<html><head><meta name="color-scheme" content="light dark"></head><body><pre style="word-wrap: break-word; white-space: pre-wrap;">\documentclass[a4paper,12pt]{article}
\usepackage{epsfig}
\newcommand{\pa}{\partial}
\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 Identities}}
\end{center}

\textbf{Question}

It is given that div$\un{F}=0$ in a domain $D$, of which any point $P$
can be joined to the origin by a straight line segment in
$D$. $\un{r}$ is a parametrization of the line segment from the origin
to $(x,y,z)$ in $D$, with
$$\un{r} = tx\un{i} +ty\un{j} +tz\un{k}, \ \ \ (0 \le t \le 1).$$
$\un{G}$ is given by 
$$\un{G}(x,y,z) = \int_0^1 t\un{F}(\un{r}(t))\times \frac{d\un{r}}{dt}
\,dt.$$

Show that curl$\un{G}=\un{F}$ throughout $D$.


\textbf{Answer}

Let $\un{v} = x\un{i} + y \un{j} + z\un{k}$. As the line segment lies
inside $D$, so div$\un{F}=0$ on the path.

We have
\begin{eqnarray*}
\un{G}(x,y,z) &amp; = &amp; \int_0^1t \un{F}(\un{r}(t)) \times \un{v} \,dt\\
&amp; = &amp; \int_0^1 t\un{F}(\xi(t), \eta(t), \zeta(t)) \times \un{v} \,dt
\end{eqnarray*}
With $\xi=tx$, $\eta=ty$ and $\zeta = tz$. So the first component of
curl$\un{G}$ is
\begin{eqnarray*}
(\textrm{curl}\un{G})_1 &amp; = &amp; \int_01^1 t
(\textrm{curl}(\un{F}\times\un{v}))_1 \,dt\\
&amp; = &amp; \int_0^1 t \left ( \frac{\pa}{\pa y}(\un{F}\times\un{v})_3 -
\frac{\pa}{\pa z}(\un{F} \times \un{v})_2 \right ) \,dt\\
&amp; = &amp; \int_0^1 t \left ( \frac{\pa}{\pa y}(F_1y-F_2x) - \frac{\pa}{\pa
z} (F_3x-F_1z) \right ) \,dt\\
&amp; = &amp; \int_0^1 \left ( t F_1 + t^2y \frac{\pa F_1}{\pa \eta} -
t^2x\frac{\pa F_2}{\pa \eta} - t^2x\frac{\pa F_3}{\pa \zeta}
\right. \\
&amp; &amp; \left. + tF_1 +t^2z \frac{\pa F_1}{\pa \zeta} \right ) \,dt\\
&amp; = &amp; \int_0^1 \left ( 2tF-1 + t^2x \frac{\pa F_1}{\pa \xi} +
t^2y\frac{\pa F_1}{\pa \eta} + t^2 z \frac{\pa F_1}{\pa \zeta} \right
) \,dt.
\end{eqnarray*}
With the last line using the fact that div$\un{F}=0$ to replace
$it^2x\frac{\pa F_2}{\pa \eta} - t^2x\frac{\pa F_3}{\pa \zeta}$ with
$t^2 x \frac{\pa F_1}{\pa \xi}$.
 
So
\begin{eqnarray*}
(\textrm{curl}\un{G})_1 &amp; = &amp; \int_0^1 \frac{d}{dt} (t^2 F_1(\xi,\eta,
\zeta))\,dy\\
&amp; = &amp; \left. t^2F_1(tx,ty,tz) \right |_0^1\\
&amp; = &amp; F_1(x,y,z)
\end{eqnarray*}
Similar arguements will show that $(\textrm{curl}\un{G})_2 = F_2)$ and
$(\textrm{curl}\un{G})_3=F_3$.
$$\Rightarrow \textrm{curl}\un{G} - \un{F}.$$

\end{document}
</pre></body></html>