If the area of the region x , y : x 2 3 + y 2 3 ≤ 1 , x + y ≥ 0 , y ≥ 0 is A , then 256 A…

Plotting the diagram of given data we have,

Now finding limit by diagram, we get

$x^{\frac{2}{3}}+x^{\frac{2}{3}}=1$

$\Rightarrow x=\pm \frac{1}{2\sqrt{2}}$

So total area will be given by,

$A={\int }_{-\frac{1}{2\sqrt{2}}}^0{\left(1-x^{\frac{2}{3}}\right)}^{\frac{3}{2}}dx-\left(\frac{1}{2}\times \frac{1}{2\sqrt{2}}\times \frac{1}{2\sqrt{2}}\right) +{\int }_0^1{\left(1-x^{\frac{2}{3}}\right)}^{\frac{3}{2}}dx$

Now for solving $\int {\left(1-x^{\frac{2}{3}}\right)}^{\frac{3}{2}}dx$

 Let $x={\sin }^3\theta$

$=\int {\left(1-{\sin }^2\theta \right)}^{\frac{3}{2}}·3{\sin }^2\theta \cos \theta d\theta$

$=\int 3{\sin }^2\theta {\cos }^4\theta d\theta$

$=3\int {\sin }^2\theta {\cos }^4\theta d\theta$

Now putting the limit and solving we get

$\Rightarrow A=\frac{9\pi }{64}-\frac{1}{16}+\frac{1}{16}=\frac{36\pi }{256}$

$\Rightarrow \frac{256 \mathrm{A}}{\pi }=36$