You still didn't get the point.
Look at the documentation of integral3:
q = integral3(fun,xmin,xmax,ymin,ymax,zmin,zmax) approximates the integral of the function z = fun(x,y,z) over the region xmin ≤ x ≤ xmax, ymin(x) ≤ y ≤ ymax(x) and zmin(x,y) ≤ z ≤ zmax(x,y)
Thus if your integration limit is a function of both other variables, it must appear as the last variable in the list ( in this case s ).
And this limit has to be a function handle, not a simple algebraic (or even symbolic) expression.
s = @(m,r) -(-1.85-1.013*(m-6)+1.496*log(sqrt(r.^2+31.025)));
fun = @(m,r,s)(r.*exp(-(s.^2/2+1.727*(m-5))))./(sqrt(r.^2-10^2));
result= integral3(fun,5,7.5,10,18,-Inf,s);
final_result=result*(2*1.727)/(sqrt(2*pi)*30)
