Острогра́дский's divergence theorem

Let $Ω$ be a solid and $𝜕 Ω$ be its oriented boundary. Let $⃗ 𝐅$ be a vector field differentiable in $Ω$ . Then #m/thm/calculus

\subset \supset \iint 𝜕 Ω ⃗ 𝐅 \cdot 𝑑 ⃗ 𝐚 = ∭ Ω ⃗ \nabla \cdot ⃗ 𝐅 𝑑 𝜏

Note the left hand side is equivalent to the flux through the surface of $Ω$ , the right hand side refers to Divergence. Heuristically, if a region has no divergence, there is no nett in-flow or out-flow, and therefore the flux through the boundary is zero.

Corollaries

$∭ Ω (⃗ \nabla \times ⃗ 𝐀) 𝑑 𝜏 = - \subset \supset \iint 𝜕 Ω ⃗ 𝐀 \times 𝑑 ⃗ 𝐚$

Proof

For any vector $⃗ 𝐜 \in ℝ 3$ , we have

⃗ 𝐜 \cdot \subset \supset \iint 𝜕 Ω ⃗ 𝐀 \times ⃗ 𝐝 𝑎 = - \subset \supset \iint 𝜕 Ω (⃗ 𝐀 \times ⃗ 𝐜) \cdot 𝑑 ⃗ 𝐚 = - ∭ Ω ⃗ \nabla \cdot (⃗ 𝐀 \times ⃗ 𝐜) 𝑑 𝜏 = ⃗ 𝐜 \cdot ∭ Ω (⃗ \nabla \times ⃗ 𝐀) 𝑑 𝜏

proving ^C1.

Practice problems

2016. Calculus, pp. 1185–1186 (§16.9 exercises)

#state/develop | #lang/en | #SemBr | #review