Calculate the measure of this set $A=[(x,y): (x^2+y^2)^2=x^3-3xy^2]$ The question is to calculate the 2-dimensional Lebesgue measure of this set : 
$$A=[(x,y): (x^2+y^2)^2=x^3-3xy^2]$$
My intuition tells me that this measure would be zero, as this is equality, but I do not have any formal tools to prove it (I do not remember any), can someone help me out? Thanks, also would be great to see an example of the set where there is equality like this, but the measure is not zero.
 A: Fubini yields
$$m_2(A) = \int_{\mathbb{R}^2} \chi_A \, \mathrm{d}m_2 = \int_{\mathbb{R}} \int_{\mathbb{R}} \chi_A(x,y) \, \mathrm{d}x \, \mathrm{d}y = \int_{\mathbb{R}} 0 \, \mathrm{d}y = 0$$
since for fixed $y$, the support of $\chi_A(\cdot,y)$ contains only finitely many points, hence is equal to $0$ almost everywhere.
A: I just realized this is actually a folium. This curve is easy to understand!
Write $x = r \cos \vartheta$ and $y = r \sin \vartheta$. Then the left hand side is now just $r^2$, and the right is $\cos ^3 \vartheta - 3 \cos \vartheta \sin^2 \vartheta$ which is the well-known triple angle formula for cosine. This makes your curve a 3-leafed folium in polar coordinates!
Now there was a good suggestion in the comments for how to kill this problem using Fubini. If you don't know Fubini, here is a geometric way of understanding what's going on. There are 3 leaves and they are all identical, so you can limit yourself to just one leaf.
Now to get a grip on the measure of the folium, you can use a shear transform to bring the leaf into position such that the furthest most point from the origin is the point $(0,1)$. Now you can stereographically project your folium to the line to get a bijection to the line, and the derivative of the stereographic projection map being bounded, together with shear transform having a bounded derivative as a map $\mathbb{R}^2 \to \mathbb{R}^2$ tells you by change of variables that the measure of the line and the measure of the leaf of the folium differ by at most a multiplicative constant, but the former is $0$ as you know, so the latter is $0$, and then multiplication by $3$ gets you that the whole thing is $0$.
I remark that Fubini is much easier, as the other answer shows, but it is nice to have visual proofs of things sometimes too.
