Question: triple integral function,NonlinearFit, integration range or variable must be specified in the second argument

Dear Friends, 

I would appreciate your help in resolving some issues. Let me describe my dummy code and the issues I am having. 

I want to know the proper value of two parameters beta and `ΔA` in a tripe integral function  by the NonlinearFit command.The triple integral function is complex.

The code of function is below:

int(tan(beta)^2*exp(-Pi*tan(beta)^2*((x-varepsilon)^2+(0-varsigma)^2)/eta^2)/eta^2, [eta = 22.83-sqrt((5.83+`ΔA`)^2-varepsilon^2) .. 22.83+sqrt((5.83+`ΔA`)^2-varepsilon^2), varepsilon = -5.83-`ΔA` .. 5.83+`ΔA`, varsigma = -1 .. 1])-(int(tan(beta)^2*exp(-Pi*tan(beta)^2*((x-varepsilon)^2+(0-varsigma)^2)/eta^2)/eta^2, [eta = 22.83-sqrt(5.83^2-varepsilon^2) .. 22.83+sqrt(5.83^2-varepsilon^2), varepsilon = -5.83 .. 5.83, varsigma = -1 .. 1])) 

And this is the complete code:

datax := [-8, -4.5, -.5, 4.5, 8, 11.5, 14.5];

datay := [0.287e-2, 0.266e-2, 0.259e-2, 0.199e-2, 0.164e-2, 0.113e-2, 0.78e-3];

f := NonlinearFit(int(tan(beta)^2*exp(-Pi*tan(beta)^2*((x-varepsilon)^2+(0-varsigma)^2)/eta^2)/eta^2, [eta = 22.83-sqrt((5.83+`ΔA`)^2-varepsilon^2) .. 22.83+sqrt((5.83+`ΔA`)^2-varepsilon^2), varepsilon = -5.83-`ΔA` .. 5.83+`ΔA`, varsigma = -1 .. 1])-(int(tan(beta)^2*exp(-Pi*tan(beta)^2*((x-varepsilon)^2+(0-varsigma)^2)/eta^2)/eta^2, [eta = 22.83-sqrt(5.83^2-varepsilon^2) .. 22.83+sqrt(5.83^2-varepsilon^2), varepsilon = -5.83 .. 5.83, varsigma = -1 .. 1])), datax, datay, x)

I try to run it and get the value of parameters beta and `ΔA` ,but I keep getting this error,

Error, (in Statistics:-NonlinearFit) integration range or variable must be specified in the second argument, got HFloat(1.0) = HFloat(16.073603031200726) .. HFloat(29.58639696879927)
Does someone how to deal with this problem?

I would sincerely appreciate any inputs in this regard.