Divergence theorem

Statement

Let $V \subseteq \mathbb{R}^3$ be compact with a piecewise smooth boundary surface $S = \partial V$, oriented by the outward unit normal $n$. If $F$ is a vector field of class $C^1$ on an open set that contains $V$, then $$\iiint_V \nabla \cdot F \, dV = \iint_S F \cdot n \, dS.$$

Remarks

The left side is the triple integral of the divergence, the total source strength in $V$. The right side is the outward flux through the boundary. Example: for $F(x, y, z) = (x, y, z)$ and the unit ball, $\nabla \cdot F = 3$. So both sides equal $3 \cdot \tfrac{4}{3} \pi = 4\pi$. With $g F$ in place of $F$, for a $C^1$ function $g$, one gets $\iiint_V (F \cdot \nabla g + g \, \nabla \cdot F) \, dV = \iint_S g \, F \cdot n \, dS$. This is a form of integration by parts. The same statement holds in $\mathbb{R}^n$. For a rectangle in the plane, the proof is a direct use of the fundamental theorem of calculus.

History

Lagrange discovered the theorem in 1762, and Gauss proved special cases from 1813. Ostrogradsky gave the first proof of the general theorem in 1826, in his work on heat flow.

Explanations

L1 human

What flows out of a closed surface must come from sources inside it. The divergence theorem makes this exact: the total divergence inside equals the net flux out through the boundary.