How to find enclosed area by the astroid $x^{\frac{2}{3}} + y^{\frac{2}{3}} = 1$? 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. 
 A: $$ A = 4 \int_{0}^{1}(1-x^{2/3})^{3/2}\,dx = 6\int_{0}^{1}u^{1/2}(1-u)^{3/2}\,du = 6\, B(3/2,5/2)=\frac{6\,\Gamma(3/2)\,\Gamma(5/2)}{\Gamma(4)}$$
leads to $A = \color{blue}{\large\frac{3\pi}{8}}$. I used Euler's Beta function and the substitution $x=u^{3/2}$.
A: By the parametric equation,
$$\frac12\oint x\,dy-y\,dx=\frac32\int_0^{2\pi}(\cos^4t\sin^2t+\sin^4t\cos^2t)\,dt=\frac32\int_0^{2\pi}\cos^2t\sin^2t\,dt\\
=\frac38\int_0^{2\pi}\sin^22t\,dt=\frac{3\pi}8.$$
The last identity is obtained from the average value of the squared sine.
A: Hint. The area enclosed by the asteroid in the first quadrant is given by
$$
\mathcal{A}=\int_0^1\left(1-x^{2/3} \right)^{3/2}dx
$$ then with the change of variable $x=\sin^3 t$, $dx=3\cos t \sin^2 t\:dt$, one gets
$$
\mathcal{A}=3\int_0^{\pi/2}\left(1-\sin^2 t \right)^{3/2}\cos t \cdot\sin^2 t\:dt=3\int_0^{\pi/2}\cos^4t \sin^2t\:dt
$$ then one may linearize the integrand to get
$$
\mathcal{A}=\frac{3\pi}{32}.
$$
A: I will make from 0 to 1 just to give you a hint 
$\int_{0}^1 (1-x^\frac{2}{3})^\frac{3}{2}dx$ 
Trig sub: $x=cos^3(u)$ 
$dx=-3cos^2(u)sin(u)$ and $\frac{\pi}{2}\le u\le 0$
So
$\int_\frac{\pi}{2}^{0} -\sin^4(u) \cos^2(u) du$
And this you can solve using trig transformantions
