Question

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

Answer 1

Answer: 36

$A=\frac{3}{2}\underset{0}{\overset{1}{\int }}{\left(1-x^{2/3}\right)}^{3/2}dx$
Let $x={\sin }^3\theta$
$A=\frac{3}{2}\underset{0}{\overset{\pi /2}{\int }}{\left(1-{\sin }^2\theta \right)}^{3/2}\cdot 3{\sin }^2\theta \cos \theta d\theta$
$=\frac{3}{2}\underset{0}{\overset{\pi /2}{\int }}3{\sin }^2\theta {\cos }^4\theta d\theta$
$=\frac{9}{2}\underset{0}{\overset{\pi /2}{\int }}{\sin }^2\theta {\cos }^4\theta d\theta$
$A=\frac{9}{2}\times \frac{\mathrm{1.3.1}}{(2+4)(4)(2)}\cdot \frac{\pi }{2}$
$\Rightarrow A=\frac{9\pi }{64}$
$\Rightarrow \frac{64A}{\pi }=9$
$\Rightarrow \frac{256A}{\pi }=36$