
# 4.9: The Divergence Theorem and a Unified Theory


When we looked at Green's Theorem, we saw that there was a relationship between a region and the curve that encloses it. This gave us the relationship between the line integral and the double integral. Now consider the following theorem:

Divergence Theorem

Let $$Q$$ be a solid region bounded by a closed surface oriented with outward pointing unit normal vector $$\text{n}$$, and let $$\textbf{F}$$ be a differentiable vector field (i.e., components have continuous partial derivatives). Then

$\iint \limits_{S} \textbf{F} \cdot \text{n} \, dS = \iiint \limits_{Q} \nabla \cdot \textbf{F}\,dv.$

Moving to three dimensions, the divergence theorem provides us with a relationship between a triple integral over a solid and the surface integral over the surface that encloses the solid.

Example $$\PageIndex{1}$$

Find

$\iint \limits_{S} \textbf{F} \cdot \textbf{N} \,ds \nonumber$

where

$\textbf{F} (x,y,z) = y^2 \hat{\textbf{i}} + e^x(1-\cos{(x^2 + z^2)}\hat{\textbf{j}} + (x + z) \hat{\textbf{k}} \nonumber$

and $$S$$ is the unit sphere centered at the point $$(1,4,6)$$ with outwardly pointing normal vector.

Solution

This seemingly difficult problem turns out to be quite easy once we have the divergence theorem. We have

$\nabla \cdot \textbf{F} = 0 + 0 + 1 = 1 \nonumber$

Now recall that a triple integral of the function $$1$$ is the volume of the solid. Since the solid is a sphere of radius $$1$$ we get $$\dfrac{4}{3}\pi$$.

Partial Proof

As usual, we will make some simplifying remarks and then prove part of the divergence theorem. We assume that the solid is bounded below by $$z = g_1(x,y)$$ and above by $$z = g_2(x,y)$$.

Notice that the outward pointing normal vector is upward on the top surface and downward for the bottom region. We also note that the divergence theorem can be written as

\begin{align*} \iint\limits_{S} \textbf{F} \cdot \textbf{N} dS &= \iint\limits_{S} (M \hat{\textbf{i}} \cdot \textbf{N} + N\hat{\textbf{j}} \cdot \textbf{N} + P \hat{\textbf{k}} \cdot \textbf{N} ) dS \\ &= \iiint\limits_{Q} \nabla \cdot \textbf{F} dv \\ &= \iiint\limits_{Q} (M_x + N_y + P_z) dv. \end{align*}

We will show that

$\iint\limits_{S} P \hat{\textbf{k}} \cdot \textbf{N} dS = \iiint \limits_{Q} P_x dv.$

We have on the top surface

$P\hat{\text{k}} \cdot \text{n}\, dS = P\hat{\text{k}} \cdot \left( (-g_2)_x \hat{\text{i}} - (g_2)_y \hat{\text{j}} + \hat{\text{k}} \right) = P(x,y,g_2(x,y)).$

On the bottom surface, we get

$P\hat{\text{k}} \cdot \text{n} dS = P\hat{\text{k}} \cdot \left( (g_1)_x \hat{\text{i}} + (g_1)_y \hat{\text{j}} - \hat{\text{k}} \right) = -P(x,y,g_1(x,y)).$

Putting these together we get

$\iint\limits_S P\hat{\text{k}} \cdot \textbf{N} dS = \iint\limits_R [P(x,y,g_2 (x,y))-P(x,y,g_1 (x,y)) ] dydx .$

For the triple integral, the Fundamental Theorem of Calculus tell us that

\begin{align*} \iiint\limits_Q P_z dzdydx &= \iint\limits_R [P(x,y,z) ]_{g_1(x,y)}^{g_2(x,y)} dydx \\ &= \iint\limits_R [P(x,y,g_2(x,y))-P(x,y,g_1(x,y))]dy\,dx. \end{align*}

$$\square$$

## An Interpretation of Divergence

We have seen that the flux is the amount fluid flow per unit time through a surface. If the surface is closed, then the total flux will equal the flow out of the solid minus the flow in. Often in the solid there is a source (such as a star when the flow is electromagnetic radiation) or a sink (such as the earth collecting solar radiation). If we have a small solid $$S(P)$$ containing a point $$P$$, then the divergence of the vector field is approximately constant, which leads to the approximation

$\iiint\limits_Q \nabla \cdot \textbf{F} \, dv \approx \nabla \cdot \textbf{F}(P) \text{Volume.}$

The divergence theorem expresses the approximation

Flux through $$S(P) \approx \nabla \cdot \textbf{F}(P)$$(Volume).

Dividing by the volume, we get that the divergence of $$\textbf{F}$$ at $$P$$ is the Flux per unit volume.

• If the divergence is positive, then the $$P$$ is a source.
• If the divergence is negative, then $$P$$ is a sink.