\(\def\ans#1{\bbox[border:1px solid green,6pt]{#1}}\)

Divergence and Curl

Introduction

Recall that Green's Theorem has two forms, the curl form, involving \(\dfrac{\partial g}{\partial x} - \dfrac{\partial f}{\partial y}\), and the divergence form, involving \(\dfrac{\partial f}{\partial x} + \dfrac{\partial g}{\partial y}\). In general, the divergence and curl can be expressed as either the dot product or cross product of the gradient vector \(\nabla = \left\langle \dfrac{\partial}{\partial x},\dfrac{\partial}{\partial y},\dfrac{\partial}{\partial z} \right\rangle\) and the vector field \(\vec{F} = \langle f,g,h \rangle\) (in three dimensions):

\[\begin{align} \textrm{Div } &= \nabla \cdot \vec{F}\\ \textrm{Curl } &= \nabla \times \vec{F} \end{align}\]

Important Note

A vector field \(\vec{F}\) is conservative if and only if \(\nabla \times \vec{F} = \vec{0}\).

Find \(curl(\vec{F})\) for the field \(\vec{F} = \langle x,y,z \rangle\) (a radial field).

Solution

\[\begin{align} \nabla \times \vec{F} &= \begin{vmatrix} \hat{\imath} & \hat{\jmath} & \hat{k}\\ \dfrac{\partial}{\partial x} & \dfrac{\partial}{\partial y} & \dfrac{\partial}{\partial z}\\ x & y & z \end{vmatrix}\\ \\ &= (0-0)\hat{\imath} - (0-0)\hat{\jmath} + (0-0)\hat{k}\\ &= \ans{\langle 0,0,0 \rangle} \end{align}\]

Find \(curl(\vec{F})\) for the field \(\vec{F} = \langle -y,x-z,y \rangle\) (a rotation field).

Solution

\[\begin{align} \nabla \times \vec{F} &= \begin{vmatrix} \hat{\imath} & \hat{\jmath} & \hat{k}\\ \dfrac{\partial}{\partial x} & \dfrac{\partial}{\partial y} & \dfrac{\partial}{\partial z}\\ -y & x-z & y \end{vmatrix}\\ \\ &= (1-(-1))\hat{\imath} - (0-0)\hat{\jmath} + (1-(-1))\hat{k}\\ &= \ans{\langle 2,0,2 \rangle} \end{align}\]

Try it yourself:

(click on a problem to show/hide its answer)

  1. Find \(curl(\vec{F})\) for the field \(\vec{F} = \langle x^2-y^2,xy,z \rangle\).
  2. \(curl(\vec{F}) = \langle 0,0,3y \rangle\)

  3. Find \(curl(\vec{F})\) for the field \(\vec{F} = \langle x^2-z^2,1,2xz \rangle\).
  4. \(curl(\vec{F}) = \langle 0,-4z,0 \rangle\)

  5. Find \(curl(\vec{F})\) for the field \(\vec{F} = \langle z^2 \sin y,xz^2 \cos y,2xz \sin y \rangle\).
  6. \(curl(\vec{F}) = \langle 0,0,0 \rangle\)

Interpretation of Curl

Suppose that \(\vec{F}\) describes the velocity field of a fluid. Then the curl of \(\vec{F}\) is the tendency of particles at \((x,y,z)\) to rotate about the axis that points in the direction of the curl vector.

If \(curl(\vec{F}) = \vec{0}\) (a conservative vector field), then the fluid is called irrotational, as in the example above of the radial field.

Examples

Determine if \(\vec{F} = \langle x^2y,xyz,-x^2y^2 \rangle\) is a conservative vector field.

Solution

\[\begin{align} \nabla \times \vec{F} &= \begin{vmatrix} \hat{\imath} & \hat{\jmath} & \hat{k}\\ \dfrac{\partial}{\partial x} & \dfrac{\partial}{\partial y} & \dfrac{\partial}{\partial z}\\ x^2y & xyz & -x^2y^2 \end{vmatrix}\\ \\ &= (-2x^2y-xy)\hat{\imath} - (-2xy^2-0)\hat{\jmath} + (yz-x^2)\hat{k}\\ &\neq \vec{0} \implies \ans{\vec{F} \textrm{ is not conservative.}} \end{align}\]

Try it yourself:

(click on the problem to show/hide the answer)

Determine if \(\vec{F} = \langle 2x,-2y,2z \rangle\) is conservative.
Yes, it is conservative.

Let \(\vec{F} = \langle yz,xz,xy \rangle\).

  1. Find \(curl(\vec{F})\).
  2. Find the work done in moving a particle from \((1,2,2)\) to \((3,4,4)\) through this force field.

Solution

\[\begin{align} \nabla \times \vec{F} &= \begin{vmatrix} \hat{\imath} & \hat{\jmath} & \hat{k}\\ \dfrac{\partial}{\partial x} & \dfrac{\partial}{\partial y} & \dfrac{\partial}{\partial z}\\ yz & xz & xy \end{vmatrix}\\ \\ &= \vec{0} \end{align}\]

Since \(\nabla \times \vec{F} = \vec{0}\), this force field is conservative, which means that we can use the fundamental theorem of line integrals to calculate this work:

\[W = \int_C \vec{F} \cdot \vec{r}\ '(t) \ dt = \phi(B)-\phi(A)\]

First, though, we need to find \(\phi(x,y,z)\):

\[\begin{align} \phi_x &= yz \implies \phi = xyz + c(y,z)\\ \phi_y &= xz+c_y (y,z) = xz \implies c_y (y,z) = 0\\ &\implies c(y,z) = d(z)\\ \\ \phi_z &= xy+d_z (z) = xy\\ &\implies d_z (z) = 0\\ \\ &\implies \phi (x,y,z) = xyz\\ \\ &\implies W = \phi(3,4,4) - \phi(1,2,2) = (3)(4)(4)-(1)(2)(2) = \ans{44} \end{align}\]

Find \(div(\vec{F})\) for the field \(\vec{F} = \langle xz,xyz,-y^2 \rangle\).

Solution

\[\begin{align} div(\vec{F}) &= \nabla \cdot \vec{F}\\ &= \left\langle \dfrac{\partial}{\partial x},\dfrac{\partial}{\partial y},\dfrac{\partial}{\partial z} \right\rangle \cdot \langle xz,xyz,-y^2 \rangle\\ &= z+xz+0 = \ans{z+xz} \end{align}\]

Find \(div(\vec{F})\) for the field \(\vec{F} = \langle x^2y,xyz,-x^2y^2 \rangle\).

Solution

\[\begin{align} div(\vec{F}) &= \nabla \cdot \vec{F}\\ &= \left\langle \dfrac{\partial}{\partial x},\dfrac{\partial}{\partial y},\dfrac{\partial}{\partial z} \right\rangle \cdot \langle x^2y,xyz,-x^2y^2 \rangle\\ &= \ans{2xy+xz} \end{align}\]

Connection Between Divergence and Curl

\[div(curl(\vec{F})) = 0\]

Verify the identity above for the field \(\vec{F} = \langle yz^2,xy,yz \rangle\).

Solution

\[\begin{align} curl(\vec{F}) &= \nabla \times \vec{F}\\ &= \begin{vmatrix} \hat{\imath} & \hat{\jmath} & \hat{k}\\ \dfrac{\partial}{\partial x} & \dfrac{\partial}{\partial y} & \dfrac{\partial}{\partial z}\\ yz^2 & xy & yz \end{vmatrix}\\ \\ &= \langle z,2yz,y-z^2 \rangle\\ \\ div(curl(\vec{F})) &= \nabla \cdot curl(\vec{F})\\ &= \left\langle \dfrac{\partial}{\partial x},\dfrac{\partial}{\partial y},\dfrac{\partial}{\partial z} \right\rangle \cdot \langle z,2yz,y-z^2 \rangle\\ &= 0+2z-2z = \ans{0} \end{align}\]

Note

The expression \(curl(div(\vec{F}))\) is meaningless, because \(div(\vec{F})\) is a scalar, and you can't take the curl of a scalar, because you can't use it in a cross product.