\( \newcommand{\abs}[1]{\left| \, {#1} \, \right| } \) \( \newcommand{\cm}{\mathrm{cm} } \) \( \newcommand{\sec}{ \, \mathrm{sec} \, } \) \( \newcommand{\vhat}[1]{\,\hat{#1}} \) \( \newcommand{\vhati}{\,\hat{i}} \) \( \newcommand{\vhatj}{\,\hat{j}} \) \( \newcommand{\vhatk}{\,\hat{k}} \) \( \newcommand{\vect}[1]{\boldsymbol{\vec{#1}}} \) \( \newcommand{\norm}[1]{\|{#1}\|} \) \( \newcommand{\arccot}{ \, \mathrm{arccot} \, } \) \( \newcommand{\arcsec}{ \, \mathrm{arcsec} \, } \) \( \newcommand{\arccsc}{ \, \mathrm{arccsc} \, } \) \( \newcommand{\sech}{ \, \mathrm{sech} \, } \) \( \newcommand{\csch}{ \, \mathrm{csch} \, } \) \( \newcommand{\arcsinh}{ \, \mathrm{arcsinh} \, } \) \( \newcommand{\arccosh}{ \, \mathrm{arccosh} \, } \) \( \newcommand{\arctanh}{ \, \mathrm{arctanh} \, } \) \( \newcommand{\arccoth}{ \, \mathrm{arccoth} \, } \) \( \newcommand{\arcsech}{ \, \mathrm{arcsech} \, } \) \( \newcommand{\arccsch}{ \, \mathrm{arccsch} \, } \)

17Calculus - Divergence of Vector Fields

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The divergence of a vector field is a measure of how a vector field diverges. The result of calculating the divergence will be a function. This function can be evaluated at a point to give a number that tells us how the vector field diverges at that point. First, let's look at the gradient to refresh our memories on the del operator.

Divergence is also called flux density.

The result of calculating the divergence of a vector field is a scalar function.

Remember from your study of gradients that the del operator is \(\displaystyle{ \nabla = \frac{\partial }{\partial x}\vhat{i} + \frac{\partial }{ \partial y}\vhat{j} + \frac{\partial }{ \partial z}\vhat{k} }\). If we are given a function \(g(x,y,z)\), the gradient of \(g\) is \(\displaystyle{ \nabla g = \frac{\partial g }{\partial x}\vhat{i} + \frac{\partial g }{ \partial y}\vhat{j} + \frac{\partial g }{ \partial z}\vhat{k} }\), which is a vector field. Sometimes \( \nabla g \) is written grad g.

Calculating Divergence

To calculate the divergence, we use the same del operator in a little different way and, as you would expect, the notation looks different. The divergence is given by the equation \( \vec{ \nabla } \cdot \vec{F} \) where \( \vec{F} \) is a vector field and '\( \cdot \)' indicates the dot product. The notation gets a bit strange but here is what this means. If we have a vector field \( \vec{F}(x,y,z) = f_i(x,y,z)\vhat{i} + \) \( f_j(x,y,z)\vhat{j} + \) \( f_k(x,y,z)\vhat{k} \), then the divergence of the vector field \(\vec{F}\) is

\(\displaystyle{ \vec{ \nabla } \cdot \vec{F} = }\) \(\displaystyle{ \left[ \frac{\partial }{\partial x} \vhat{i} + \frac{\partial }{ \partial y}\vhat{j} + \frac{\partial }{ \partial z}\vhat{k} \right] \cdot }\) \(\displaystyle{ \left[ f_i(x,y,z)\vhat{i} + \right. }\) \(\displaystyle{ \left. f_j(x,y,z)\vhat{j} +f_k(x,y,z)\vhat{k} \right] = }\) \(\displaystyle{ \frac{\partial f_i }{\partial x} + \frac{\partial f_j }{ \partial y} + \frac{\partial f_k }{ \partial z} }\)

Things To Notice
1. We do not have a true dot product in the above equation since \( \vec{ \nabla } \) is not a vector, it is an operator. However, we stretch the notation here to think of the del operator as a vector. In actuality, we can think of the dot product here as a way to 'distribute' the partial derivatives to each term in \(\vec{F}\).
2. When writing the divergence, mathematicians often write the del operator with the vector sign above it to emphasize that we need to think of the del operator as a vector. We have done this here too. However, \(\vec{ \nabla }\) and \(\nabla \) both refer to the same del operator shown above. [As usual, check with your instructor to see what they expect.]
3. In the gradient equation \(\nabla g\), there is no dot for a dot product. It would be incorrect to have a dot for a gradient since a dot product is an operation on two vectors.
4. Another way to write the divergence is to write \(div ~ \vec{F}\), so \(div ~ \vec{F} = \vec{ \nabla } \cdot \vec{F} \).

Okay, so let's watch a video clip that explains the divergence in more detail.

Dr Chris Tisdell - What is the divergence? (Part 1) [6mins-29secs]

video by Dr Chris Tisdell

Properties of Divergence

It may not be obvious from the equations we use to calculate divergence but divergence is a linear operator. That means most of the rules of algebra and calculus that you already know apply also to divergence. Here is a list of a few of them.

In the table below, \(\vec{F}\) and \(\vec{G}\) are vector fields, \(a\) is a scalar and \( \varphi \) is a scalar function.

equation

notes

\( \vec{ \nabla } \cdot \left( a\vec{F} \right) = a \left( \vec{ \nabla } \cdot \vec{F} \right) \)

\( \vec{ \nabla } \cdot \left( \vec{F} + \vec{G} \right) = \vec{ \nabla } \cdot \vec{F} + \vec{ \nabla } \cdot \vec{G} \)

\( \vec{ \nabla } \cdot \left( \varphi \vec{F} \right) = \left( \nabla \varphi \right) \cdot \vec{F} + \varphi \left( \vec{ \nabla } \cdot \vec{F} \right) \)

product rule involving a scalar function

\( \vec{ \nabla } \cdot \left( \vec{F} \times \vec{G} \right) = \left( \vec{ \nabla } \times \vec{F} \right) \cdot \vec{G} - \) \( \vec{F} \cdot \left( \vec{ \nabla } \times \vec{G} \right) \)

product rule involving the curl

\( \vec{ \nabla } \cdot \left( \nabla \varphi \right) = \vec{ \nabla }^2 \varphi \)

Laplacian of a scalar function

\( \vec{ \nabla } \cdot \left( \vec{ \nabla } \times \vec{F} \right) = 0 \)

identity

Here is a quick video clip discussing a few of these properties.

Dr Chris Tisdell - What is the divergence? (Part 2) [1min-56secs]

video by Dr Chris Tisdell

Meaning of Divergence

Whew! That is a lot of math. Now that you know how to calculate divergence of a vector field, you may be asking yourself, what does it mean and how do I use this? This video clip gives great examples and explanation on how to understand the result of a divergence calculation.

Dr Chris Tisdell - What is the divergence? (Part 3) [13mins-38secs]

video by Dr Chris Tisdell

Big Picture

The divergence measures the net outflow of a vector field.

If the divergence is positive everywhere, then there is a net outflow over every closed curve/surface. This is sometimes referred to as a source.

If the divergence is negative everywhere, then there is a net inflow over every closed curve/surface. This is sometimes referred to as a sink.

A vector field with zero divergence everywhere is called 'incompressible' with zero net outflow over every closed curve/surface.

Okay, so after all that math and stuff, your head is probably spinning. Lets get an idea of the big picture. The list above is adapted from the notes of Dr Chris Tisdell. In this video clip, he discusses these items. [We highly recommend that you go to his website and download the notes for this topic.]

Dr Chris Tisdell - Divergence of vector fields [1min-34secs]

video by Dr Chris Tisdell

This is a lot to assimilate. So, it is time to work some practice problems. Once you are done with those, the next topic is conservative vector fields.

Practice

Unless otherwise instructed, calculate the divergence of these vector fields. If a point is given, find the divergence at that point also.

Basic

Calculate the divergence of the vector field \( \vec{F} = xyz\vhat{i} + \cos(xyz)\vhat{j} + xy^2z^3\vhat{k} \)

Problem Statement

Calculate the divergence of the vector field \( \vec{F} = xyz\vhat{i} + \cos(xyz)\vhat{j} + xy^2z^3\vhat{k} \)

Solution

832 video

video by Dr Chris Tisdell

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Unless otherwise instructed, calculate the divergence of the vector field \( \vec{F} = \langle xy^2, 2xz, 4-z^2y \rangle \) and at the point \( (1,1,2) \).

Problem Statement

Unless otherwise instructed, calculate the divergence of the vector field \( \vec{F} = \langle xy^2, 2xz, 4-z^2y \rangle \) and at the point \( (1,1,2) \).

Solution

833 video

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Calculate the divergence of the vector field \( \vec{F}(x,y) = x^3y\vhat{i} + yx^2\vhat{j} \)

Problem Statement

Calculate the divergence of the vector field \( \vec{F}(x,y) = x^3y\vhat{i} + yx^2\vhat{j} \)

Solution

834 video

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Calculate the divergence of the vector field \( \cos(z)\vhat{i} + \sin(y)\vhat{j}+ \tan(x)\vhat{k} \)

Problem Statement

Calculate the divergence of the vector field \( \cos(z)\vhat{i} + \sin(y)\vhat{j}+ \tan(x)\vhat{k} \)

Solution

835 video

video by Dr Chris Tisdell

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Calculate the divergence of the vector field \( \vec{F} = 2xz \vhat{i} - xy\vhat{j} - z\vhat{k} \)

Problem Statement

Calculate the divergence of the vector field \( \vec{F} = 2xz \vhat{i} - xy\vhat{j} - z\vhat{k} \)

Solution

836 video

video by Dr Chris Tisdell

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Calculate the divergence of the vector field \( \vec{G}(x,y) = x\vhat{j} \)

Problem Statement

Calculate the divergence of the vector field \( \vec{G}(x,y) = x\vhat{j} \)

Solution

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Unless otherwise instructed, calculate the divergence of the vector field \( \vec{F} = (x^2-y)\vhat{i} + (y+z)\vhat{j} + (z^2-x)\vhat{k} \) and at that point \( (1,2,3) \).

Problem Statement

Unless otherwise instructed, calculate the divergence of the vector field \( \vec{F} = (x^2-y)\vhat{i} + (y+z)\vhat{j} + (z^2-x)\vhat{k} \) and at that point \( (1,2,3) \).

Solution

837 video

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Calculate the divergence of the vector field \( \vec{F}(x,y,z) = xye^z\vhat{i} + yze^x\vhat{j} + xze^z\vhat{k} \)

Problem Statement

Calculate the divergence of the vector field \( \vec{F}(x,y,z) = xye^z\vhat{i} + yze^x\vhat{j} + xze^z\vhat{k} \)

Final Answer

\( ye^z + ze^x + xe^z(z+1) \)

Problem Statement

Calculate the divergence of the vector field \( \vec{F}(x,y,z) = xye^z\vhat{i} + yze^x\vhat{j} + xze^z\vhat{k} \)

Solution

2203 video

video by Michel vanBiezen

Final Answer

\( ye^z + ze^x + xe^z(z+1) \)

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Calculate the divergence of the vector field \( \vec{F} = \langle 6z\cos(x), 7z\sin(x),5z \rangle \)

Problem Statement

Calculate the divergence of the vector field \( \vec{F} = \langle 6z\cos(x), 7z\sin(x),5z \rangle \)

Final Answer

\( -6z \sin(x) + 5 \)

Problem Statement

Calculate the divergence of the vector field \( \vec{F} = \langle 6z\cos(x), 7z\sin(x),5z \rangle \)

Solution

2133 video

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Final Answer

\( -6z \sin(x) + 5 \)

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Calculate the divergence of the curl of \( \langle x+y+z, xyz, 2x+3y+4z \rangle \).

Problem Statement

Calculate the divergence of the curl of \( \langle x+y+z, xyz, 2x+3y+4z \rangle \).

Solution

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Intermediate

Compute the divergence and the curl of \( \vec{F}(x,y,z) = (\sinh x)\vhat{i} + (\cosh y)\vhat{j} -xyz\vhat{k} \) and verify that \( \vec{\nabla} \cdot ( \vec{\nabla} \times \vec{F}) = 0\).

Problem Statement

Compute the divergence and the curl of \( \vec{F}(x,y,z) = (\sinh x)\vhat{i} + (\cosh y)\vhat{j} -xyz\vhat{k} \) and verify that \( \vec{\nabla} \cdot ( \vec{\nabla} \times \vec{F}) = 0\).

Solution

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\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\)

\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\)

\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\)

\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\)

Set 2 - squared identities

\( \sin^2t + \cos^2t = 1\)

\( 1 + \tan^2t = \sec^2t\)

\( 1 + \cot^2t = \csc^2t\)

Set 3 - double-angle formulas

\( \sin(2t) = 2\sin(t)\cos(t)\)

\(\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }\)

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\(\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }\)

\(\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }\)

 

\(\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }\)

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\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }\)

 

\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }\)

\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\)

 

\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }\)

\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }\)

 

\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }\)

Trig Integrals

\(\int{\sin(x)~dx} = -\cos(x)+C\)

 

\(\int{\cos(x)~dx} = \sin(x)+C\)

\(\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C\)

 

\(\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C\)

\(\int{\sec(x)~dx} = \) \( \ln\abs{\sec(x)+\tan(x)}+C\)

 

\(\int{\csc(x)~dx} = \) \( -\ln\abs{\csc(x)+\cot(x)}+C\)

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Practice Instructions

Unless otherwise instructed, calculate the divergence of these vector fields. If a point is given, find the divergence at that point also.

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