I'm trying to calculate the volume form $dx\wedge dy\wedge dz$ on $\mathbb{T}^3=\mathbb{R}^3/\mathbb{Z}^3$:
$\int_{\mathbb{T}^3}dx\wedge dy\wedge dz$
I've been told that it's simply the volume of the 'integral domain' $[0,1)^3$, i.e $1$. Though I'm having trouble understanding why this is. Going by definition (that I learned), we have to find a partition of unity subordinate to an atlas on $\mathbb{T}^3$ and integrate those locally. Is there any way to see this easily?
As a followoup question, in general can one say something about a integrating on a quotient of a smooth manifold where we know how to integrate on the original space?