Divergence and curl are two measurements of vector fields that arevery useful in a variety of applications. Both are most easilyunderstood by thinking of the vector field as representing a flow of aliquid or gas;that is, each vector in the vector field should be interpreted as avelocity vector. Roughly speaking, divergencemeasures the tendency of the fluid to collect or disperse at a point, and curl measures thetendency of the fluid to swirl around the point. Divergence is ascalar, that is, a single number, while curl is itself a vector. Themagnitude of the curl measures how much the fluid is swirling, thedirection indicates the axis around which it tends to swirl. Theseideas are somewhat subtle in practice, and are beyond the scope ofthis course. You can find additional information on the web, forexample at
and
and inmany books including Div, Grad, Curl, and All That: An Informal Text on Vector Calculus,by H. M. Schey.
Recall that if $f$ is a function, the gradient of $f$is given by $$\nabla f=\left\langle {\partial f\over\partial x},{\partial f\over\partial y},{\partial f\over\partial z}\right\rangle.$$A useful mnemonic for this (and for the divergence and curl, as itturns out) is to let$$\nabla = \left\langle{\partial \over\partial x},{\partial \over\partial y},{\partial \over\partial z}\right\rangle,$$that is, we pretend that $\nabla$ is a vector with rather odd lookingentries. Recalling that $\langle u,v,w\rangle a=\langle ua,va,wa\rangle$,we can then think of the gradient as$$\nabla f=\left\langle{\partial \over\partial x},{\partial \over\partial y},{\partial \over\partial z}\right\rangle f = \left\langle {\partial f\over\partial x},{\partial f\over\partial y},{\partial f\over\partial z}\right\rangle,$$that is, we simply multiply the $f$ into the vector.
The divergence and curl can now be defined in terms of this same oddvector $\nabla$ by using the cross product and dot product.The divergence of a vector field ${\bf F}=\langle f,g,h\rangle$ is$$\nabla \cdot {\bf F} =\left\langle{\partial \over\partial x},{\partial \over\partial y},{\partial \over\partial z}\right\rangle\cdot\langle f,g,h\rangle= {\partial f\over\partial x}+{\partial g\over\partial y}+{\partial h\over\partial z}.$$The curl of $\bf F$ is$$\nabla\times{\bf F} = \left|\matrix{{\bf i}&{\bf j}&{\bf k}\cr{\partial \over\partial x}&{\partial \over\partial y}&{\partial \over\partial z}\crf&g&h\cr}\right| = \left\langle {\partial h\over\partial y}-{\partial g\over\partial z},{\partial f\over\partial z}-{\partial h\over\partial x},{\partial g\over\partial x}-{\partial f\over\partial y}\right\rangle.$$
Here are two simple but useful facts about divergence and curl.
Theorem 16.5.1 $\nabla\cdot(\nabla\times{\bf F})=0$.$\qed$
In words, this says that the divergence of the curl is zero.
Theorem 16.5.2 $\nabla\times(\nabla f) = {\bf 0}$.$\qed$
That is, the curl of a gradient is the zero vector. Recalling thatgradients are conservative vector fields, this says that the curl of aconservative vector field is the zero vector. Under suitableconditions, it is also true that if the curl of $\bf F$ is $\bf 0$then $\bf F$ is conservative. (Note that this is exactly the same testthat we discussedin section 16.3.)
Example 16.5.3 Let ${\bf F} = \langle e^z,1,xe^z\rangle$. Then $\nabla\times{\bf F} = \langle 0,e^z-e^z,0\rangle = {\bf 0}$.Thus, $\bf F$ is conservative, and we can exhibit this directly byfinding the corresponding $f$.
Since $f_x=e^z$, $f=xe^z+g(y,z)$. Since $f_y=1$, it must be that$g_y=1$, so $g(y,z)=y+h(z)$. Thus $f=xe^z+y+h(z)$ and$$xe^z = f_z = xe^z + 0 + h'(z),$$so $h'(z)=0$, i.e., $h(z)=C$, and $f=xe^z+y+C$.$\square$
We can rewrite Green's Theorem using these new ideas; these rewrittenversions in turn are closer to some later theorems we will see.
Suppose we write a two dimensional vector field in theform ${\bf F}=\langle P,Q,0\rangle$, where $P$ and $Q$ are functionsof $x$ and $y$. Then $$\nabla\times {\bf F} =\left|\matrix{{\bf i}&{\bf j}&{\bf k}\cr{\partial \over\partial x}&{\partial \over\partial y}&{\partial \over\partial z}\crP&Q&0\cr}\right|=\langle 0,0,Q_x-P_y\rangle,$$and so $(\nabla\times {\bf F})\cdot{\bf k}=\langle 0,0,Q_x-P_y\rangle\cdot\langle 0,0,1\rangle = Q_x-P_y$. So Green's Theorem says$$\eqalignno{\int_{\partial D} {\bf F}\cdot d{\bf r}&=\int_{\partial D} P\,dx +Q\,dy = \dint{D} Q_x-P_y \,dA=\dint{D}(\nabla\times {\bf F})\cdot{\bf k}\,dA.&(16.5.1)\cr}$$Roughly speaking, the right-most integral adds up the curl (tendencyto swirl) at each point in the region; the left-most integral adds upthe tangential components of the vector field around the entireboundary. Green's Theorem says these are equal, or roughly, that thesum of the "microscopic'' swirls over the region is the same as the"macroscopic'' swirl around the boundary.
Next, suppose that the boundary $\partial D$ has a vector form${\bf r}(t)$, so that ${\bf r}'(t)$ is tangent to the boundary, and${\bf T}={\bf r}'(t)/|{\bf r}'(t)|$ is the usual unit tangent vector.Writing ${\bf r}=\langle x(t),y(t)\rangle$ we get$${\bf T}={\langle x',y'\rangle\over|{\bf r}'(t)|}$$and then$${\bf N}={\langle y',-x'\rangle\over|{\bf r}'(t)|}$$is a unit vector perpendicular to $\bf T$, that is, a unit normal tothe boundary. Now$$\eqalign{\int_{\partial D} {\bf F}\cdot{\bf N}\,ds&=\int_{\partial D} \langle P,Q\rangle\cdot{\langle y',-x'\rangle\over|{\bf r}'(t)|} |{\bf r}'(t)|dt=\int_{\partial D} Py'\,dt - Qx'\,dt\cr&=\int_{\partial D} P\,dy - Q\,dx=\int_{\partial D} - Q\,dx+P\,dy.\cr}$$So far, we've just rewritten the original integral using alternatenotation. The last integral looks just like the right side of Green'sTheorem (16.4.1) except that $P$ and $Q$ havetraded places and $Q$ has acquired a negative sign. Then applyingGreen's Theorem we get $$\int_{\partial D} - Q\,dx+P\,dy=\dint{D} P_x+Q_y\,dA=\dint{D} \nabla\cdot{\bf F}\,dA.$$Summarizing the long string of equalities, $$\eqalignno{\int_{\partial D} {\bf F}\cdot{\bf N}\,ds&=\dint{D} \nabla\cdot{\bf F}\,dA.&(16.5.2)\cr}$$ Roughly speaking, the first integral adds up the flow across theboundary of the region, from inside to out, and the second sums thedivergence (tendency to spread) at each point in the interior. Thetheorem roughly says that the sum of the "microscopic'' spreads isthe same as the total spread across the boundary and out of the region.
Exercises 16.5
Sage knows how to compute divergence and curl.
Ex 16.5.1Let ${\bf F}=\langle xy,-xy\rangle$ and let $D$ be given by $0\le x\le 1$, $0\le y\le 1$.Compute $\ds\int_{\partial D} {\bf F}\cdot d{\bf r}$ and$\ds\int_{\partial D} {\bf F}\cdot{\bf N}\,ds$.(answer)
Ex 16.5.2Let ${\bf F}=\langle ax^2,by^2\rangle$ and let $D$ be given by $0\le x\le 1$, $0\le y\le 1$.Compute $\ds\int_{\partial D} {\bf F}\cdot d{\bf r}$ and$\ds\int_{\partial D} {\bf F}\cdot{\bf N}\,ds$.(answer)
Ex 16.5.3Let ${\bf F}=\langle ay^2,bx^2\rangle$ and let $D$ be given by $0\le x\le 1$, $0\le y\le x$.Compute $\ds\int_{\partial D} {\bf F}\cdot d{\bf r}$ and$\ds\int_{\partial D} {\bf F}\cdot{\bf N}\,ds$.(answer)
Ex 16.5.4Let ${\bf F}=\langle \sin x\cos y,\cos x\sin y\rangle$ and let $D$ be given by $0\le x\le \pi/2$, $0\le y\le x$.Compute $\ds\int_{\partial D} {\bf F}\cdot d{\bf r}$ and$\ds\int_{\partial D} {\bf F}\cdot{\bf N}\,ds$.(answer)
Ex 16.5.5Let ${\bf F}=\langle y,-x\rangle$ and let $D$ be given by $x^2+y^2\le 1$.Compute $\ds\int_{\partial D} {\bf F}\cdot d{\bf r}$ and$\ds\int_{\partial D} {\bf F}\cdot{\bf N}\,ds$.(answer)
Ex 16.5.6Let ${\bf F}=\langle x,y\rangle$ and let $D$ be given by $x^2+y^2\le 1$.Compute $\ds\int_{\partial D} {\bf F}\cdot d{\bf r}$ and$\ds\int_{\partial D} {\bf F}\cdot{\bf N}\,ds$.(answer)
Ex 16.5.7Prove theorem 16.5.1.
Ex 16.5.8Prove theorem 16.5.2.
Ex 16.5.9If $\nabla \cdot {\bf F}=0$, $\bf F$ is said to be incompressible. Show that any vector fieldof the form ${\bf F}(x,y,z) = \langle f(y,z),g(x,z),h(x,y)\rangle$ isincompressible. Give a non-trivial example.
