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