# Divergence theorem on special unbounded domains

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 $$\int_{B_\epsilon} \nabla\cdot \mathbf{F}=\int_{\partial B_\epsilon}\mathbf{F}\cdot \mathbf{n}\qquad\qquad\qquad$$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 $$$$ holds on the unbounded domain $$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 $$\int_{D_\epsilon} \nabla\cdot \mathbf{F}=\int_{\partial D_\epsilon}\mathbf{F}\cdot \mathbf{n}\qquad ?\qquad\qquad\qquad$$

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

• $\frac{1}{x^2+y^2}(x,y)^T$ is a divergence-free vector field on $D_\epsilon$, disproving . Oct 28 '19 at 12:31
• Thanks! Does $$ hold when we also suppose $\mathbf{F}$ with compact support? Oct 28 '19 at 18:02
• 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. Oct 28 '19 at 18:27
• 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. Oct 28 '19 at 19:16
• Am I correct @JosefE.Greilhuber? Oct 28 '19 at 19:19