Ellipses 101.

I chose the ellipses $(\dfrac{x^2}{9} + y^2 = 1$ above the line $y = 0)$ and $(\dfrac{x^2}{9} + (y - 1)^2 = 1$ below the line $y = 1)$ .

Solving for $y$ in both ellipses we want:

$f(x) = \dfrac{1}{3}\sqrt{9 - x^2}$ and $g(x) = 1 - \dfrac{1}{3}\sqrt{9 - x^2}$ .

$f(x) = g(x) \implies x = \pm\dfrac{3\sqrt{3}}{2} \implies$ the area $A = \displaystyle\int_{-\frac{3\sqrt{3}}{2}}^{\frac{3\sqrt{3}}{2}} f(x) - g(x) dx = \dfrac{2}{3}\displaystyle\int_{-\frac{3\sqrt{3}}{2}}^{\frac{3\sqrt{3}}{2}} \sqrt{9 - x^2} dx - 3\sqrt{3}$

For $I = \dfrac{2}{3}\displaystyle\int_{-\frac{3\sqrt{3}}{2}}^{\frac{3\sqrt{3}}{2}} \sqrt{9 - x^2} dx$

Let $x = 3\sin(\theta) \implies dx = 3\cos(\theta) \implies I = 6\displaystyle\int_{-\frac{\pi}{3}}^{\frac{\pi}{3}} \cos^2(\theta) d\theta = 3\displaystyle\int_{-\frac{\pi}{3}}^{\frac{\pi}{3}} 1 + \cos(2\theta) d\theta =$ $3(\theta + \dfrac{1}{2}\sin(2\theta))|_{-\frac{\pi}{3}}^{\frac{\pi}{3}} = 2\pi + \dfrac{3\sqrt{3}}{2} \implies A = 2\pi - \dfrac{3\sqrt{3}}{2} \approx \boxed{3.685109}$

It is given that the centre of the ellipses are seperated at 1 unit distance. So considering $(0,\frac{1}{2})$ as the centre of one ellipse (say ${\color{#D61F06}A}$ ) and $(0,\frac{-1}{2})$ as the centre of the other ellipse (say ${\color{#3D99F6}B}$ ), we can censtruct two congruent ellipses in the coordinate axis who meet at two points which lie on the $x$ axis (as a consequence of symmetry).

It is given that the length of the major axis and the minor axis of the ellipses are $3$ units and $1$ unit respectively. So the equation of the ellipses are -

$\Large{{\color{#D61F06}A}: \frac{x^2}{9} + \frac{(y-\frac{1}{2})^2}{1} = 1}$

$\Large{{\color{#3D99F6}B}: \frac{x^2}{9} + \frac{(y+\frac{1}{2})^2}{1} = 1}$

They are constructed in the coordinate axis as below

We notice that the enclosed area between the curves in twice this area -

So what we can do is that we can find the integeral of the ellipse ${\color{#3D99F6}B}$ from $x = \frac{{-3}{\sqrt{3}}}{2}$ to $x = \frac{{3}{\sqrt{3}}}{2}$ and then multiply it by 2.

The integeral we get is $\large { I = \displaystyle\int_{-\frac{3\sqrt{3}}{2}}^{\frac{3\sqrt{3}}{2}} (\sqrt{1 - \frac{x^2}{9}} + \frac{1}{2}) dx }$

And hence the total area is $\large {A = 2I = 2\int_{-\frac{3\sqrt{3}}{2}}^{\frac{3\sqrt{3}}{2}} (\sqrt{1 - \frac{x^2}{9}} + \frac{1}{2}) dx }$

Which after calculation gives the answer $A = 3.685109$

NOTE: The enclosed area could also be divided into four equal parts by dividing it along the $x$ and $y$ axis and in that case, the same integral will be calculated but with the lower limit as $x=0$ . Multiplying 4 to the integral gives us the enclosed area