Desired Area

Using the symmetry about the $y-axis$ we only need the right side $(x \geq 0)$ of the graph.

For $x \geq 0 \implies f(x) = \dfrac{4}{5}(\sqrt{x} - \sqrt{1 - x^2}), g(x) = \dfrac{4}{5}(\sqrt{x} - 3\sqrt{1 - x^2})$ , and $x = \dfrac{25}{64}y^2 \implies y = \pm\dfrac{8}{5}\sqrt{x}$ .

Let $h(x) = \dfrac{8}{5}\sqrt{x}$ and $j(x) = \dfrac{-8}{5}\sqrt{x}$ .

$I_{1} = \int_{0}^{\dfrac{\sqrt{5} - 1}{2}} (f(x) - h(x)) dx = \dfrac{4}{5}\int_{0}^{\dfrac{\sqrt{5} - 1}{2}} (\sqrt{1 - x^2} - \sqrt{x}) dx$

Let $x = \sin(\theta) \implies dx = \cos(\theta)$

$\implies I_{1} = \dfrac{4}{5}(\dfrac{1}{2}\arcsin(\dfrac{\sqrt{5} - 1}{2}) + \dfrac{1}{2}(\dfrac{\sqrt{5} - 1}{2})^{\dfrac{3}{2}} - \dfrac{2}{3}(\dfrac{\sqrt{5} - 1}{2})^{\dfrac{3}{2}}) =$ $\dfrac{2}{5}(\arcsin(\dfrac{\sqrt{5} - 1}{2}) - \dfrac{1}{3}(\dfrac{\sqrt{5} - 1}{2})^{\dfrac{3}{2}})$ .

$I_{2} = \int_{0}^{\dfrac{\sqrt{5} - 1}{2}} (j(x) - g(x)) dx = \dfrac{12}{5}\int_{0}^{\dfrac{\sqrt{5} - 1}{2}} (\sqrt{1 - x^2} - \sqrt{x}) dx = 3I_{1} \implies$

The desired area $A = 2(4I_{1}) = 8I_{1} = \dfrac{16}{5}(\arcsin(\dfrac{\sqrt{5} - 1}{2}) - \dfrac{1}{3}(\dfrac{\sqrt{5} - 1}{2})^{\dfrac{3}{2}})$ .

Letting $\phi = \dfrac{\sqrt{5} + 1}{2} \implies A = \dfrac{16}{5}(\arcsin(\phi - 1) - \dfrac{1}{3}(\phi - 1)^{\dfrac{3}{2}}) = \dfrac{a}{b}(\arcsin(\phi - 1) - \dfrac{1}{c}(\phi - 1)^{\dfrac{c}{d}}) \implies a + b +c+ d = \boxed{26}$