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

Question

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

Answer 1

In maxima using the adjoint curves method

solve((y^2-(-1/10)^2)*(2*(-1/10)^2-5*(-1/10)+3)-4*((-1/10)^2-2*(-1/10)+y^2)^2); # [y = -sqrt(23)/10,y = sqrt(23)/10]
solve([a5,a0+a1+a2+a3+a4+a5,a0-a1+a2+a3-a4+a5,(-1/10)*(-1/10)*a0+(-1/10)*(sqrt(23)/10)*a1+(-1/10)*a2+(sqrt(23)/10)*(sqrt(23)/10)*a3+(sqrt(23)/10)*a4+a5],[a0,a1,a2,a3,a4,a5]); # [[a0 = %r1,a1 = -%r2,a2 = (sqrt(23)*%r2-2*%r1)/3, a3 = -(sqrt(23)*%r2+%r1)/3,a4 = %r2,a5 = 0]]
factor(resultant((y^2-x^2)*(2*x^2-5*x+3)-4*(x^2-2*x+y^2)^2,t*x^2-x*y+(sqrt(23)-2*t)/3*x-(sqrt(23)+t)/3*y^2+y,x)); # 18*(x-1)^4*x^2*(10*x+1)*(20*t^4*x+20*sqrt(23)*t^3*x+234*t^2*x+64*sqrt(23)*t*x+158*x+2*t^4+2*sqrt(23)*t^3-87*t^2-56*sqrt(23)*t-187)
factor(resultant((y^2-x^2)*(2*x^2-5*x+3)-4*(x^2-2*x+y^2)^2,t*x^2-x*y+(sqrt(23)-2*t)/3*x-(sqrt(23)+t)/3*y^2+y,x)); # 18*(y-1)^2*y^2*(y+1)^2*(200*t^4*y^2+200*sqrt(23)*t^3*y^2+2340*t^2*y^2+640*sqrt(23)*t*y^2+1580*y^2-660*t^3*y-624*sqrt(23)*t^2*y-4602*t*y-498*sqrt(23)*y-46*t^4+20*sqrt(23)*t^3+897*t^2+313*sqrt(23)*t+782)
solve(200*t^4*y^2+200*sqrt(23)*t^3*y^2+2340*t^2*y^2+640*sqrt(23)*t*y^2+1580*y^2-660*t^3*y-624*sqrt(23)*t^2*y-4602*t*y-498*sqrt(23)*y-46*t^4+20*sqrt(23)*t^3+897*t^2+313*sqrt(23)*t+782,y); # [y = -(sqrt(9200*t^8+5200*sqrt(23)*t^7-54860*t^6-53440*sqrt(23)*t^5-159328*t^4+105664*sqrt(23)*t^3+1013749*t^2+150878*sqrt(23)*t+190463)-330*t^3-312*sqrt(23)*t^2-2301*t-249*sqrt(23))/(200*t^4+200*sqrt(23)*t^3+2340*t^2+640*sqrt(23)*t+1580), y = (sqrt(9200*t^8+5200*sqrt(23)*t^7-54860*t^6-53440*sqrt(23)*t^5-159328*t^4+105664*sqrt(23)*t^3+1013749*t^2+150878*sqrt(23)*t+190463)+330*t^3+312*sqrt(23)*t^2+2301*t+249*sqrt(23))/(200*t^4+200*sqrt(23)*t^3+2340*t^2+640*sqrt(23)*t+1580)]
solve(20*t^4*x+20*sqrt(23)*t^3*x+234*t^2*x+64*sqrt(23)*t*x+158*x+2*t^4+2*sqrt(23)*t^3-87*t^2-56*sqrt(23)*t-187,x); # [x = -(2*t^4+2*sqrt(23)*t^3-87*t^2-56*sqrt(23)*t-187)/(20*t^4+20*sqrt(23)*t^3+234*t^2+64*sqrt(23)*t+158)]
solve(-(2*t^4+2*sqrt(23)*t^3-87*t^2-56*sqrt(23)*t-187)/(20*t^4+20*sqrt(23)*t^3+234*t^2+64*sqrt(23)*t+158)-1,t),float; # [t = -2.444534971313947,t = -1.453380790342413,t = -1.205651455247495, t = 0.3077356935911356]

In geogebra; check the parametrization

In maple; get the area $0.057$

int((sqrt(9200*t^8+5200*sqrt(23)*t^7-54860*t^6-53440*sqrt(23)*t^5-159328*t^4+105664*sqrt(23)*t^3+1013749*t^2+150878*sqrt(23)*t+190463)+330*t^3+312*sqrt(23)*t^2+2301*t+249*sqrt(23))/(200*t^4+200*sqrt(23)*t^3+2340*t^2+640*sqrt(23)*t+1580)*diff(-(2*t^4+2*sqrt(23)*t^3-87*t^2-56*sqrt(23)*t-187)/(20*t^4+20*sqrt(23)*t^3+234*t^2+64*sqrt(23)*t+158),t),t=-1.205651455247495..0.3077356935911356); # -0.057

In geogebra

Answer 2

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.