JEE Main · 2022 · 27 Jun 2022 · Shift 2 · Area Under Curves

The curve $x^{2/3}+y^{2/3}=1$ is an astroid. In the first quadrant with $x+y\ge 0$ and $y\ge 0$, we need the area under the curve from $x=0$ to $x=1$.

From $x^{2/3}+y^{2/3}=1$, we get $y=(1-x^{2/3}{)}^{3/2}$.

The area is:
$A=\frac{3}{2}{\int }_0^1(1-x^{2/3}{)}^{3/2}dx$

Substituting $x={\sin }^3\theta$, $dx=3{\sin }^2\theta \cos \theta d\theta$:

$A=\frac{9}{2}{\int }_0^{\pi /2}{\sin }^2\theta {\cos }^4\theta d\theta$

Using the beta function formula:
$A=\frac{9}{2}\cdot \frac{1\cdot 3\cdot 1}{2\cdot 4\cdot 6}\cdot \frac{\pi }{2}=\frac{9\pi }{64}$

Therefore $\frac{256A}{\pi }=\frac{256\cdot 9\pi }{64\pi }=36$.