Going threedimensional (my first solution for the elliptic integral) Creating a field of velocities with several arbitrary functions of z, we can write the circulation along an ellipse be proportional to the ellipse arc. Then we create a rotational tub, from the speed field having horizontal ellipses, as M(z)x2+N(z) y2= 1
(1)
On their surface, circulation along a closed line will be zero (in the case of a triangle, the sum of 3 circulations) and we shall put the ellipse arc AB as function of this circulation ar BC amd CA. Choosing planes passing by z-ax, y=L(z) 路x with L defined by x1,y1 and x2,y2 extremes of arc.
We choose CD for the circulation be integrable, a surface made with straight lines y=L(z) 路x ( passing by z-ax). In the lines BC and CA the calculation is elemental because with L=constant, we will have from (2)