I know how one can define the surface area via the charts of a surface in $\mathbb{R}^3.$
Now, I read that the canonical surface area form for such a surface with surface normal $n$ is given by $\omega( \xi,\eta ) = \det(n,\xi,\eta).$
My question is now: How can I see that both definitions coincide? So why does the integral over $\int_{A} \omega$ the surface area now? The thing is that I am completely new to differential forms and maybe it is more or less the definition, but my problem is that I don't know how it follows now.
If anything is unclear, please let me know.
Let $r(u,v)$ denote the parameterized surface, then $r_u$ and $r_v$ are the tangent vectors, and the unit normal vector is $$n=\frac{r_u\times r_v}{\left\| r_u\times r_v \right\|}$$ Now according to the reference the surface area element is $\sigma=\left\|r_u\times r_v\right\|du\wedge dv$, we want to show that $\sigma=\omega$, that is, given any two vectors $\xi$ and $\eta$, we want to show that $$\left\|r_u\times r_v\right\|du\wedge dv(\xi, \eta)=\det(n, \xi, \eta)$$ Since both functions are multilinear and skew-symmetric we need only to verify the equality on the basis $r_u\wedge r_v$. That is plug in $\xi=r_u$ and $\eta=r_v$ we get $$LHS=\left\|r_u\times r_v\right\|$$ and $$RHS=\det(\frac{r_u\times r_v}{\left\| r_u\times r_v \right\|}, r_u, r_v)$$ $$=\frac{r_u\times r_v}{\left\| r_u\times r_v \right\|}\cdot (r_u\times r_v)=\frac{\left\|r_u\times r_v\right\|^2}{\left\| r_u\times r_v \right\|}$$ $$=LHS$$ Refer to Triple Product to see how the determinant turns to triple product.