= 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> by
$$
\omega_{g,x}=\sqrt{\det G_x}\,dx^1\wedge\cdots\wedge dx^n.
$$
If $y$ is another positive <coordinate chart> and $J=\partial x/\partial y$, the transformation rule for the <metric tensor> is $G_y=J^TG_xJ$. Taking <determinants> gives
$$
\det G_y=(\det J)^2\det G_x,
\qquad \sqrt{\det G_y}=|\det J|\sqrt{\det G_x}.
$$
The <oriented atlas> has $\det J>0$, so the <wedge product of differential forms> transformation from part (a) gives
$$
\sqrt{\det G_y}\,dy^1\wedge\cdots\wedge dy^n
=\sqrt{\det G_x}\,dx^1\wedge\cdots\wedge dx^n.
$$
The local expressions therefore glue to a global smooth top-degree <differential form>. Its local coefficient never vanishes, so it is a <volume form>. \b[The global answer is $\boxed{\omega_g=\sqrt{\det(g_{ij})}\,dx^1\wedge\cdots\wedge dx^n}$ 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.
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