The parametric representation of that asteroid is $x=2\cos^3\theta$, $y=2\sin^3\theta$. We hit the point $(2,0)$ when $\theta=0$, and the point $(1/\sqrt2,1/\sqrt2)$ when $\theta=\pi/4$. The area is between the curve and the $x$-axis, so equals the sum of infinitesimal vertical strips of width $|dx|=|x'(\theta)\,d\theta|$ and height $|y(\theta)|$. Therefore the area is
The minus sign comes from the fact that we are moving from right to left as the parameter $\theta$ grows (IOW $x'(\theta)<0$ in this interval).
I leave the calculation of that integral to you.