Curl and Divergence

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Objectives:

1.: Know what the curl of a vector field is, and how it is computed.
2.: Know what the curl of a vector field represents geometrically.
3.: Be able to compute the divergence of a vector field.
4.: Know what the divergence represents geometrically.
5.: Know what $\text {div}(\text {curl}(F))$ and $\text {curl}(\nabla f)$ are.

Recap Video

Take a look at the following video which goes over the basics of the curl and the divergence.

To summarize:

If $\textbf {F} = \left \langle F_1,F_2,F_3 \right \rangle$ is a three-dimensional vector field, then the curl of $\textbf {F}$ , denoted $\text {curl}(\textbf {F})$ , is a vector given by $\text {curl}(F) = \left | \begin {array}{ccc} \textbf {i} & \textbf {j} & \textbf {k}\\ \frac {\partial }{\partial x} & \frac {\partial }{\partial y} & \frac {\partial }{\partial z} \\ F_1 & F_2 & F_3 \end {array}\right | = \left \langle \frac {\partial F_3}{\partial y}-\frac {\partial F_2}{\partial z},\frac {\partial F_1}{\partial z}-\frac {\partial F_3}{\partial x},\frac {\partial F_2}{\partial x}-\frac {\partial F_1}{\partial y}\right \rangle .$
If $\textbf {F}$ is a two-dimensional vector field, then we make the $z$ -component $0$ and then take the curl as above.
If $\textbf {F} = \left \langle F_1,F_2,F_3 \right \rangle$ is a vector field, then the divergence of $\textbf {F}$ , denoted $\text {div}(\textbf {F})$ , is $\text {div}(\textbf {F}) = \frac {\partial F_1}{\partial x}+\frac {\partial F_2}{\partial y}+\frac {\partial F_3}{\partial z}.$

Problems

If $\textbf {F} = \left \langle xy,e^x,y+z \right \rangle$ , then

$\begin{eqnarray*} \text{curl}(\textbf{F}) &=&\left| \begin{array}{ccc} \textbf{i} & \textbf{j} & \textbf{k}\\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ F_1 & F_2 & F_3 \end{array}\right| \\ &=& \left| \begin{array}{ccc} \textbf{i} & \textbf{j} & \textbf{k}\\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ \answer{xy} & \answer{e^x} & \answer{y+z} \end{array}\right| \\ &=& \left\langle \answer{1},\answer{0},\answer{e^x-x} \right\rangle. \end{eqnarray*}$

If $\textbf {F} = \left \langle x^2,x+y \right \rangle$ , then

$\begin{eqnarray*} \text{curl}(\textbf{F}) &=& \text{curl}(\left\langle x^2,x+y,0 \right\rangle) \\ &=& \left| \begin{array}{ccc} \textbf{i} & \textbf{j} & \textbf{k}\\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ \answer{x^2} & \answer{x+y} & \answer{0} \end{array}\right| \\ &=& \left\langle \answer{0},\answer{0},\answer{1} \right\rangle \end{eqnarray*}$

As we saw in the video, we have the following theorem.

If $f(x,y,z)$ is an function whose second partials are continuous everywhere, then $\text {curl}(\nabla f) = \textbf {0}.$

Let’s look at this theorem through an example.

Suppose $f(x,y,z) = xy+y^2+2z$ . Then $\nabla f = \left \langle \answer {y},\answer {x+2y},\answer {2} \right \rangle .$ We can now take the curl of $\nabla f$ , since this is a vector field. We compute

$\begin{eqnarray*} \text{curl}(\textbf{F}) &=& \text{curl}(\left\langle y,x+2y,2 \right\rangle) \\ &=& \left| \begin{array}{ccc} \textbf{i} & \textbf{j} & \textbf{k}\\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ \answer{y} & \answer{x+2y} & \answer{2} \end{array}\right|\\ &=& \left\langle \answer{0},\answer{0},\answer{0} \right\rangle \end{eqnarray*}$

If $\textbf {F} = \left \langle xy,e^x,z \right \rangle$ , then $\text {div}(\textbf {F})= \frac {\partial }{\partial x}(\answer {xy})+\frac {\partial }{\partial y}(\answer {e^x})+\frac {\partial }{\partial z}(\answer {z}) = \answer {y+1}.$

If $\textbf {F} = \left \langle xy,e^x,y+z\right \rangle$ , then we saw above that $\text {curl}(\textbf {F}) = \left \langle 1,0,e^x-x \right \rangle$ . In this case $\text {div}(\text {curl}(\textbf {F})) = \frac {\partial }{\partial x}(\answer {1})+\frac {\partial }{\partial y}(\answer {0})+\frac {\partial }{\partial z}(\answer {e^x-x}) = \answer {0}.$

This is an example of the following theorem.

If $\textbf {F}$ is a vector field whose components have second partials which are continuous everywhere, then $\text {div}(\text {curl}(\textbf {F})) = 0$ .

We can use this theorem to answer the question of when a vector field $\textbf {F}$ is the curl of another vector field, i.e. when there is a vector field $\textbf {G}$ such that $\text {curl}(\textbf {G}) = \textbf {F}$ .

If $\textbf {F} = \left \langle x\sin (y),\cos (y),z-xy \right \rangle$ , is there a vector field $\textbf {G}$ such that $\text {curl}(\textbf {G}) = \textbf {F}$ ?

We can use the theorem above to help us. The vector field $\textbf {F}$ is defined everywhere. We compute: $\text {div}(\textbf {F}) = \answer {1} \neq 0.$ What does this tell you about the vector field $\textbf {F}$ ?

It is the curl of a vector field $\textbf {G}$ . It is not the curl of a vector field $\textbf {G}$ .

True/False Questions

If $f(x,y,z)$ is a function whose second partials are continuous everywhere, then $\text {div}(\nabla f) = 0$ .

True False

If $f(x,y,z) = x^2+y^2+z^2$ , then $\text {div}(\nabla f) = 6 \neq 0$ .

If $\textbf {F}$ is a conservative three-dimensional vector field whose components are defined everywhere, then $\text {curl}(\textbf {F}) = \textbf {0}$ .

True False

This is the first theorem in the curl section.

If $\textbf {F}$ is a three-dimensional vector field whose components are defined everywhere, then $\text {curl}(\text {curl}(\textbf {F})) = \textbf {0}$ .

True False

If $\textbf {F} = \left \langle 0,x,xz \right \rangle$ , then $\text {curl}(\text {curl}(\textbf {F})) \neq \textbf {0}$ .