Let α be the area of the larger region bounded by the curve y ^ 2 =...

We have,

${y}^{2} = 8x$

$y = x$

Point of intersection of curve and a line is $A\left(8, 8\right)$.

Required area

$\alpha = {∫}_{2}^{8}\left(\sqrt{8x} - x\right)dx$

$⇒ \alpha = {\left[\dfrac{4\sqrt{2}{x}^{\dfrac{3}{2}}}{3} - \dfrac{{x}^{2}}{2}\right]}_{2}^{8}$

$⇒ \alpha = \left[\left(\dfrac{4\sqrt{2}{\left(2\sqrt{2}\right)}^{3}}{3} - \dfrac{64}{2}\right) - \left(\dfrac{4\sqrt{2}{\left(\sqrt{2}\right)}^{3}}{3} - \dfrac{4}{2}\right)\right]$

$⇒ \alpha = \dfrac{112}{3} - 30$

$⇒ \alpha = \dfrac{22}{3}$

So, $3\alpha = 22$