# Is the area enclosed by p(x,y) always irrational?

Take a polynomial $$p \in \mathbb{Q}[X,Y]$$. Now draw the graph of $$p(x,y)=0$$. If, like $$X^2-Y^2-1$$, this turns out to enclose a finite area, is the area enclosed always irrational?

There are some answers that I regard as 'cheating', like combining $$4$$ lines to make an enclosed square, as this doesn't feel similar to what we are doing with $$X^2+Y^2-1$$: $$p=(X-1)(X+1)(Y-1)(Y+1)$$ I think cases like this can be eliminated by requiring that $$p$$ is irreducible, so more precisely my question is:

Is there an irreducible $$p \in \mathbb{Q}[X,Y]$$ such that a finite nonzero area enclosed by the graph $$p(x,y)=0$$ is rational?

I've found it quite difficult to even find a polynomial that encloses an area that isn't of the form $$a(x)+b(y)-c$$, and all of those intuitively feel like they should be irrational, though proving any individual case is presumably extremely difficult, based on how much work it takes to prove $$\pi$$ is irrational.

I would also be interested in an answer for polynomials in $$n>2$$ variables. I don't know the answer for $$n=1$$ either, as things like $$X^2-1$$ are reducible, though I suspect the answer here might be obvious if I knew a bit of Galois Theory.

• en.wikipedia.org/wiki/… Commented Jul 5 at 23:55
• This question seems to give an example as well. Commented Jul 5 at 23:56
• @user51547 and Greg Martin these are both really nice examples, thank you! When I posed this questions what I had in mind were things like $X^4+Y^4-1$ where it's just a loop disconnected from anything else, so I still wonder if it's possible to find something like that with a rational nonzero area. Though the lemniscate looks tantalisingly close to such a curve. Commented Jul 6 at 0:13