we have a Riemannian manifold (M, g), a smooth, proper function $u$ and a smooth function $f$.
I am stuck with computing the derivative along $r$ of the following quantity:
$$ I(r)=\int_{\{u=r\}} f d\sigma. $$
I guess that actually a normal derivative of $f$ has to appear, as well as the mean curvature $H$ of $\{u=r\}$, coming from the derivative of the area element.
Anyway, I am not able to figure out a correct formula.
Thank you in advance.
It is a bit more convenient to define $$ F(r) = \int_{\{u=r\}} f \, \frac{d\sigma}{|\nabla u|}. $$ That is because $|\nabla u|^{-1} \,d\sigma$ is a more natural surface measure for level sets, since it takes the thickness into account. That is large gradient gives a thin level set and small gradient gives a thick one.
In this case, it holds $$ F'(r)= \int_{\{u=r\}} \operatorname{div}\left( f \frac{\nabla u}{|\nabla u|^2}\right) \, \frac{d\sigma}{|\nabla u|} . $$ By using the chain rule for the divergence, one recovers two contributions $$ \operatorname{div}\left( f \frac{\nabla u}{|\nabla u|^2}\right) = \nabla f \cdot \frac{\nabla u}{|\nabla u|^2} + f \operatorname{div}\left( \frac{\nabla u}{|\nabla u|^2}\right) . $$ Hence, we see the two contribution you conjecture. The first is related to the normal derivative of $f$ and the second is related to the curvature of the level set itself.
To proof the formula, we note that for any smooth function $g:\mathbb{R}\to \mathbb{R}$ it holds by the co-area formula $$ \int_{\mathbb{R}} F(r) g(r) \, dr = \int_{\mathbb{R}} \int_{\{u=r\}} f\ g(r) \frac{d\sigma}{|\nabla u|} \, dr = \int_M f \ g\circ u \ d{vol} . $$ Then, we can calculate with $g'$ instead of $g$ inside of the above formula $$\begin{aligned} \int_{\mathbb{R}} F(r) g'(r) \, dr &= \int_M f\ g'\circ u \, d{vol}\\ &= \int_M f \ \frac{\nabla u}{|\nabla u|^2} \cdot \nabla (g\circ u) \, d{vol}\\ &= - \int_M g\circ u \ \operatorname{div}\left( f \ \frac{\nabla u}{|\nabla u|^2} \right) \, d{vol} \\ &= - \int_{\mathbb{R}} g(r) \int_{\{u=r\}} \operatorname{div}\left( f \ \frac{\nabla u}{|\nabla u|^2} \right) \, \frac{d\sigma}{|\nabla u|} \, dr \end{aligned}$$ Now, letting vary $g$ over all smooth functions, we identified the formula in the distributional sense.