How to calculate the area of one of the smaller loops of the curve: $(y^2−x^2)(2x^2−5x+3) =4(x^2−2x+y^2)^2$? How to calculate the area of one of the smaller loops of the curve: $$(y^2−x^2)(2x^2−5x+3) =4(x^2−2x+y^2)^2$$?

 A: According to Wolfram, The equation for the blue curve is
$$
y = \frac{\sqrt{-6 x^2-\sqrt{3} \sqrt{-(x-1)^2 \left(20 x^2-28 x-3\right)}+11 x+3}}{2
   \sqrt{2}} := g_1(x)
$$
and the equation for the orange curve is
$$
y = \frac{\sqrt{-6 x^2+\sqrt{3} \sqrt{-(x-1)^2 \left(20 x^2-28 x-3\right)}+11 x+3}}{2
   \sqrt{2}}:= g_2(x)
$$
and the points in the smaller loop satisfy $1 \leq x \leq \frac 32$. Using Green's theorem with $Q=x, P = 0$, you have that
$$
\iint_D \underbrace{\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)}_{=1}dx dy  = \int_{\gamma} P dx +Q dy
$$
and so
$$
Area = \int_{\gamma} x \, dy = \int_{\gamma_1} x dy + \int_{\gamma_2} x dy.
$$
Where $\gamma_1$ corresponds to the blue curve, travelled from left to ro right and $\gamma_2$ to the orange curve, travelled from right to left. The line integrals over $\gamma_1, \gamma_2$ can be computed using the standard parametrizations associated to the expressions above ($y = g_1(x),\quad y=g_2(x)$):
$$\int_{\gamma_1} x dy = \int_1^{3/2} x g_1 '(x) dx \approx -0.146917$$
$$\int_{\gamma_2} x dy = -\int_1^{3/2} x g_2 '(x) dx \approx 0.204086$$
So, the estimate for the area would be 0.0571698, as pointed out in other answers. I did not try to compute the integrals exactly.

