Question

If the area of the region $\left\{( x , y ): x ^{\frac{2}{3}}+ y ^{\frac{2}{3}} \leq 1 x + y \geq 0, y \geq 0\right\}$ is $A$, then $\frac{256 A }{\pi}$ is

Answer 1

$A =\frac{3}{2} \int\limits_{0}^{1}\left(1- x ^{2 / 3}\right)^{3 / 2} dx$
Let $x =\sin ^{3} \theta$
$A =\frac{3}{2} \int\limits_{0}^{\pi / 2}\left(1-\sin ^{2} \theta\right)^{3 / 2} \cdot 3 \sin ^{2} \theta \cos \theta d \theta$
$=\frac{3}{2} \int\limits_{0}^{\pi / 2} 3 \sin ^{2} \theta \cos ^{4} \theta d \theta$
$=\frac{9}{2} \int\limits_{0}^{\pi / 2} \sin ^{2} \theta \cos ^{4} \theta d \theta$
$A =\frac{9}{2} \times \frac{1.3 .1}{(2+4)(4)(2)} \cdot \frac{\pi}{2}$
$\Rightarrow A =\frac{9 \pi}{64}$
$\Rightarrow \frac{64 A }{\pi}=9$
$\Rightarrow \frac{256 A }{\pi}=36$