What role does the volume form take in the definition of the integral of an $n$-form?

Question

In Differential Geometry class I have been given the following definition for the integral of an $n$-form:

Let $(M,\nu)$ be an oriented manifold of dimension $n$. Let $\mathfrak{U} = \{(U_\alpha, \phi_\alpha)\}_{\alpha \in A}$ be an atlas of $M$ whose transition functions have positive jacobians and such that $\nu|_{U_\alpha} = f_\alpha dx^1 \wedge \dots \wedge dx^n$ with $f_\alpha > 0$ in $C^\infty(U_\alpha)$ for each local chart. Let $\{h_\alpha:\alpha \in A\}$ be a smooth, locally finite partition of unity such that the support of $h_\alpha$ is contained in $U_\alpha$ for each $\alpha \in A$. If $\omega \in \mathcal{A}^n(M)$, then $\omega = \sum_\alpha h_\alpha \omega$. It is not hard to see that the sum $\sum_\alpha \int_{U_\alpha}(h_\alpha \omega)$ does not depend on the partition of unity that is used. Therefore, we define the integral of the $n$-form $\omega$ over $M$ (relative to the orientation given) as follows:

$$\int_M \omega := \sum_\alpha \int_{U_\alpha} (h_\alpha \omega).$$

The issue I have with this definition is that I don't understand what role $\nu$ is playing in here. This is supposed to be a volume form in $M$, i.e., a nowhere-vanishing $n$-form. But it is not clear to me how things change if I use a different volume form. For example, if I change $\nu$ for $-\nu$, I know this is supposed to change the sign of the integral, but I don't understand how, where is $\nu$ being used in this definition?

Answer 1

Pages 204-206 of Nakahara's book on Geometry, Topology and Physics might help you understand more explicitly how the volume element affects the integral.

Answer 2

The only thing that matters about $\nu$ is its sign, i.e., any other n-form $\eta$ that satisfies $\eta=f\nu$ for some smooth function $f>0$ will induce the same orientation. Considering this orientation, one may tell if the charts from your atlas are oriented with respect to $\nu$ or not.

The definition of the integrals on the right hand side involves the orientation of the charts from the atlas. This is exactly the role that $\nu$ is playing. If you flip the sign and change $\nu$ by $-\nu$, the sign of every integral on the right hand side changes as well, therefore switching the sign of the whole integral at the left hand side.

Chapter 16 of John M. Lee's Introduction to Smooth Manifolds probably explains this much better than me at the subsection Integration on Manifolds, but the idea is this.