I have been trying to understand the numerical integration approach employed by Charles Hutton in calculating the gravitational attraction of the Schiehallion mountain on a plumb line. The paper is available here: https://royalsocietypublishing.org/doi/10.1098/rstl.1778.0034. In the paper, he provides a geometric argument and I'm trying to formulate it in terms of calculus. The relevant pages (if you are interested) are 750-755.
The basic approach can be seen in the following diagram:
The plumb line is at A in the figure. He is using cylindrical coordinates to do the integration. He takes the meridian going toward north to be the positive X-axis and computes the X-component of the gravitational attraction of each mass element on the plumb line. If you have a mass element enclosed in the ring between $r = r_1$ and $r = r_2$, and the sector between $\theta = \theta_1$ and $\theta = \theta_2$ and planes $z = 0$ and $z = h$, the X-component of its gravitational attraction would be:
$$\int_{\theta_1}^{\theta_2}\int_{0}^{h}\int_{r_1}^{r_2} Gm\rho\;\frac{r^2}{(r^2 + z^2)^\frac{3}{2}}\;\cos{\theta}\;dr\;d\theta\;dz$$
Here, $G$ is the gravitational constant, $m$ is the mass of the plumb bob and $\rho$ is the density of the mountain.
The integral (if you factor out $Gm\rho$) evaluates to:
$$h\;(\sin{\theta_2} - \sin{\theta_1})\;\Biggl\{\log\Biggl(\sqrt{\frac{r_2^2}{h^2} + 1} + \frac{r_2}{h}\Biggr) - \log\Biggl(\sqrt{\frac{r_1^2}{h^2} + 1} + \frac{r_1}{h}\Biggr)\Biggr\}$$
(I hope I have calculated the integral correctly.)
He seems to be assuming that $r_2$ is close to $r_1$ and $h \ll \frac{r_1 + r_2}{2}$. With these assumptions, it seems to me that the following approximation should hold:
$$\Biggl\{\log\Biggl(\sqrt{\frac{r_2^2}{h^2} + 1} + \frac{r_2}{h}\Biggr) - \log\Biggl(\sqrt{\frac{r_1^2}{h^2} + 1} + \frac{r_1}{h}\Biggr)\Biggr\} \approx \frac{r_2 - r_1}{\sqrt{(\frac{r_1 + r_2}{2})^2 + h^2}}$$
But I'm unable to see how to derive this approximation. I would greatly appreciate any help you can provide in helping me understand this. Thanks!!
In your integral development of Hutton's gravitational attraction of the mountain, you have the attraction in the $X$-direction as $$ \text{Attraction}=\int_{\theta_{1}}^{\theta_{2}}\int_{0}^{h}\int_{r_{1}}^{r_{2}} Gm\rho \,\dfrac{r^2}{\left(r^2+z^2\right)^{3/2}}\, \cos\theta\, {\rm d}r\,{\rm d}\theta\,{\rm d}z. $$ On setting $Gm\rho=1$, this integrates to $h\left(\sin\theta_{2}-\sin\theta_{1}\right)I$, where Hutton's mean radius to the center of the "finite element" is $r_{m}=\left(r_{1}+r_{2}\right)/2$ and $I\,$ is the expression that you wish to approximate. We may now relate the inner and outer radii of the element by means of the relations $r_{1}=r_{m}-\varepsilon$ and $r_{2}=r_{m}+\varepsilon$, where, from the last relation, $$ \varepsilon=r_{2}-r_{m}=r_{2}-\left(r_{1}+r_{2}\right)/2=\left(r_{2}-r_{1}\right)/2. $$
With these expressions, together with the relation $\log(a)-\log(b)=\log(a/b)$, we can now write your Hutton $X$-attraction integral as $$ I=\log\left(\dfrac{H\left(r_{2}\right)}{H\left(r_{1}\right)}\right) = \log\left(\dfrac{H\left(r_{m}+\varepsilon\right)}{H\left(r_{m}-\varepsilon\right)}\right) $$ where $$ H(r)=\dfrac{1}{h}\left\{\sqrt{r^2+h^2}+r\right\}. $$ Expanding $H\left( r_{m}+\varepsilon\right)$ in a Taylor series and keeping only the linear term in $\varepsilon$, that is we are assuming that $r_{2}-r_{1}\ll 2\, r_{m}$, gives the approximate relation $$ H\left(r_{m}+ \varepsilon\right)\approx H\left(r_{m}\right) +\varepsilon \left( \dfrac{{\rm d}H}{{\rm d}r} \right)_{r=r_{m}}. $$ The integral logarithm term then becomes $$I\approx\log\left( \dfrac{H\left(r_{m}\right)+\varepsilon \left(\dfrac{{\rm d}H}{{\rm d}r}\right)_{r=r_{m}}}{H\left(r_{m}\right)-\varepsilon \left(\dfrac{{\rm d}H}{{\rm d}r}\right)_{r=r_{m}}} \right), $$ and by dividing the numerator and denominator by $H\left(r_{m}\right)$, we get $$ I\approx\log\left( \dfrac{1+\dfrac{\varepsilon}{H(r)} \dfrac{{\rm d}H(r)}{{\rm d}r}}{1-\dfrac{\varepsilon}{H(r)}\dfrac{{\rm d}H(r)}{{\rm d}r}} \right)_{r=r_{m}}. $$ If we now use the binomial expansion of the relation $(1+\varepsilon x)/(1-\varepsilon\, x)\approx 1+2\varepsilon\, x$, this relation becomes $$\, I\approx\log\left( 1+\dfrac{2\varepsilon}{H(r)}\dfrac{{\rm d}H(r)}{{\rm d}r} \right)_{r=r_{m}}\approx \left( \dfrac{2\varepsilon}{H(r)}\dfrac{{\rm d}H(r)}{{\rm d}r}\right)_{r=r_{m}} $$ where we have used the series expansion $\log(1+\varepsilon\, x)\approx \varepsilon\, x$. We have now to differentiate $H(r)$ to get $$ \dfrac{{\rm d}H(r)}{{\rm d}r}=\dfrac{1}{h}\left[ \dfrac{r}{\sqrt{r^2+h^2}} +1 \right]. $$ The approximate expression for the integral, namely, $$I\approx\left( \dfrac{2\varepsilon}{H(r)}\dfrac{{\rm d}H(r)}{{\rm d}r}\right)_{r=r_{m}},$$ then reduces to $\dfrac{2\varepsilon}{\sqrt{r_{m}^2+h^2}+r_{m}}\left[\dfrac{r_{m}+\sqrt{r_{m}^2+h^2}}{\sqrt{r_{m}^2+h^2}}\right]$, and on cancelling terms in the numerator and denominator, we recover the expression you were looking to obtain, namely, $$ I\approx\dfrac{r_{2}-r_{1}}{\sqrt{r_{m}^2+h^2}}, $$ when we use the defined quantities $\varepsilon = r_{2}-r_{1}$ and $r_{m}=\left(r_{1}+r_{2}\right)/2$.
This has now prompted me to read Hutton's paper. I guess Playfair also provided input into Maskelyne's problem a few years later. Has anyone retraced Hutton's approximations with the more accurate logarithmic expressions to see how accurate his results were?
Addendum The simple result for the OP's calculus result of Hutton's approximation, used in the calculation of the mountain Schiehallion's attraction on a plumb line, suggests that a more economic derivation was lurking in the background.
The integral $I\,$ can be written as $$I=\left[F(r+\varepsilon)-F(r-\varepsilon)\right]_{r=r_{m}},\quad\text{where}$$ $$F(r)=\log[H(r)]\quad\text{ and }\quad H(r)=\dfrac{1}{h}\left\{\sqrt{r^2+h^2}+r\right\}. $$ If the integral is written as $$ I=2\varepsilon\left[\dfrac{F(r+\varepsilon)-F(r-\varepsilon)}{2\varepsilon}\right]_{r=r_{m}}, $$ the quantity in brackets is the central difference representation of the differential ${\rm d}F(r)/{\rm d}r$ when $\varepsilon\ll r_{m}$. Thus $I\approx 2\varepsilon\left[{\rm d}F(r)/{\rm d}r\right]_{r=r_{m}}$, and using the definition $F(r) =\log\left[H(r)\right]$ we recover the OP's result. The central difference operator is known to be accurate to the second order in $\varepsilon$, and if we expand the function $F$ in the numerator in a Taylor series, we get $$I\approx 2\varepsilon\left[\dfrac{\left(F+\varepsilon F' +\tfrac{1}{2}\varepsilon^2 F''+\tfrac{1}{6}\varepsilon^3 F'''\right)- \left(F-\varepsilon F' +\tfrac{1}{2}\varepsilon^2 F''-\tfrac{1}{6}\varepsilon^3 F'''\right)}{2\varepsilon}\right]_{r=r_{m}},$$ or $$I\approx 2\varepsilon F'\left(r_{m}\right) +\tfrac{1}{3}\varepsilon^3 F'''\left(r_{m}\right)\approx \dfrac{2\varepsilon}{\left(r_{m}^2+h^2\right)^{1/2}} +\dfrac{\varepsilon^3 \left(2 r_{m}^2-h^2\right)}{3\left(r_{m}^2+h^2\right)^{5/2}}. $$ This shows that Hutton's geometric and the OP's calculus representations of the plumb bob attraction are accurate to the second order in $\varepsilon$, since third order terms are required in order to affect the result. If the mean radius to the element is chosen as $r_{m}=h/\sqrt{2}$, the preceding result shows that Hutton's calculation would have been correct to the fourth order of small quantities, since then the next term, $2\varepsilon^{5}F^{'''''}/5!\,$, would be needed to affect Hutton's result.
Edit
I have just found an expansion agreeing with the preceding result on page 597 of the 1856 paper by Henry James, On the deflection of the plumb-line at Arthur's Seat, and the mean density of the earth, Proceedings of the Royal Society, Vol. 146, pages 591-606, https://royalsocietypublishing.org/doi/pdf/10.1098/rstl.1856.0029, where R.E. Clarke computes the attraction formula as $$A=\rho h\left(\sin\theta_{2}-\sin\theta_{1} \right)\left[ \dfrac{\left(r_{2}-r_{1}\right)}{\left(r_{m}^2+h^2\right)^{1/2}}+\dfrac{\left(r_{2}-r_{1}\right)^3}{24}\dfrac{2r_{m}^2-h^2}{\left(r_{m}^2+h^2\right)^{5/2}} \right], $$ which agrees with preceding result on putting $\varepsilon=\left(r_{2}-r_{1}\right)/2.$