Volume Of My Heart.

For the $x$ intercept of $y_{1} = \dfrac{4}{5} (\sqrt{x} - \sqrt{1 - x^2})$ we obtain:

$x^2 + x - 1 = 0 \implies x = \dfrac{-1 \pm \sqrt{5}}{2}$ , since $x = -\dfrac{1 + \sqrt{5}}{2} = -\phi$ results in a complex valued square root $\implies x = \dfrac{\sqrt{5} - 1}{2}$ is the x intercept of $y_{1} = \dfrac{4}{5} (\sqrt{x} - \sqrt{1 - x^2})$ .

The equation of the line passing thru $A: (0,\dfrac{4}{5})$ and $B: (\dfrac{\sqrt{5} - 1}{2},0)$ is:

$y = \dfrac{4}{5} (\dfrac{-2x}{\sqrt{5} - 1} + 1) \implies$

Let $\phi = \dfrac{\sqrt{5} + 1}{2}$

$\implies$

$V_{R} = 2\pi (\dfrac{4}{5}) \int_{0}^{\phi - 1}$ $x(1 - \dfrac{x}{\phi - 1} + \sqrt{1 - x^2} - \sqrt{x}) dx$

$= \dfrac{8\pi}{5} \int_{0}^{\phi - 1} (x - \dfrac{x^2}{\phi - 1} + x(1 - x^2)^{\frac{1}{2}} - x^{\frac{3}{2}}) dx$

$= \dfrac{8\pi}{5} (\dfrac{x^2}{2} - \dfrac{x^3}{3(\phi - 1)} - \dfrac{1}{3}(1 - x^2)^{\frac{3}{2}} - \dfrac{2}{5}x^{\frac{5}{2}})|_{0}^{\phi - 1}$

$= \dfrac{8\pi}{5}(\dfrac{(\phi - 1)^2}{6} + \dfrac{1}{3} - \dfrac{1}{3}(1 - (\phi - 1)^2)^\frac{3}{2} - \dfrac{2}{5}(\phi - 1)^{\frac{5}{2}})$

$= \dfrac{8\pi}{5}(\dfrac{1}{3} - \dfrac{1}{3}(\phi - 1)^{\frac{3}{2}} + \dfrac{1}{6}(\phi - 1)^2 - \dfrac{2}{5}(\phi - 1)^{\frac{5}{2}})$

$= \dfrac{2^3\pi}{5}(\dfrac{1}{3} - \dfrac{1}{3}(\phi - 1)^{\frac{3}{2}} + \frac{1}{2 * 3}(\phi - 1)^2 - \dfrac{2}{5}(\phi - 1)^{\frac{5}{2}}) =$

$= \dfrac{a^3\pi}{b}(\dfrac{1}{c} - \dfrac{1}{c}(\phi - 1)^{\frac{c}{a}} + \frac{1}{a * c}(\phi - 1)^{a} -\dfrac{a}{b}(\phi - 1)^{\frac{b}{a}})$

$\implies a + b + c = \boxed{10}$ .

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Note: $1 - (\phi - 1)^2 = 2\phi - \phi^2 = \sqrt{5} + 1 - \dfrac{5 + 2\sqrt{5} + 1}{4} = \dfrac{2\sqrt{5} - 2}{4} =$ $\dfrac{\sqrt{5} - 1}{2} = \phi - 1$ .