 Solution

Use coordinate charts from the chosen oriented atlas. In each, the matrix $G_x=(g_{ij})$ of the Riemannian metric is smooth and positive-definite, so $\det G_x>0$ and its positive square root is smooth. Define the local Riemannian volume form byIf $y$ is another positive coordinate chart and $J=\partial x/\partial y$, the transformation rule for the metric tensor is $G_y=J^T{G}_{x}J$. Taking determinants givesThe oriented atlas has $\det J>0$, so the wedge product of differential forms transformation from part (a) givesThe local expressions therefore glue to a global smooth top-degree differential form. Its local coefficient never vanishes, so it is a volume form. The global answer is $\boxed{{\omega }_g=\sqrt{\det (g_{ij})}d{x}^1\wedge \cdots \wedge d{x}^n}$ in positive charts. In an arbitrary negatively oriented chart the corresponding expression needs a minus sign; the orientation is essential to the form, although a positive metric volume density exists without it.