OBJECTIVES

When you finish this section, you should be able to:

1. Find the divergence of a vector field (p. 1038)
2. Use the Divergence Theorem (p. 1038)
3. Interpret the divergence of \(\mathbf{F}\) (p. 1042)

spanDEFINITIONspan Divergence of \(\mathbf{F}\)

Let \(\mathbf{F}=\mathbf{F}(x,y,z)=P(x,y,z)\mathbf{i}+Q(x,y,z)\mathbf{j }+R(x,y,z)\mathbf{k}\) be a vector field, where \(\dfrac{\partial P}{ \partial x}\), \(\dfrac{\partial Q}{\partial y}\), and \(\dfrac{\partial R}{ \partial z}\) exist. The divergence of \(\mathbf{F}\), denoted by \(\rm{div}\mathbf{F}\), is defined as the function* \[\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt]{ {\rm div} \mathbf{F}=\dfrac{\partial P}{ \partial x}+\dfrac{\partial Q}{\partial y}+ \dfrac{\partial R}{\partial z} }} \]

Finding the Divergence of \(\mathbf{F}\)

The divergence of the vector field \[ \mathbf{F}(x,y,z)=x^{2}yz^{2}\mathbf{i}+(2xz+y^{3})\mathbf{j}+x^{2}y^{3}z \mathbf{k} \]

is \[ {\rm div}\mathbf{F}=\dfrac{\partial }{\partial x}(x^{2}yz^{2})+\dfrac{ \partial }{\partial y}(2xz+y^{3})+\dfrac{\partial }{\partial z} (x^{2}y^{3}z)=2xyz^{2}+3y^{2}+x^{2}y^{3} \]

NOW WORK

Problem 5.

ORIGINS

Although Carl Friedrich Gauss (1777 – 1855) was neither the first to discover nor the first to prove the Divergence Theorem, the theorem is named in honor of him. The Divergence Theorem was proved by the Russian mathematician Mikhail Ostrogradsky (1801 – 1862), whose name is also associated with the theorem. Gauss is considered among the greatest mathematicians of all time. He was the mentor of many other notable mathematicians including Riemann, Bessel, Dedekind, and Möbius.

THEOREM Divergence Theorem

Let \(S\) be a positively oriented surface with outer unit normal \(\mathbf{n}\) that encloses a closed, bounded solid \(E\) in space. Let \(\mathbf{ F}=\mathbf{F}(x,y,z)=P(x,y,z)\mathbf{i}+Q(x,y,z)\mathbf{j}+R(x,y,z)\mathbf{k}\) be a vector field for which the functions \(P\), \(Q\), and \(R\) have continuous first-order partial derivatives on an open set containing \(E\). Then \[\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt]{ \iiint\limits_E {{\rm div}}\ \mathbf{F} \,dV=\iiint\limits_S{\mathbf{F}}\,{\cdot}\, \mathbf{n}\,dS }} \]

If \(\mathbf{n}=\cos \alpha \mathbf{i}+\cos \beta \mathbf{j}+\cos \gamma \mathbf{k}\), then \[\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt]{ \iiint\limits_E \left( \dfrac{\partial P}{\partial x}+\dfrac{\partial Q}{ \partial y}+\dfrac{\partial R}{\partial z}\right) dV=\iint\limits_S (P\cos \alpha +Q\cos \beta +R\cos \gamma )\,dS }} \]

Proof

We assume that surfaces \(S_{1}\) and \(S_{2}\) are defined by the equations \[ S_{1}{:} \quad z=f_{1}(x,y) \qquad \hbox{and}\qquad S_{2}{:}\quad z=f_{2}(x,y) \]

where \(f_{1}(x,y)<f_{2}(x,y)\) on \(D\), \(f_{1}\) and \(f_{2}\) are continuous on \(D\), and \(f_{1}\) and \(f_{2}\) have continuous partial derivatives on \(D\). The surface \(S_{3}\) is the portion of a cylindrical surface between \(S_{1}\) and \(S_{2}\) formed by lines parallel to the \(z\)-axis and along the boundary of the region \(D\) in the \(xy\)-plane. The solid \(E\) is enclosed on the bottom by \(S_{1}\), on the top by \(S_{2}\), and on the sides by the lateral surface \(S_{3}\).

We now show that \[ \begin{equation} \tag{4} \iiint\limits_E \dfrac{\partial R}{\partial z}\,dV=\iint\limits_S R\cos \gamma \,dS \end{equation} \]

We start with the right side of (4). Then using the fact that \(S\) consists of the three surfaces \(S_{1}\), \(S_{2}\), and \(S_{3}\), we have \[ \iint\limits_S R\cos \gamma \,dS=\iint\limits_{S_{1}}R\cos \gamma \,dS+\iint\limits_{S_{2}}R\cos \gamma \,dS+\iint\limits_{S_{3}}R\cos \gamma \,dS \]

For the bottom surface \(S_{1}\): \(z=f_{1}(x,y)\), the outer unit normal \(\mathbf{n}_{1}\) equals the downward-pointing unit normal vector \(-\mathbf{n}\) to \(S_{1}\). So, \(\cos \gamma\) is given by \[ \cos \gamma =\mathbf{n}_{1}\,{\bf\cdot}\, \mathbf{k}=\dfrac{-1}{\sqrt{ [(f_{1})_x(x,y)]^{2}+[(f_{1})_y(x,y)]^{2}+1}} \] See Example 5 (page 1032).

Then \[ \iint\limits_{S_{1}}R\cos \gamma \,dS=\iint\limits_{S_{1}}R(x,y,f_{1}(x,y)) \dfrac{-1}{\sqrt{[(f_{1})_x(x,y)]^{2}+[(f_{1})_y(x,y)]^{2}+1}}\,dS \]

We can express the surface integral as a double integral \[ \begin{eqnarray} \iint\limits_{S_{1}}R\cos \gamma \,dS&=&-\iint\limits_D R(x,y,f_{1}(x,y)) \dfrac{\sqrt{[(f_{1})_x(x,y)]^{2}+[(f_{1})_y(x,y)]^{2}+1}\,}{\sqrt{ [(f_{1})_x(x,y)]^{2}+[(f_{1})_y(x,y)]^{2}+1}}\,dx\,dy\nonumber \\ &=&-\iint\limits_D R(x,y,f_{1}(x,y))\,dx\,dy \tag{5} \end{eqnarray} \]

For the top surface \(S_{2}\): \(\ z=f_{2}(x,y)\), the outer unit normal \(\mathbf{n}_{2}\) equals the upward-pointing unit normal vector \(\mathbf{n}\) to \(S_{2}\). So, \(\cos \gamma\) is given by \[ \cos \gamma =\mathbf{n}_{2}\,{\bf\cdot}\, \mathbf{k}=\dfrac{1}{\sqrt{ [(f_{2})_x(x,y)]^{2}+[(f_{2})_y(x,y)]^{2}+1}} \]See Example 5 (page 1032).

and \[ \iint\limits_{S_{2}}R\cos \gamma \,dS=\iint\limits_{S_{2}}R(x,y,f_{2}(x,y)) \dfrac{1}{\sqrt{[(f_{2})_x(x,y)]^{2}+[(f_{2})_y(x,y)]^{2}+1}}\,dS \]

Again, we can express the surface integral as a double integral. \[ \begin{eqnarray} \iint\limits_{S_{2}}R\cos \gamma \,dS&=&\iint\limits_{D}R(x,y,f_{2}(x,y))\dfrac{ \sqrt{[(f_{2})_x(x,y)]^{2}+[(f_{2})_y(x,y)]^{2}+1}\,}{\sqrt{ [(f_{2})_x(x,y)]^{2}+[(f_{2})_y(x,y)]^{2}+1}}\,dx\,dy \nonumber \\ &=&\iint\limits_{D}R(x,y,f_{2}(x,y))\,dx\,dy \tag{6} \end{eqnarray} \]

On the lateral surface \(S_{3}\), the angle between \(\mathbf{n}_{3}\) and \(\mathbf{k}\) is \(\gamma =\dfrac{\pi }{2}\), so \(\cos \gamma =0\). Therefore, \[ \begin{equation*} \iint\limits_{\kern-3ptS_{3}}R\cos \gamma \,dS=0 \end{equation*} \tag{7} \]

We now obtain \(\iint\limits_S R\cos \gamma \,dS\) by summing (5), (6), and (7): \[ \begin{eqnarray*} \iint\limits_{S} R\cos \gamma \,dS& =&\iint\limits_{S_{1}}R\cos \gamma \,dS+\iint\limits_{S_{2}}R\cos \gamma \,dS+\iint\limits_{S_{3}}R\cos \gamma \,dS \\ &=&-\iint\limits_{D}R(x,y,f_{1}(x,y))\,dx\,dy+\iint\limits_{D}R(x,y,f_{2}(x,y))\,dx\,dy+0 \\ & =&\iint\limits_{D}[R(x,y,f_{2}(x,y))-R(x,y,f_{1}(x,y))]\,dx\,dy \\ & =&\iint\limits_{D}\left[ \int_{f_{1}(x,y)}^{f_{2}(x,y)}\dfrac{\partial R}{ \partial z}\,dz\right] \,dx\,dy=\iiint\limits_E \dfrac{\partial R}{\partial z }\,dV \end{eqnarray*} \]

which proves (3).

Using the Divergence Theorem to Find Flux

Find the mass of a fluid of constant mass density \(\rho\) flowing across the cube enclosed by the planes \(x=0\), \(x=1\), \(y=0\), \(y=1\), \(z=0\), and \(z=1\) in a unit of time, in the direction of the outer unit normal vectors if the velocity of the fluid at any point on the cube is \(\mathbf{F}=\mathbf{F} (x,y,z)=4xz\mathbf{i}-y^{2}\mathbf{j}+yz\mathbf{k}\). (This is Example 8 on page 1035.)

Solution Suppose \(E\) is the solid cube and \(S\) is its surface. The mass of the fluid is \(\iint\limits_S \rho \mathbf{F}\,{\bf\cdot}\, \mathbf{n} \,dS\), where \(\mathbf{n}\) is the outer unit normal vector of \(S\). Since the velocity of the fluid is \(\mathbf{F}=4xz\mathbf{i}-y^{2}\mathbf{j}+yz\mathbf{ k}\), we have \({\rm{div}}\mathbf{F}=4z-2y+y=4z-y\). Now we use the Divergence Theorem. \[ \begin{eqnarray*} \iint\limits_{S}\rho \mathbf{F}\,{\bf\cdot}\, \mathbf{n}\,dS &=&\rho \iint\limits_S \mathbf{F}\,{\bf\cdot}\, \mathbf{n}\,dS=\rho \iiint\limits_E {\rm{div}}\mathbf{F} \,dV=\rho \int_{0}^{1}\int_{0}^{1}\int_{0}^{1}(4z-y)\,dz\,dy\,dx \\ &=&\rho \int_{0}^{1}\int_{0}^{1}(2-y)\,dy\,dx=\rho \int_{0}^{1}\dfrac{3}{2} \,dx=\dfrac{3}{2}\rho \end{eqnarray*} \]

which is the same as the result found in Example 8, Section 15.7.

NOW WORK

Problem 9.

NEED TO REVIEW?

Triple integrals using cylindrical coordinates are discussed in Section 14.7, pp. 950-954.

Using the Divergence Theorem to Find a Surface Integral

Let \(S\) be the surface of a cylindrical solid \(E\) whose boundary is \(x^{2}+y^{2}=4\), \(z=0\), and \(z=1\). Let \(\mathbf{F}=x^{3}\mathbf{i}+y^{3} \mathbf{j}+z^{2}\mathbf{k}\) and let \(\mathbf{n}\) be the outer unit normal to \(S\). Use the Divergence Theorem to find \(\iint\limits_S \mathbf{F}\,{\bf\cdot}\, \mathbf{n}\,dS\).

Figure 68

Solution Figure 68 illustrates the cylindrical surface \(S\). We begin by finding \({\rm{div}}\mathbf{F}=3x^{2}+3y^{2}+2z\). Then we use the Divergence Theorem, obtaining \[ \iint\limits_{S}\mathbf{F}\,{\bf\cdot}\, \mathbf{n}\,dS=\iiint \limits_{E}(3x^{2}+3y^{2}+2z)\,dV \]

Because \(E\) is a cylindrical solid, we use cylindrical coordinates: \[ 3x^{2}+3y^{2}+2z=3r^{2}\cos ^{2}\theta +3r^{2}\sin ^{2}\theta+ 2z=3r^{2}( \cos ^{2}\theta +\sin ^{2}\theta ) +2z=3r^{2}+2z \]

The solid \(E\) is given by \(0\leq r\leq 2\), \(0\leq \theta \leq 2\pi\) , \(0\leq z\leq 1\). Then \[ \begin{eqnarray*} \iint\limits_{S}\mathbf{F}\,{\cdot}\, \mathbf{n}\,ds& =&\iiint\limits_E(3x^{2}+3y^{2}+2z)\,dV \underset{\underset{\color{#0066A7}{dV=r\,dr\,d\theta \,dz}}{\color{#0066A7}{\uparrow}}} {=} \iiint\limits_E(3r^{2}+2z)\,r\,dr\,d\theta dz \\ &=&\int_{0}^{2\pi }\int_{0}^{2}\int_{0}^{1}(3r^{3}+2rz)\,dz\,dr\,d\theta \\ & =&\int_{0}^{2\pi }\int_{0}^{2}\big[ 3r^{3}z+r z^{2}\big] _{0}^{1}\,dr\,d\theta =\int_{0}^{2\pi }\int_{0}^{2}(3r^{3}+r)\,dr\,d\theta\\ &=&\int_{0}^{2\pi }\left[ \dfrac{3r^{4}}{4}+\dfrac{r^{2}}{2}\right] _{0}^{2}\,d\theta =\int_{0}^{2\pi }14\,d\theta =28\pi \end{eqnarray*} \]

NOW WORK

Problem 13.

Using the Divergence Theorem to Find a Surface Integral

Let \(\mathbf{F}(x,y,z)=x^{3}\mathbf{i}+y^{3}\mathbf{j}+z^{3}\mathbf{k}\). If \(S\) is the surface of the spherical solid \(E\) in the first octant enclosed by \(z=\sqrt{ 4-x^{2}-y^{2}}\) and \(z=0\), and \(\mathbf{n}\) is the outer unit normal to \(S\), use the Divergence Theorem to find \(\iint\limits_S \mathbf{F}\,{\bf\cdot}\, \mathbf{n}\,dS\).

Figure 69 \(dV=\rho ^{2}\sin \phi \,d\rho \,d\theta \,d\phi\)

Solution See Figure 69. Since \({\rm{div}}\mathbf{F} =3x^{2}+3y^{2}+3z^{2}=3(x^{2}+y^{2}+z^{2})\), we use the Divergence Theorem, obtaining \[ \iint\limits_{S}\mathbf{F}\,{\bf\cdot}\, \mathbf{n}\,dS=3\iiint \limits_E(x^{2}+y^{2}+z^{2})\,dV \]

We use spherical coordinates to find the triple integral. Then \(x^{2}+y^{2}+z^{2}=\rho ^{2}\) and \(dV=\rho ^{2}\sin \phi \,d\rho \,d\theta \,d\phi\). The solid \(E\) is given by \(0\leq \rho \leq 2\), \(0\leq \theta \leq \dfrac{\pi}{2}\), \(0\leq \phi \leq \dfrac{\pi }{2}\). Then \[ \begin{eqnarray*} \iint\limits_{S}\mathbf{F}\,{\cdot}\, \mathbf{n}\,dS &= &3\iiint\limits_E \rho ^{2}( \rho ^{2}\sin \phi \,d\rho \,d\theta \,d\phi ) = 3\int_{0}^{\pi /2}\int_{0}^{\pi /2 }\int_{0}^{2}\rho ^{4}\,\sin \phi \,d\rho \,d\theta \,d\phi \notag \\ &=&3\int_{0}^{\pi /2}\int_{0}^{\pi /2}\left[ \dfrac{\rho ^{5}}{5}\right] _{0}^{2}\sin \phi\, d\theta \,d\phi =3\int_{0}^{\pi /2}\int_{0}^{\pi /2 }\dfrac{ 32}{5}\sin \phi\, d\theta \,d\phi \\ &=&\dfrac{96}{5}\int_{0}^{\pi /2}\big[ \theta \big] _{0}^{\pi /2}\sin \phi \,d\phi \notag \\ &=&\dfrac{48\pi }{5}\int_{0}^{\pi /2}\sin \phi \,d\phi =\dfrac{48\pi }{5} \big[ -\cos \phi \big] _{0}^{\pi /2}=\dfrac{48\pi }{5} \end{eqnarray*} \]

NEED TO REVIEW?

Triple integrals using spherical coordinates are discussed in Section 14.8, pp. 956-960.

NOW WORK

Problem 15.

NEED TO REVIEW?

The Mean Value Theorem for integrals is discussed in Section 5.4, pp. 372-374.

Applying the Divergence Theorem to an Electric Force Field

Let \(\mathbf{F}=\mathbf{F}(x,y,z)\) be an electric force field \[ \mathbf{F}=\mathbf{F}(x,y,z)=\frac{\varepsilon qQ}{x^{2}+y^{2}+z^{2}}\mathbf{u} \]

where \(\varepsilon \) is a constant that depends on the units used; \(q\) and \(Q \) are the charges of two objects, one located at the point \((0,0,0)\) and the other at a point \((x,y,z)\); and \(\mathbf{u}\) is the unit vector in the direction from \((0,0,0)\) to \((x,y,z)\). Let \(S\) be a closed surface with positive orientation that encloses a solid \(E \) in space.

Show that the following statements are true under the assumptions of the Divergence Theorem:

1. If neither \(S\) nor its interior contains the point \((0,0,0)\), then \(\iint\limits_S \mathbf{F}\,{\bf\cdot}\, \mathbf{n}\,dS=0\). That is, the flux of \(\mathbf{F}\) across \(S\) equals \(0\).
2. If the interior of \(S\) contains the point \((0,0,0)\), then \(\iint\limits_S \mathbf{F}\,{\bf\cdot}\, \mathbf{n}\,dS=4\pi \varepsilon qQ\). That is, the flux of \(\mathbf{F}\) across \(S\) equals \(4\pi \varepsilon qQ\).

Solution The unit vector \(\mathbf{u}=\dfrac{x\mathbf{i}+y\mathbf{j }+z\mathbf{k}}{\sqrt{x^{2}+y^{2}+z^{2}}}\), so that \[ \mathbf{F}=\mathbf{F}(x,y,z)=\varepsilon qQ\dfrac{x\mathbf{i}+y\mathbf{j}+z \mathbf{k}}{(x^{2}+y^{2}+z^{2})^{3/2}} \]

The electric force field \(\mathbf{F}\) is continuous everywhere except at the point \((0,0,0)\).

(a) The surface \(S\) that forms the boundary of the solid \(E \) satisfies the assumptions of the Divergence Theorem. We find \({\rm{div}} \mathbf{F}\). \[ \begin{eqnarray*} {\rm{div}}\mathbf{F}&= &\varepsilon qQ\left\{\dfrac{\partial }{\partial x}\left[ \dfrac{x}{ (x^{2}+y^{2}+z^{2})^{3/2}}\right] +\dfrac{\partial }{\partial y}\left[ \dfrac{y }{(x^{2}+y^{2}+z^{2})^{3/2}}\right] +\dfrac{\partial }{\partial z}\left[ \dfrac{z}{(x^{2}+y^{2}+z^{2})^{3/2}}\right] \right\} \\ &=&\varepsilon qQ\bigg\{\dfrac{(x^{2}+y^{2}+z^{2})^{3/2}-3x^{2} \sqrt{x^{2}+y^{2}+z^{2}}}{(x^{2}+y^{2}+z^{2})^{3}}+\dfrac{ (x^{2}+y^{2}+z^{2})^{3/2}-3y^{2}\sqrt{x^{2}+y^{2}+z^{2}}}{ (x^{2}+y^{2}+z^{2})^{3}} \\ &&+\,\dfrac{(x^{2}+y^{2}+z^{2})^{3/2}-3z^{2}\sqrt{x^{2}+y^{2}+z^{2}}}{ (x^{2}+y^{2}+z^{2})^{3}} \bigg\} \\ &=&\varepsilon qQ\dfrac{3(x^{2}+y^{2}+z^{2})^{3/2}-3\sqrt{x^{2}+y^{2}+z^{2}}(x^{2}+y^{2}+z^{2})}{(x^{2}+y^{2}+z^{2})^{3}}=0 \end{eqnarray*} \]

Therefore, by the Divergence Theorem, \[ \iint\limits_{S}\mathbf{F}\,{\cdot}\, \mathbf{n}\,dS=\iiint\limits_E {\rm{div}} \mathbf{F}\,dV=0 \]

Figure 70

(b) Since the interior of \(S\) contains the origin, \(\mathbf{F}\) is not continuous at the origin, so we cannot use the Divergence Theorem. We use the following argument to prove (b). Let \(E \) be the closed solid enclosed by two separate surfaces: the surface \(S\) and a sphere \(S_{a}\) of radius \(a\), with its center at \((0,0,0)\), as shown in Figure 70. The outer surface is \(S\) and the inner surface is \(S_{a}\). Now \(\mathbf{F}\) is continuous throughout \(E\), so the Divergence Theorem can be used. \[ \iiint\limits_E {\rm{div}}\mathbf{F}\,dV=\iint\limits_{S}\mathbf{F}\,{\cdot}\, \mathbf{n}\,dS+\iint\limits_{S_{a}}\mathbf{F}\,{\bf\cdot}\, \mathbf{n}\,dS \]

From (a), we know that \(\iiint\limits_{E}{\rm{div}}\mathbf{F}\,dV=0\). So, \[ \iint\limits_{S}\mathbf{F}\,{\bf\cdot}\, \mathbf{n}\,dS=-\iint\limits_{S_{a}}\mathbf{F }\,{\bf\cdot}\, \mathbf{n}\,dS \]

On the inner surface \(S_{a}\), a sphere of radius \(a\), the outer unit normal is \[ \mathbf{n}=-\dfrac{x\mathbf{i}+y\mathbf{j}+z\mathbf{k}}{a}=-\dfrac{x\mathbf{i}+ y\mathbf{j}+z\mathbf{k}}{\sqrt{x^{2}+y^{2}+z^{2}}} \]

Then \[ \begin{eqnarray*} \mathbf{F}\,{\bf\cdot}\, \mathbf{n\,} &=&\dfrac{\varepsilon qQ}{x^{2}+y^{2}+z^{2}} \mathbf{u}\,{\bf\cdot}\, \left( -\dfrac{x\mathbf{i}+y\mathbf{j}+z\mathbf{k}}{\sqrt{ x^{2}+y^{2}+z^{2}}}\right)\\ &=&-\dfrac{\varepsilon qQ}{x^{2}+y^{2}+z^{2}}\dfrac{x \mathbf{i}+y\mathbf{j}+z\mathbf{k}}{\sqrt{x^{2}+y^{2}+z^{2}}}\,{\bf\cdot}\, \dfrac{x \mathbf{i}+y\mathbf{j}+z\mathbf{k}}{\sqrt{x^{2}+y^{2}+z^{2}}} \\ &=&-\varepsilon qQ\dfrac{x^{2}+y^{2}+z^{2}}{a^{4}}=-\dfrac{\varepsilon qQ}{ a^{2}} \end{eqnarray*} \]

So,