# Second derivative of integral function

Let $$f: \mathbb R^2 \rightarrow \mathbb R$$ be a smooth strictly convex function with unique minimum at $$0$$ such that all level sets $$A_x:=\left\{z ; f(z) \le x \right\}$$ are compact. Imagine something like $$f(z)=\Vert z \Vert^2.$$

Define the integral function

$$F(x):=\int_{A_x} g(z) dz$$

where $$g$$ is as smooth as you like.

Question: Is there a way to analytically determine an expression for $$F''(x)$$?

Let $$B_u:=\{z\colon f(z)=u\}$$ for $$u>u_0:=\min f$$. Let $$[0,2\pi)\ni t\mapsto(x_u(t),y_u(t))$$ be any smooth parametrization of $$B_u$$, so that $$B_u=\{(x_u(t),y_u(t))\colon t\in[0,2\pi)\}$$.
(For instance, one may take $$(x_u(t),y_u(t))=(\rho_u(t)\cos t,\rho_u(t)\sin t)$$, where $$\rho_u(t):=f_t^{-1}(u)$$ and $$f_t^{-1}\colon(u_0,\infty)\to(0,\infty)$$ is the function inverse to the function $$f_t\colon(0,\infty)\to(u_0,\infty)$$ defined by the formula $$f_t(r):=f(r\cos t,r\sin t)$$.)
Then $$$$F'(u)=\int_0^{2\pi}dt\,\sqrt{x'_u(t)^2+y'_u(t)^2}\frac{g(x_u(t),y_u(t))}{|(\nabla f)(x_u(t),y_u(t))|}, \tag{\ast}$$$$ and hence $$$$F''(u)=\int_0^{2\pi}dt\,\frac d{du}\Big(\sqrt{x'_u(t)^2+y'_u(t)^2}\frac{g(x_u(t),y_u(t))}{|(\nabla f)(x_u(t),y_u(t))|}\Big).$$$$
Here, $$dt\,\sqrt{x'_u(t)^2+y'_u(t)^2}$$ is the infinitesimal length element of the curve $$B_u$$, and $$\frac{du}{|(\nabla f)(x_u(t),y_u(t))|}$$ is the infinitesimal distance between the curves $$B_u$$ and $$B_{u+du}$$ near the point $$(x_u(t),y_u(t))$$.
We can verify formula $$(\ast)$$ for e.g. $$f(x,y)=x^2+y^4$$ and $$g(x,y)=y^2$$, in which case we have the following closed-form expressions: for $$u>0$$ $$$$F(u)=\frac{2 \sqrt{\pi } \Gamma \left(\frac{7}{4}\right)}{3 \Gamma \left(\frac{9}{4}\right)}\,u^{5/4} ,\quad F'(u)=\frac{5 \sqrt{\pi } \Gamma \left(\frac{7}{4}\right)}{6 \Gamma \left(\frac{9}{4}\right)}\,u^{1/4},$$$$ and the latter expression coincides with the integral in $$(\ast)$$, with the parametrization $$(x_u(t),y_u(t))=(u^{1/2}\cos t,u^{1/4}(\sin t)^{[1/2]})$$ of $$B_u$$, where $$w^{[1/2]}:=|w|^{1/2}\,\text{sign}\, w$$.