\documentclass[a4paper,12pt]{article}
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\newcommand{\dci}{\rlap{$\displaystyle\int\!\!\!\int_{B_{a,b,c}}$}
{\hspace{2.2pt}\bigcirc} \ \ }
\newcommand{\un}{\underline}
\begin{document}
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\begin{center}
\textbf{Vector Calculus}

\textit{\textbf{Grad, Div and Curl}}
\end{center}

\textbf{Question}

$\un{F}$ is a 3-dimensional smooth vector field.

$B_{a,b,c}$ is the surface of the box defined by
\begin{eqnarray*}
-a \le & x & \le a\\
-b \le & y & \le b\\
-c \le & z & \le c
\end{eqnarray*}
with outward normal $\un{\hat{N}}$.

Show that
$$\lim_{a,b,c \to 0^+} \frac{1}{8abc} \dci
\un{F}\bullet\un{\hat{N}} \,dS = \un{\nabla}\bullet\un{F}(0,0,0)$$


\textbf{Answer}

Use the Maclaurin expansion of $\un{F}$:
$$\un{F}=\un{F}_0+  \un{F}_1x+ \un{F}_2y + \un{F}_3z+ \cdots$$
with
\begin{eqnarray*}
 \un{F}_0 & = &  \un{F}(0,0,0)\\
 \un{F}_1 & = & \left. \frac{\partial}{\partial x} \un{F}(x,y,z)
 \right |_{(0,0,0)} = \left. \left ( \frac{\partial F_1}{\partial x}
 \un{i} + \frac{\partial F_2}{\partial x} \un{j} + \frac{\partial
 F_3}{\partial x}\un{k} \right ) \right |_{(0,0,0)}\\
 \un{F}_2 & = & \left. \frac{\partial}{\partial y} \un{F}(x,y,z)
 \right |_{(0,0,0)} = \left. \left ( \frac{\partial F_1}{\partial y}
 \un{i} + \frac{\partial F_2}{\partial y} \un{j} + \frac{\partial
 F_3}{\partial y}\un{k} \right ) \right |_{(0,0,0)}\\
 \un{F}_3 & = & \left. \frac{\partial}{\partial z} \un{F}(x,y,z)
 \right |_{(0,0,0)} = \left. \left ( \frac{\partial F_1}{\partial z}
 \un{i} + \frac{\partial F_2}{\partial z} \un{j} + \frac{\partial
 F_3}{\partial z}\un{k} \right ) \right |_{(0,0,0)}
\end{eqnarray*}
$\cdots$ represents terms in $x$, $y$ and $z$ that are of degree two
or above.

On the top of the box: $z=c$, $\un{\hat{N}}=\un{k}$.

On the bottom of the box: $z=-c$, $\un{\hat{N}}=-\un{k}$

On both of these: $dS =dxdy$

So
\begin{eqnarray*}
& & \left ( \int\!\!\!\int_{top} + \int\!\!\!\int_{top} \right )
\un{F} \bullet \un{\hat{N}} dS\\
& = & \int_{-a}^a dx \int_{-b}^b dy (c\un{F}_3\bullet \un{k}
-c\un{F}\bullet (-\un{k}+ \cdots \\
& = & 8abc\un{F}_3\bullet\un{k} + \cdots + 8abc
\left. \frac{\partial}{\partial z} F_3(x,y,z) \right |_{0,0,0} +
\cdots
\end{eqnarray*}
Here, $\cdots$ represented terms in $a$, $b$ and $c$ that are of degree
4 or higher.

Similar formulas can be used for the other two face pairs.

Combining the three formulas gives

$$\dci \un{F} \bullet \un{\hat{N}} \,dS = 8abc
\textrm{div} \un{F}(0,0,0) + \cdots$$

So it can be seen that
$$\lim_{a,b,c \to 0^+} \frac{1}{8abc} \dci
\un{F}\bullet\un{\hat{N}} \,dS = \textrm{div}\un{F}(0,0,0)$$

\end{document}