Plotting $(x^2 + y^2 - 1)^3 - x^2 y^3 = 0$

Question

I have no idea how this equation: \begin{equation} (x^2 + y^2 - 1)^3 - x^2 y^3 = 0 \end{equation}

Produces this picture:

Can someone provide a general explanation of plotting this function?

Answer 1

The solution set is obviously symmetric with respect to the $y$-axis. Therefore we may assume $x\geq 0$. In the domain $\{(x,y)\in {\mathbb R}^2\ |\ x\geq0\}$ the equation is equivalent with $$x^2+ y^2 -1=x^{2/3} y\ ,$$ which can easily be solved for $y$: $$y={1\over2}\bigl(x^{2/3}\pm\sqrt{x^{4/3}+4(1-x^2)}\bigr)\ .$$ Now plot this, taking both branches of the square root into account. You might have to numerically solve the equation $x^{4/3}+4(1-x^2)=0$ in order to get the exact $x$-interval.

Answer 2

It's easier to be understood in polar coordinates.

(x^2 + y^2 - 1)^3 - x^2*y^3 == 0 /. {x -> r*Cos[\[Theta]], 
   y -> r*Sin[\[Theta]]} // FullSimplify

% // ApplySides[#/r^5 &, #] &

$$ \left(r^2-1\right)^3=r^5 \sin ^3(\theta ) \cos ^2(\theta ) \\ \frac{\left(r^2-1\right)^3}{r^5}=\sin ^3(\theta ) \cos ^2(\theta ) $$

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Plot[Sin[\[Theta]]^3*Cos[\[Theta]]^2, {\[Theta], 0, 2 Pi}]

Plot[(r^2 - 1)^3/r^5, {r, 0, 5}]