It's all Symmetry.

Let $f(x) = \sqrt{|x|}$ and $g(x) = -x^2$ and $A: (x_{0},y_{0})$ .

$R_{(0,0)} \circ R_{y = x}(x_{0},y_{0}) = (-y_{0},-x_{0}) = B$

and $R_{y - axis}(-y_{0},-x_{0}) = (y_{0},-x_{0}) = D$ and $R_{y - axis}(x_{0},y_{0}) = (-x_{0},y_{0}) = C$

$\implies m_{AB} = 1$ and $m_{CD} = -1$

$m_{AB} = 1 \implies f'(x_{0}) = \dfrac{1}{2\sqrt{x_{0}}} = 1 \implies x_{0} = \dfrac{1}{4} \implies y_{0} = \dfrac{1}{2} \implies y = x + \dfrac{1}{4}$

and $m_{CD} = -1 \implies y = -x + \dfrac{1}{4}$

To find the $x$ coordinate of the find point of intersection $E$ of $y = -x + \dfrac{1}{4}$ and $f(x) = \sqrt{x}$

let $-x + \dfrac{1}{4} = \sqrt{x} \implies 1 - 4x = 4\sqrt{x} \implies 1 - 8x + 16x^2 = 16x \implies 16x^2 - 24x + 1 = 0 \implies x = \dfrac{3 \pm 2\sqrt{2}}{4}$

We drop $x = \dfrac{3 + 2\sqrt{2}}{4}$ which doesn't satisfy the initial equation and choose $x = \dfrac{3 - 2\sqrt{2}}{4}$ which satisfies the initial equation.

For $A_{R_{1}}$ we have:

$A_{R_{1}} = \displaystyle\int_{0}^{\frac{1}{4}} x + \dfrac{1}{4} - x^{\frac{1}{2}} \:\ dx =$ $\dfrac{1}{2}x^2 + \dfrac{1}{4}x - \dfrac{2}{3}x^{\frac{3}{2}}|_{0}^{\frac{1}{4}} =$ $\boxed{\dfrac{1}{96}}$

For $A_{R_{2}}$ we have:

$A_{R_{2}} = \displaystyle\int_{0}^{\frac{3 - 2\sqrt{2}}{4}} x^{\frac{1}{2}} + x^{2} \:\ dx =$ $\dfrac{2}{3}x^{\frac{3}{2}} + \dfrac{1}{3}x^{3}|_{0}^{\frac{3 - 2\sqrt{2}}{4}} =$

$\boxed{\dfrac{2}{3}(\dfrac{3 - 2\sqrt{2}}{4})^{\frac{3}{2}} + \dfrac{1}{3}(\dfrac{3 - 2\sqrt{2}}{4})^{3}}$ .

For $A_{R_{3}}$ we have:

$A_{R_{3}} = \displaystyle\int_{\frac{3 - 2\sqrt{2}}{4}}^{\frac{1}{2}} (-x + \dfrac{1}{4} + x^2) \:\ dx =$

$-\dfrac{1}{2}x^2 + \dfrac{1}{4}x + \dfrac{1}{3}x^3|_{\frac{3 - 2\sqrt{2}}{4}}^{\frac{1}{2}} =$

$\dfrac{1}{24} - (-\dfrac{1}{2}(\dfrac{3 - 2\sqrt{2}}{4})^2 + \dfrac{1}{4}(\dfrac{3 - 2\sqrt{2}}{4}) + \dfrac{1}{3}(\dfrac{3 - 2\sqrt{2}}{4})^{3})$

$A_{R_{2}} + A_{R_{3}} = (\dfrac{3 - 2\sqrt{2}}{4})(\dfrac{8\sqrt{3 - 2\sqrt{2}} + 3 - 6\sqrt{2}}{24}) + \dfrac{1}{24}$

The Area $A^{*} = A_{R_{1}} + A_{R_{2}} + A_{R_{3}} =$ $\dfrac{1}{96}[5 + (3 - 2\sqrt{2})(8\sqrt{3 - 2\sqrt{2}} + 3 - 6\sqrt{2})]$

Using the symmetry about the $y$ axis the total area $A = 2A^{*} =$

$\dfrac{2}{96}[5 + (3 - 2\sqrt{2})(8\sqrt{3 - 2\sqrt{2}} + 3 - 6\sqrt{2})] \approx \boxed{0.09640452}$