Divergence and Curl

Introduction

Recall that Green's Theorem has two forms, the curl form, involving ∂g∂x−∂f∂y\dfrac{\partial g}{\partial x} - \dfrac{\partial f}{\partial y}, and the divergence form, involving ∂f∂x+∂g∂y\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 ∇=⟨∂∂x,∂∂y,∂∂z⟩\nabla = \left\langle \dfrac{\partial}{\partial x},\dfrac{\partial}{\partial y},\dfrac{\partial}{\partial z} \right\rangle and the vector field F⃗=⟨f,g,h⟩\vec{F}=⟨f,g,h⟩ (in three dimensions):

Div =∇⋅F⃗Curl =∇×F⃗\begin{aligned} \textrm{Div } &= \nabla \cdot \vec{F}\ \textrm{Curl } &= \nabla \times \vec{F} \end{aligned}

Important Note

A vector field F⃗\vec{F} is conservative if and only if ∇×F⃗=0⃗.\nabla \times \vec{F} = \vec{0}.

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

Solution

∇×F⃗=∣ı^ȷ^k^∂∂x∂∂y∂∂zxyz∣=(0−0)ı^−(0−0)ȷ^+(0−0)k^=⟨0,0,0⟩\begin{aligned} \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{aligned}

Find curl(F⃗)curl(\vec{F}) for the field F⃗=⟨−y,x−z,y⟩\vec{F}=⟨−y,x−z,y⟩ (a rotation field).

Solution

∇×F⃗=∣ı^ȷ^k^∂∂x∂∂y∂∂z−yx−zy∣=(1−(−1))ı^−(0−0)ȷ^+(1−(−1))k^=⟨2,0,2⟩\begin{aligned} \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{aligned}

  1. Find curl(F⃗)curl(\vec{F}) for the field F⃗=⟨x2−y2,xy,z⟩.\vec{F} = \langle x^2-y^2,xy,z \rangle.

    2. curl(F⃗)=⟨0,0,3y⟩curl(\vec{F}) = \langle 0,0,3y \rangle

  2. Find curl(F⃗)curl(\vec{F}) for the field F⃗=⟨x2−z2,1,2xz⟩.\vec{F} = \langle x^2-z^2,1,2xz \rangle.

    3. curl(F⃗)=⟨0,−4z,0⟩curl(\vec{F}) = \langle 0,-4z,0 \rangle

  3. Find curl(F⃗)curl(\vec{F}) for the field F⃗=⟨z2sin⁡y,xz2cos⁡y,2xzsin⁡y⟩.\vec{F} = \langle z^2 \sin y,xz^2 \cos y,2xz \sin y \rangle.

    4. curl(F⃗)=⟨0,0,0⟩curl(\vec{F}) = \langle 0,0,0 \rangle

Interpretation of Curl

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

If curl(F⃗)=0⃗curl(\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 F⃗=⟨x2y,xyz,−x2y2⟩\vec{F} = \langle x^2y,xyz,-x^2y^2 \rangle is a conservative vector field.

Solution

∇×F⃗=∣ı^ȷ^k^∂∂x∂∂y∂∂zx2yxyz−x2y2∣=(−2x2y−xy)ı^−(−2xy2−0)ȷ^+(yz−x2)k^≠0⃗  ⟹  F⃗ is not conservative.\begin{aligned} \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{aligned}

  1. Determine if F⃗=⟨2x,−2y,2z⟩\vec{F}=⟨2x,−2y,2z⟩ is conservative.

    2. Yes, it is conservative.

Let F⃗=⟨yz,xz,xy⟩\vec{F}=⟨yz,xz,xy⟩.

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

Solution

∇×F⃗=∣ı^ȷ^k^∂∂x∂∂y∂∂zyzxzxy∣=0⃗\begin{aligned} \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{aligned}

Since ∇×F⃗=0⃗∇×\vec{F}=0⃗, this force field is conservative, which means that we can use the fundamental theorem of line integrals to calculate this work: W=∫CF⃗⋅r⃗ ′(t) dt=ϕ(B)−ϕ(A)W = \int_C \vec{F} \cdot \vec{r}\ '(t) \ dt = \phi(B)-\phi(A)

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

ϕx=yz  ⟹  ϕ=xyz+c(y,z)ϕy=xz+cy(y,z)=xz  ⟹  cy(y,z)=0  ⟹  c(y,z)=d(z)ϕz=xy+dz(z)=xy  ⟹  dz(z)=0  ⟹  ϕ(x,y,z)=xyz  ⟹  W=ϕ(3,4,4)−ϕ(1,2,2)=(3)(4)(4)−(1)(2)(2)=44\begin{aligned} \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{aligned}

Find div(F⃗)div(\vec{F}) for the field F⃗=⟨xz,xyz,−y2⟩\vec{F}=⟨xz,xyz,−y^2⟩.

Solution

div(F⃗)=∇⋅F⃗=⟨∂∂x,∂∂y,∂∂z⟩⋅⟨xz,xyz,−y2⟩=z+xz+0=z+xz\begin{aligned} 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{aligned}

Find div(F⃗)div(\vec{F}) for the field F⃗=⟨x2y,xyz,−x2y2⟩\vec{F}=⟨x2y,xyz,−x2y2⟩.

Solution

div(F⃗)=∇⋅F⃗=⟨∂∂x,∂∂y,∂∂z⟩⋅⟨x2y,xyz,−x2y2⟩=2xy+xz\begin{aligned} 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{aligned}

  1. Find div(F⃗)div(\vec{F}) for the field F⃗=⟨−y,x,z⟩\vec{F} = \langle -y,x,z \rangle.

    2. div(F⃗)=1div(\vec{F}) = 1

  2. Find div(F⃗)div(\vec{F}) for the field F⃗=⟨2x,4y,−3z⟩\vec{F} = \langle 2x,4y,-3z \rangle.

    3. div(F⃗)=3div(\vec{F}) = 3

  3. Find div(F⃗)div(\vec{F}) for the field F⃗=⟨12x,−6y,−6z⟩\vec{F} = \langle 12x,-6y,-6z \rangle.

    4. div(F⃗)=0div(\vec{F}) = 0

  4. Find div(F⃗)div(\vec{F}) for the field F⃗=⟨x,y,z⟩1+x2+y2\vec{F} = \dfrac{\langle x,y,z \rangle}{1+x^2+y^2}.

    5. div(F⃗)=x2+y2+3(1+x2+y2)2div(\vec{F}) = \dfrac{x^2+y^2+3}{(1+x^2+y^2)^2}

Connection Between Divergence and Curl

div(curl(F⃗))=0div(curl(\vec{F})) = 0

Verify the identity above for the field F⃗=⟨yz2,xy,yz⟩\vec{F} = \langle yz^2,xy,yz \rangle.

Solution

curl(F⃗)=∇×F⃗=∣ı^ȷ^k^∂∂x∂∂y∂∂zyz2xyyz∣=⟨z,2yz,y−z2⟩div(curl(F⃗))=∇⋅curl(F⃗)=⟨∂∂x,∂∂y,∂∂z⟩⋅⟨z,2yz,y−z2⟩=0+2z−2z=0\begin{aligned} 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{aligned}

Note

The expression curl(div(F⃗))curl(div(\vec{F})) is meaningless, because div(F⃗)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.