Editing Apps/MaclaurinSpheroids/GoogleBooks (section)

==Prolate Spheroid==
As one specific example, Maclaurin discusses the gravitational attraction at the pole of a uniform-density, prolate spheroid in his &sect;647 of Volume II (pp. 125-127).  While his discussion of this topic necessarily refers back to earlier sections &#8212; most importantly, &sect;643 (spheres) and &sect;644 (oblate spheroids) &#8212; it frequently refers to Figure 291, No. 2, which we have reproduced in the following figure.  

<div align="center" id="Paragraph630">
<table border="1" cellpadding="5" width="80%">
<tr><td align="center" colspan="2">
Figure extracted<sup>&dagger;</sup> from [https://books.google.com/books?id=xfQ7AQAAMAAJ&printsec=frontcover&source=gbs_ge_summary_r&cad=0#v=onepage&q&f=false Colin Maclaurin (1742)]<br />
"''A Treatise of Fluxions''"<br />
Volume II, Figure Plate XXXIII
</td></tr>
<tr>
  <td align="center">
[[File:Figure291Pt2.png|300px|center|Maclaurin (1742)]]
  </td>
  <td align="center">
<!-- [[File:MyFig291Pt2C.png|380px|center|Our Construction]] -->
[[File:Maclaurin01.gif|380px|center|Our Construction]]
  </td>
</tr>
<tr>
  <td align="left">
<sup>&dagger;</sup>The .  
  </td>
  <td align="left">
In this animation sequence, the blue ellipse has an eccentricity, <math>e = 0.55</math>, and the red line-segment is drawn at an angle to the horizontal axis that varies over the range,  <math>\tfrac{3\pi}{8} \ge \alpha \ge \tfrac{\pi}{8}</math>.  As is described in the accompanying text, the orange curve &#8212; Maclaurin's "AKG" curve &#8212; is mapped out as the angle, <math>\alpha</math>, is varied.
  </td>
</tr>
</table>
</div>

While trying to decipher Maclaurin's geometrically based derivation, we have found it useful to re-draw key elements of this figure, deriving relevant algebraic or trigonometric relations along the way.  Because the "oblong" ellipse in Maclaurin's figure is oriented such that the axis of revolution &#8212; the symmetry axis &#8212; is horizontal, and because we prefer to associate the <math>z</math> axis with the symmetry axis of the configuration, we will rather unconventionally refer to the horizontal axis as the <math>z</math> (symmetry) axis.  Adopting a cylindrical coordinate system furthermore means that, in our notation, the vertical axis will be the <math>\varpi</math> axis.

The blue ellipse centered on the origin of our plot represents Maclaurin's "ADBE" ellipse; in the expressions that follow, we will use <math>a_1</math> and <math>a_3</math> to refer, respectively, to the length of the semi-major axis (Maclaurin's line-segment "CA") and semi-minor axis (Maclaurin's line-segment "CD") of the ellipse; and, following Maclaurin, we will refer to the leftmost point on the ellipse (Maclaurin's point "A") as the "pole" which is positioned at coordinate <math>(z, \varpi) = (-a_1, 0)</math>.   (Note that in our plot, we have set <math>a_1 = 1</math>, but otherwise we will leave its specification arbitrary.)   In our plot, Maclaurin's "CNH" circle is represented by the green quarter circle centered on the pole.  The blue ellipse is defined by the relation,

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

<tr>
  <td align="right">
<math>\frac{z^2}{a_1^2} + \frac{\varpi^2}{a_3^2}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>1 </math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~~ \frac{z^2}{a_1^2} + (1-e^2)^{-1}\frac{\varpi^2}{a_1^2}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>1 \, ,</math>
  </td>
</tr>
</table>

and, because it is [[Appendix/Ramblings/ToroidalCoordinates#Off-center_Circle|off-center]], the green quarter circle is defined by the relation,

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

<tr>
  <td align="right">
<math>(a_1 + z)^2 + \varpi^2</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>a_1^2 </math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~~ \biggl(1 + \frac{z}{a_1} \biggr)^2 + \frac{\varpi^2}{a_1^2}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>1 \, .</math>
  </td>
</tr>
</table>

The red line-segment in our plot is what Maclaurin refers to as "AM" which, in his words, is ''any right line from the pole A.''  We will use the variable, <math>\alpha</math>, to identify the angle between this line-segment and the horizontal axis; in determining the gravitational attraction at pole A, Maclaurin ultimately sweeps this angle through the range, <math>0 \le \alpha \le \tfrac{\pi}{2}</math>.  Points lying along this line are defined by the relation,

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

<tr>
  <td align="right">
<math>\varpi</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>(a_1+z)\tan\alpha</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~~ \frac{\varpi}{a_1}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\biggl(1+\frac{z}{a_1}\biggr)\tan\alpha \, .</math>
  </td>
</tr>
</table>


===Intersection With Circle===

Hence, the red line-segment intersects the green quarter-circle when,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~1- \biggl(1 + \frac{z}{a_1} \biggr)^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl(1+\frac{z}{a_1}\biggr)^2\tan^2\alpha </math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~~\biggl(1+\frac{z}{a_1}\biggr)^2(1+\tan^2\alpha)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~1 </math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~~\frac{z}{a_1}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~(1+\tan^2\alpha)^{-1/2} -1 </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\cos\alpha -1 \, .</math>
  </td>
</tr>
</table>
</div>

Correspondingly,
<div align="center">
<math>\frac{\varpi}{a_1} = (\cos\alpha)\tan\alpha = \sin\alpha \, .</math>
</div>

===Intersection With Ellipse===

Similarly, the red line-segment intersects the blue ellipse when,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~(1-e^2)\biggl[1- \biggl(\frac{z}{a_1} \biggr)^2\biggr]</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl(1+\frac{z}{a_1}\biggr)^2\tan^2\alpha </math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~~ (1-e^2) - (1-e^2)\biggl(\frac{z}{a_1} \biggr)^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[1+2\biggl(\frac{z}{a_1}\biggr)+\biggl(\frac{z}{a_1}\biggr)^2\biggr]\tan^2\alpha </math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~~ 0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[\tan^2\alpha - (1-e^2)\biggr]+2\biggl(\frac{z}{a_1}\biggr)\tan^2\alpha
+\biggl(\frac{z}{a_1}\biggr)^2\biggl[\tan^2\alpha+ (1-e^2) \biggr] \, .</math>
  </td>
</tr>
</table>
</div>

The roots of this quadratic equation are,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{z}{a_1}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{b}{2a}\biggl\{ -1 \pm \biggl[ 1 - \frac{4ac}{b^2} \biggr]^{1/2} \biggr\}</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{\tan^2\alpha}{(\tan^2\alpha+ 1-e^2 )}\biggl\{ -1 
\pm \biggl[ 1 - \frac{[\tan^2\alpha - (1-e^2)][ \tan^2\alpha+ (1-e^2) ]}{\tan^4\alpha} \biggr]^{1/2} \biggr\}</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{\tan^2\alpha}{(\tan^2\alpha+ 1-e^2 )}\biggl\{ -1 
\pm \frac{1}{\tan^2\alpha}\biggl[ \tan^4\alpha - [\tan^4\alpha - (1-e^2)^2] \biggr]^{1/2} \biggr\}</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-~\frac{[\tan^2\alpha \mp (1-e^2) ]}{[\tan^2\alpha+ (1-e^2 )]}  \, .</math>
  </td>
</tr>
</table>
</div>
The "inferior" root gives, 

<div align="center">
<math>~\frac{z}{a_1}\biggr|_- = -1 \, ,</math>
</div>
while the "superior" root gives,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{z}{a_1}\biggr|_+</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~- ~\frac{[\tan^2\alpha - (1-e^2) ]}{[\tan^2\alpha+ (1-e^2) ]} \, .</math>
  </td>
</tr>
</table>
</div>

Correspondingly,
<div align="center">
<math>\frac{\varpi}{a_1}\biggr|_- = 0 \, ,</math>
</div>
and,
<div align="center">
<math>\frac{\varpi}{a_1}\biggr|_+ = \tan\alpha\biggl\{  1
- ~\frac{[\tan^2\alpha - (1-e^2) ]}{[\tan^2\alpha+ (1-e^2) ]}
\biggr\} \, .</math>
</div>

The "inferior" root clearly identifies the intersection of the red line-segment with the pole of the prolate spheroid.  The "superior" root identifies the point at which the red line-segment intersects and crosses the blue ellipse, away from the pole.

===Generate Curve AKG===

In &sect;644 (p. 122 of Volume II) of his Treatise, Maclaurin states that "the gravity at the pole A towards the spheroid ADBE will be measured by [the ratio],"
<div align="center">
<math>\frac{\mathrm{AKGC}}{AC} \, ,</math>
</div>
where the numerator, "AKGC," is the area under the curve, "AKG," and the denominator is the length of the semi-major axis, <math>~a_1</math>.  So a key element of Maclaurin's derivation is the proper construction of the curve, "AKG."  The key sentence from his &sect;644 appears to be the following:

[[File:Vol2Par644KeyPhrase.png|400px|thumb|center|Extracted directly from &sect;644 of Maclaurin's Book 1, as digitized by Google]]

We interpret this phrase to mean that, as the angle, <math>~\alpha</math>, varies from <math>~\tfrac{\pi}{2}</math> to <math>~0</math>, the curve "AKG" is generated by plotting, as the ordinate, the length of the line-segment "AQ" and, as the abscissa, the <math>~z</math>-coordinate location of the vertical line, "RN" &#8212; we'll label this, <math>~z_\mathrm{RN}</math>.  In our terminology, as [[Apps/MaclaurinSpheroids/GoogleBooks#Intersection_With_Circle|derived above]], <math>~z_\mathrm{RN}</math> is the <math>~z</math>-coordinate of the intersection of the red line-segment with the green quarter-circle, that is,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~z_\mathrm{RN}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\cos\alpha - 1 \, .</math>
  </td>
</tr>
</table>
</div>
And the magnitude of "AQ" is obtained from the <math>~z</math>-coordinate location of the vertical line, "MQ" &#8212; that is, the <math>~z</math>-coordinate of the intersection of the red line-segment with the blue ellipse &#8212; as [[Apps/MaclaurinSpheroids/GoogleBooks#Intersection_With_Ellipse|derived above]].  Specifically, we interpret Maclaurin's phrase to mean that,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{\mathrm{AQ}}{a_1} \equiv 1 + \frac{z}{a_1}\biggr|_+</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~1 - ~\frac{[\tan^2\alpha - (1-e^2) ]}{[\tan^2\alpha+ (1-e^2) ]} \, .</math>
  </td>
</tr>
</table>
</div>

Using the first of these two expressions, we can rewrite the tangent function directly in terms of the ordinate, <math>~z_\mathrm{RN}</math>; specifically,
<div align="center">
<math>\tan^2\alpha = \frac{1- \cos^2\alpha}{\cos^2\alpha} = \frac{1- (1+z_\mathrm{RN})^2}{(1+z_\mathrm{RN})^2}  \, .</math>
</div>
Hence,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{\mathrm{AQ}}{a_1} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~1 - ~\frac{(1+z_\mathrm{RN})^2[\tan^2\alpha - (1-e^2) ]}{(1+z_\mathrm{RN})^2[\tan^2\alpha+ (1-e^2) ]}</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~1 - ~\frac{[1-(1+z_\mathrm{RN})^2 - (1-e^2)(1+z_\mathrm{RN})^2 ]}{[1-(1+z_\mathrm{RN})^2+ (1-e^2) (1+z_\mathrm{RN})^2]}</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~1 - ~\frac{[1+(1+z_\mathrm{RN})^2 (-1 - (1-e^2) ]}{[1+(1+z_\mathrm{RN})^2(-1+1-e^2)]}</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~1 - ~\frac{[1-(2-e^2)(1+z_\mathrm{RN})^2 ]}{[1-e^2(1+z_\mathrm{RN})^2]}</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{[1-e^2(1+z_\mathrm{RN})^2]-[1-(2-e^2)(1+z_\mathrm{RN})^2 ]}{[1-e^2(1+z_\mathrm{RN})^2]}</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{[2(1-e^2)(1+z_\mathrm{RN})^2 ]}{[1-e^2(1+z_\mathrm{RN})^2]} \, .</math>
  </td>
</tr>
</table>
</div>

In our above figure, this function, <math>~\mathrm{AQ}(z_\mathrm{RN})</math> normalized to <math>~a_1</math>, is delineated by the orange triangles over the coordinate range, <math>~-1 \le z_\mathrm{RN} \le 0</math>.  If our interpretation of Maclaurin's discussion is correct, this is precisely the curve in Maclauin's Figure 291, No. 2 that is labeled, "AKG."

===Area Under the Curve===
The area, "AKGC" &#8212; which, according to Maclaurin, gives the gravitational acceleration at the pole A &#8212; can be determined by carrying out the integral,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{\mathrm{AKGC}}{a_1} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\int_{-1}^0 \biggl[\frac{\mathrm{AQ}}{a_1}\biggr]dz_\mathrm{RN} \, .</math>
  </td>
</tr>
</table>
</div>

This integral can be more cleanly evaluated if we make the variable substitution, 
<div align="center">
<math>u_\mathrm{RN} \equiv 1 + z_\mathrm{RN} = \cos\alpha \, ,</math>
</div> 
with a corresponding shift in the limits of integration from <math>~(-1 \rightarrow 0)</math>, for <math>~z_\mathrm{RN}</math>, to <math>~(0 \rightarrow 1)</math>, for <math>~u_\mathrm{RN}</math>.  Specifically, we find,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{\mathrm{AKGC}}{a_1}  </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2(1-e^2) \int_{0}^1 \frac{u_\mathrm{RN}^2 }{1-e^2 u_\mathrm{RN}^2} ~du_\mathrm{RN}</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-2(1-e^2) \biggl[ \frac{u_\mathrm{RN}}{e^2} \biggr]_{0}^1 + \frac{1}{e^4} \int_{0}^1 \frac{du_\mathrm{RN}}{e^{-2}- u_\mathrm{RN}^2} </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-\frac{2(1-e^2)}{e^2}  + \frac{2(1-e^2)}{e^4} \biggl\{ \frac{e}{2} \ln \biggl[ \frac{e^{-2} +u_\mathrm{RN} }{e^{-2} -u_\mathrm{RN} } \biggr] \biggr\}_{0}^1 </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{(1-e^2)}{e^3} \ln \biggl[ \frac{1 + e^{2} }{1-e^{2}} \biggr]-\frac{2(1-e^2)}{e^2}   \, .</math>
  </td>
</tr>
</table>
</div>
This is precisely the expression for the coefficient, <math>~A_1</math>, that is [[ThreeDimensionalConfigurations/HomogeneousEllipsoids#Prolate_Spheroids_.28a1_.3E_a2_.3D_a3.29|traditionally used in the expression for the gravitational potential inside homogeneous, prolate spheroids]].  


{{SGFworkInProgress}}

===Comparison With More Traditional Derivation===

In our [[ThreeDimensionalConfigurations/HomogeneousEllipsoids#Acceleration_at_the_Pole|separate derivation of the gravitational acceleration at the pole of a prolate spheroid]] that uses a more traditional approach, we obtained this key coefficient by integrating over a different function, namely,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\mathcal{Z}</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~\frac{(1-\zeta)}{ [ (2-e^2) - 2\zeta + e^2\zeta^2 ]^{1/2}} -1 </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{(1-\zeta)^{1/2}}{ [ 2-e^2 - e^2\zeta  ]^{1/2}} -1 \, .</math>
  </td>
</tr>
</table>
</div>

By comparing Maclaurin's approach to the more traditional one, we conclude that,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\int_{-1}^0 \biggl[\frac{\mathrm{AQ}}{a_1}\biggr]dz_\mathrm{RN} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\int_{-1}^1 \mathcal{Z} d\zeta \, .</math>
  </td>
</tr>
</table>
</div>

We should be able to demonstrate explicitly that these two integrands are the same.  We begin by recognizing that the relationship between the dimensionless coordinate, <math>~z/a_1</math>, that we have used to decipher Maclaurin's presentation and the dimensionless coordinate, <math>~\zeta</math>, that was [[ThreeDimensionalConfigurations/HomogeneousEllipsoids#Acceleration_at_the_Pole|used in our separate analysis]], is,
<div align="center">
<math>\frac{z}{a_1} = -\zeta \, .</math>
</div>

Returning to the first expression in the above subsection that details the [[Apps/MaclaurinSpheroids/GoogleBooks#Intersection_With_Ellipse|intersection of the red line-segment with the blue ellipse]], we can therefore write,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~(1-e^2) (1-\zeta^2)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~(1-\zeta)^2\tan^2\alpha </math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~~ \tan^2\alpha </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{(1-e^2) (1+\zeta)}{(1-\zeta)} \, .</math>
  </td>
</tr>
</table>
</div>

Recalling that,
<div align="center">
<math>u_\mathrm{RN} = \cos\alpha = [1+\tan^2\alpha]^{-1/2} \, ,</math>
</div>
we see that,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{\mathrm{AQ}}{a_1} = \frac{2(1-e^2)u_\mathrm{RN}^2}{1-e^2 u_\mathrm{RN}^2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{2(1-e^2)[1+\tan^2\alpha]^{-1}}{1-e^2[1+\tan^2\alpha]^{-1}}</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{2(1-e^2)}{1-e^2 +\tan^2\alpha}</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{2(1-e^2)(1-\zeta)}{(1-e^2)(1-\zeta) +(1-e^2)(1+\zeta)}</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~(1-\zeta) \, .</math>
  </td>
</tr>
</table>
</div>

Furthermore, we see that the transformation from <math>~u_\mathrm{RN}</math> to <math>~\zeta</math> is,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~u_\mathrm{RN} = [1+\tan^2\alpha ]^{-1/2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[1 + \frac{(1-e^2) (1+\zeta)}{(1-\zeta)}  \biggr]^{-1/2}</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[\frac{(1-\zeta) + (1-e^2) (1+\zeta)}{(1-\zeta)}  \biggr]^{-1/2}</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~(1-\zeta)^{1/2} (2-e^2 -e^2\zeta)^{-1/2}</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~~ du_\mathrm{RN}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~[
-2^{-1}(1-\zeta)^{-1/2} (2-e^2 -e^2\zeta)^{-1/2}
-2^{-1}(1-\zeta)^{1/2} (2-e^2 -e^2\zeta)^{-3/2}(-e^2)]d\zeta
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-2^{-1}(1-\zeta)^{-1/2}  (2-e^2 -e^2\zeta)^{-3/2}[
(2-e^2 -e^2\zeta)-e^2(1-\zeta) ]d\zeta
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-(1-e^2)(1-\zeta)^{-1/2}  (2-e^2 -e^2\zeta)^{-3/2} d\zeta \, .</math>
  </td>
</tr>
</table>
</div>

Putting these two expressions together, we conclude that the integrand is,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\biggl[\frac{\mathrm{AQ}}{a_1}\biggr] du_\mathrm{RN}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-\biggl[\frac{(1-e^2)(1-\zeta)^{1/2} }{ (2-e^2 -e^2\zeta)^{3/2}}\biggr] d\zeta </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-\biggl[\frac{(1-e^2)}{(2-e^2 -e^2\zeta)}  \biggr]\biggl[\frac{(1-\zeta)^{1/2} }{ (2-e^2 -e^2\zeta)^{1/2}}\biggr] d\zeta \, .</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-\biggl[\frac{(1-e^2)}{(2-e^2 -e^2\zeta)}  \biggr]\biggl[1 + \mathcal{Z}\biggr] d\zeta \, .</math>
  </td>
</tr>
</table>
</div>

This does not appear to match the other integrand.  I must be misinterpreting something!

To be done:  I suspect that Maclaurin's approach will look more familiar if I redo the "traditional" derivation from the point of view of a spherical coordinate system (positioning pole A at the origin), in which an axisymmetric volume element takes the form,
<div align="center">
<math>~dV = 2\pi r^2 \sin\alpha dr d\alpha = 2\pi r^2 dr du_\mathrm{RN} \, ,</math>
</div>
where, from above, <math>u_\mathrm{RN} \equiv \cos\alpha \, .</math>