Given four non-planar 3D points, these points define a skew quadrilateral ABCD.
Example:
A = (0,0,0); B = (0,3,0); C = (3,3,2); D = (3,0,-1);
The bilinear interpolation of these vertices define a patch for a doubly-ruled surface that is an hyperbolic paraboloid. My question is:
Is there an analytical formula to compute the surface area of this patch?
(Note: this is not the minimal surface solved by Schwarz using abelian integrals.)
In a trivial case, with the surface given by:
z = x y
Parametric equations: x(u,v) = u; y(u,v) = v; z(u,v) = u v
And points given by:
A = (-1, -1, 1)
B = ( 1, -1, -1)
C = ( 1, 1, 1)
D = (-1, 1, -1)
The surface area can be found by integrating dS over this domain:
$dS = \sqrt{1 + u^2 + v^2}$
I've found the surface area for this trivial surface patch to be:
$S = 2/9 (-\pi + 6 (\sqrt{3} + \log{(7 + 4 \sqrt{3})})) \approx 5.12316$
It can be extended to other simple cases, such as A = (0, 0, 0), B = (0, b, 0), C = (a, b, a b), D = (a, 0, 0), but I could not extend it to the generic case (i.e., any A, B, C, D points). Note that the generic case is just an affine transformation of the trivial case.
