발산 정리

Let $S'$ be the disk $x^2+y^2\le 3,\ z=0$ (with $\hat{\mathbf{n}}$ the downward pointing normal) and let $V$ be the portion of the ball $x^2+y^2+(z-1)^2\le 4$ with $z\ge 0$. Then, by the divergence theorem,

$$\begin{aligned} \iint _{S}\mathbf{F}\cdot\hat{\mathbf{n}}\,\mathrm{d}S &=\iiint _{V}{ \boldsymbol{\nabla} }\cdot\mathbf{F}\, \mathrm{d}V -\iint _{S'}\mathbf{F}\cdot(-\hat{\mathbf{k}})\,\mathrm{d}S\\ &=\iiint _{V}(2x+2y)\, \mathrm{d}V +\iint _{S'}(4+5x)\,\mathrm{d}x\mathrm{d}y \end{aligned}$$

Because $x$ is odd under $x\rightarrow-x$ and $y$ is odd under $y\rightarrow -y$,

$$\iiint _{V}x\, \mathrm{d}V=\iiint _{V}y\, \mathrm{d}V=\iint _{S'}x\,\mathrm{d}x\mathrm{d}y=0$$

so that

$$\begin{aligned} \iint _{S}\mathbf{F}\cdot\hat{\mathbf{n}}\,\mathrm{d}S =4\iint _{S'}\,\mathrm{d}x\mathrm{d}y =4\,\text{Area}(S') =4\times \pi\big(\sqrt{3}\big)^2=12\pi \end{aligned}$$