Divergence and curl are two measurements of vector fields that are very useful in a variety of applications. Both are most easily understood by thinking of the vector field as representing a flow of a liquid or gas; that is, each vector in the vector field should be interpreted as a velocity vector. Roughly speaking, divergence measures the tendency of the fluid to collect or disperse at a point, and curl measures the tendency of the fluid to swirl around the point. Divergence is a scalar, that is, a single number, while curl is itself a vector. The magnitude of the curl measures how much the fluid is swirling, the direction indicates the axis around which it tends to swirl. These ideas are somewhat subtle in practice, and are beyond the scope of this course. You can find additional information on the web, for example at

*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 it turns 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 looking entries. 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 odd vector $\nabla$ 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 $$\nabla\times{\bf F} = \left|\matrix{{\bf i}&{\bf j}&{\bf k}\cr {\partial \over\partial x}&{\partial \over\partial y}&{\partial \over\partial z}\cr f&g&h\cr}\right| = \left\langle {\partial h\over\partial y}-{\partial g\over\partial z}, {\partial f\over\partial z}-{\partial h\over\partial x}, {\partial g\over\partial x}-{\partial f\over\partial y}\right\rangle.$$

Here are two simple but useful facts about divergence and curl.

In words, this says that the divergence of the curl is zero.

That is, the curl of a gradient is the zero vector. Recalling that gradients are conservative vector fields, this says that the curl of a conservative vector field is the zero vector. Under suitable conditions, 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 test that we discussed in 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 by finding 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 $f=xe^z+y+C$.

We can rewrite Green's Theorem using these new ideas; these rewritten versions in turn are closer to some later theorems we will see.

Suppose we write a two dimensional vector field in the form ${\bf F}=\langle P,Q,0\rangle$, where $P$ and $Q$ are functions of $x$ and $y$. Then $$\nabla\times {\bf F} = \left|\matrix{{\bf i}&{\bf j}&{\bf k}\cr {\partial \over\partial x}&{\partial \over\partial y}&{\partial \over\partial z}\cr P&Q&0\cr}\right|= \langle 0,0,Q_x-P_y\rangle,$$ 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 (tendency to swirl) at each point in the region; the left-most integral adds up the tangential components of the vector field around the entire boundary. Green's Theorem says these are equal, or roughly, that the sum 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 to the 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 alternate notation. The last integral looks just like the right side of Green's Theorem (16.4.1) except that $P$ and $Q$ have traded places and $Q$ has acquired a negative sign. Then applying Green'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 the boundary of the region, from inside to out, and the second sums the divergence (tendency to spread) at each point in the interior. The theorem roughly says that the sum of the "microscopic'' spreads is the same as the total spread across the boundary and out of the region.

## Exercises 16.5

**Ex 16.5.1**
Let ${\bf F}=\langle xy,-xy\rangle$ 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 16.5.2**
Let ${\bf F}=\langle ax^2,by^2\rangle$ 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 16.5.3**
Let ${\bf F}=\langle ay^2,bx^2\rangle$ 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 16.5.4**
Let ${\bf F}=\langle \sin x\cos y,\cos x\sin y\rangle$ 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 16.5.5**
Let ${\bf F}=\langle y,-x\rangle$ 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 16.5.6**
Let ${\bf F}=\langle x,y\rangle$ 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 16.5.7**
Prove theorem 16.5.1.

**Ex 16.5.8**
Prove theorem 16.5.2.

**Ex 16.5.9**
If $\nabla \cdot {\bf F}=0$, $\bf F$ is said to be **incompressible**. Show that any vector field
of the form ${\bf F}(x,y,z) = \langle f(y,z),g(x,z),h(x,y)\rangle$ is
incompressible. Give a non-trivial example.