Estimating the volume of a region bounded by polynomial inequalities

Question

Let $Q(x,y,z)$ be a geometrically irreducible quadratic form in $x,y,z$ with real coefficients, such that $z^2$ appears with non-zero coefficient. Define the region $\mathcal{R}(X)$ by

$$\displaystyle \mathcal{R}(X) = \{(x,y,z) \in \mathbb{R}^3 : |x|, |y|, |Q(x,y,z)| \geq 1, 1 \leq |x||y|Q(x,y,z)^2 \leq X\}.$$

This region is compact, since the lower bound for $|x|, |y|, |Q(x,y,z)|$ implies that each of these quantities is bounded, and since $|x|, |y|$ are bounded and $Q(x,y,z)^2$ is bounded, it follows that $|z|$ is also bounded. Therefore $\mathcal{R}(X)$ has a finite volume.

Does anyone know how to estimate the order of magnitude of $\mathcal{R}(X)$? Heuristically, the polynomial $xyQ(x,y,z)^2$ has degree 6 and has three variables, therefore the volume should be something like $X^{3/6} (\log X)^\rho$ for some non-negative real number $\rho$. Methods using elementary calculus seems to be nightmarish, so hopefully a higher browed approach will be better.

Answer 1

Since we are restricted now to $Q(x,y,z)=z^2-4xy$, it makes sense to use the symmetry and consider $x,y\ge 1$ once with the same $Q$ and once with $Q(x,y,z)=z^2+4xy$. Let now $w=xy$. Then, if we switch to the new variable $w$, we shall get the regions $1\le w\le X, 1\le |z^2\pm 4w|\le \sqrt{X/w}$ with the measure that, for $w$ not too close to the endpoints, is approximately $(\log w)\,dw\,dz$. If $4w$ is not much greater than $\sqrt{X/w}$, i.e., if $w^3\le CX$, then we can estimate the measure of $z$ that satisfy the condition by $CX^{1/4}w^{-1/4}$ resulting in the integral $X^{1/4}\int_1^{CX^{1/3}}w^{-1/4}\log w\,dw=O(\sqrt X\log X)$. In the case when we have $z^2+4w$ that's it. But when we have $z^2-4w$, we can continue beyond $w=CX^{1/3}$ using the observation that $z\mapsto z^2-a^2$ is approximately linear near $a$ with the coefficient $2a$, so the length of the interval where $|z^2-a^2|\le b$ is almost $b/a$, provided that $b$ is small compared to $a^2$. Thus, we end up with the integral $$ \int_{\approx X^{1/3}}^{\approx X} \frac{\sqrt{X/w}}{2\sqrt w}\log w\,dw=\frac 14\left.\sqrt X\log^2 w\right|_{w\approx X^{1/3}}^{w\approx X}\approx \frac 29\sqrt X\log^2 X\,. $$

This has to be doubled to account for the symmetric region $x,y\le -1$ finally giving $\frac 49\sqrt X\log^2 X$ (twice less than what I got during my midnight computation attempt; I hope it is correct now).