Least surface of volume with constraints

Question

We know that in 2D/3D the shape with the least surface of a certain volume is a circle/sphere (e.g. soap bubbles). Now Imagine we have a flat surface (tabletop) that can be used as part of the surface of the volume. Let's say the given volume is 1. Which shape (with the usage) has now the least surface (when you ignore the area of the 'flat' surface part)? When you look at soap bubbles that stick to a surface it looks like it was just a half sphere, is that true respectively do you know a proof?

Answer 1

Bubbles attempt to minimise their surface area (because that way the surface free energy is minimised).

It follows from the isoperimetric inequality that the shape will be either a sphere or a spherical cap.

A sphere of radius $R_0$ has surface area $4\pi R_0^2$ and volume $\frac{4}{3}\pi R_0^3$.

A hemisphere (with no base) of radius $R$ has surface area $2\pi R^2$ and volume $\frac{2}{3}\pi R^3$.

To achieve a constant volume (and therefore equilibrated pressure) we require $\frac{2}{3}\pi R^3=\frac{4}{3}\pi R_0^3$ or $R=R_02^{1/3}$. The surface area of the hemisphere is therefore $2\pi R_0^2 2^{2/3}$ which is less than the area of the complete sphere $4\pi R_0^2$. A spherical bubble can therefore reduce its surface area by becoming a hemisphere on a surface (with a 25% larger radius).

If you compute the surface area for a fixed volume as a function of contact angle then you will be able to show that $\theta=90^\circ$ is indeed the minimum. (Note that the contact angle is seldom $90^\circ$ for water droplets because they have an interfacial energy).