Skip to main content
\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)
Mathematics LibreTexts

16.9: The Divergence Theorem

The third version of Green's Theorem we saw was:

$$\int_{\partial D} {\bf F}\cdot{\bf N}\,ds=\iint\limits_{D} \nabla\cdot{\bf F}\,dA.$$

With minor changes this turns into another equation, the Divergence Theorem:

Theorem: Divergence Theorem

Under suitable conditions, if \(E\) is a region of three dimensional space and \(D\) is its boundary surface, oriented outward, then

$$\iint\limits_{D} {\bf F}\cdot{\bf N}\,dS=\iiint\limits_{E} \nabla\cdot{\bf F}\,dV.$$

Proof

Again this theorem is too difficult to prove here, but a special case is easier. In the proof of a special case of Green's Theorem, we needed to know that we could describe the region of integration in both possible orders, so that we could set up one double integral using \(dx\,dy\) and another using \(dy\,dx\). Similarly here, we need to be able to describe the three-dimensional region \(E\) in different ways.

We start by rewriting the triple integral:

$$\iiint\limits_{E} \nabla\cdot{\bf F}\,dV = \iiint\limits_{E} (P_x+Q_y+R_z)\,dV =\iiint\limits_{E} P_x\,dV + \iiint\limits_{E} Q_y\,dV + \iiint\limits_{E} R_z\,dV.$$

The double integral may be rewritten:

$$\iint\limits_{D} {\bf F}\cdot{\bf N}\,dS=\iint\limits_{D} (P{\bf i}+Q{\bf j}+R{\bf k})\cdot{\bf N}\,dS=\iint\limits_{D} P{\bf i}\cdot{\bf N}\,dS+\iint\limits_{D} Q{\bf j}\cdot{\bf N}\,dS+\iint\limits_{D} R{\bf k}\cdot{\bf N}\,dS.$$

To prove that these give the same value it is sufficient to prove that

$$ \iint\limits_{D} P{\bf i}\cdot{\bf N}\,dS=\iiint\limits_{E} P_x\,dV,\;
\iint\limits_{D} Q{\bf j}\cdot{\bf N}\,dS=\iiint\limits_{E} Q_y\,dV,\;\hbox{and}\;
 \iint\limits_{D} R{\bf k}\cdot{\bf N}\,dS=\iiint\limits_{E} R_z\,dV.$$

Not surprisingly, these are all pretty much the same; we'll do the first one.

We set the triple integral up with \(dx\) innermost:

$$\iiint\limits_{E} P_x\,dV=\iint\limits_{B}\int_{g_1(y,z)}^{g_2(y,z)} P_x\,dx\,dA= \iint\limits_{B} P(g_2(y,z),y,z)-P(g_1(y,z),y,z)\,dA,$$

where \(B\) is the region in the \(y\)-\(z\) plane over which we integrate. The boundary surface of \(E\) consists of a "top'' \(x=g_2(y,z)\), a "bottom'' \(x=g_1(y,z)\), and a "wrap-around side'' that is vertical to the \(y\)-\(z\) plane. To integrate over the entire boundary surface, we can integrate over each of these (top, bottom, side) and add the results. Over the side surface, the vector \(\bf N\) is perpendicular to the vector \(\bf i\), so

$$\iint\limits_{\hbox{\text{ sevenpoint side}}} P{\bf i}\cdot{\bf N}\,dS =\iint\limits_{\hbox{\text{sevenpoint side}}}0\,dS=0.$$

Thus, we are left with just the surface integral over the top plus the surface integral over the bottom. For the top, we use the vector function \({\bf r}=\langle g_2(y,z),y,z\rangle\) which gives \({\bf r}_y\times{\bf r}_z=\langle 1,-g_{2y},-g_{2z}\rangle\); the dot product of this with \({\bf i}=\langle 1,0,0\rangle\) is 1. Then

$$\iint\limits_{\hbox{\sevenpoint top}} P{\bf i}\cdot{\bf N}\,dS=\iint\limits_{B} P(g_2(y,z),y,z)\,dA.$$

In almost identical fashion we get

$$\iint\limits_{\hbox{\sevenpoint bottom}} P{\bf i}\cdot{\bf N}\,dS=-\iint\limits_{B} P(g_1(y,z),y,z)\,dA,$$

where the negative sign is needed to make \(\bf N\) point in the negative \(x\) direction. Now

$$\iint\limits_{D} P{\bf i}\cdot{\bf N}\,dS=\iint\limits_{B} P(g_2(y,z),y,z)\,dA-\iint\limits_{B} P(g_1(y,z),y,z)\,dA,$$

which is the same as the value of the triple integral above.

Example \(\PageIndex{1}\)

Let \({\bf F}=\langle 2x,3y,z^2\rangle\), and consider the three-dimensional volume inside the cube with faces parallel to the principal planes and opposite corners at \((0,0,0)\) and \((1,1,1)\). We compute the two integrals of the divergence theorem.

The triple integral is the easier of the two:

$$\int_0^1\int_0^1\int_0^1 2+3+2z\,dx\,dy\,dz=6.$$

The surface integral must be separated into six parts, one for each face of the cube. One face is \(z=0\) or \({\bf r}=\langle u,v,0\rangle\), \(0\le u,v\le 1\). Then \({\bf r}_u=\langle 1,0,0\rangle\), \({\bf r}_v=\langle 0,1,0\rangle\), and \({\bf r}_u\times{\bf r}_v= \langle 0,0,1\rangle\). We need this to be oriented downward (out of the cube), so we use \(\langle 0,0,-1\rangle\) and the corresponding integral is

$$\int_0^1\int_0^1 -z^2\,du\,dv=\int_0^1\int_0^1 0\,du\,dv=0.$$

Another face is \(y=1\) or \({\bf r}=\langle u,1,v\rangle\). Then \({\bf r}_u=\langle 1,0,0\rangle\), \({\bf r}_v=\langle 0,0,1\rangle\), and \({\bf r}_u\times{\bf r}_v= \langle 0,-1,0\rangle\). We need a normal in the positive \(y\) direction, so we convert this to \(\langle 0,1,0\rangle\), and the corresponding integral is

$$\int_0^1\int_0^1 3y\,du\,dv=\int_0^1\int_0^1 3\,du\,dv=3.$$

The remaining four integrals have values 0, 0, 2, and 1, and the sum of these is 6, in agreement with the triple integral.

Example \(\PageIndex{2}\)

Let \({\bf F}=\langle x^3,y^3,z^2\rangle\), and consider the cylindrical volume \(x^2+y^2\le9\), \(0\le z\le2\). The triple integral (using cylindrical coordinates) is

$$\int_0^{2\pi}\int_0^3\int_0^2 (3r^2+2z)r\,dz\,dr\,d\theta=279\pi.$$

For the surface we need three integrals. The top of the cylinder can be represented by \({\bf r}=\langle v\cos u,v\sin u,2\rangle\); \({\bf r}_u\times{\bf r}_v=\langle 0,0,-v\rangle\), which points down into the cylinder, so we convert it to \(\langle 0,0,v\rangle\). Then

$$\int_0^{2\pi}\int_0^3 \langle v^3\cos^3u,v^3\sin^3u,4\rangle\cdot\langle 0,0,v\rangle\,dv\,du=\int_0^{2\pi}\int_0^3 4v\,dv\,du=36\pi.$$

The bottom is \({\bf r}=\langle v\cos u,v\sin u,0\rangle\); \({\bf r}_u\times{\bf r}_v=\langle 0,0,-v\rangle\) and

$$\int_0^{2\pi}\int_0^3 \langle v^3\cos^3u,v^3\sin^3u,0\rangle\cdot\langle 0,0,-v\rangle \,dv\,du= \int_0^{2\pi}\int_0^3 0\,dv\,du=0.$$

The side of the cylinder is \({\bf r}=\langle 3\cos u,3\sin u,v\rangle\); \({\bf r}_u\times{\bf r}_v=\langle 3\cos u,3\sin u,0\rangle\) which does point outward, so

$$\eqalign{\int_0^{2\pi}\int_0^2 &\langle 27\cos^3 u,27\sin^3 u,v^2\rangle\cdot\langle 3\cos u,3\sin u,0\rangle \,dv\,du\cr &=\int_0^{2\pi}\int_0^2 81\cos^4 u+81\sin^4u\,dv\,du=243\pi.\cr}$$

The total surface integral is thus \(36\pi+0+243\pi=279\pi\).

Contributors

David Guichard (Whitman College)

  • Integrated by Justin Marshall.