It's All Areas!

$x^2 + xy + y^2 = 7$

$x^2 - xy + y^2 = 5$

The coefficient( $x^2$ ) = coefficient( $y^2$ ) $= 1 \implies \theta = 45^{\circ}$ and the equations of rotation are :

$x = \dfrac{1}{\sqrt{2}}x' - \dfrac{1}{\sqrt{2}}y'$

$y = \dfrac{1}{\sqrt{2}}x' + \dfrac{1}{\sqrt{2}}y'$

Replacing $(x,y)$ in the initial equations we obtain:

$x'^2 + 3y'^2 = 10$

$3x'^2 + y'^2 = 14$

Solving the above system we obtain $x' = \pm 2$ (We only need $x'$ here) and

$y_{2}(x') = \sqrt{14 - 3x'^2}$ and $y_{1}(x') = \dfrac{\sqrt{10 - x'^2}}{\sqrt{3}} \implies A_{1} = \displaystyle\int_{-2}^{2} y_{2}(x') - y_{1}(x') dx' = A_{2}$ .

$I_{1} = \displaystyle\int_{-2}^{2} y_{2}(x') dx' = \displaystyle\int_{-2}^{2} \sqrt{14 - 3x'^2} dx'$

Letting $\sqrt{3}x' = \sqrt{14}\sin(\theta) \implies \sqrt{3} dx' = \sqrt{14}\cos(\theta) d\theta \implies$

$I_{1} = \dfrac{7}{\sqrt{3}}(\arcsin(\dfrac{\sqrt{3}}{\sqrt{14}}x') + \dfrac{\sqrt{3}x'\sqrt{14 - 3x'^2}}{14})|_{-2}^{2} =$ $\dfrac{14}{\sqrt{3}}(\arcsin(\sqrt{\dfrac{6}{7}}) + \dfrac{\sqrt{6}}{7})$

and

$I_{2} = \displaystyle\int_{-2}^{2} y_{1}(x') dx' = \dfrac{1}{\sqrt{3}}\displaystyle\int_{-2}^{2} \sqrt{10 - x'^2} dx'$

Letting $x' = \sqrt{10}\sin(\theta) \implies dx' = \sqrt{10}\cos(\theta) d\theta \implies$

$I_{2} = \dfrac{5}{\sqrt{3}}(\arcsin(\dfrac{x'}{\sqrt{10}}) + \dfrac{x'\sqrt{10 - x'^2}}{10})|_{-2}^{2} =$ $\dfrac{10}{\sqrt{3}}(\arcsin(\sqrt{\dfrac{2}{5}}) + \dfrac{\sqrt{6}}{5})$

$\implies A_{!} = I_{1} - I_{2} = \dfrac{2}{\sqrt{3}}(7\arcsin(\sqrt{\dfrac{6}{7}}) - 5\arcsin(\sqrt{\dfrac{2}{5}})) = A_{2}$

$\implies A = A_{1} + A_{2} = \dfrac{4}{\sqrt{3}}(7\arcsin(\sqrt{\dfrac{6}{7}}) - 5\arcsin(\sqrt{\dfrac{2}{5}})) =$

$\dfrac{\alpha^{\alpha}}{\sqrt{\beta}}(\lambda\arcsin(\sqrt{\dfrac{\omega}{\lambda}}) - \gamma\arcsin(\sqrt{\dfrac{\alpha}{\gamma}})) \implies \alpha + \beta + \lambda + \omega + \gamma = \boxed{23}$