How is the differential form $dx_1 \wedge …\wedge dx_k$ “intuitively” an oriented $k$-dimensional volume element?

I am reading a paper on differential forms. They define it as follows:

A differential $$k$$-form on a domain $$U$$ is a continuous, infinitely differentiable map $$\omega \colon U \to \Lambda^k (\Bbb{R}^{n \ast})$$ where $$\Bbb{R}^{n\ast}$$ is the dual of $$\Bbb{R}^n$$ as a vector space.

In particular we may let $$x_1,...x_n$$ be a basis for $$\Bbb{R}^n$$ and let $$dx_i\in \Bbb{R}^{n \ast}$$ be the unique linear map $$\Bbb{R}^n\to \Bbb{R}$$ that satisfies $$dx_i(x_j)=\delta_{ij}$$

It then goes on to make a comment:

Intuitively, we may consider $$dx_1 \wedge ... \wedge dx_k$$ to be an oriented, $$k$$ dimensional volume element

I don't understand this remark. I know $$dx$$ from calculus is supposed to be an infinitesimal change and I also understand that wedge product is intuitively a way of turning vectors into higher dimensional objects.

But forget about wedge products. I don't understand how $$dx$$ defined as $$\delta_{ij}$$ has anything to do with an infinitesimal change.

What is going on here? Why is this abstract definition related to an infinitesimal change?

If you feed a list of $k$ vectors $v_1,\dots,v_k\in\Bbb R^n$ into the $k$-form $dx_1\wedge\dots\wedge dx_k$ you'll get the signed volume of the parallelepiped formed by $v_1',\dots,v_k'$, where $v_j'$ is the projection of $v_j$ into $x_1\dots x_k$-space.
• Actually, I have one really stupid question. Where does the "infinitesimal" come in? You feed in $k$ vectors to the $k$ form and you get a volume, where does infinitesimal come in? – user223391 Jul 1 '17 at 0:54
• "Infinitesimal" is a relic from the old-style, or physics-engineering, way of thinking of integrals. If you chop a $k$-dimensional surface into small pieces, the vectors $v_j$ you feed in here will be small vectors :) – Ted Shifrin Jul 1 '17 at 1:02