I'm reading about differential forms and I have the following example. Let $M= \{(x,y,z) \in \Bbb R^3 \mid x^2+y^2+z^2 < 1 \}$ and consider the $2$-form $\omega = x dy \wedge dz - y dx \wedge dz+zdx \wedge dy.$ We can now integrate $\omega$ over $\partial M$ as follows.
We have that $$\begin{align*}dx &= \cos(u)\cos(v)du - \sin(v)\sin(u) dv \\ dy &= \cos(u)\sin(v)du + \cos(v)\sin(u) dv \\ dz &= -\sin(u)du \end{align*}$$
and the wedge products are $$\begin{align*}dx \wedge dy &= \cos(u)\sin(u)du \wedge dv \\ dx \wedge dz &= -\sin(v)\sin^2(u) du \wedge dv \\ dy \wedge dz &= \sin^2(u)\cos(v)du \wedge dv \end{align*}$$
and we get that $\omega = \sin(u) du \wedge dv.$ Thus integrating $$\int_{\partial M} \omega = \int_{0}^{2\pi}\int_{0}^\pi \sin(u) dudv = 4\pi.$$
What I don't get here is that what is the interpretation of this $2$-form $\omega$. I can see that the integral checks out by looking at the definitions, but I have no understanding what is $\omega$ representing.
What is the use of integrating over something that one cannot understand. Even in this case we're just integrating something over the boundary of the unit disc in $\Bbb R^3$ and yes the output is some number, but I have no idea what can I conclude from this.
I have been watching this lecture series by Keenan Crane and he does a really nice job on visualizing these things, but it's still not clicking for me quite yet.
He's given the following visualizations and what I'm trying to do is to figure out how they apply for example in the integral above. Can this even be done?
You have the volume 3-form $dV$ on the $\mathbb{R}^{3}$ which finds the signed volume of the parallelpiped spanned by a 3-vector (the determinant formula). Feeding it a unit vector makes the resulting 2-form essentially measure areas orthogonal to that unit vector.
For a surface $i:S^{2}\rightarrow \mathbb{R}^{3}$ the 2-form measuring areas is given by $$ds=i^{*}(n\lrcorner dV),$$ with interior product (contraction) and the pull-back (restriction) operations. The normals on the unit-sphere are $n=(x,y,z)$; the linearity of interior product gives
$$ds=x\ \partial_{x} \lrcorner dV+y\ \partial_{y} \lrcorner dV+z\ \partial_{z} \lrcorner dV.$$
Now remember that $dV=dx\wedge dy\wedge dz=dy\wedge dz\wedge dx=dz\wedge dx\wedge dy$; so contracting $dV$ with $\partial_{x^{i}}$ just removes the $dx^{i}$. Thus $$ds=x\ dy\wedge dz+y\ dz\wedge dx+z\ dx\wedge dy.$$
Integration intuitively is the limit process of feeding this 2-form with infinitesimal tangent 2-vectors and summing up all the measured infinitesimal areas.