We fix $\epsilon>0$ and consider the open ball in $\mathbb{R}^3$ $$B_\epsilon=\left\{x\in \mathbb{R}^3:\lVert \mathbf{x} \rVert<{\epsilon}\right\}.$$

If $\mathbf{F}$ is a vector field of class $C^1$ on the closure of $B_\epsilon$, then by the Divergence Theorem we know that $$\int_{B_\epsilon} \nabla\cdot \mathbf{F}=\int_{\partial B_\epsilon}\mathbf{F}\cdot \mathbf{n}\qquad\qquad\qquad[1] $$where $\mathbf{n}$ is the outward pointing unit normal field of the boundary $\partial B_\epsilon=\left\{x\in \mathbb{R}^3:\lVert \mathbf{x} \rVert={\epsilon}\right\}$. The left side of the previous is a volume integral, the right side is a surface integral.

I would like to know if $[1]$ holds on the unbounded domain $D_\epsilon:={\overline{B_\epsilon}}^c$. More precisely:

Lets define $$D_\epsilon=\left\{x\in \mathbb{R}^3:\lVert \mathbf{x} \rVert>{\epsilon}\right\},$$ and suppose $\mathbf{F}$ is a vector field of class $C^1$ on $\overline{D_\epsilon}=\left\{x\in \mathbb{R}^3:\lVert \mathbf{x} \rVert\geq{\epsilon}\right\}$. Here, obviously, $\partial D_\epsilon =\partial B_\epsilon$.

If the improper Riemann integral of $\nabla \cdot \mathbf{F}$ on $D_\epsilon$ is convergent, can we state that $$\int_{D_\epsilon} \nabla\cdot \mathbf{F}=\int_{\partial D_\epsilon}\mathbf{F}\cdot \mathbf{n}\qquad ?\qquad\qquad\qquad[2] $$

Any hint for a proving/disproving $[2]$ would be really appreciated.

  • $\begingroup$ $\frac{1}{x^2+y^2}(x,y)^T$ is a divergence-free vector field on $D_\epsilon$, disproving [2]. $\endgroup$ Oct 28 '19 at 12:31
  • $\begingroup$ Thanks! Does $[2]$ hold when we also suppose $\mathbf{F}$ with compact support? $\endgroup$
    – eleguitar
    Oct 28 '19 at 18:02
  • $\begingroup$ Yes it does. You can use the divergence theorem on the annulus $D_\epsilon^c \cap D_R$ for large enough $R$ to show it. $\endgroup$ Oct 28 '19 at 18:27
  • $\begingroup$ If exist $R>0$ such that $\mathbf{F}=0$ on $\left\{x\in \mathbb{R}^3:\lVert \mathbf{x} \rVert\geq R \right\}$, then $\int_{D_\epsilon} (\nabla \cdot \mathbf{F})^{+}=\lim_{r\to +\infty}\int_{{D_\epsilon}\cap B_r(0)} (\nabla \cdot \mathbf{F})^{+}=\int_{\{\epsilon<|x|<R\}} (\nabla \cdot \mathbf{F})^{+}$ and also $\int_{D_\epsilon} (\nabla \cdot \mathbf{F})^{-}=\int_{\{\epsilon<|x|<R\}} (\nabla \cdot \mathbf{F})^{-}$. So $\int_{D_\epsilon} (\nabla \cdot \mathbf{F})=\int_{\{\epsilon<|x|<R\}} (\nabla \cdot \mathbf{F})=\int_{\partial D_\epsilon}\mathbf{F}\cdot n$ by the divergence theorem. $\endgroup$
    – eleguitar
    Oct 28 '19 at 19:16
  • $\begingroup$ Am I correct @JosefE.Greilhuber? $\endgroup$
    – eleguitar
    Oct 28 '19 at 19:19

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