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I'd like to know if there's a way to find the equation of a surface, like the one shown in this image: enter image description here

In this image, there is a series of red lines plotted in $x,y$ space. Each red line is a line of equal $z$ value, such that the red lines define a surface, and that z is a function of x and y: $z = f(x, y)$ (imagine the red lines extend to infinity and do not cut off at the graph edges). Arbitrarily, suppose the red line closest to the origin is a low number, and the value of each red line (that is, the value of $z$) increases towards the "northeast". Conceptually, it would look something like this in $x, y, z$ space:

surface in 3-d

With a graph like this, how do I find an equation that describes, or even just approximates, the shape of the surface in terms of $x$ and $y$?

Edit:

This is what I've tried so far. Since all the red lines are linear, I can easily compute the slope and intercept of each. I can then compile a table of numerous $x, y$ pairs for every $z$ value. In this case, I made a table of about $n=1600$. I have some software (XLSTAT) that can do multi-variate non-linear regressions, so I plugged $z$ in as a dependent variable, and $x$ and $y$ as independent variables, to come up with a 4th order equation that approximates the shape of the surface. Is this a sound method? The result is below. I've superimposed a colored Excel plot of the surface in plan view, and a chart with the red lines that I am trying to approximate mathematically. You can see that the charts match in places but diverge in places as well:

superimposition of non-linear regression solution

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    $\begingroup$ If those red lines are straight lines and not parallel, they will intersect. What does that say about your assertion that $z = f(x,y)$? $\endgroup$ Commented Dec 3, 2014 at 23:31
  • $\begingroup$ I suppose it means that $f(x,y)$ has non-unique solutions. Does that then mean that it isn't possible to approximate the shape of the surface with a function? What about if the surface has limits? $\endgroup$
    – dagrha
    Commented Dec 3, 2014 at 23:58

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