Divergence Theorem

Divergence Theorem

Theorem 1. Let $\mathbf{F}$ be a vector field whose components have continuous first partial derivatives, and let $S$ be a piecewise smooth oriented closed surface. The flux of $\mathbf{F}$ across $S$ in the direction of the surface’s outward unit normal field $\mathbf{n}$ equals the triple integral of the divergence $\nabla \cdot \mathbf{F}$ over the region $D$ enclosed by the surface: $$\iint_S \mathbf{F} \cdot \mathbf{n} d \sigma = \iint_S \mathbf{F} \cdot d \mathbf{a} = \iiint_D \nabla \cdot \mathbf{F} dV.$$

Divergence theorem, 즉 발산 정리는 그린 정리의 두번째 버전을 확장한 정리이다. 즉 폐곡면에서 벡터장 $\mathbf{F}$를 적분한 값은 그 폐곡면으로 둘러쌓인 영역에서 $\mathbf{F}$의 divergence를 적분한 값과 같다. 폐곡면 내부 영역에서 작은 piece들의 divergence는 맞닿아 있는 부분에서 서로 상쇄되므로 겉면의 정보만 알면 영역 내부의 divergence를 파악할 수 있다는 것이다. 

Proof. Let $\mathbf{F}(x, y, z) = \langle M(x, y, z), N(x, y, z), P(x, y, z) \rangle$. We suppose that $D$ is a convex region with no holes or bubbles and that any line perpendicular to the $xy$-plane at an interior point of the region $R_{xy}$ intersects the surface $S$ in exactly two points, producing surfaces $$S_1: z = f_1(x, y), \\ S_2: z = f_2(x, y),$$ with $f_1 \leq f_2$. We make similar assumptions about $R_{yz}$ and $R_{xz}$. 
Note that for $\mathbf{n} = \langle n_1, n_2, n_3 \rangle$, $$n_1 = \mathbf{n} \cdot \mathbf{i} = \cos \alpha \\ n_2 = \mathbf{n} \cdot \mathbf{j} = \cos \beta \\ n_3 = \mathbf{n} \cdot \mathbf{k} = \cos \gamma.$$ Then $\mathbf{F} \cdot \mathbf{n} = M \cos \alpha + N \cos \beta + P \cos \gamma$. 
We need to show that $$\iint_S (M \cos \alpha + N \cos \beta + P \cos \gamma) d \sigma = \iiint_D \left( \frac{\partial M}{\partial x} + \frac{\partial N}{\partial y} + \frac{\partial P}{\partial z} \right) dx dy dz \cdots (*).$$ We prove the theorem by establish the following equation: $$\iint_S P \cos \gamma d \sigma = \iiint_D \frac{\partial P}{\partial z} dx dy dz.$$ Because $\gamma$ is the angle between $\mathbf{n}$ and $\mathbf{k}$, we have $\cos \gamma d\sigma = dA = dx dy$ on $S_2$. Similarly, $\cos \gamma d\sigma = -dx dy$ on $S_1$. Therefore, $$\iint_S P \cos \gamma d\sigma = \iint_{S_2} P \cos \gamma d\sigma + \iint_{S_1} P \cos \gamma d\sigma \\ = \iint_{R_{xy}} P(x, y, f_2(x, y)) dx dy - \iint_{R_{xy}} P(x, y, f_1(x, y)) dx dy \\ = \iint_{R_{xy}} [P(x, y, f_2(x, y)) - P(x, y, f_1(x, y))] dx dy \\ = \iint_{R_{xy}} \left[ \int_{f_1(x, y)}^{f_2(x, y)} \frac{\partial P}{\partial z} dz \right] dx dy = \iiint_D \frac{\partial P}{\partial z} dx dx dy.$$ This procedure is true for other two parts of $(*)$. This proves the theorem for these special regions. $\blacksquare$