Calculus on manifolds - limit of an integral

[meiji1]

Homework Statement

Let $$\phi \in C^{\infty}_{0}(\mathbb{R}^2)$$ and $$f: \mathbb{R}^2 \to \mathbb{R}$$ a smooth, non-negative function. For $$c > 0$$, let $$< F_c, \phi > := \int_{\{f(x,y) \le c\}} \phi(x,y)\mbox{dx dy}$$. Supposing the gradient of $$\frac{\partial f}{\partial x}$$ is nonzero everywhere on $$M = \{(x,y) : f(x,y) = c \}$$, prove that the pointwise limit $$lim_{\epsilon \to 0^+} \frac{<F_{c+\epsilon},\phi> - <F_c,\phi>}{\epsilon}$$ exists for every $$\phi \in C^{\infty}_{0}(\mathbb{R}^2)$$.

Homework Equations

N/A

The Attempt at a Solution

I know nothing about calculus on manifolds, which I'm sure would help here. The exercise is given in the context of distribution theory.

I get that the idea is to use the implicit function theorem within $$\frac{F_{c+\epsilon} - F_c}{\epsilon} = \int_{c < f(x,y) \le c+\epsilon} \frac{\phi(x,y)}{\epsilon}\mbox{dy dx}$$, somehow. This poses two major problems I can't get past: first, I need to resolve the epsilon in the denominator of the integrand, I would suspect through a substitution of the form $$y' = \epsilon y$$ (or analogously with x, of course).

Second, I can't see how to extend the implicit function theorem to the set $$M_\epsilon = \{(x,y): c \le f(x,y) \le c+\epsilon\}$$. The implicit function $$g(y)$$ given by the theorem is defined and differentiable on some ambiguous open neighborhood of a point $$(x,y) \in M$$ only, with $$f(g(y),y) = c$$ in that neighborhood; it's not clear how to extend g past M to obtain $$f(g(y),y) = f(x,y)$$ for all points $$(x,y) \in M_\epsilon$$, if that's the proper approach.

 

[Nino MMP]

Edit: I found the answer: the integral formulation is incorrect, and this is an application of the fundamental theorem of calculus of variations. The limit is \int_{M} \frac{\partial f}{\partial x} \phi\mbox{dy dx}.