# Divergence theorem: Wikis

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# Encyclopedia

In vector calculus, the divergence theorem, also known as Gauss' theorem (Carl Friedrich Gauss), Ostrogradsky's theorem (Mikhail Vasilievich Ostrogradsky), or Gauss–Ostrogradsky theorem is a result that relates the flow (that is, flux) of a vector field through a surface to the behavior of the vector field inside the surface.

More precisely, the divergence theorem states that the outward flux of a vector field through a closed surface is equal to the volume integral of the divergence of the region inside the surface. Intuitively, it states that the sum of all sources minus the sum of all sinks gives the net flow out of a region.

The divergence theorem is an important result for the mathematics of engineering, in particular in electrostatics and fluid dynamics.

The theorem is a special case of the more general Stokes' theorem, which generalizes the fundamental theorem of calculus.

## Intuition

If a fluid is flowing in some area, and we wish to know how much fluid flows out of a certain region within that area, then we need to add up the sources inside the region and subtract the sinks. The fluid flow is represented by a vector field, and the vector field's divergence at a given point describes the strength of the source or sink there. So, integrating the field's divergence over the interior of the region should equal the integral of the vector field over the region's boundary. The divergence theorem says that this is true.

The divergence theorem is thus a conservation law which states that the volume total of all sinks and sources, the volume integral of the divergence, is equal to the net flow across the volume's boundary.

## Mathematical statement

A region V bounded by the surface S=∂V with the surface normal n

Suppose V is a subset of Rn (in the case of n = 3, V represents a volume in 3D space) which is compact and has a piecewise smooth boundary. If F is a continuously differentiable vector field defined on a neighborhood of V, then we have

$\iiint\limits_V\left(\nabla\cdot\mathbf{F}\right)dV=\iint\limits_{S}\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\;\;\;\subset\!\supset \mathbf{F}\cdot\mathbf{n}\,dS .$

The left side is a volume integral over the volume V, the right side is the surface integral over the boundary of the volume V. The closed manifold ∂V is quite generally the boundary of V oriented by outward-pointing normals, and n is the outward pointing unit normal field of the boundary ∂V. (dS may be used as a shorthand for ndS.) By the symbol within the two integrals it is stressed once more that ∂V is a closed surface. In terms of the intuitive description above, the left-hand side of the equation represents the total of the sources in the volume V, and the right-hand side represents the total flow across the boundary ∂V.

### Corollaries

By applying the divergence theorem in various contexts, other useful identities can be derived (cf. vector identities).

• Applying the divergence theorem to the product of a scalar function g and a vector field F, the result is
$\iiint\limits_V\left(\mathbf{F}\cdot \left(\nabla g\right) + g \left(\nabla\cdot \mathbf{F}\right)\right) dV = \iint\limits_{S}\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\;\;\;\subset\!\supset g\mathbf{F}\cdot d\mathbf{S}.$
A special case of this is $\scriptstyle \mathbf{F}=\nabla f$, in which case the theorem is the basis for Green's identities.
• Applying the divergence theorem to the cross-product of two vector fields $\scriptstyle \mathbf{F}\times \mathbf{G}$, the result is
$\iiint\limits_V \left(\mathbf{G}\cdot\left(\nabla\times\mathbf{F}\right) - \mathbf{F}\cdot \left( \nabla\times\mathbf{G}\right)\right)\, dV = \iint\limits_{S}\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\;\;\;\subset\!\supset \mathbf F\times\mathbf{G}{\cdot}d\mathbf{S}.$
• Applying the divergence theorem to the product of a scalar function, f, and a non-zero constant vector, the following theorem can be proven:[1]
$\iiint\limits_V \nabla f\, dV =\iint\limits_{S}\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\;\;\;\subset\!\supset{f}d \mathbf{S}$.
• Applying the divergence theorem to the cross-product of a vector field F and a non-zero constant vector, the following theorem can be proven:[1]
$\iiint\limits_V \nabla\times\mathbf{F}\, dV = \iint\limits_{S}\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\;\;\;\subset\!\supset{d}\mathbf{S}\times \mathbf{F}.$

## Example

A vector field on a sphere (this is not the field in the example).

Suppose we wish to evaluate

$\iint\limits_{S}\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\;\;\;\subset\!\supset \mathbf{F}\cdot\mathbf{n}\,{d}S\,,$

where S is the unit sphere defined by $\scriptstyle x^2+y^2+z^2=1$ and F is the vector field

$\mathbf{F} = 2 x\mathbf{i}+y^2\mathbf{j}+z^2\mathbf{k}.$

The direct computation of this integral is quite difficult, but we can simplify the derivation of the result using the divergence theorem:

\begin{align} \iint\limits_{S}\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\;\;\;\subset\!\supset \mathbf{F} \cdot \mathbf{n}\,{d}S &= \iiint\limits_W\left(\nabla\cdot\mathbf{F}\right) \, dV\ &= 2\iiint\limits_W\left(1+y+z\right) \, dV\ &= 2\iiint\limits_W \,dV + 2\iiint\limits_W y \,dV + 2\iiint\limits_W z \,dV. \end{align}

Since the functions $\scriptstyle y$ and $\scriptstyle z$ are odd on $\scriptstyle S$ (which is a symmetric set with respect to the coordinate planes), one has

$\iiint\limits_W y\, dV = \iiint\limits_W z\, dV = 0.$

Therefore,

$\iint\limits_{S}\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\;\;\;\subset\!\supset \mathbf{F}\cdot\mathbf{n}\,{d}S = 2\iiint\limits_W \,dV = \frac{8\pi}{3}\,,$

because the unit sphere W has volume $\scriptstyle \frac{4\pi}{3}$.

## Applications

### Differential form and integral form of physical laws

As a result of the divergence theorem, a host of physical laws can be written in both a differential form (where one quantity is the divergence of another) and an integral form (where the flux of one quantity through a closed surface is equal to another quantity). Three examples are Gauss's law (in electrostatics), Gauss's law for magnetism, and Gauss's law for gravity.

#### Continuity equations

Continuity equations offer more examples of laws with both differential and integral forms, related to each other by the divergence theorem. In fluid dynamics, electromagnetism, quantum mechanics, and a number of other fields, there are continuity equations that describe the conservation of mass, momentum, energy, probability, or other quantities. Generically, these equations state that the divergence of the flow of the conserved quantity is equal to the distribution of sources or sinks of that quantity. The divergence theorem states that any such continuity equation can be written in a differential form (in terms of a divergence) and an integral form (in terms of a flux).

### Inverse-square laws

Any inverse-square law can instead be written in a Gauss' law-type form (with a differential and integral form, as described above). Two examples are Gauss' law (in electrostatics), which follows from the inverse-square Coulomb's law, and Gauss' law for gravity, which follows from the inverse-square Newton's law of universal gravitation. The derivation of the Gauss' law-type equation from the inverse-square formulation (or vice-versa) is exactly the same in both cases; see either of those articles for details.

## History

The theorem was first discovered by Joseph Louis Lagrange in 1762, then later independently rediscovered by Carl Friedrich Gauss in 1813, by George Green in 1825 and in 1831 by Mikhail Vasilievich Ostrogradsky, who also gave the first proof of the theorem. Subsequently, variations on the Divergence theorem are called Gauss's Theorem, Green's theorem, and Ostrogradsky's theorem.

## Examples

Verify the planar variant of the divergence theorem for a region R, with F(x,y)=2yi + 5xj, where R is the region bounded by the circle

$\scriptstyle x^2 + y^2 = 1.\$

Solution: The boundary of R is the unit circle, C, that can be represented parametrically by:

$x\, = \cos(s);\ y\, = \sin(s)\$

such that $\scriptstyle 0\, \le \,s\, \le \,2\pi$ where s units is the length arc from the point s = 0 to the point P on C. Then a vector equation of C is

$\mathbf{C}(s) = \cos(s)\mathbf{i} + \sin(s)\mathbf{j}.\$

At a point $\scriptstyle\mathbf{P}\, = \,(\cos(s),\, \sin(s))$ on C, $\scriptstyle\mathbf{F}\, = \,2\sin(s)\mathbf{i} + 5\cos(s)\mathbf{j}$. Therefore,

\begin{align}\oint_C \mathbf{F} \cdot \mathbf{N}\, ds &= \,\int_{0}^{2 \pi} ( 2 \sin s \mathbf{i} + 5 \cos s \mathbf{j}) \cdot (\cos s \mathbf{i} + \sin s \mathbf{j})\, ds\ &= \,\int_{0}^{2 \pi} (2 \sin s \cos s + 5 \sin s \cos s)\, ds\ &= \,7\int_{0}^{2 \pi} \sin s \cos s\, ds\ &= \,0.\end{align}\

Because $\scriptstyle M\, = \,2y\,$, $\scriptstyle \frac{\partial M}{\partial x} = \,0$, and because $\scriptstyle N\, = \,5x\,$, $\scriptstyle \frac{\partial N}{\partial y} = \,0$. Thus

\begin{align}\iint\limits_R \, \operatorname{div} \mathbf{F} \,dA\, &= \iint\limits_R \left (\frac{\partial M}{\partial x} + \frac{\partial N}{\partial y} \right) dA \\&= \,0.\end{align}\