Area enclosed by curve What is the area enclosed by the curve $$(x^2+y^2-1)^3 = x^2y^3$$
I am guessing this is solvable via double integration or perhaps polar but I am unsure of where to begin. I have tried isolating $y$ but this yields a complicated function which is difficult to integrate. Is it possible there is a way to transform a graph with known area such as a circle to yield the figure given by the equation from which the Jacobian can be calculated to find the scaling factor?
Do the sharp points at $(0,1)$ and $(0,-1)$ cause any problems?
 A: The idea espoused in this answer works here too.
$$(x^2+y^2-1)^3=x^2y^3\implies y=\frac{x^{2/3}\pm\sqrt{x^{4/3}-4(x^2-1)}}2\,dx$$
$$A=2\int_0^{a^{3/2}}\sqrt{x^{4/3}-4x^2+4}\,dx$$
Substitute $t=x^{2/3}$:
$$A=3\int_0^a\sqrt{t(t^2-4t^3+4)}\,dt=3.66197272581827\dots$$
Here $a=1.09066\dots$ is the only real root of $t^2-4t^3+4$.
While this can theoretically be expressed through complete elliptic integrals (since the integral runs from zero to zero of the quartic), the resulting form is so complicated that it's not worth it. I did it anyway, and this is the result as Mathematica code:
Print[3 NIntegrate[Sqrt[t(t^2-4t^3+4)], {t, 0, Root[t^2-4t^3+4,1]}, WorkingPrecision->50]]
r = Sqrt[Root[x^3+3x^2-1732, 1]]
m = 1/2 + AlgebraicNumber[r, {0, -1/96, 0, -863/83136, 0, 1/83136}]
n = 1/2 + AlgebraicNumber[r, {0, -7/288, 0, -3461/249408, 0, 1/249408}]
Sqrt[r/2]/32 * (3*EllipticE[m] - AlgebraicNumber[r, {2, 48, 5, 0, 2}]/3*EllipticK[m] +
AlgebraicNumber[r, {-69280, 2661218, 138560, 4043, 55424, 1925}]/(3*64*433)*EllipticPi[n, m]) // N[#,50]& // Print

