In one line
The divergence theorem states that the total outward flux of a vector field through a closed surface equals the volume integral of its divergence over the volume enclosed.
Explained simply
Think of a room full of people where some doors let people out. You can count people leaving in two ways: stand at every door and count those walking out (surface integral), or go around inside counting how many each little region "produces" (volume integral of divergence). Both totals must match. Inside the room, flows between neighbouring regions cancel, leaving only what crosses the outer walls.
How it works
- Divide the volume into tiny cubes. For each cube, outward flux \(\approx(\nabla\cdot\vec{A})\,\Delta V\) by definition of divergence.
- Add over all cubes. On each shared internal face, flux leaving one cube enters its neighbour, so these cancel.
- Only flux through the outer boundary survives, giving \(\oint_S\vec{A}\cdot d\vec{S} = \int_V(\nabla\cdot\vec{A})\,dV\).
- Use: it converts surface integrals into volume integrals. Applied to Gauss's law it gives \(\nabla\cdot\vec{E} = \rho/\varepsilon_0\); applied to \(\oint\vec{B}\cdot d\vec{S} = 0\) it gives \(\nabla\cdot\vec{B} = 0\).
Key formula
\(\oint_S\vec{A}\cdot d\vec{S} = \int_V(\nabla\cdot\vec{A})\,dV\). \(S\): closed surface, \(d\vec{S}\): outward area element (m2), \(V\): enclosed volume, \(dV\): volume element (m3).
Quick example
Verify for \(\vec{A} = x\hat{i} + y\hat{j} + z\hat{k}\) over the unit cube \(0 \le x, y, z \le 1\). Volume side: \(\nabla\cdot\vec{A} = 3\), so \(\int 3\,dV = 3\times1 = 3\). Surface side: on face \(x = 1\), \(\vec{A}\cdot\hat{i} = 1\), flux = 1; on \(x = 0\), \(A_x = 0\), flux = 0; similarly for y and z faces. Total = 3. Both sides agree.
Common mistakes
- Applying it to an open surface; the surface must be closed.
- Using inward normals; \(d\vec{S}\) must point outward.
- Mixing it up with Stokes' theorem (line integral to surface integral).
How MU asks it
Asked as "State Gauss divergence theorem" (3 marks), or as the key step in "Derive the differential form of Gauss's law" (5 marks). Write the statement in words and symbols, give the physical idea of cancelling internal fluxes, and show where it is used in Maxwell's equations.