Q:

Derivative of an integral quantity involving a moving level set

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.

A:

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. 

