A 3D surface can be written like $z = f(x,y)$. In my case, the domain of $f(x,y)$ is $x \ge 0$ and $y \ge 0$. Furthermore, $z>0$ in all the domain, and it is finite, continuous and derivable.
Given an arbitrary function $y = g(x)$, with domain $x \ge 0$ and such that $y \ge 0$, we can calculate the integral
$$ I_g = \int_0^\infty f(x, g(x)) dx $$
Now I don't really know the shape of $f(x,y)$, but for physical reasons it must be that $I_g=1$ for any arbitrary $g(x)$ (i.e. the integral domain). My intuition is that this condition ($I_g=1$) can be easily fulfilled if $f(x,y) = f(x)$ (i.e. $f$ does not depend on $y$), because any such function that integrates to 1 will integrate to 1 for any $g(x)$.
The question is: can there exist a function $f(x,y)$ that does depend on $y$ and fulfills the condition $I=1$? Or how can I prove its non-existence?
I have an idea, but I'm not sure it's correct. If $f(x,y)$ fulfills the condition $I_g=1$ for a specific domain $g(x)$, we can construct a different domain
$$g_2(x) = g(x) + \begin{cases} a & \text{if $x = b$} \\ 0 & \text{otherwise} \\ \end{cases}$$
where $a, b$ are two parameters. In this case, $g_2(x) = g(x)$ in all the domain, except for $x=b$, where $g_2(b)-g(b) = a$. Then, in order for $I_{g_2}$ to be 1 (i.e. the integral with $g_2(x)$ as domain), it must be that $f(b, g_2(b)) = f(b, g(b))$. Repeat for all possible values of $a$ and $b$, and we proved that $f$ must be constant along $y$.
Am I correct? Maybe the problem is that $g_2(x)$ isn't continuous?
Thanks