so that $dx_1, dx_2, dx_3$ form the basis of one forms, and therefore any three form is written as $$\rho \cdot dx_1 \wedge dx_2 \wedge dx_3.$$