There are two tools, generalizing a concept of a volume to the case of submanifolds in $\mathbb{R}^n$, namely the Hausdorff measure $H^k$ and the volume form. The question is how to show that if $M$ is an orientable $k$-submanifold in $\mathbb{R}^n$ with a volume form $dV$ then
$$
 \int\limits\_{M} f(x) dV = \int\limits\_{M} f(x) H^k(dx)
$$
if the integrals exist.


P.S. Maybe the question is too silly for MathOverflow and more suitable for Mathematics Stack Exchange, but I have 2 reasons to post it here:


1. In books on geometric integration theory (Krantz, Parks; Federer) I failed to find an answer.
2. I've already posted the question on Mathematics Stack Exchange and on one more forum, but I didn't receive any response.



Hausdorff measure and the volume form
I guess you know that it is true in $\mathbb R^k$.


Without loss of generality we can assume that $f\ge 0$.
Fix $\varepsilon>0$ and cover your manifold by $(1\mp\varepsilon)$-Lipschitz charts.
Break your integrals into pieces using subordinate partition of unity and put these pieces back together.
Since in $\mathbb{R}^{k}$ you have equality, you will get
$$
\int\limits\_{M} f(x)\cdot dV\ \ \lessgtr\ \ (1\pm \varepsilon)^k\cdot\int\limits\_{M} f(x)\cdot H^k(dx)
$$
Since $\varepsilon>0$ is arbitrary your statement follows.


