Of course, this is an elliptic integral. But I don't know much about elliptic integrals, and besides what good is that information? It doesn't give me a number. (I didn't actually need to know this for anything, but somehow the knowledge that the answer is, say, 2π+ε (it's not) is more satisfying than the fact that it can be written in terms of some special function that I don't know that well.
It turns out the answer is (2+ε)π.
Here's a sketch of a proof. This ellipse can be parametrized by x = (1+ε) cos t, y = sin t, with t varying from 0 to 2π. So the length is
Expand, and let ε2 = 0, to get
which simplifies to (recalling cos2 t + sin2 t = 1)
Now, since ε is "small", 2ε cos2 t is also "small". In general, for small h, (1+h)1/2 ~ 1 + (h/2); thus this is approximately
and now we recall that cos2 t has average value 1/2. So this is (2+ε)π, which is the answer.
More formally, if f(a) is the length of an ellipse with axis lengths 1 and a, this shows (modulo a couple leaps of faith) that f'(1) = π -- this can easily be checked with a computer algebra system, or if you actually know more about elliptic integrals than I do. This has the nice interpretation that if the lengths of the two axes of an ellipse are close to each other, the length of the ellipse is just π times the mean of the axis lengths -- which is true for a circle, where it just says that the circumference of a circle is π times its diameter. The Taylor series for the elliptic integral which gives the length is, according to Maple,
and from numerical data, for, say, ε = 0.1, this method predicts an ellipse length of 2.1π = 6.5973; the actual length is 6.6011. Even for ε = 0.5 the error's only about one percent (7.85 versus 7.93).
The real question that I started out with, though, still isn't answered -- is there some nice geometric way to see this?