Note

More advanced treatments of vector calculus use the theory of “differential forms” to formulate a general version of Stokes’ Theorem that is valid in all dimensions and includes each of our main theorems (Green’s, Stokes’, Divergence) as a special case.

EXAMPLE 1

Evaluate the divergence of \({\bf F} = \langle e^{xy}, xy,z^4\rangle\) at \(P=(1,0,2)\).

Solution \begin{align*} \hbox{div}({\bf F}) &= \dfrac{\partial{}}{\partial{x}}\, e^{xy} + \dfrac{\partial}{\partial{y}}\, xy + \dfrac{\partial}{\partial{z}}\,z^4 = ye^{xy}+x+4z^3\\ \hbox{div}({\bf F})(P) &= \hbox{div}({\bf F})(1,0,2) = 0\cdot e^0+1+4\cdot 2^3 = 33 \end{align*}

THEOREM 1 Divergence Theorem

Let \(S\) be a closed surface that encloses a region \(W\) in \({\bf R}^3\). Assume that \(S\) is piecewise smooth and is oriented by normal vectors pointing to the outside of \(W\). Let \({\bf F}\) be a vector field whose domain contains \(W\). Then \begin{equation} \iint_{S\,} {\bf F} \cdot d{\bf S} = \iiint_{W\,} \hbox{div}({\bf F})\, dV\tag{2} \end{equation}

Proof

We prove the Divergence Theorem in the special case that \(W\) is a box \([a,b]\times[c,d]\times[e,f]\) as in Figure 17.57. The proof can be modified to treat more general regions such as the interiors of spheres and cylinders.

We write each side of Eq. (2) as a sum over components: \begin{eqnarray*} \iint_{\partial W}(F_1{\bf i}+F_2{\bf j}+F_3{\bf k})\cdot d{\bf S} &=& \iint_{\partial W}\! F_1{\bf i}\cdot d{\bf S} + \iint_{\partial W}\! F_2{\bf j}\cdot d{\bf S} + \iint_{\partial W}\! F_3{\bf k}\cdot d{\bf S} \\ \iiint_{W} \hbox{div}(F_1{\bf i}+F_2{\bf j}+F_3{\bf k})\, dV &=& \iiint_{W} \hbox{div}(F_1{\bf i})\, dV + \iiint_{W} \hbox{div}(F_2{\bf j})\, dV \\ &&{} + \iiint_{W} \hbox{div}(F_3{\bf k})\, dV \end{eqnarray*}

As in the proofs of Green’s and Stokes’ Theorems, we show that corresponding terms are equal. It will suffice to carry out the argument for the \({\bf i}\)-component (the other two components are similar). Thus we assume that \({\bf F} = F_1{\bf i}\).

The surface integral over boundary \(S\) of the box is the sum of the integrals over the six faces. However, \({\bf F}=F_1{\bf i}\) is orthogonal to the normal vectors to the top and bottom as well as the two side faces because \(\displaystyle{ {\bf F}\cdot{\bf j} ={\bf F}\cdot{\bf k} = 0}\). Therefore, the surface integrals over these faces are zero. Nonzero contributions come only from the front and back faces, which we denote \(S_f\) and \(S_b\) (Figure 17.58): \[ \iint_{S\,} {\bf F}\cdot d{\bf S} = \iint_{S_f\,} {\bf F}\cdot d{\bf S} + \iint_{S_b\,} {\bf F}\cdot d{\bf S} \]

To evaluate these integrals, we parametrize \(S_f\) and \(S_b\) by \begin{align*} G_f(y,z) &=(b,y,z),&\qquad& c \le y \le d,\ e \le z \le f\\ G_b(y,z) &=(a,y,z),&\qquad&c \le y \le d,\ e \le z \le f \end{align*}

The normal vectors for these parametrizations are \begin{align*} \dfrac{\partial{G_f}}{\partial{y}}\times\dfrac{\partial{G_f}}{\partial{z}} &= {\bf j}\times{\bf k} = {\bf i}\\ \dfrac{\partial{G_b}}{\partial{y}}\times\dfrac{\partial{G_b}}{\partial{z}} &= {\bf j}\times{\bf k} = {\bf i} \end{align*}

However, the outward-pointing normal for \(S_b\) is \(-{\bf i}\), so a minus sign is needed in the surface integral over \(S_b\) using the parametrization \(G_b\): \begin{align*} \iint_{S_f\,} {\bf F}\cdot d{\bf S} +\iint_{S_b\,} {\bf F}\cdot d{\bf S} &= \int_e^f\int_c^d F_1(b,y,z)\,dy\,dz -\int_e^f\int_c^d F_1(a,y,z) \,dy\,dz\\ &= \int_e^f\int_c^d \Big(F_1(b,y,z)-F_1(a,y,z)\Big)\,dy\,dz \end{align*}

By the FTC in one variable, \[ F_1(b,y,z)-F_1(a,y,z) = \int_a^b\,\dfrac{\partial{F_1}}{\partial{x}}(x,y,z)\,dx \]

Since \(\displaystyle{\hbox{div}({\bf F})=\hbox{div}(F_1{\bf i})=\dfrac{\partial{F_1}}{\partial{x}}}\), we obtain the desired result: \begin{align*} \iint_{S\,} {\bf F}\cdot d{\bf S} = \int_e^f\int_c^d\int_a^b \dfrac{\partial{F_1}}{\partial{x}}(x,y,z) \,dx\,dy\,dz = \iiint_{W\,} \hbox{div}({\bf F})\,dV \end{align*}

EXAMPLE 2 Verifying the Divergence Theorem

Verify Theorem 1 for \({\bf F} = \langle y,yz,z^2 \rangle\) and the cylinder in Figure 17.59.

Figure 17.59: Cylinder of radius 2 and height 5.

Solution We must verify that the flux \(\displaystyle{\iint_{S\,} {\bf F} \cdot d{\bf S}}\), where \(S\) is the boundary of the cylinder, is equal to the integral of \(\hbox{div}(W)\) over the cylinder. We compute the flux through \(S\) first: It is the sum of three surface integrals over the side, the top, and the bottom.

Step 1. Integrate over the side of the cylinder

We use the standard parametrization of the cylinder: \begin{align*} G(\theta,z) &=(2\cos\theta,2\sin\theta,z),\qquad 0\le\theta>2\pi,\quad 0 \le z\le 5 \end{align*}

The normal vector is \begin{align*} {\bf n} &= {\bf T}_{\theta} \times {\bf T}_z = \langle -2\sin\theta,2\cos\theta,0 \rangle \times \langle 0,0,1 \rangle = \langle 2\cos\theta,2\sin\theta,0 \rangle \end{align*} and \({\bf F}(G(\theta,z)) = \langle y,yz,z^2 \rangle =\langle 2\sin\theta,2z\sin\theta,z^2 \rangle\). Thus \begin{align} {\bf F} \cdot d{\bf S}& = \langle 2\sin\theta,2z\sin\theta,z^2 \rangle \cdot \langle 2\cos\theta,2\sin\theta,0 \rangle\,d\theta\,dz\notag \\[4pt] &= 4\cos\theta\sin\theta + 4z\sin^2\theta\,d\theta\,dz\notag \\[4pt] \iint_{\textrm{side}} {\bf F} \cdot d{\bf S} &= \int_0^5\int_0^{2\pi} (4\cos\theta\sin\theta + 4z\sin^2\theta)\,d\theta\,dz\notag \\[4pt] &= 0 + 4\pi\int_0^5 z \,dz = 4\pi\left(\frac{25}2\right) = 50\pi\tag{3} \end{align}

Step 2. Integrate over the top and bottom of the cylinder

The top of the cylinder is at height \(z=5\), so we can parametrize the top by \(G(x,y)=(x, y,5)\) for \((x,y)\) in the disk \(D\) of radius \(2\): \[ D = \{(x,y):x^2+y^2\le 4\} \]

Then \begin{align*} {\bf n} &= {\bf T}_x \times {\bf T}_y = \langle 1,0,0 \rangle \times \langle 0,1,0 \rangle = \langle 0,0,1 \rangle \end{align*} and since \({\bf F}(G(x,y)) ={\bf F}(x,y,5) = \langle y,5y,5^2 \rangle\), we have \begin{align*} {\bf F}(G(x,y)) \cdot {\bf n} &= \langle y,5y,5^2 \rangle \cdot \langle 0,0,1 \rangle = 25\\ \iint_{\textrm{top}} {\bf F}\cdot d{\bf S} &= \iint_{D\,} 25\,dA = 25\,\mathrm{Area}(D) = 25(4\pi)=100\pi \end{align*}

Along the bottom disk of the cylinder, we have \(z=0\) and \({\bf F}(x, y, 0) = \langle y, 0 , 0\rangle\). Thus \({\bf F}\) is orthogonal to the vector \(-{\bf k}\) normal to the bottom disk, and the integral along the bottom is zero.

Step 3. Find the total flux \[ \iint_{S\,} {\bf F} \cdot d{\bf S} = \textrm{sides \(+\) top \(+\) bottom} = 50\pi+100\pi +0=\boxed{150\pi} \]

Step 4. Compare with the integral of divergence \[ \hbox{div}({\bf F}) = \hbox{div}\bigl(\langle y, yz, z^2\rangle\bigr)=\dfrac{\partial}{\partial{x}}y+\dfrac{\partial}{\partial{y}}(yz)+\dfrac{\partial}{\partial{z}}z^2 = 0+z+2z=3z \]

The cylinder \(W\) consists of all points \((x,y,z)\) for \(0\le z\le 5\) and \((x,y)\) in the disk \(D\). We see that the integral of the divergence is equal to the total flux as required: \begin{align*} \iiint_{W\,}\hbox{div}({\bf F})\,dV &= \iint_{D\,}\int_{z=0}^5 3z\,dV = \iint_{D\,} \frac{75}2\,dA \\ &= \left(\frac{75}2\right)(\textrm{Area}(D)) = \left(\frac{75}2\right)(4\pi) = \boxed{150\pi} \end{align*}

EXAMPLE 3 Using the Divergence Theorem

Use the Divergence Theorem to evaluate \(\displaystyle{\iint_{S\,} \langle x^2,z^4,e^z\rangle\cdot d{\bf S}}\), where \(S\) is the boundary of the box \(W\) in Figure 17.60.

Solution First, compute the divergence: \[ \hbox{div}\bigl(\langle x^2,z^4,e^z\rangle\bigr) = \dfrac{\partial}{\partial{x}}\,x^2 + \dfrac{\partial}{\partial{y}}\,z^4 + \dfrac{\partial}{\partial{z}}\,e^z = 2x+e^z \]

Then apply the Divergence Theorem and use Fubini’s Theorem: \begin{align*} \iint_{S\,} \langle x^2,z^4,e^z\rangle\cdot d{\bf S} &= \iiint_{W\,} (2x+e^z)\,dV= \int_0^2\int_0^3\int_0^1(2x+e^z)\,dz\,dy\,dx\\ &= 3\int_0^2 2x \,dx + 6\int_0^1 e^z\,dz = 12+6(e-1)=6e+6 \end{align*}

EXAMPLE 4 A Vector Field with Zero Divergence

Compute the flux of \[ {\bf F} = \langle z^2+xy^2, \cos(x+z), e^{-y}-zy^2\rangle \] through the boundary of the surface \(S\) in Figure 17.61.

Solution Although \({\bf F}\) is rather complicated, its divergence is zero: \[ \hbox{div}({\bf F}) = \dfrac{\partial{}}{\partial{x}}(z^2+xy^2)+ \dfrac{\partial{}}{\partial{y}}\cos(x+z)+\dfrac{\partial{}}{\partial{z}}(e^{-y}-zy^2) = y^2-y^2 = 0 \]

The Divergence Theorem shows that the flux is zero. Letting \(W\) be the region enclosed by \(S\), we have \[ \iint_{S\,} {\bf F} \cdot d{\bf S} = \iiint_{W\,} \hbox{div}({\bf F})\, dV = \iiint_{W\,} 0\, dV = 0 \]

Figure 17.62: For a velocity field, the flux through a surface is the flow rate (in volume per time) of fluid across the surface.

GRAPHICAL INSIGHT Interpretation of Divergence

Let’s assume again that \({\bf F}\) is the velocity field of a fluid (Figure 17.62). Then the flux of \({\bf F}\) through a surface \(S\) is the flow rate (volume of fluid passing through \(S\) per unit time). If \(S\) encloses the region \(W\), then by the Divergence Theorem, \begin{equation} \textrm{Flow rate across \(S\)} = \iiint_{W\,}\hbox{div}({\bf F})\,dV\tag{4} \end{equation}

Now assume that \(S\) is a small surface containing a point \(P\). Because \(\displaystyle{\hbox{div}({\bf F})}\) is continuous (it is a sum of derivatives of the components of \({\bf F}\)), its value does not change much on \(W\) if \(S\) is sufficiently small and to a first approximation, we can replace \(\hbox{div}({\bf F})\) by the constant value \(\hbox{div}({\bf F})(P)\). This gives us the approximation \begin{equation} \textrm{Flow rate across \(S\)} = \iiint_{W\,}\hbox{div}({\bf F})\,dV \approx \hbox{div}({\bf F})(P)\,\cdot\,\textrm{Vol}(W)\tag{5} \end{equation}

In other words, the flow rate through a small closed surface containing \(P\) is approximately equal to the divergence at \(P\) times the enclosed volume, and thus \(\hbox{div}({\bf F})(P)\) has an interpretation as “flow rate (or flux) per unit volume”:

• If \(\hbox{div}({\bf F})(P)>0\), there is a net outflow of fluid across any small closed surface enclosing \(P\), or, in other words, a net “creation” of fluid near \(P\).
• If \(\hbox{div}({\bf F})(P)<0\), there is a net inflow of fluid across any small closed surface enclosing \(P\), or, in other words, a net “destruction” of fluid near \(P\).

Because of this, \(\hbox{div}({\bf F})\) is sometimes called the source density of the field.

• If \(\hbox{div}({\bf F})(P)=0\), then to a first-order approximation, the net flow across any small closed surface enclosing \(P\) is equal to zero.

A vector field such that \(\hbox{div}({\bf F}) = 0\) everywhere is called incompressible.

To visualize these cases, consider the two-dimensional situation, where we define \[\displaystyle \hbox{div}(\langle F_1, F_2\rangle)=\dfrac{\partial{F_1}}{\partial{x}}+\dfrac{\partial{F_2}}{\partial{y}}\]

In Figure 17.63, field (A) has positive divergence. There is a positive net flow of fluid across every circle per unit time. Similarly, field (B) has negative divergence. By contrast, field (C) is incompressible. The fluid flowing into every circle is balanced by the fluid flowing out.

Note

Do the units match up in Eq. (5)? The flow rate has units of volume per unit time. On the other hand, the divergence is a sum of derivatives of velocity with respect to distance. Therefore, the divergence has units of “distance per unit time per distance,” or unit time\(^{-1}\), and the right-hand side of Eq. (5) also has units of volume per unit time.

REMINDER

\[ r=\sqrt{x^2+y^2+z^2} \] For \(r\ne 0\), \[ {\bf e}_r = \frac{\langle x, y, z\rangle}{r} = \frac{\langle x, y, z\rangle}{\sqrt{x^2+y^2+z^2}} \]

EXAMPLE 5 The Inverse-Square Vector Field

Verify that \(\displaystyle{{\bf F}_{\textrm{i-sq}}=\frac{{\bf e}_r}{r^2}}\) has zero divergence: \[ \hbox{div}\left(\frac{{\bf e}_r}{r^2}\right) = 0 \]

Solution Write the field as \[ {\bf F}_{\textrm{i-sq}} = \langle F_1, F_2, F_3\rangle = \frac1{r^2}\langle \frac{x}r,\frac{y}r, \frac{z}r\rangle = \langle xr^{-3},yr^{-3}, zr^{-3}\rangle \]

We have \begin{align*} \dfrac{\partial{r}}{\partial{x}} &= \dfrac{\partial}{\partial{x}}(x^2+y^2+z^2)^{1/2} = \frac12(x^2+y^2+z^2)^{-1/2}(2x)=\frac{x}r \\ \dfrac{\partial{F_1}}{\partial{x}} &= \dfrac{\partial}{\partial{x}} xr^{-3} = r^{-3} - 3xr^{-4}\dfrac{\partial{r}}{\partial{x}} = r^{-3} -(3xr^{-4})\frac{x}r = \frac{r^2-3x^2}{r^{5}} \end{align*}

The derivatives \(\displaystyle{\dfrac{\partial{F_2}}{\partial{y}}}\) and \(\displaystyle{\dfrac{\partial{F_3}}{\partial{z}}}\) are similar, so \[ \hbox{div}({\bf F}_{\textrm{i-sq}}) = \frac{r^2-3x^2}{r^{5}}+\frac{r^2-3y^2}{r^{5}}+\frac{r^2-3z^2}{r^{5}} = \frac{3r^2-3(x^2+y^2+z^2)}{r^{5}} = 0 \]

THEOREM 2 Flux of the Inverse-Square Field

The flux of \(\displaystyle{{\bf F}_{\textrm{i-sq}}=\frac{{\bf e}_r}{r^2}}\) through closed surfaces has the following remarkable description: \[\boxed{ \iint_{S}\left(\frac{{\bf e}_r}{r^2}\right)\cdot\,d\,{\bf S} = \begin{cases} 4\pi \quad & \textrm{if \(S\) encloses the origin}\\ 0& \textrm{if \(S\) does not enclose the origin} \end{cases} }\]

Proof

First, assume that \(S\) does not contain the origin (Figure 17.65). Then the region \(W\) enclosed by \(S\) is contained in the domain of \({\bf F}_{\textrm{i-sq}}\) and we can apply the Divergence Theorem. By Example 5, \(\hbox{div}({\bf F}_{\textrm{i-sq}})=0\) and therefore \[ \iint_{S}\left(\frac{{\bf e}_r}{r^2}\right)\cdot\,d\,{\bf S} = \iiint_{W}\hbox{div}({\bf F}_{\textrm{i-sq}})\,dV = \iiint_{W}0\,\,dV =0 \]

Figure 17.65: \(S\) is contained in the domain of \({\bf F}_{\textrm{i-sq}}\) (away from the origin).

Next, let \(S_R\) be the sphere of radius \(R\) centered at the origin (Figure 17.66). We cannot use the Divergence Theorem because \(S_R\) contains a point (the origin) where \({\bf F}_{\textrm{i-sq}}\) is not defined. However, we can compute the flux of \({\bf F}_{\textrm{i-sq}}\) through \(S_R\) using spherical coordinates. Recall from Section 16.4 [Eq. (5)] that the outward-pointing normal vector in spherical coordinates is \[ {\bf n} = {\bf T}_\phi\times {\bf T}_\theta = (R^2\sin\phi) {\bf e}_r \]

The inverse-square field on \(S_R\) is simply \({\bf F}_{\textrm{i-sq}} = R^{-2}{\bf e}_r\), and thus \[{\bf F}_{\textrm{i-sq}}\cdot {\bf n} = (R^{-2}{\bf e}_r)\cdot (R^2\sin\phi{\bf e}_r) = \sin\phi({\bf e}_r\cdot{\bf e}_r)=\sin\phi\] \begin{align*} \iint_{S_R\,}{\bf F}_{\textrm{i-sq}}\cdot d{\bf S} &= \int_0^{2\pi}\int_0^{\pi}{\bf F}_{\textrm{i-sq}}\cdot{\bf n}\,d\phi\,d\theta\\ &= \int_0^{2\pi}\int_{0}^{\pi}\sin\phi\,d\phi\,d\theta\\ & = 2\pi\int_{0}^{\pi}\sin\phi\,d\phi = 4\pi \end{align*}

To extend this result to any surface \(S\) containing the origin, choose a sphere \(S_R\) whose radius \(R>0\) is so small that \(S_R\) is contained inside \(S\). Let \(W\) be the region between \(S_R\) and \(S\) (Figure 17.67). The oriented boundary of \(W\) is the difference \[ \partial W = S-S_R \]

Figure 17.67: \(W\) is the region between \(S\) and the sphere \(S_R\).

This means that \(S\) is oriented by outward-pointing normals and \(S_R\) by inward-pointing normals. By the Divergence Theorem, \begin{align*} \iint_{\partial W}\,{\bf F}_{\textrm{i-sq}}\cdot\,d{\bf S} &=& \iint_{S\,}{\bf F}_{\textrm{i-sq}}\cdot d{\bf S} - \iint_{S_R\,}{\bf F}_{\textrm{i-sq}}\cdot d{\bf S}&\\ &= & \iiint_{W\,}\hbox{div}({\bf F}_{\textrm{i-sq}})\,dV\qquad&\textrm{(Divergence Theorem)} \\ &=& \iiint_{W\,} 0 \,dV = 0\qquad&\textrm{(Because \(\hbox{div}({\bf F}_{\textrm{i-sq}})=0\))} \end{align*}

This proves that the fluxes through \(S\) and \(S_R\) are equal, and hence both equal \(4\pi\).

Notice that we just applied the Divergence Theorem to a region \(W\) that lies between two surfaces, one contained in the other. This is a more general form of the theorem than the one we stated formally in Theorem 1 above. The marginal comment explains why this is justified.

THEOREM 3 Uniformly Charged Sphere

The electric field due to a uniformly charged hollow sphere \(S_R\) of radius \(R\), centered at the origin and of total charge \(Q\), is \begin{equation} {\bf E} = \begin{cases} \dfrac{Q}{4\pi\epsilon_0 r^2}{\bf e}_r& \textrm{if \(r >R\)}\\ \mathbf{0} & \textrm{if \(r < R\)}\tag{7} \end{cases} \end{equation} where \(\displaystyle{\epsilon_0=8.85\times 10^{-12} \mathrm{C^2\text{/}N\hbox{-}m}^2}\).

Proof

By symmetry (Figure 17.68), the electric field \({\bf E}\) must be directed in the radial direction \({\bf e}_r\) with magnitude depending only on the distance \(r\) to the origin. Thus, \({\bf E}=E(r){\bf e}_r\) for some function \(E(r)\). The flux of \({\bf E}\) through the sphere \(S_r\) of radius \(r\) is \[ \iint_{S_r\,} {\bf E} \cdot d{\bf S} = E(r)\underbrace{\iint_{S_r\,}{\bf e}_r\cdot d{\bf S}}_{\textrm{Surface area of sphere}} = 4\pi r^2 E(r) \]

Figure 17.68: The electric field due to a uniformly charged sphere.

By Gauss’s Law, this flux is equal to \(C/\epsilon_0\), where \(C\) is the charge enclosed by \(S_r\). If \(r<R\), then \(C=0\) and \({\bf E}=\mathbf{0}\). If \(r>R\), then \(C=Q\) and \(4\pi r^2 E(r)=Q/\epsilon_0\), or \(\displaystyle{E(r)=Q/(\epsilon_04\pi r^2)}\). This proves Eq. (7).

Note

We proved Theorem 3 in the analogous case of a gravitational field (also a radial inverse-square field) by a laborious calculation in Exercise 48 of Section 16.4. Here, we have derived it from Gauss’s Law and a simple appeal to symmetry.

CONCEPTUAL INSIGHT

Here is a summary of the basic operations on functions and vector fields: \[ \boxed{ \begin{array} \(f & \overset{\nabla}{\longrightarrow} & {\bf F} & \overset{\textrm{curl}}{\longrightarrow} & {\bf G} & \overset{\textrm{div}}{\longrightarrow} & g\\ \hbox{function}&&\hbox{vector field}&&\hbox{vector field}&&\hbox{function} \end{array}}\]

One basic fact is that the result of two consecutive operations in this diagram is zero: \[ \boxed{ {\bf curl}(\nabla(f)) = {\bf 0},\qquad\hbox{div}({\bf curl}({\bf F})) = 0} \]

We verified the first identity in Example 1 of Section 17.2. The second identity is left as an exercise (Exercise 6).

An interesting question is whether every vector field satisfying \({\bf curl}({\bf F})={\bf 0}\) is necessarily conservative—that is, \({\bf F}=\nabla V\) for some function \( V\). The answer is yes, but only if the domain \(D\) is simply connected (every path can be drawn down to a point in \(D\)). We saw in Section 16.3 that the vortex vector satisfies \({\bf curl}({\bf F})={\bf 0}\) and yet cannot be conservative because its circulation around the unit circle is nonzero (the circulation of a conservative vector field is always zero). However, the domain of the vortex vector field is \({\bf R}^2\) with the origin removed, and this domain is not simply-connected.

The situation for vector potentials is similar. Can every vector field \({\bf G}\) satisfying \(\hbox{div}({\bf G}) = \textbf{0}\) be written in the form \({\bf G} = {\bf curl}({\bf A})\) for some vector potential \({\bf A}\)? Again, the answer is yes—provided that the domain is a region \(W\) in \({\bf R}^3\) that has “no holes,” like a ball, cube, or all of \({\bf R}^3\). The inverse-square field \({\bf F}_{\textrm{i-sq}} = {\bf e}_r/r^2\) plays the role of vortex field in this setting: Although \(\hbox{div}({\bf F}_{\textrm{i-sq}})=0\), \({\bf F}_{\textrm{i-sq}}\) cannot have a vector potential because its flux through the unit sphere is nonzero as shown in Theorem 2 (the flux over a closed surface of a vector field with a vector potential is always zero by Theorem 2 of Section 17.2). In this case, the domain of \({\bf e}_r/r^2\) is \({\bf R}^3\) with the origin removed, which “has a hole.”

These properties of the vortex and inverse-square vector fields are significant because they relate line and surface integrals to “topological” properties of the domain, such as whether the domain is simply-connected or has holes. They are a first hint of the important and fascinating connections between vector analysis and the area of mathematics called topology.

Note

This is not just mathematical elegance\(\dots\) but beauty. It is so simple and yet it describes something so complex.

Francis Collins (1950– ), leading geneticist and former director of the Human Genome Project, speaking of the Maxwell Equations.