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 \(\displaystyle{ grad ~ g = \nabla g = \frac{\partial g }{\partial x}\vhat{i} + \frac{\partial g }{ \partial y}\vhat{j} + \frac{\partial g }{ \partial z}\vhat{k} }\), which is a vector field.

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

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.

Before we go on, let's work a few practice problems.

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.

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

Now it's time for a practice problem.

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.

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. ]

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