# The Divergence Theorem

We will now look at a counterpart to both Green’s Theorem and Stoke’s Theorem known as **The Divergence Theorem**.

Theorem 1 (The Divergence Theorem): Let $E$ be a regular domain in $\mathbb{R}^3$ and let $S$ be the closed surface boundary of $E$ that is oriented outward and with unit normal field $\hat{N}$. Let $\mathbf{F}(x, y, z) = P(x, y, z) \vec{i} + Q(x, y, z) \vec{j} + R(x, y, z) \vec{k}$ be a vector field on $\mathbb{R}^3$ where the partial derivatives of $P$, $Q$, and $R$ are all continuous on an open region containing $E$. Then $\iint_{\delta} \mathbf{F} \cdot d \vec{S} = \iiint_D \mathrm{div} (\mathbf{F}) \: dV$ |

*An alternative name for The Divergence Theorem is Gauss’s Theorem.*

**Proof:**Since $D$ is a regular domain in $\mathbb{R}^3$, then $E$ is the union of a finite number of non-overlapping domains $D_1$, $D_2$, …, $D_n$ that are $x$-simple, $y$-simple, and $z$-simple, and so it is sufficient to show that for each of these subdomains the theorem above holds.

- To show this, suppose that the region $D$ is split by the surface $\delta^*$ into two subregions $D_1$ and $D_2$ such that $D = D_1 \cup D_2$ so that then the boundary surface $\delta$ is split into two surfaces $\delta_1$ and $\delta_2$ such that $\delta = \delta_1 \cup \delta_2$ and that $\delta_1 \cup \delta^*$ is the boundary surface of $D_1$ and $\delta_2 \cup \delta^*$ is the boundary surface of $D_2$.

- Note that the normals $\hat{N}_1$ and $\hat{N}_2$ of the subdomains $D_1$ and $D_2$ point in opposite directions on the boundary surface $\delta^*$. If we suppose that $\iiint_{D_1} \mathrm{div} (\mathbf{F}) \: dV = \iint_{S_1 \cup S^*} \mathbf{F} \cdot d \vec{S}$ and $\iiint_{D_2} \mathrm{div} (\mathbf{F}) \: dV = \iint_{S_2 \cup S^*} \mathbf{F} \cdot d \vec{S}$ and then add these equations, then we get that this $\hat{N_1} = – \hat{N_2}$, the contributions from the surface $\delta^*$ cancel out and:

(1)

\begin{align} \quad \iiint_D \mathrm{div} (\mathbf{F}) \: dV =\iint_{\delta_1 \cup \delta_2} \mathbf{F} \cdot d \vec{S} \end{align}

- Let $D$ be an $x$-simple, $y$-simple, and $z$-simple domain, and let $\mathbf{F}(x, y, z) = P(x, y, z) \vec{i} + Q(x, y, z) \vec{j} + R(x, y, z) \vec{k}$ be a vector field. Then:

(2)

\begin{align} \quad \iiint_D \mathrm{div} (\mathbf{F}) \: dV = \iiint_D \left ( \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z} \right ) \: dV \\ \quad \iiint_D \mathrm{div} (\mathbf{F}) \: dV = \iiint_D \frac{\partial P}{\partial x} \: dV + \iiint_D \frac{\partial Q}{\partial y} \: dV + \iiint_D \frac{\partial R}{\partial z} \: dV \end{align}

- Since $\hat{N}$ is the unit normal outward, then we have that:

(3)

\begin{align} \quad \iint_{\delta} \mathbf{F} \cdot d \vec{S} = \iint_{\delta} \mathbf{F} \cdot \hat{N} dS \\ \quad \iint_{\delta} \mathbf{F} \cdot d \vec{S} = \iint_{\delta} (P(x, y, z) \vec{i} + Q(x, y, z) \vec{j} + R(x, y, z) \vec{k}) \cdot \hat{N} \: dS \\ \quad \iint_{\delta} \mathbf{F} \cdot d \vec{S} = \iint P(x, y, z) \vec{i} \cdot \hat{N} dS + \iint_{\delta} Q(x, y, z) \vec{j} \cdot \hat{N} \: dS + \iint_{\delta} R(x, y, z) \vec{k} \cdot \hat{N} \: dS \end{align}

- To prove The Divergence Theorem all we need to show is that $\iint_{\delta} P(x, y, z) \vec{i} \cdot \hat{N} \: dS = \iiint_D \frac{\partial P}{\partial x} \: dV$, $\iint_{\delta} Q(x, y, z) \vec{j} \cdot \hat{N} \: dS = \iiint_D \frac{\partial Q}{\partial y} \: dV$ and $\iint_{\delta} R(x, y, z) \vec{k} \cdot \hat{N} \: dS = \iiint_D \frac{\partial R}{\partial z} \: dV$.

- We will first show that $\iint_{\delta} R(x, y, z) \vec{k} \cdot \hat{N} \: dS = \iiint_D \frac{\partial R}{\partial z} \: dV$. Since $D$ is a $z$-simple domain then if $B$ is the projection of the region $D$ onto the $xy$ plane then we can write $D$ as:

(4)

\begin{align} \quad D = \{ (x, y, z) \in \mathbb{R}^3 : (x, y) \in P : u_1(x, y) ≤ z ≤ u_2(x, y) \} $]] \end{align}

- Now we look at one side applying the Fundamental Theorem of Calculus we can greatly simplify the righthand side of the equation containing the triple integral:

(5)

\begin{align} \quad \iiint_D \frac{\partial R}{\partial z} \: dV = \iint_B \left [ \int_{u_1(x, y)}^{u_2(x, y)} \frac{\partial R}{\partial z} \: dz \right ] \: dA \\ \quad \iiint_D \frac{\partial R}{\partial z} \: dV = \iint_B \left [ R(x, y, u_2(x, y)) – R(x, y, u_1(x, y)) \right ] \: dA \end{align}

- Now we will show that the lefthand side (the one containing the double integral) is equal to this one. Now, the boundary of $\delta$ contains at most three important surfaces – the bottom surface which we denoted by $\delta_1$ which is generated by the function $z = u_1(x, y)$, the top surface, $\delta_2$, which is generated by the function $z = u_2(x, y)$ and possibly an upright surface $\delta_3$. Therefore we have that:

(6)

\begin{align} \quad \quad \iint_{\delta} R(x, y, z) \vec{k} \cdot \hat{N} \: dS = \iint_{\delta_1} R(x, y, z) \vec{k} \cdot \hat{N} \: dS + \iint_{\delta_2} R(x, y, z) \vec{k} \cdot \hat{N} \: dS + \iint_{\delta_3} R(x, y, z) \vec{k} \cdot \hat{N} \: dS \end{align}

- On $\delta_3$ notice that $\vec{k} \cdot \hat{N} = 0$ though since $\vec{k}$ is a vertical unit vector and $\hat{N}$ is a horizontal unit vector and so these vectors are perpendicular so their dot product is zero and so:

(7)

\begin{align} \quad \iint_{\delta_3} R(x, y, z) \vec{k} \hat{N} \: dS = \iint_{\delta_3} 0 \: dS = 0 \end{align}

- Therefore we can simplify the equation from earlier for the lefthand side of the equation:

(8)

\begin{align} \quad \iint_{\delta} R(x, y, z) \vec{k} \cdot \hat{N} \: dS = \iint_{\delta_1} R(x, y, z) \vec{k} \cdot \hat{N} \: dS + \iint_{\delta_2} R(x, y, z) \vec{k} \cdot \hat{N} \: dS \end{align}

- Now since $\delta_1$ is generated by the function $z = u_1(x, y)$ we will have that the outward unit normal $\hat{N}$ will point downwards, and so:

(9)

\begin{align} \quad \iint_{\delta_1} R(x, y, z) \vec{k} \cdot \hat{N} \: dS = – \iint_B R(x, y, u_1(x, y)) \: dA \end{align}

- Similarly since $\delta_2$ is generated by $z = u_2(x, y)$ we will have that the outward unit normal $\hat{N}$ will point upwards, and so:

(10)

\begin{align} \quad \iint_{\delta_2} R(x, y, z) \vec{k} \cdot \hat{N} \: dS = \iint_B R(x, y, u_2(x, y)) \: dA \end{align}

- Therefore we have that:

(11)

\begin{align} \quad \iint_{\delta} R(x, y, z) \vec{k} \cdot \hat{N} \: dS = – \iint_B R(x, y, u_1(x, y)) \: dA + \iint_B R(x, y, u_2(x, y)) \: dA \\ \quad \iint_{\delta} R(x, y, z) \vec{k} \cdot \hat{N} \: dS = \iint_B \left [ R(x, y, u_2(x, y)) – R(x, y, u_1(x, y)) \right ] \: dA \end{align}

- This proves that $\iint_{\delta} R(x, y, z) \vec{k} \cdot \hat{N} \: dS = \iiint_D \frac{\partial R}{\partial z} \: dV$. We can then repeat the proof by treating $D$ as $y$-simple and $z$-simple regions to show that the other equalities hold and we thus:

(12)

\begin{align} \quad \iint_{\delta} \mathbf{F} \cdot d \vec{S} = \iiint_D \mathrm{div} (\mathbf{F}) \: dV \quad \blacksquare \end{align}

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