very strange phenomenon $f(x,y)=x^4-6x^2y^2+y^4$ integral goes wild I am going over my lecture's notes in preparation for exam and I saw something a bit strange I would like someone to explain how it is possible.
Look at the function $f(x,y) = x^4-6x^2y^2+y^4$
if we convert it to polar coordinates, we will get $f(r,\theta)=r^4\cos(4\theta)$
the weird thing about this function, is that it has 2 different integrals over the same area.
What I mean: if we calculate the integral $$ \lim_{n \to \infty}\int_{0}^{2\pi} \int_{0}^{n} r^5\cos(4\theta)dr d\theta$$ we will see it is equal to zero.
However, if we calculate $$\lim_{n \to \infty}\int_{-n}^{n} \int_{-n}^{n} x^4-6x^2y^2+y^4dxdy = \lim_{n \to \infty}\int_{-n}^{n} \frac{2n^5}{5}-4n^3y^2+2ny^4 dy=\lim_{n \to \infty} -\frac{16}{15}n^6 = -\infty$$
How come the integral is so different based just on how we choose to represent the same area? (it is $\mathbb R^2$ in both cases. first it is a giant circle, second case it is a giant square)
So if asked what is $$\iint_{\mathbb R^2} x^4-6x^2y^2+y^4$$ is the correct thing to say that it does not exist?

This integral does not converge. To illustrate, we can see that $$\lim_{n \to \infty} \int_{-n}^{n}x \, \text dx = \lim_{n \to \infty} \left( \frac{n^2}{2} - \frac{(-n)^2}{2} \right) = \lim_{n \to \infty} 0 = 0$$ even though $\int_{-\infty}^{\infty} x \, \text dx$ does not exist. What was just computed is called the Cauchy principal value, which assigns values to integrals which would otherwise be undefined.
Your case is similar, except that in 2D the limit of the expanding domain can be taken in many ways, which give different answers. Furthermore, the fact that the integral taken in two equivalent ways give different results is sufficient to state that it is undefined.
