Chhaer Sothun Your equation seems like a dY/dS-type gradient, so it shouldn't be hard to do an integral to obtain an analytical formula for the surface.

Chhaer Sothun Maybe you meant (Y2-Y1)/sqrt(pow(X2-X1,2)+pow(Y2-Y1,2)). This expression can be treated as sin(Th) in the local cylindrical coordinate system (R,Th,Z) with the origin at the point (X1,Y1) and the polar angle Th. Coordinate system can be created in the same dialog box as analytical field.

Zherui Martinez-Guo Thank you for your answer. But I did not get what you mean by that.

Victor J. Аdlucky Thank you for your answer. But this expression is not for cylindrical coordinates, I use it for rectangular coordinates (X, Y). And in the analytic field, as default, I can input X and Y parameters only (could not express as X2-X1 or Y2-Y1).