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The divergence of a vector field in the space.

Definition 6.37   Let $\overrightarrow{F} (x,y,z)= M(x,y,z) \overrightarrow{i} +
N((x,y,z) \overrightarrow{j} + P(x,y,z) \overrightarrow{k} $ be a vector filed in the 3-dimensional space. The divergence of the vector field $\overrightarrow{F} $ is the scalar function

\begin{displaymath}\text{div} \overrightarrow{F} = \frac {\partial M}{\partial x...
...rac {\partial N}{\partial y} +\frac {\partial P}{\partial z} .
\end{displaymath}

It has the same physical interpretation as the divergence of a vector field in the plane (v.s. Remark  6.34).

Remark 6.38   The divergence of the vector field $\overrightarrow{F} $ can be expressed as the formal scalar product

\begin{displaymath}\text{div} \overrightarrow{F} = \nabla \cdot \overrightarrow{F}\end{displaymath}

where

\begin{displaymath}\nabla = \frac {\partial }{\partial x} \overrightarrow{i} +\f...
...ghtarrow{j} +\frac {\partial }{\partial z}\overrightarrow{k} .
\end{displaymath}

Remark 6.39   Let f(x,y,z) be a function of three real variables defined on an open domain $\mathcal{D}$ in $\mathbb{R} ^3$. Then ,as we saw in  6.18, the gradient of f is:

\begin{displaymath}\overrightarrow{\nabla f} = \frac {\partial f}{\partial x} \o...
...htarrow{j} +\frac {\partial f}{\partial z}\overrightarrow{k} .
\end{displaymath}

Apply the divergence operator to this vector field (=the gradient field); we have:

\begin{displaymath}\nabla \cdot ( \overrightarrow{\nabla f} )=
\frac {\partial^...
...rtial^2 f }{\partial y^2} +\frac {\partial^2 f }{\partial z^2}
\end{displaymath}

i.e. $\nabla^2$ is Laplace's operator.

More on this topic will be studied in Section  5, about the Divergence Theorem.


next up previous contents
Next: Multiple integrals. Up: Applications of the derivative. Previous: The curl of a
Noah Dana-Picard
2001-05-30