The Divergence Theorem

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}