\documentclass[a4paper,12pt]{article}
\usepackage{epsfig}
\newcommand{\dci}{\rlap{$\displaystyle\int\!\!\!\int_{B_{a,b,c}}$}
{\hspace{2.2pt}\bigcirc} \ \ }
\newcommand{\un}{\underline}
\begin{document}
\parindent=0pt

\begin{center}
\textbf{Vector Calculus}

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

\textbf{Question}

$\un{F}$ is a 2-dimensional smooth vector field. 

$C_{\epsilon}$ is a circle of radius $\epsilon$ centred at the origin.

$\un{\hat{N}}$ is the unit outward normal to $C_{\epsilon}$.

Show that $$\lim_{\epsilon \to O^+} \frac{1}{\pi \epsilon^2}
\oint_{C_{\epsilon}} \un{F}\bullet\un{\hat{N}} \,ds = \textrm{div}\un{F}(0,0)$$

\textbf{Answer}

Use the Maclaurin expansion of $\un{F}$,
$$\un{F} = \un{F}_0 + \un{F}_1x +\un{F}_2y + \cdots$$
with
\begin{eqnarray*}
\un{F}_0 & = & \un{F}(0,0)\\
\un{F}_1 & = & \frac{\partial}{\partial x} \left. \un{F}(x,y) \right
|_{(0,0)} = \left. \left( \frac{\partial F_1}{\partial x} \un{i}
+\frac{\partial F_2}{\partial x} \un{j} \right ) \right |_{(0,0)}\\
\un{F}_2 & = & \frac{\partial}{\partial y} \left. \un{F}(x,y) \right
|_{(0,0)} = \left. \left( \frac{\partial F_1}{\partial 2} \un{i}
+\frac{\partial F_2}{\partial 2} \un{j} \right ) \right |_{(0,0)}
\end{eqnarray*}
Here, $\cdots$ represent terms in $x$ and $y$ of degree 2 and higher.

On the curve $C_{\epsilon}$ of radius $\epsilon$ centered at the
origin, $\un{\hat{N}}=\frac{1}{\epsilon}(x\un{i}+y\un{j})$.

$\Rightarrow$
\begin{eqnarray*}
\un{F}\bullet\un{\hat{N}} & = & \frac{1}{\epsilon} (\un{F}_0\bullet
\un{i} x + \un{F}_0\bullet\un{j} y + \un{F}_1\bullet\un{i} x^2\\
& & + \un{F}_1\bullet\un{j} xy + \un{F}_2\bullet\un{i} xy + \un{F}_2
\bullet\un{j} y^2 + \cdots
\end{eqnarray*}
Here $\cdots$ represents terms in $x$ and $y$ of degree 3 or higher.

Since
\begin{eqnarray*}
\oint_{C_{\epsilon}} x \,ds & = & \oint_{C_{\epsilon}}y \,ds =
\oint_{C_{\epsilon}} xy \,ds = 0\\
\oint_{C_{\epsilon}} x^2 \,ds & = & \oint_{C_{\epsilon}} y^2 \,ds =
\int_{0}^{2\pi} \epsilon^2 \cos^2\theta \epsilon \,d\theta =
\pi\epsilon^3
\end{eqnarray*}

This gives
\begin{eqnarray*}
\frac{1}{\pi\epsilon^2} \oint_{C_{\epsilon}} \un{F}\bullet
\un{\hat{N}} \,ds & = & \frac{1}{\pi\epsilon^2}
\frac{\pi\epsilon^3}{\epsilon} (\un{F}_1\bullet\un{i} +
\un{F}_2\bullet\un{j}) + \cdots\\
& = & \textrm{div}\un{F}(0,0) + \cdots
\end{eqnarray*}
Here $\cdots$ represents terms in $\epsilon$ of degree 1 or higher.

So taking the limit as $\epsilon \to 0$ gives
$$\lim_{\epsilon \to O^+} \frac{1}{\pi \epsilon^2}
\oint_{C_{\epsilon}} \un{F}\bullet\un{\hat{N}} \,ds = \textrm{div}\un{F}(0,0)$$

\end{document}

