Leibnitz rule multiple integral

Question

Let $$h(z) = \iint_{\{(x,y)\in\mathbb{R}:\; g(x,y)\leq z\}}f(x,y)dxdy.$$

How to find the derivative of $h$? Should $g$ be smooth? Is there a generalization of the Leibnitz rule to this case?

Answer 1

Not a complete answer, but \begin{align*} h(r)=\iint_{B_{r}(0)}f(x,y)dxdy \end{align*} is such that \begin{align*} h'(r)=\int_{\partial B_{r}(0)}f(x,y)d\mathcal{H}^{1}(x,y) \end{align*} a.e. $r>0$, this is due to Coarea formula.

Answer 2

It seems that we have a theorem called coarea formula which states that if $\Omega$ is an open subset of $\mathbb{R}^n$ and that both $f$ and $u$ are nice functions on $\Omega$, then

$$ \int_{\Omega} f(\mathrm{x})|\nabla u(\mathrm{x})| \, d\mathrm{x} = \int_{\mathbb{R}} \left( \int_{u^{-1}(t)} f(\mathrm{x}) \, \mathcal{H}^{n-1}(d\mathrm{x}) \right) \, d\mathrm{t} $$

Here $\mathcal{H}^{n-1}$ is the $(n-1)$-dimensional Hausdorff measure (or simply you may consider it the $(n-1)$-dimensional surface measure). In fact, it only requires $u$ is Lipschitz and $f$ is integrable, but probably this is an overkill.

In our case, this tells that

$$ h(z) = \int_{\{ g(\mathrm{x}) \leq z \}} f(\mathrm{x}) \, d\mathrm{x} = \int_{-\infty}^{z} \left( \int_{\{ g(\mathrm{x}) = t\}} \frac{f(\mathrm{x})}{|\nabla g(\mathrm{x})|} \, \mathcal{H}^{1}(dx) \right) \, dt $$

and hence we have

$$ h(z) = \int_{\{ g(\mathrm{x}) = z\}} \frac{f(\mathrm{x})}{|\nabla g(\mathrm{x})|} \, \mathcal{H}^{1}(dx) = \int_{\{ g(\mathrm{x}) = z\}} \frac{f(\mathrm{x})}{|\nabla g(\mathrm{x})|} \, ds. $$

Here, the last integral is a line integral and the last equality holds when the level set for $g(\mathrm{x}) = z$ is a nice curve.