# Difference between revisions of "Divergence Theorem"

Fountain Flux
Field: Calculus
Image Created By: Brendan John
Website: [1]

Fountain Flux

The water flowing out of a fountain demonstrates an important theorem for vector fields, the Divergence Theorem.

# Basic Description

Consider the top layer of the fountain pictured. The rate that water flows out of the fountain's spout is directly related to the amount of water that flows off the top layer. Because something like water isn't easily compressed like air, if more water is pumped out of the spout, then more water will have to flow over the boundaries of the top layer. This is essentially what The Divergence Theorem states: the total the fluid being introduced into a volume is equal to the total fluid flowing out of the boundary of the volume if the quantity of fluid in the volume is constant.

# A More Mathematical Explanation

Note: understanding of this explanation requires: *Some multivariable calculus

The Divergence Theorem in its pure form applies to Vector Fields. Flowing water can be considere [...]

The Divergence Theorem in its pure form applies to Vector Fields. Flowing water can be considered a vector field because at each point the water has a velocity vector. Faster moving water is represented by a larger vector in our field. The divergence of a vector field at a point is the net flow of water going outwards from that point. Analytically divergence of a field $F$ is expressed in partial derivatives:

$\nabla\cdot\mathbf{F} =\partial{F_x}/\partial{x} + \partial{F_y}/\partial{y} + \partial{F_z}/\partial{z}$,

where $F _i$ is the component of $F$ in the $i$ direction. Intuitively, if F has a large positive rate of change in the x direction, the partial derivative with respect to x in this direction will be large, increasing total divergence.

The divergence theorem is formally stated as:

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

This theorem requires that we sum divergence over an entire volume, as shown in the left side of this equation.. If this sum is positive, then the field must indicate some movement out of the volume through its boundary, while if this sum is negative, the field must indicate some movement into the volume through its boundary. We use the notion of flux, the flow through a surface, to quantify this movement through the boundary, which itself is a surface. The right side of this equation is the sum of the field perpendicular to the volume's boundary at the boundary, which is the total flux through the boundary.

Summing up divergence over the entire volume means we sum the flow into or out of each infinitely subregion. A flow into one infinitesimal subregion means flow out of an adjacent subregion, which effects the next adjacent subregion, and so on until the boundary of the entire volume is reached. The total sum of divergence over the volume is thus equal to the flow at the boundary, as the theorem states.

A volume can be broken into infinitely small subregions, each of whose divergence effects the adjacent regions' divergence, up to the volume's boundary.

### Example of Divergence Theorem Verification

The following example verifies that given a volume and a vector field, the Divergence Theorem is valid.

Cutaway view of the cube used in the example. The purple lines are the vectors of the vector field F.

Consider the vector field $F = \begin{bmatrix} x^2 \\ 0\\ 0\\ \end{bmatrix}$.

For a volume, we will use a cube of edge length two, and vertices at (0,0,0), (2,0,0), (0,2,0), (0,0,2), (2,2,0), (2,0,2), (0,2,2), (2,2,2). This cube has a corner at the origin and all the points it contains are in positive regions.

• We begin by calculating the left side of the Divergence Theorem.
Step 1: Calculate the divergence of the field:
$\nabla\cdot F = 2x$
Step 2: Integrate the divergence of the field over the entire volume.
$\iiint\nabla\cdot F\,dV =\int_0^2\int_0^2\int_0^2 2x \, dxdydz$
$=\int_0^2\int_0^2 4\, dydx$
$=16$
• We now turn to the right side of the equation, the integral of flux.
Step 3: We first parametrize the parts of the surface which have non-zero flux.
Notice that the given vector field has vectors which only extend in the x-direction, since each vector has zero y and z components. Therefore, only two sides of our cube can have vectors normal to them, those sides which are perpendicular to the x-axis. Furthermore, the side of the cube perpendicular to the x axis with all points satisfying x = 0 cannot have any flux, since all vectors on this surface are zero vectors.
We are thus only concerned with one side of the cube since only one side has non-zero flux. This side is parametrized using
$X=\begin{bmatrix} x \\ y\\ z\\ \end{bmatrix} = \begin{bmatrix} 2 \\ u\\ v\\ \end{bmatrix}\, , u \in (0,2)\, ,v \in (0,2)$
Step 4: With this parametrization, we find a general normal vector to our surface.
To find this normal vector, we find two vectors which are always tangent to (or contained in) the surface, and are not collinear. The cross product of two such vectors gives a vector normal to the surface.
The first vector is the partial derivative of our surface with respect to u: $\frac{\part{X}}{\part{u}} = \begin{bmatrix} 0\\ 1\\ 0\\ \end{bmatrix}$
The second vector is the partial derivative of our surface with respect to v: $\frac{\part{X}}{\part{v}} = \begin{bmatrix} 0\\ 0\\ 1\\ \end{bmatrix}$
The normal vector is finally the cross product of these two vectors, which is simply $N = \begin{bmatrix} 1\\ 0\\ 0\\ \end{bmatrix}.$
Step 5: Integrate the dot product of this normal vector with the given vector field.
The amount of the field normal to our surface is the flux through it, and is exactly what this integral gives us.
$\iint\limits_{\partial V}\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\;\;\;\subset\!\supset \mathbf F\;\cdot\mathbf n\,{d}S .$
$= \int_0^2 \int_0^2 F \cdot N \,dsdt$
$= \int_0^2 \int_0^2 \begin{bmatrix} x^2 \\ 0\\ 0\\ \end{bmatrix} \cdot \begin{bmatrix} 1 \\ 0\\ 0\\ \end{bmatrix} \,dsdt = \int_0^2 \int_0^2 x^2dsdt = \int_0^2 \int_0^2 4 \,dsdt$
$=16$
• Both sides of the equation give 16, so the Divergence Theorem is indeed valid here. ■

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