# Why does expanding the radius of a drop in contact with a surface by $\mathrm{d}x$ increase the surface area by $2\pi r \mathrm{d}x \cos\theta$?

Suppose we have a drop of water sitting on a solid surface in air.

The drop forms a portion of a sphere and intersects the surface in a circle. The radius of the circle, $$x$$ depends on the surface tension of the water-air, water-surface, and air-surface interfaces.

Suppose we increase the radius of the circle by a small amount $$\mathrm{d}x$$ while keeping the volume of the water fixed. This is illustrated in Capillarity and Wetting Phenomena by De Gennes, pp 17.

The first claim is that the water-solid surface area increases by $$2\pi x \mathrm{d}x$$. That makes sense to me.

The second claim is that, to first order, the change in area of the water-air surface is $$2\pi x \mathrm{d}x\cos\theta_E$$. Why is that true?

From the picture, I see that this is the area of a strip near the bottom of the dotted line, but I don't see why the rest of the dotted line should have the same area as the solid line.

I've shown that this is true, although not via the picture I included in the question. I still don't understand why the picture's argument works.

Wikipedia has these definitions for a spherical cap:

The volume is

$$V = \frac16 \pi h(3a^2 + h^2)$$

the area is

$$A = \pi(a^2 + h^2)$$

The volume of water is fixed, so $$\mathrm{d}V = 0$$. This implies

$$\mathrm{d}h = \frac{-2ah}{a^2 + h^2} \mathrm{d}a$$

Taking the differential of the area and plugging in the above expression, we get

$$\mathrm{d}A = 2\pi a\left(1 - \frac{2h^2}{a^2 + h^2}\right)\mathrm{d}a$$

If we define $$\phi$$ such that $$\tan\phi = \frac{h}{a},$$ then recalling $$2\cos^2\alpha - 1 = \cos(2\alpha)$$, the above is equivalent to

$$\mathrm{d}A = 2\pi a \cos(2\phi) \mathrm{d}a$$

A little geometry using the inscribed angle theorem can show that $$2\phi = \theta,$$ so

$$\mathrm{d}A = 2\pi a \cos\theta \mathrm{d}a,$$

which is what I wanted to show.

The new circumference is $$2\pi (x+dx)$$ and that is multiplied by the length of the inclined bit $$dx\, \cos \theta _{\rm E}$$ to give the required area if $$(dx)^2$$ is neglected as a second order term.

• The entire question is why, though. Why is the inclined bit the only extra area we need to worry about? Oct 18, 2018 at 12:50