How to find enclosed area by astroid
$$x^{\frac{2}{3}} + y^{\frac{2}{3}} = 1$$ for $-1≤x≤1$ and $-1≤y≤1$ ?
This curve is also given parametericallyly by $x(t) = (\cos(t))^{3}, y(t) = (\sin(t))^{3} $.
I have tried calculating it with formula for volume but my attempts were unsuccessful. How to solve this or at least hints would be helpful.
Best Answer
$$ A = 4 \int_{0}^{1}(1-x^{2/3})^{3/2}\,dx = 6\int_{0}^{1}u^{1/2}(1-u)^{3/2}\,du = 6\, B(3/2,5/2)=\frac{6\,\Gamma(3/2)\,\Gamma(5/2)}{\Gamma(4)}$$ leads to $A = \color{blue}{\large\frac{3\pi}{8}}$. I used Euler's Beta function and the substitution $x=u^{3/2}$.