Editing 2DStructure/ToroidalCoordinates (section)

====Volume with Cropped Top====

[[File:CropTopB.png|300px|right|Diagram of "Cropped Top" Torus]]
During my development of a computer program to integrate over the volume of a circular torus while using toroidal coordinates, I decided that it would be useful to determine the volume of a torus that has a "cropped top" as illustrated in the accompanying diagram, shown here on the right.  Let's evaluate this "cropped-top" volume using cylindrical coordinates so that we will know what the correct answer is when developing an integration scheme using toroidal coordinates.  Specifically, let's determine the volume of the "green" portion for a specified value of <math>~h < r_t</math>.

The radial integration will be evaluated between the limits: (lower) <math>~(\varpi_t - b) </math> and (upper) <math>~(\varpi_t + b) </math>, where,
<div align="center">
<math>~b = \sqrt{r_t^2-h^2} \, .</math>
</div>
And the limits on the vertical integration will be:  (lower) <math>~h</math> and, as before when integrating over the entire volume, (upper) <math>~\sqrt{r_t^2 - (\varpi_t - \varpi)^2} \, .</math>  With these limits, the volume integration gives,

<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~V_\mathrm{green}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2\pi \int\limits_{(\varpi_t - b)}^{(\varpi_t + b)} \varpi d\varpi  \int\limits_{h}^{\sqrt{r_t^2 - (\varpi_t - \varpi)^2}} dz </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2\pi \int\limits_{(\varpi_t - b)}^{(\varpi_t + b)} \varpi d\varpi \biggl[\sqrt{r_t^2 - (\varpi_t - \varpi)^2} - h \biggr] </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-2\pi h \int\limits_{(\varpi_t - b)}^{(\varpi_t + b)} \varpi d\varpi  
+ 2\pi \int\limits_{(\varpi_t - b)}^{(\varpi_t + b)} \varpi \sqrt{X} d\varpi \, ,
</math>
  </td>
</tr>
</table>
</div>
where,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~X</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~a_0 + a_1\varpi + a_2\varpi^2 \, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~a_0</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~-( \varpi_t^2 - r_t^2) \, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~a_1</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~2\varpi_t \, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~a_2</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~-1 \, .</math>
  </td>
</tr>
</table>
</div>
Carrying out this pair of integrations gives,

<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~V_\mathrm{green}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-\pi h \biggl[ (\varpi_t + b)^2 -  (\varpi_t - b)^2  \biggr]  
+ 2\pi \biggl[\frac{X\sqrt{X}}{3a_2} - \frac{a_1(2a_2\varpi + a_1)}{8a_2^2} \sqrt{X} \biggr]_{(\varpi_t - b)}^{(\varpi_t + b)} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
- \frac{2\pi a_1 ( 4a_0 a_2 - a_1^2)}{(4a_2)^2} \int\limits_{(\varpi_t - b)}^{(\varpi_t + b)} \frac{d\varpi}{\sqrt{X}} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-\pi h \biggl[\varpi_t^2 + 2b\varpi_t + b^2 - (\varpi_t^2 - 2b\varpi_t + b^2)  \biggr]
+ \frac{\pi}{12} \biggl\{-8[ r_t^2 - (\varpi_t - \varpi)^2]^{3/2} + 6\varpi_t(2\varpi - 2\varpi_t) [r_t^2 - (\varpi_t - \varpi)^2 ]^{1/2} \biggr\}_{(\varpi_t - b)}^{(\varpi_t + b)} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
- \frac{2\pi a_1 ( 4a_0 a_2 - a_1^2)}{(4a_2)^2} 
\biggl\{ - \frac{1}{\sqrt{-a_2}} ~\sin^{-1}\biggl[ \frac{2a_2 \varpi + a_1}{\sqrt{a_1^2 - 4a_0 a_2}} \biggr] \biggr\}_{(\varpi_t - b)}^{(\varpi_t + b)}  
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-4\pi h b\varpi_t  
+ \frac{\pi}{3} \biggl[3\varpi_t b (r_t^2 - b^2 )^{1/2} -2( r_t^2 - b^2)^{3/2} \biggr]
+ \frac{\pi}{3} \biggl[3\varpi_t b (r_t^2 - b^2 )^{1/2} +2( r_t^2 - b^2)^{3/2} \biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
- \pi \varpi_t ( a_0  + \varpi_t^2)
\biggl\{ \sin^{-1}\biggl[ \frac{\varpi_t - \varpi }{\sqrt{\varpi_t^2 + a_0}} \biggr] \biggr\}_{(\varpi_t - b)}^{(\varpi_t + b)}  
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-4\pi h b\varpi_t  + 2\pi\varpi_t b (r_t^2 - b^2 )^{1/2} 
- \pi \varpi_t  r_t^2  \biggl\{ \sin^{-1}\biggl[ \frac{-b}{r_t} \biggr] - \sin^{-1}\biggl[ \frac{b }{r_t} \biggr] \biggr\}  
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-4\pi h \varpi_t (r_t^2 -h^2 )^{1/2} + 2\pi\varpi_t h (r_t^2 -h^2 )^{1/2} 
+2 \pi \varpi_t  r_t^2  \sin^{-1}\biggl[1 - \biggl(\frac{h }{r_t}\biggr)^2\biggr]^{1/2}   
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
2 \pi \varpi_t  r_t^2  \biggl\{ \sin^{-1}\biggl[1 - \biggl(\frac{h }{r_t}\biggr)^2\biggr]^{1/2}   
- \biggl(\frac{h}{r_t}\biggr) \biggl[ 1 - \biggl(\frac{h}{r_t}\biggr)^2 \biggr]^{1/2} \biggr\}\, .
</math>
  </td>
</tr>
</table>
</div>

Hence, the ''fractional'' volume is,

<div align="center" id="GreenAnalytic">
<table border="1" cellpadding="8"><tr><th align="center" colspan="2">
Analytic Expression for Green Volume</th></tr>
<tr><td align="center">

<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{V_\mathrm{green}}{V_\mathrm{torus}}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{\pi}  \biggl\{ \sin^{-1}\biggl[1 - \biggl(\frac{h }{r_t}\biggr)^2\biggr]^{1/2}   
- \biggl(\frac{h}{r_t}\biggr) \biggl[ 1 - \biggl(\frac{h}{r_t}\biggr)^2 \biggr]^{1/2} \biggr\}\, .
</math>
  </td>
</tr>
</table>

</td>
<td align="center">
[[File:CropTopB.png|150px|right|Diagram of "Cropped Top" Torus]]
</td></tr>
</table>
</div>


REALITY CHECK:  This should give a zero (green) volume if <math>~h = r_t</math>; and the fractional volume should be one-half if <math>~h = 0</math>.  In the former case, our expression gives,

<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{V_\mathrm{green}}{V_\mathrm{torus}}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{\pi} \sin^{-1} (0) = 0 \, ,  
</math>
  </td>
</tr>
</table>
</div>
as expected.  And in the latter case we have,

<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{V_\mathrm{green}}{V_\mathrm{torus}}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{\pi}  \sin^{-1} (1) =  \frac{1}{2}  \, ,  
</math>
  </td>
</tr>
</table>
</div>
which means that <math>~V_\mathrm{green}</math> is indeed half of the total torus volume, [[2DStructure/ToroidalCoordinates#Total_Volume|as derived earlier]].