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.
 A: 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.
A: 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.
