Divergence Theorem

_ V F· d S = _V ( · F) dV — outward flux equals total divergence inside.

The idea

Theorem (Divergence theorem (Gauss)).

Let $V$ be a bounded solid region in $\mathbb{R}^3$ whose boundary surface $\partial V$ is closed, piecewise smooth, and oriented with its normal pointing outward. For every smooth vector field $\mathbf{F}$ on $V$, $\oiint_{\partial V} \mathbf{F} \cdot d\mathbf{S} = \iiint_V (\nabla \cdot \mathbf{F})\, dV.$

The left side is the total flux of $\mathbf{F}$ outward through the boundary surface. The right side adds up, over every point inside, the divergence $\nabla \cdot \mathbf{F}$ — the local rate at which $\mathbf{F}$ spreads away from that point. The theorem says the two totals agree: whatever flows out through the surface was produced somewhere within.

The reason is the same cancellation that drives Green's theorem, one dimension up. Cut $V$ into two pieces with an interior wall. Flux leaving the left piece through that wall is flux entering the right piece through the same wall, so the wall's two contributions have opposite signs and cancel when the pieces are added, and only the outer boundary survives. Slicing $V$ into arbitrarily many small cells cancels every internal face the same way, leaving the flux through $\partial V$ on one side and a sum of per-cell contributions on the other — in the limit, the volume integral.

The outward orientation is part of the statement: flipping to inward normals negates the left side and leaves the right side alone.

Ways to work on it

• Walkthrough. Unpack the statement (flux vs. triple integral, divergence operator, outward orientation), then compute on a cube.
• Practice. Constant-divergence fields on cubes and balls: divergence × volume.
• Hardest. Compute the flux of a variable-divergence field through a closed surface.

Not sure where to start? Take the ten-question placement test.