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\large\centerline{\textbf{Calculating Line Elements in Cylindrical and Spherical Coordinates}}\normalsize

\bigskip\bigskip
\leftline{\bf Rectangular Coordinates:}

The arbitrary infinitesimal displacement vector in Cartesian coordinates is:
$$d\rr=dx\,\ii + dy\,\jj +dz\,\kk$$
Given the cube shown below, find $d\rr$ on each of the three
paths, leading from $a$ to $b$.

\vspace{0.4in}
Path 1:  $d\rr=$
\bigskip

Path 2:  $d\rr=$
\bigskip

Path 3:  $d\rr=$
\bigskip\bigskip

\centerline{\includegraphics[width=4.5 in]{vfdrvectorcurvihandfig1.jpg}}
\bigskip
\newpage
The first expression above for $d\rr$ is valid for any path in rectangular
coordinates.  Find the appropriate expression for $d\rr$ for the path which
goes directly from $a$ to $c$ as drawn below.

\vspace{0.6in}
Path 4:  $d\rr=$

\vspace{-0.5in}
\centerline{\includegraphics[width=4.5 in]{vfdrvectorcurvihandfig2.png}}
\bigskip\bigskip

However, Cartesian coordinates would be a {\bf poor} choice to describe a path
on a cylindrically or spherically shaped surface.  Next we will find an
appropriate expression in these cases.

\newpage

\leftline{\bf Cylindrical Coordinates:}

You will now derive the general form for $d\rr$ in cylindrical coordinates
by determining $d\rr$ along the specific paths below.

Note that an infinitesmial element of length in the $\rhat$ direction is
simply $dr$, just as an infinitesimal element of length in the $\ii$
direction is $dx$.  {\bf But}, an infinitesimal element of length in the
$\phat$ direction is {\bf not} just $d\phi$, since this would be an angle
and does not even have the units of length.

Geometrically determine the length of the three paths leading from $a$
to $b$ and write these lengths in the corresponding boxes on the diagram.

Now, remembering that $d\rr$ has both magnitude and direction, write down
below the infinitesimal displacement vector $d\rr$ along the three paths
from $a$ to $b$.  Notice that, along any of these three paths, only one
coordinate $r$, $\phi$, or $z$ is changing at a time. (i.e.\ along path 1,
$dz\ne0$, but $d\phi=0$ and $dr=0$).
\bigskip

\vspace{1truein}

Path 1:  $d\rr=$
\bigskip

Path 2:  $d\rr=$
\bigskip

Path 3:  $d\rr=$
\bigskip

\vspace{-2.25truein}

\centerline{\includegraphics[width=4.5 in]{vfdrvectorcurvihandfig3.png}}

\bigskip
If all three coordinates are allowed to change simultaneously, by an
infinitesimal amount, we could write this $d\rr$ for any path as:

\bigskip
\qquad$d\rr$=
\bigskip

This is the general line element in cylindrical coordinates.

\newpage


\centerline{\includegraphics[width=4.5 in]{vfdrvectorcurvihandfig4.png}}

\leftline{\bf Spherical Coordinates:}

You will now derive the general form for $d\rr$ in spherical coordinates by
determining $d\rr$ along the specific paths below.  As in the cylindrical
case, note that an infinitesimal element of length in the $\that$ or $\phat$
direction is {\bf not} just $d\theta$ or $d\phi$.  You will need to be more
careful.  Geometrically determine the length of the three paths leading from
$a$ to $b$ and write these lengths in the corresponding boxes on the diagram.
Now, remembering that $d\rr$ has both magnitude and direction, write down
below the infinitesimal displacement vector $d\rr$ along the three paths from
$a$ to $b$.  Notice that, along any of these three paths, only one coordinate
$r$, $\theta$, or $\phi$ is changing at a time. (i.e.\ along path 1,
$d\theta\ne0$, but $dr=0$ and $d\phi=0$).
\medskip

Path 1:  $d\rr=$
\smallskip

Path 2:  $d\rr=$
\hfill{\small (Be careful, this is the tricky one.)}
\smallskip

Path 3:  $d\rr=$
\medskip

If all 3 coordinates are allowed to change simultaneously, by an infinitesimal
amount, we could write this $d\rr$ for any path as:
\medskip
\qquad$d\rr$=
\medskip

This is the general line element in spherical coordinates.

\vfill
\leftline{\it by Corinne Manogue and Katherine Meyer}
\leftline{\copyright 1997 \& 2006 Corinne A. Manogue}
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