Intuition Behind Volume Forms on Surfaces I saw the theory of scalar integration on manifolds, where we define with respect to a riemann metric the volume form $$\omega_g = \text{det}[g_{ij}]^{\frac{1}{2}}\text{d}x^1\wedge\cdots\wedge\text{d}x^n$$ and using it defining scalar integration as just $\int_Mf\omega_g $. This writing is using local coordinates on an open subset of $M$ of course.
What I am looking for is some intuition on why this makes sense. I know that $\omega_g$ as an $n$-form accepts $n$ vectors and spits out volumes of parallelepipeds in a way that is compatible with $g$ (it gives $1$ to orthonormal frames). However, I don't see what the $$\text{det}[g_{ij}]^{\frac{1}{2}}=\omega_g\left ( \frac{\partial}{\partial x^1},\dots,\frac{\partial}{\partial x^n} \right )$$ is supposed to represent, besides just making $\omega_g$ independant of the chart. Is it some sort of infinitesimal volume?
 A: The metric tensor $g$ dictates the geometry on $M$. In a more direct way as an inner product it lets us have lengths and angles. But in a way having an inner product also gives us volume.
The origin of the term is how $g$-orthonormal bases relate to the coordinate basis.
Take some positively oriented orthonormal basis $(e_1,\dots,e_n)$ of $T_p M$. we have that $\omega_g(e_1,\dots,e_n)=1$. Now take any other basis $(v_1,\dots,v_n)$, with coordinates
$$v_i=\sum_{j=1}^n a_{ij}e_j$$
and denote $A=(a_{ij})$.
What do we expect $\omega_g(v_1,\dots,v_n)$ to be? Well this is scalling the cube created by $e_1,\dots,e_n$ using the linear transformation $A$, so if it were any like the euclidian case we would expect $\det A$! This could be shown formally by plugging in and using the permutation propety of determinants. But what is $\det A$?
We see that computing $g(v_i,v_j)$ translates to the usual dot product in components since
$$g(v_i,v_j) = \sum_\alpha \sum_\beta a_{i\alpha}a_{j\beta}g(e_\alpha,e_\beta)=\sum_\alpha \sum_\beta a_{i\alpha}a_{j\beta}\delta_{\alpha\beta}=\sum_\alpha a_{i\alpha}a_{j\alpha}=(A^\top A)_{ij}$$
So we have that for $G=(g(v_i,v_j))$
$$\sqrt{\det G}=\sqrt{\det(A^\top A)}=|\det A|$$
This tells us that in general, $\omega_g$ does infact as you said measures volume. The square root factor is what happenes when you feed $\omega_g$ the coordinate basis instead of the orthonormal one and it is just the determinant of the transition matrix but expressed in a different way (in the case $v_i = \frac{\partial}{\partial x^i}$).
A: For example think of spherical coordinates in $\mathbb{R}^3$. Then you have $\omega_g \equiv dV=r^2 \sin^2(\theta) dr \wedge d\theta \wedge d\phi$. The formula you wrote is just the generalization of the Jacobian in $\mathbb{R}^d$ to manifolds with a metric $g$.
