Let $(M,g_{ij})$ be a smooth Riemannian manifold. We introduce a new metric $\tilde{g}_{ij}: = (e^{\phi} g)_{ij}$ where $\phi:M \to \mathbb{R}$ is a smooth function. Now we want to calculate the volume form w.r.t. the new metric in terms of the old volume form. All of this in local coordinates. I found two ways to do this:
Let $(U,x_1,...,x_n)$ be the local coordinates on M, such that the frame $(\partial x_1,..., \partial x_n)$ is orthonormal w.r.t. $g_{ij}$.
the "old" volume form is then just $vol = dx_1 \wedge dx_2 \wedge ... \wedge dx_n$.
an orthonormal frame $(f_1,...,f_n)$ w.r.t. $\tilde{g}_{ij}$ is then given by $f_i := e^{-\frac{\phi}{2}} \partial{x_i}$. So that we get the new volume form: $vol_{new} = e^{-\frac{\phi}{2}} d{x_1} \wedge ... \wedge e^{-\frac{\phi}{2}}d{x_n} = e^{-n\frac{\phi}{2}} vol$
$vol_{new} = \sqrt{det(\tilde{g}_{ij}})dx_1 \wedge ... \wedge dx_n = e^{n \frac{\phi}{2}}\sqrt{(det(g_{ij})}dx_1 \wedge...\wedge dx_n = e^{n \frac{\phi}{2}} vol$.
Now why don't I get the same result ?