Understanding integration of k-forms I try to wrap my head around the idea of integrating a k-form over a manifold. Loosely speaking, my intuition is, that for  each point of the manifold, we do a projection onto the tangent space and get the "area". For example the 3 dimensional unit-sphere is a 2-dimensional manifold, consequently the tanget space is also 2-dimensional. To this point my intuition was that, we do an integral over 2-dimensial-volumes that were measured in the tanget space. 
This also fits together with the idea that a k-form spits out the volume of the parallelepiped, that was spanned by its k-input vectors. In case of two dimensions, this would be a plane. This idea leads me to choosing the following 2-form to measure the surface of the unitball in 3 dimensions:
$$\omega = dx_1\wedge dx_2 + dx_1\wedge dx_3 + dx_3\wedge dx_2$$
Each 2-form in the sum can measure a surface. For the innocent purpose of just measuring the surface, all the coefficients are set to 1. The integral would look like this:
$$\int_{S} \omega= \int_{S}dx_1\wedge dx_2 + dx_1\wedge dx_3 + dx_3\wedge dx_2$$
But this makes no sense to me, because those 2-forms are the derivatives of the canonical projections that are wedged together. When I plugin a point $p\in S $ into $\omega$, all I get is a sum of functionals that still need an input: $$\omega(p) = x_1\wedge x_2 + x_1\wedge dx_3 + x_3\wedge x_2$$
Where does it take the additional input from? Propably from the vectors that span the tangent space. I could do a parametrization of the manifold to get them as the columns in the jacobian-matrix.  
 A: A more natural form for the calculation of the area of the unit sphere is
$$\omega = \sin\varphi \, d\varphi \wedge d\theta$$
where $\varphi \in [0, \pi]$ and $\theta \in [0, 2\pi)$ are the angles shown here.
To transform this into Cartesian coordinates we start from
$$\begin{cases}
x = \sin\varphi \, \cos\theta \\
y = \sin\varphi \, \sin\theta \\
z = \cos\varphi
\end{cases}$$
which gives
$$\begin{cases}
dx = \cos\varphi \, \cos\theta \, d\varphi - \sin\varphi \, \sin\theta \, d\theta \\
dy = \cos\varphi \, \sin\theta \, d\varphi + \sin\varphi \, \cos\theta \, d\theta \\
dz = -\sin\varphi \, d\varphi
\end{cases}$$
or
$$\begin{cases}
d\varphi = \frac{1}{z} \left( \frac{x}{\sqrt{x^2+y^2}} dx + \frac{y}{\sqrt{x^2+y^2}} dy \right) \\
d\theta = \frac{x}{x^2+y^2} dy - \frac{y}{x^2+y^2} dx
\end{cases}$$
Thus,
$$\omega = \sqrt{x^2+y^2} \, \frac{1}{z} \left( \frac{x}{\sqrt{x^2+y^2}} dx + \frac{y}{\sqrt{x^2+y^2}} dy \right) \wedge \left( \frac{x}{x^2+y^2} dy - \frac{y}{x^2+y^2} dx \right) \\
= \frac{1}{z} \left( \frac{x^2}{x^2+y^2} dx \wedge dy - \frac{y^2}{x^2+y^2} dy \wedge dx \right) \\
= \frac{1}{z} dx \wedge dy
$$
