Integral theorems · Definition

Divergence Theorem

multivariable calculusvector calculus

Before you start you need to know

It turns a flux through a closed surface into a triple integral, which is often easier, and vice versa.

Definition

The divergence theorem relates the flux of a field \(\mathbf F\) through a closed surface \(S\), with outward normal, to the triple integral of its divergence over the solid \(E\) that \(S\) encloses.

\[\iint_S\mathbf F\cdot d\mathbf S=\iiint_E\nabla\cdot\mathbf F\,dV\]

The surface \(S\) must be closed, piecewise smooth and outward oriented, and \(\mathbf F\) must have continuous partial derivatives on an open set containing \(E\) and \(S\). The physical idea is a balance: \(\nabla\cdot\mathbf F\) measures how much fluid is created per unit volume at each point, and the net outward flux is the sum of all those sources inside \(E\).

Guiding example: for \(\mathbf F=\langle x,y,z\rangle\) we have \(\nabla\cdot\mathbf F=3\), so the outward flux through any closed surface is \(3\) times the enclosed volume; for the unit sphere it is \(4\pi\). If \(\nabla\cdot\mathbf F=0\) throughout \(E\), the flux through \(S\) is zero: what goes in comes out.

An open surface can be closed with a simple lid: apply the theorem to the closed surface and subtract the flux through the lid, as with the hemisphere closed by its disk. The most frequent mistake is applying the theorem when \(\mathbf F\) is not smooth at some interior point, like the field \(\mathbf r/|\mathbf r|^3\) at the origin, or forgetting that the normal must point outward.

In the plane, Green's theorem in flux form is the analogous case. Together with Stokes' theorem, the divergence theorem is an instance of one principle: the integral of a derivative over a region is an integral of the field over its boundary.

Think about it

If \(\nabla\cdot\mathbf F=5\) in a solid of volume \(2\), what is the outward flux through its boundary?

\(10\): the flux is the constant divergence times the volume.

Why can't the theorem be used for \(\mathbf r/|\mathbf r|^3\) on the unit sphere even though its divergence is \(0\)?

Because the field is not smooth at the origin, which lies inside the sphere; the actual flux is \(4\pi\), not \(0\).

Exercises 10

References

  • James Stewart, Daniel K. Clegg, Saleem Watson, Calculus: Early Transcendentals, 9th edition, Cengage, 2021. §16.9
  • Joel Hass, Christopher Heil, Przemyslaw Bogacki, Maurice D. Weir, George B. Thomas Jr., Thomas' Calculus, 15th edition, Pearson, 2023. §16.8
  • Jerrold E. Marsden, Anthony Tromba, Vector Calculus, 6th edition, W. H. Freeman, 2012. §8.3