More of Ellipses 101.

I chose the ellipses $(\dfrac{x^2}{a^2} + y^2 = 1$ above the line $y = 0)$ and $(\dfrac{x^2}{a^2} + (y - 1)^2 = 1$ below the line $y = 1)$ .

Solving for $y$ in both ellipses we want:

$f(x) = \dfrac{1}{a}\sqrt{a^2 - x^2}$ and $g(x) = 1 - \dfrac{1}{a}\sqrt{a^2 - x^2}$ .

$f(x) = g(x) \implies x = \pm\dfrac{\sqrt{3}}{2}a \implies$ the area $A = \displaystyle\int_{-\frac{\sqrt{3}}{2}a}^{\frac{\sqrt{3}}{2}a} f(x) - g(x) dx = \dfrac{2}{a}\displaystyle\int_{-\frac{\sqrt{3}}{2}a}^{\frac{3\sqrt{3}}{2}a} \sqrt{a^2 - x^2} dx - \sqrt{3}a$

For $I = \dfrac{2}{a}\displaystyle\int_{-\frac{\sqrt{3}}{2}a}^{\frac{\sqrt{3}}{2}a} \sqrt{a^2 - x^2} dx$

Let $x = a\sin(\theta) \implies dx = a\cos(\theta) \implies I = 2a\displaystyle\int_{-\frac{\pi}{3}}^{\frac{\pi}{3}} \cos^2(\theta) d\theta = a\displaystyle\int_{-\frac{\pi}{3}}^{\frac{\pi}{3}} 1 + \cos(2\theta) d\theta =$ $a(\theta + \dfrac{1}{2}\sin(2\theta))|_{-\frac{\pi}{3}}^{\frac{\pi}{3}} = a(\dfrac{2\pi}{3} + \dfrac{\sqrt{3}}{2}) \implies A = a(\dfrac{2\pi}{3} - \dfrac{\sqrt{3}}{2}) = a(\dfrac{\alpha\pi}{\beta} - \dfrac{\sqrt{\beta}}{\alpha}) \implies \alpha + \beta = \boxed{5}$