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

Eric Wofsey
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piteer
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  • What is the context of the problem and please add more detail. – Тyma Gaidash Oct 17 '21 at 13:48
  • do you have a final solution? – mark Oct 17 '21 at 13:49
  • Wolfram Alpha cannot find a general anti derivative for a branch of the function here, so good luck. – Тyma Gaidash Oct 17 '21 at 13:53
  • Parametrizing the curve via adjoint curves I get the integral

    $\int_{-1.205651455247495}^{0.3077356935911356} \frac{\sqrt{9200t^8+5200sqrt{23}t^7-54860t^6-53440\sqrt{23}t^5-159328t^4+105664\sqrt{23}t^3+1013749t^2+150878\sqrt{23}t+190463}+330*t^3+312\sqrt{23}t^2+2301t+249\sqrt{23}}{200t^4+200\sqrt{23}t^3+2340t^2+640\sqrt{23}t+1580} (-\frac{2t^4+2\sqrt{23}t^3-87t^2-56\sqrt{23}t-187}{20t^4+20\sqrt{23}t^3+234t^2+64\sqrt{23}t+158})'dt$

     – Jan-Magnus Økland Oct 17 '21 at 18:35
  • Which maple evaluates to about $0.057$ – Jan-Magnus Økland Oct 18 '21 at 02:25
  • Can you differentiate and apply Green's thm with line integral $ 2 A= \int ( x dy - y dx) ?$ – Narasimham Nov 02 '21 at 14:02
  • @piteer I find it strange that, not only here, but in many of your questions, you don't dialog with people, beginning by answering to the questions they place in comments. Try to do it, you will see, it' rewarding... – Jean Marie Nov 20 '21 at 22:34

2 Answers2

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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

eq

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

wingpoly

Jan-Magnus Økland
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  • To go back from the parametrization to the implicit form (in M2) S=QQ[a,s,t,w,x,y,z,MonomialOrder=>Lex];I=ideal(x+(2*t^4+2*a*t^3*s-87*t^2*s^2-56*a*t*s^3-187*s^4),10*y-w+330*t^3*s+312*a*t^2*s^2+2301*t*s^3+249*a*s^4,z-(20*t^4+20*a*t^3*s+234*t^2*s^2+64*a*t*s^3+158*s^4),w^2-(9200*t^8+5200*a*t^7*s-54860*t^6*s^2-53440*a*t^5*s^3-159328*t^4*s^4+105664*a*t^3*s^5+1013749*t^2*s^6+150878*a*t*s^7+190463*s^8),a^2-23) gens gb I factor oo_0_0 and set $z=1.$ – Jan-Magnus Økland Oct 18 '21 at 13:09
  • What is the closed form area or at least a series representation? – Тyma Gaidash Oct 18 '21 at 20:48
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    @TymaGaidash Excuse my ignorance, is this a usual request for the integration/area tags? I'm in it for the curves tag. The integral given can be approximated any way you want. The integral limits are also approximations, the exact values are (-sqrt(((-23^(3/2))-(8*(4*23^(1/2)))/11+(294*23^(1/2))/11)/6+18/11)/2)+((-1)*23^(1/2))/4+3/4, sqrt(((-23^(3/2))-(8*(4*23^(1/2)))/11+(294*23^(1/2))/11)/6+18/11)/2+((-1)*23^(1/2))/4+3/4. – Jan-Magnus Økland Oct 19 '21 at 10:52
  • @Jan-MagnusØkland In this case it becomes a bit simpler because $y$ can be explicitly written in terms of $x$ for points over the curve. – PierreCarre Nov 02 '21 at 15:44
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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.

enter image description here

PierreCarre
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