The area of the loop of the curve x^2+(y-1)y^2=0 is equal to:

$x=±y\sqrt{1-y}$. Defined for y ≤ 1.

$A=\int\limits_0^12y\sqrt{1-y}dy$ put 1 - y = t²

$⇒A=-\int\limits_1^04t^2(1-t^2)dt=\int\limits_0^1(4t^2-4t^4)dt=\frac{4t^3}{3}-\frac{4t^5}{t}|_0^1=\frac{8}{15}$