Let $F$ be a continuous vector field on $R^3$ such that for $x\in \mathbb{R}^3$ $$|F(x)|\leq \frac{1}{1+|x|^3}$$ Prove that $$\int_{\mathbb{R}^3}\operatorname{div}{F}\,dx=0$$
I was given a hint: Consider a large ball $B_R$ and then take $R\to \infty$.
I think one is supposed to apply divergence theorem, but I'm not quite sure how to get started. Can someone tell me how to solve the problem using the hint and what the conclusion indicates in physics (this is from our PDE course and, without solid background knowledge in physics, I always find myself really confused when trying to understand those differential equations and other related topics involving physics)? Any help is appreciated.
Here is an outline. By the divergence theorem,
$$\int_{B_R} \operatorname{div}(\mathbf{F})\, d\mathbf{x} = \int_{\partial B_R} \mathbf{F}\cdot \mathbf{n}\, d\sigma$$
Since $\lvert \mathbf{F}\cdot \mathbf{n}\rvert \le (1 + R^3)^{-1}$ on $\partial B_R$, the integral $\int_{\partial B_R} \mathbf{F}\cdot \mathbf{n}\, d\sigma$ is bounded by $\pi R^2(1 + R^3)^{-1}$, which tends to $0$ as $R\to \infty$. Therefore $\int_{B_R}\operatorname{div}(\mathbf{F})\, d\mathbf{x} \to 0$ as $R\to \infty$. In particular, $\int_{\Bbb R^3} \operatorname{div}(\mathbf{F})\, d\mathbf{x} = \lim\limits_{R\to \infty} \int_{B_{R}} \operatorname{div}(\mathbf{F})\, d\mathbf{x} = 0$.