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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.

enter image description here

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.

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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:

enter image description here

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.

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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.

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

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