Volume of my Valentine 3

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:

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

$V_{R} = \dfrac{8\pi}{5} \int_{0}^{\frac{\sqrt{5} - 1}{2}}$ $(a - x)(1 - \dfrac{2x}{\sqrt{5} - 1} + \sqrt{1 - x^2} - \sqrt{x})) dx =$ $\dfrac{8\pi}{5} \int_{0}^{\frac{\sqrt{5} - 1}{2}} (a - x)\sqrt{1 - x^2} + (a - x) - \dfrac{2ax}{\sqrt{5} -1} + \dfrac{2x^2}{\sqrt{5} - 1} - a\sqrt{x} + x^{\frac{3}{2}}) dx$

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

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

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

$\implies \int_{0}^{\beta} x\sqrt{1 - x^2} dx =\dfrac{a}{2}\arcsin(\beta) + \dfrac{a}{2}(\beta)^{\frac{3}{2}} + \dfrac{1}{3}(\beta)^{\frac{3}{2}} - \dfrac{1}{3} \implies$

$V_{R} = \dfrac{8\pi}{5} (\dfrac{a}{2}\arcsin(\beta) + \dfrac{a}{2}(\beta)^{\frac{3}{2}} + \dfrac{1}{3}(\beta)^{\frac{3}{2}} - \dfrac{1}{3} +$ $(ax - \dfrac{x^2}{2} - \dfrac{ax^2}{\sqrt{5} - 1}$ + $\dfrac{2}{3(\sqrt{5} - 1)} x^3 - \dfrac{2a}{3}x^{\frac{3}{2}} + \dfrac{2}{5}x^{\frac{5}{2}})|_{0}^{\beta}) =$ $\dfrac{8\pi}{5}((\dfrac{1}{2}\arcsin(\beta) + \dfrac{1}{2}(\beta) - \dfrac{1}{6}(\beta)^{\frac{3}{2}})a +$ $(\dfrac{1}{3}(\beta)^{\frac{3}{2}} - \dfrac{1}{6}(\beta)^2 + + \dfrac{2}{5}(\beta)^{\frac{5}{2}} - \dfrac{1}{3}))$

$V_{R} = \dfrac{8\pi}{5}(\arcsin(\beta) + \beta - \dfrac{1}{6}(\beta)^2 + \dfrac{2}{5}(\beta)^{\frac{5}{2}} - \dfrac{1}{3}) \implies$

$(\dfrac{3\arcsin(\beta) + 3\beta - (\beta)^{\frac{3}{2}}}{6})a = \dfrac{3\arcsin(\beta) + 3\beta - (\beta)^{\frac{3}{2}}}{3} \implies \boxed{a = 2}$ .