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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 $\mathbb{R}^3$.

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up vote 3 down vote accepted

Assume that $M = \{x \in \mathbb{R}^n \mid g(x) = 0 \}$ and $\partial_1 g > 0$. Then by the area formula we have $$ I = \int\limits_{M}f(x)dH^{n-1}(dx) = \int\limits_{\mathbb{R}^{n-1}}f(y) J_[\varphi](y) dL^{n-1}(y) $$ where $J_{n-1}[\varphi]$ is a $(n-1)$-dimensional Jacobian of the mapping $\varphi(y_1,...,y_{n-1}) = (x_1(y),y_1,...,y_{n-1})$, where $x_1(y)$ is such that $g(x_1(y),y_1,...,y_{n-1})=0$. So $J_{n-1}(y) = \frac{|\nabla g|}{\partial_1 g}$. By the other hand $\frac{|\nabla g|}{\partial_1 g} y_1 \wedge ... \wedge y_{n-1}$ is a pullback of the form $\omega$ such that $\frac{dg}{| \nabla g |} \wedge \omega$ = dx. The volume form $dS$ on $M$ satisfies the same equation. Then $$ I = \int\limits_{M} f(x)dS $$ If M is arbitraty we must use a partition of unity.

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