The Divergence Theorem

From local sources to boundary flow

For a bounded solid $V$ with a piecewise smooth closed boundary, and a continuously differentiable field on a neighborhood of the solid,

\iint_{\partial V}\mathbf F\cdot\mathbf n\,dS=\iiint_V\nabla\cdot\mathbf F\,dV.

The boundary normal points outward. Flux through interior faces of adjacent small boxes cancels; only the exterior remains. That cancellation explains why integrating local divergence gives total outward flow.

Visual guide

VISUAL GUIDEOpposite internal fluxes cancel

Two adjacent boxes share an internal face. Flux leaving one enters the other, so those contributions cancel in the total. The divergence theorem sums only the remaining exterior flux over a closed boundary.

Check the surface first

The theorem applies to a closed surface. If a problem gives only a roof or curved wall, add the missing caps, apply the theorem to the resulting solid, then subtract their fluxes with the outward orientations of that solid. Also inspect the field for singularities inside the volume.

Worked example: a box

For $\mathbf F=(2x,-y,3z)$ on $[0,1]^3$ , divergence is $4$ and volume is $1$ , so total flux is $4$ . The far faces contribute $2,-1,3$ respectively; the three faces at coordinate zero contribute zero. Negative flux through the $y=1$ face represents entry, not an error.

Worked example: a hemisphere and its cap

For $\mathbf F=(0,0,z+1)$ on the upper half of a unit ball, divergence is $1$ . Total outward flux is its volume $2\pi/3$ . The flat bottom disk has outward normal $(0,0,-1)$ , so its flux is $-\pi$ . The curved hemisphere therefore has flux $2\pi/3-(-\pi)=5\pi/3$ .

A singular source

The field $\mathbf r/\|\mathbf r\|^3$ is not defined at the origin. Its divergence away from the origin is zero, but that does not justify zero flux through a sphere enclosing the origin. Exclude a small inner sphere and include both boundaries to use the theorem correctly.

Practice

Find outward flux of $(x,y,z)$ through a sphere of radius $2$ .
Find total outward flux of a constant field through any closed smooth surface.
Why can a divergence-free field still cross parts of the boundary?

Show worked solutions

Divergence $3$ times volume $32\pi/3$ gives $32\pi$ .
Divergence is zero, so total flux is zero.
Entry and exit can cancel. The theorem constrains the net flux, not each patch separately.

Further study

MIT OpenCourseWare: Multivariable Calculus provides a full university course with additional lectures and exercises.