A geometric property about certain polynomials in two variables

Question

Assume that $p(x,y)$ is a polynomial in $\mathbb{R}[x, y]$ in the form $$ p=p_{2n}+ p_{2n-1}+\ldots +p_1+p_0$$

where $p_i$ is a homogenous polynomial of degree $i$. Moreover we assume that the last term $p_{2n}$ is always nonnegative which has a unique root at origin $(0,0)$ and has no any critical point except origin.

We consider the polynomial function $p(x,y)$ as a function from $\mathbb{R}^2 \to \mathbb{R}$.

Is it true to say that for all $z$ sufficiently large, all curves $p^{-1}(z)$ are simple closed curves whose centroid $c(z)\in \mathbb{R}^2$ has a limit $L\in \mathbb{R}^2$, as $z$ goes to $+\infty$ ?

The Motivation:

In lower dimension, assume that we have an even degree polynomial $p(x)=a_{2n}x^{2n}+a_{2n-1}x^{2n_1}+\ldots+a_1x+a_0$. Then for $y$ sufficiently large $p^{-1}(y)$ consists two points $\{A(y), B(y)\}$. Then $$\lim_{y\to \infty} (A(y)+B(y)) =\frac{-a_{2n-1}}{na_{2n}}$$

Note: The centroid $c(z)$ is actually the centroid of closed region in the plane which is surrounded by $p^{-1}(z)$.

Answer 1

If we work in polar coordinates, for large enough $z$ the solution $(x,y)=(r\cos\theta,r\sin\theta)$ of $f(x,y)=z^{2n}$ with argument $\theta$ is of the form $$r=z-\frac{p_{2n-1}(\cos \theta,\sin \theta)}{2np_{2n}(\cos \theta,\sin \theta)}+o(1)$$ for all $\theta$ since the conditions of the problem assure that $p_{2n}(\cos\theta,\sin\theta)$ is always nonzero. This means that the centroid of the level set $f^{-1}(z^{2n})$ is given by $$\frac{1}{2\pi}\int_{0}^{2\pi}re^{i\theta}d\theta=\frac{1}{2\pi}\int_0^{2\pi}\left(-\frac{p_{2n-1}(\cos \theta,\sin \theta)}{2np_{2n}(\cos \theta,\sin \theta)}+o(1)\right)e^{i\theta} d\theta$$ since $\int_0^{2\pi} ze^{i\theta}d\theta=0$. Therefore as $z\to +\infty$ the centroid of the level set $f^{-1}(z^{2n})$ converges to the point $$c_1+ic_2=\frac{1}{2\pi}\int_0^{2\pi}\left(-\frac{p_{2n-1}(\cos \theta,\sin \theta)}{2np_{2n}(\cos \theta,\sin \theta)}\right)e^{i\theta} d\theta.$$