I'm currently reading a paper about incompressible Euler's equation, and I don't understand how the surface element expand. So here comes the question.
Let $\Omega$ be a Riemannian manifold with metric $g_{ab}$ and $N^{a}$ be the unit normal to $\partial\Omega$. We know the volume element is $$ dV_{g}=(det(g))^{\frac{1}{2}}\,dV $$ where $dV$ is the volume element under Eucidean meansure. Now we define the induced measure $\gamma_{ab}=g_{ab}-N_{a}N_{b}$ on the tangent space to the boundary $\partial\Omega$.Then the author claims that $$ dS_{\gamma}=(det(g))^{\frac{1}{2}}(\sum_{n}N_{n}^{2})^{-\frac{1}{2}}\,dS $$ I have no idea how the formula is derived. Since, intuitively, the measure change under different metric should be the square of the determinant of the metric and it seems not the case for the induced surface measure.
Let $(u,v)$ be an orthonormal basis (with respect to $g$) for the tangent space at some point of $\partial M$.
Then $$dS_{\gamma} = |u\times v|\,dS = (u\times v)\cdot \tilde{N}\,dS$$ where $\tilde{N}$ is the normal, under the Euclidean metric, to the surface. $N$ and $\tilde{N}$ are parallel; in particular $\tilde{N} = N/\|N\|$. Therefore $$dS_{\gamma} = \frac{(u\times v)\cdot N}{\|N\|}\,dS = \frac{dV_g}{\|N\|dV}\,dS = \frac{\sqrt{\det g}}{{\|N\|}}\,dS.$$