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George Weigt 2008-12-15 19:03:38 -07:00
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@ -84,6 +84,59 @@ $$L=1.47894$$
\subsection{Line integrals}
There are two kinds of line integrals, one for scalar fields and the other
for vector fields.
Both are closely related to arc length, as the following table shows.
\begin{center}
\begin{tabular}{|llll|}
\hline
& & & \\
& $\quad$
& Abstract form
& Computable form
\\
& & & \\
Arc length
&
& $\displaystyle{\int_C ds}$
& $\displaystyle{\int_a^b |g'(t)|\,dt}$
\\
& & & \\
Line integral, scalar field
&
& $\displaystyle{\int_C f\,ds}$
& $\displaystyle{\int_a^b f(g(t))\,|g'(t)|\,dt}$
\\
& & & \\
Line integral, vector field
&
& $\displaystyle{\int_C(F\cdot u)\,ds}$
& $\displaystyle{\int_a^b F(g(t))\cdot g'(t)\,dt}$
\\
& & & \\
\hline
\end{tabular}
\end{center}
\medskip
\noindent
We have
$$ds=|g'(t)|\,dt$$
For the vector field, $u$ is the unit tangent vector
$$u={g'(t)\over|g'(t)|}$$
The length of the tangent vector cancels wth $ds$
as follows.
$$\int_C(F\cdot u)\,ds
=\int_a^b\bigg(F(g(t))\cdot{g'(t)\over|g'(t)|}\bigg)\,\bigg(|g'(t)|\,dt\bigg)
=\int_a^b F(g(t))\cdot g'(t)\,dt
$$
\newpage
\noindent
Line integrals are easily computed by
converting the coordinates $x$, $y$ and $z$ into functions of $t$.
This has the effect of changing the measure as well.