# 16.8: The Divergence Theorem and a Unified Theory

- Page ID
- 2642

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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

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

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*.

### Contributors

- Larry Green (Lake Tahoe Community College)
Integrated by Justin Marshall.