Volume of My Valentine 2

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:

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

$V_{R} = 2\pi (\dfrac{4}{5}) \int_{0}^{\frac{\sqrt{5} - 1}{2}}$ $x((1 - \dfrac{2x}{\sqrt{5} - 1} + \sqrt{1 - x^2} - \sqrt{x})) dx =$ $\dfrac{8\pi}{5} \int_{0}^{\frac{\sqrt{5} - 1}{2}} (x - \dfrac{2x^2}{\sqrt{5} - 1} + x\sqrt{1 - x^2} - x\sqrt{x}) dx$

For $\int x\sqrt{1 - x^2} dx$

Let $x = \sin(\theta) \implies dx = \cos(\theta) \implies \int x\sqrt{1 - x^2} dx = -\int (\cos(\theta))^2 (-\sin(\theta)) d\theta = -\dfrac{1}{3}(\cos(\theta))^3$ .

$\therefore \int_{0}^{\frac{\sqrt{5} - 1}{2}} x\sqrt{1 - x^2} dx =$ $\dfrac{1}{3}(1 - (\dfrac{\sqrt{5} - 1}{2})^{\frac{3}{2}}) \implies$

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

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

$V_{R} = \dfrac{8\pi}{5}(\dfrac{1}{3} - \dfrac{1}{3}(\phi - 1)^{\frac{3}{2}} + \frac{1}{6}(\phi - 1)^2 - \dfrac{2}{5}(\phi - 1)^{\frac{5}{2}}) =$ $\dfrac{2^3\pi}{5}(\dfrac{1}{3} - \dfrac{1}{3}(\phi - 1)^{\frac{3}{2}} + \frac{1}{2 * 3}(\phi - 1)^2 - \dfrac{2}{5}(\phi - 1)^{\frac{5}{2}}) = \dfrac{a^3\pi}{b}(\dfrac{1}{c} - \dfrac{1}{c}(\phi - 1)^{\frac{c}{a}} + \frac{1}{a * c}(\phi - 1)^{a} -\dfrac{a}{b}(\phi - 1)^{\frac{b}{a}})$

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