To calculate the area delimited by the closed curve called asteroid, with parametric form: $$$\gamma(t)=\big( a\cos^3(t),a\sin^3(t) \big) \quad t\in[0,2\pi]$$$
Development:
The area that we want to find is drawn in the following graph:
This is a clear example of how useful Green's theorem can be to calculate areas of regions limited by curves. The problem already gives us the parametric curve. As a vector field we will take $$F(x,y)=(0,x)$$.
Thus: $$$\text{Area}= \iint_D 1 \ dx \ dy=\iint_D \big( Q_x-P_y\big) \ dx \ dy= \int_C F \ dr = \int_0^{2\pi} F(\gamma(t))\cdot\gamma'(t) \ dt $$$
Let's calculate,
$$$\gamma'(t)=\big( -3a\cos^2(t)\sin(t),3a\sin^2(t)\cos(t) \big)$$$
Then:
$$$ \begin{array}{rl} \int_C F \ dr =& \int_0^{2\pi} F(\gamma(t))\cdot\gamma'(t) \ dt\\ =& \int_0^{2\pi} \big(0,a\cos^3(t)\big)\cdot\big(-3a\cos^2(t)\sin(t),3a\sin^2(t)\cos(t)\big) \ dt \\ =& \int_0^{2\pi}3a^2\sin^2(t)\cos^4(t) \ dt= 3a^2\int_0^{2\pi}\Big(\dfrac{1-\cos(2t)}{2}\Big)\Big(\dfrac{1+\cos(2t)}{2}\Big)^2 \ dt \\ =& \dfrac{3a^2}{8}\int_0^{2\pi}(1+\cos(2t))(1-\cos^2(2t)) \ dt \\ =& \dfrac{3a^2}{8}\int_0^{2\pi} (1-\cos^2(2t)+\cos(2t)-\cos^3(2t)) \ dt \\ =& \dfrac{3a^2}{8}\int_0^{2\pi} \Big( 1-\dfrac{1+\cos(4t)}{2}+\cos(2t)-\cos(2t)+\cos(2t)\sin^2(2t) \Big) \ dt \\ =& \dfrac{3a^2}{8}\int_0^{2\pi} \dfrac{1}{2} \ dt =\dfrac{3a^2\pi}{8} \end{array}$$$
Solution:
$$\text{Area}=\dfrac{3a^2\pi}{8}$$