TITLE: Volume form and Hausdorff measure
QUESTION [8 upvotes]: Let $M$ be a smooth orientable $(n-1)$-dimensional submanifold in $\mathbb{R}^n$, $dS$ be its volume form and $dH^{n-1}(x)$ be an $(n-1)$-dimensional Hausdorff measure. How to show than that
$$
  \int\limits_{M} f(x) dS = \int\limits_{M} f(x) dH^{n-1}(x)
$$
In fact, it is a generaliation of an equality formula for surface integrals of first and second kind in $\mathbb{R}^3$.

REPLY [2 votes]: You can use the area formula. The following argument appears as section 3.3.4, example D in Evans and Gariepy's Measure Theory and Fine Properties of Functions. Let $f:V\to M$ for $V \subset \mathbb{R}^{n-1}$ be a coordinate map, then the area formula (section 3.3.2, theorem 1) says that for $\eta\in C^\infty(M)$, $\int_V \eta Jf \,dLeb_{n-1}=\int_{\mathbb{R}^n} \sum_{x \in f^{-1}(\{y\})} \eta(x) \, d\mathscr{H}^{n-1}(y)=\int_M \eta \, dS$. But writing $g_{ij}=\frac{\partial f}{\partial x^i} \cdot \frac{\partial f}{\partial x^j}$ for the Euclidean inner product $\cdot$, we have $(Df)^* \circ Df=((g_{ij}))$, so $Jf=\sqrt{\det(g)}$. By the standard result (e.g. Prop. 15.31 in Lee's Introduction to Smooth Manifolds), we have  $\int_V\sqrt{\det(g)} \, dLeb_{n-1}=\int_M\,dS$.