Острогра́дский's divergence theorem

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

Note the left hand side is equivalent to the flux through the surface of , 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

Proof

For any vector , we have

proving ^C1.

Practice problems

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

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