If surface tension uses perimeter and length to define, how can we use it to predict the surface a water droplet must take in 3D space? Often we say a water droplet takes spherical shape in zero gravity due to surface tension. However, to my knowledge surface tension is related to the length of the boundary over which the surface of the liquid is formed; yet a sphere is a closed surface, hence it has no boundary line. So how do we apply the concept of surface tension here?
 A: If you cut open a droplet from middle you can see

Here the boundary length is the circumference of the sphere=2πr.
You can cut anywhere, but it is generally cut through middle for symmetry and easy calculations.
A: There is a way of defining surface tension $\Sigma$ as a force per unit length on the boundary of a surface, but that turns out not to be the most useful way of thinking about $\Sigma$.  This is similar to how pressure $P$ can be conceived in mechanics as a force per unit area, but in thermodynamics, we find that it is usually more useful to think of pressure as the quantity that appears in the First Law of Thermodynamics, $dU=dQ-P\,dV$.  This makes it possible to redefine pressure as a partial derivative,
$$P=-\left(\frac{\partial U}{\partial V}\right)_{S}.$$
(The appearance of entropy $S$. here is because we need to specify that this is derivative is taken during an adiabatic process, with no heat transfer $dQ=T\,dS=0$.)
Surface tension plays the same role as $-P$ in a two-dimensional system like a surface.  The differential work associated with a surface tension is $dW=\Sigma\,dA$, where $dA$ is the differential change in surface area.  (The difference in sign between this equation and $dW=-P\,dV$ for a fluid arises from the fact that pressure wants to expand the volume of a fluid, but surface tension wants to shrink the area of a surface.  For a one-dimensional system, like a rope of spring, with a tension ${\cal T}$, stretched by a distance $d\ell$, the differential work is $dW={\cal T}\,d\ell$, which is nothing more than the Work-Energy Theorem.)  So we can think of the $\Sigma$ as the rate at which the energy increases as he increase the surface area adiabatically,
$$\Sigma=\left(\frac{\partial U}{\partial A}\right)_{S}.$$
With this notion of surface tension, it is easy to see that if you start with a spherical drop of water and start to deform it (while keeping the volume constant), the area will necessarily increase by an amount $dA$.  This is associated with an energy cost $\Sigma\,dA$, so the sphere (since it is the most compact configuration, with the smallest surface area to volume) is the lowest-energy, hence equilibrium, state of the drop.
