Consider a smooth convex/compact domain $D\subset \mathbb{R}^n$ and a smooth, concave function $F:D\to \mathbb{R}$. Then we can define the function that simply takes the volume of the upper contour sets determined by the argument:
$$G(t) = \int_{\{x\in D \; : \; F(x) \ge t\}} d\lambda$$
where $\lambda$ denotes the Lebesgue measure. I'm trying to figure out an expression for $\frac{d}{dt}G(t)$.
This seems like nothing more than a special case of a higher-dimensional Leibniz Integral Rule, but wikipedia gives me a substantially more general formula than I suspect I need for this case (for definitions of terms see the link):
$$\frac{d}{dt} \int_{\Omega(t)} \omega = \int_{\Omega(t)} i_{\vec{v}}(d_x \omega) + \int_{\partial \Omega(t)} i_{\vec{v}}\omega + \int_{\Omega(t)} \dot{\omega}.$$
I have almost no background in differential forms, but immediately I know, for starters, the volume form I'm integrating is time invariant so the last term drops out here. Moreover, given I'm just concerned with a uniform density, I'd imagine the first term should be zero too? (This corresponding to the intuition that all that really matters here is how much 'volume bleeds out of the bag $\Omega(t)$' as I cinch it shut by increasing $t$, and hence I need only be concerned with the incremental flow of volume across the boundary.) But that may be wildly incorrect.
Ideally if someone could help guide me (ideally both intuitively and analytically) to be able to understand and describe this derivative I'd be very grateful! In particular an expression for what the Leibniz rule reduces to in this case would be most welcome.