The arc-lenght integral can be expressed as the inner product:
$$ \int \sqrt{\mathbf{g}_\tau \cdot \mathbf{g}_\tau}d\tau \tag{1} $$
where $\mathbf{g}_\tau$ is a vector of the tangent space of a manifold M. Then using the replacement procedure $\mathbf{g}_\tau =\frac{\partial}{\partial \tau} \gamma(\tau) $ we get
$$ \int \sqrt{\frac{\partial \gamma_\mu(\tau)}{\partial \tau}\frac{\partial \gamma^\nu(\tau)}{\partial \tau}} d\tau $$
which is the familiar arclength integral.
Then, reading this article, https://web.science.uu.nl/itf/Teaching/2009/Mischa%20Spelt.pdf, the author recovers the Nambu-goto action using geometric algebra by defining a surface integral:
$$ S=\iint \sqrt{(\mathbf{g}_\tau \wedge \mathbf{g}_\sigma)^2}d\tau d\sigma \tag{2} $$
where $\sqrt{(\mathbf{g}_\tau \wedge \mathbf{g}_\sigma)^2}$ is the area associated with the vectors $\mathbf{g}_\tau$ and $\mathbf{g}_\sigma$. Taking these vectors to be elements of some tangent space of $M$, the replacement $\mathbf{g}_\tau=\frac{\partial}{\partial \tau }x(\tau,\sigma)$ and $\mathbf{g}_\tau=\frac{\partial}{\partial \sigma }x(\tau,\sigma)$ and with some causual algebraic manipulations, eventually produces the Nambu-Goto action (Page 25):
$$ S=\int \sqrt{ \left( \frac{\partial x^\mu(\tau,\sigma)}{\partial \tau}\frac{\partial x_\nu(\tau,\sigma)}{\partial \sigma} \right)^2 - \left( \frac{\partial x^\mu(\tau,\sigma)}{\partial \tau}\frac{\partial x_\nu(\tau,\sigma)}{\partial \sigma} \right) \left( \frac{\partial x^\mu(\tau,\tau)}{\partial \sigma}\frac{\partial x_\nu(\tau,\sigma)}{\partial \sigma} \right)}d\tau d\sigma \tag{3} $$
One can identify a progression from arc-lengh to area to volume to 4-volume by jamming more wedge products. For instance one can write the volume form as :
$$ \iiint \sqrt{(\mathbf{g}_x \wedge \mathbf{g}_y \wedge \mathbf{g}_z)^2} dxdydz $$
and the 4-volume:
$$ \iiiint \sqrt{(\mathbf{g}_x \wedge \mathbf{g}_y \wedge \mathbf{g}_z \wedge \mathbf{g}_t)^2} dxdydzdt $$
My question relates to the 4-volume and why the replacements from $\mathbf{g}$ to their derivative formulation play no role, while they do in 1 or 2 dimensions. For example, consider the following integral:
$$ \iiiint \sqrt{|\det g_{\mu\nu}|} d^4x $$
This integral is the consequence of taking
$$ \iiiint \sqrt{(\mathbf{e}_0\wedge \mathbf{e}_1 \wedge \mathbf{e}_2 \wedge \mathbf{e}_0 )^2}d^4x=\iiiint \sqrt{(\sqrt{|\det g_{\mu\nu} |}\gamma_0\wedge \gamma_1 \wedge \gamma_2 \wedge \gamma_0 )^2}d^4x=\iiiint (\sqrt{|\det g_{\mu\nu} |}d^4x \tag{4} $$
So now, I am questioning why are we not using the same definition for lower dimensions. For instance, in the case of (2), can we not just write:
$$ \iint \sqrt{\mathbf{g}_\tau \wedge \mathbf{g}_\sigma}d\tau d \sigma=\iint \sqrt{|\det g^{\mu\nu}|}d \tau \sigma \tag{5} $$
What do the derivatives give us that the determinant doesn't? What does (3) give us that (5) doesn't?
Or, form the other direction, why would I not define the volume form (4) by replacing the vectors $\mathbf{e}_u$ with derivatives, and what would I gain/lose by doing so? Why (4) and not have a 4-volume form as:
$$ \iiiint \sqrt{\left( \frac{\partial \phi}{\partial x} \wedge \frac{\partial \phi}{\partial y} \wedge \frac{\partial \phi}{\partial z} \wedge \frac{\partial \phi}{\partial t} \right)^2}dxdydzdt $$