If the area of the region \ (x, y): x(2)/(3) +y(2)/(3) ≤ 1, x + y ≥...

Plotting the diagram of given data we have,

Now finding limit by diagram, we get

$x\dfrac{2}{3} + x\dfrac{2}{3} = 1$

$⇒ x = \pm \dfrac{1}{2\sqrt{2}}$

So total area will be given by,

$A = {∫}_{- \dfrac{1}{2\sqrt{2}}}^{0}\left(1 - x\dfrac{2}{3}\right)\dfrac{3}{2} d x - \left(\dfrac{1}{2} \times \dfrac{1}{2\sqrt{2}} \times \dfrac{1}{2\sqrt{2}}\right) + {∫}_{0}^{1}\left(1 - x\dfrac{2}{3}\right)\dfrac{3}{2}dx$

Now for solving $∫ \left(1 - x\dfrac{2}{3}\right)\dfrac{3}{2}dx$

Let $x = \sin^{3}\theta$

$= ∫ \left(1 - \sin^{2}\theta\right)\dfrac{3}{2} \cdot 3\sin^{2}\theta cos\thetad\theta$

$= ∫ 3\sin^{2}\theta\cos^{4}\thetad\theta$

$= 3 ∫ \sin^{2}\theta\cos^{4}\thetad\theta$

Now putting the limit and solving we get

$⇒ A = \dfrac{9\pi}{64} - \dfrac{1}{16} + \dfrac{1}{16} = \dfrac{36\pi}{256}$

$⇒ \dfrac{256A}{\pi} = 36$