= Solution
For a spherical system, use the <spherical shell theorem> with shell mass $dm=4\pi\rho(s)s^2ds$. Inner shells contribute $-Gdm/r$ to the <gravitational potential>, while an outer shell contributes its constant interior potential $-Gdm/s$. With $\Phi(\infty)=0$, assuming the required integrals converge,
$$
\boxed{\Phi(r)=-4\pi G\left[\frac1r\int_0^r\rho(s)s^2\,ds+\int_r^\infty\rho(s)s\,ds\right].}
$$
On differentiation, the two terms containing $\rho(r)r$ cancel. Hence $\Phi'=GM(r)/r^2$, where $M(r)=4\pi\int_0^r\rho(s)s^2ds$. Radial balance for a <circular orbit> gives \b[$v_c^2=r\Phi'=GM(r)/r$].
For a razor-thin axisymmetric <astrophysical disk>, the element of mass is $dm=\Sigma(s)s\,ds\,d\phi'$, where $\Sigma$ is the <surface density>. Superposing <Newtonian gravitational potentials> therefore gives
$$
\boxed{\Phi(R)=-G\int_0^\infty\Sigma(s)s\,ds\int_0^{2\pi}\frac{d\phi'}{\sqrt{R^2+s^2-2Rs\cos(\phi'-\phi)}}.}
$$
Axisymmetry permits $\phi=0$. Unlike the spherical case, an exterior annulus generally exerts a radial force: <enclosed mass does not determine a disc rotation curve>.
For the <Legendre expansion of thin-disk gravity>, split the radial integral at $s=R$. In the inner part the kernel expands in $s/R$, while in the outer part it expands in $R/s$. The angular integrals of odd <Legendre polynomials> vanish, and those of even degree $n$ equal $2\alpha_n$, where
$$
\alpha_n=\pi\left[\frac{n!}{2^n((n/2)!)^2}\right]^2\quad(n\text{ even}).
$$
Introduce $I_n(R)=\int_0^R\Sigma(s)s^{n+1}ds$ and $J_n(R)=\int_R^\infty\Sigma(s)s^{-n}ds$. The <gravitational potential> becomes
$$
\Phi(R)=-2G\sum_{k=0}^\infty\alpha_{2k}\left[R^{-(2k+1)}I_{2k}(R)+R^{2k}J_{2k}(R)\right].
$$
When differentiating each bracket, the moving-limit terms cancel: $R^{-(n+1)}I_n'=\Sigma(R)$ and $R^nJ_n'=-\Sigma(R)$. Since $v_c^2=R\Phi'$, this gives
$$
v_c^2=2G\sum_{k=0}^\infty\alpha_{2k}\left[(2k+1)R^{-(2k+1)}I_{2k}-2kR^{2k}J_{2k}\right].
$$
The zeroth term has $\alpha_0=\pi$ and $M(R)=2\pi I_0$. Separating it proves
$$
\boxed{v_c^2=\frac{GM(R)}R+2G\sum_{k=1}^\infty\alpha_{2k}\left[(2k+1)R^{-(2k+1)}\int_0^R\Sigma(s)s^{2k+1}ds-2kR^{2k}\int_R^\infty\Sigma(s)s^{-2k}ds\right].}
$$
The inner correction is inward, while the exterior correction is outward. For a smooth <surface density>, the original potential singularity at coincident points is integrable. Its radial force is understood through a symmetric <Cauchy principal value> or a vanishing-thickness regularization; the paired interior and exterior terms above retain the cancellation at $s=R$. This avoids treating the two singular local force contributions separately.
For an <exponential galactic disk>, write $\Sigma(s)=\Sigma_0e^{-s/R_d}$, so $M_\infty=2\pi\Sigma_0R_d^2$ and $M(R)=M_\infty[1-(1+R/R_d)e^{-R/R_d}]$. At large $R$, the missing mass and exterior-ring terms are exponentially small. The leading nonspherical interior term is $k=1$, with $\alpha_2=\pi/4$ and $I_2(\infty)=6\Sigma_0R_d^4$. Consequently the <exponential-disk Keplerian asymptotic> is
$$
v_c^2(R)=\frac{GM_\infty}R+\frac{9\pi G\Sigma_0R_d^4}{R^3}+O(R^{-5})=\frac{GM_\infty}R\left[1+\frac92\left(\frac{R_d}R\right)^2+O\left((R_d/R)^4\right)\right].
$$
The positive leading correction shows that \b[the rotation curve approaches the Keplerian limit from above]. The finite-order large-radius expansion is sufficient here; an infinite moment expansion need not converge for a disk extending to infinity.
A useful special example is the <Mestel disk>, with <surface density> $\Sigma(s)=C/s$ for $s>0$, $C>0$. It has $M(R)=2\pi CR$. For every positive even $n$,
$$
I_n=\frac{CR^{n+1}}{n+1},\qquad J_n=\frac{CR^{-n}}n,\qquad(n+1)R^{-(n+1)}I_n-nR^nJ_n=C-C=0.
$$
All the nonspherical corrections cancel, giving
$$
\boxed{v_c^2=\frac{GM(R)}R=2\pi GC=v_M^2.}
$$
Thus the <Mestel disk> has a perfectly <flat galaxy rotation curve>.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2015/iii/paper-62-mestel-rotation.png]
{title=Flat rotation curve of a Mestel disk with surface density inversely proportional to radius}
{height=320}
This example has infinite total mass and a singular central <surface density>. The absolute <gravitational potential> cannot be set to zero at infinity, but the radial force exists as a limit of disks with increasing outer cutoff. Potential differences are $\Phi(R)-\Phi(R_0)=v_M^2\log(R/R_0)$. The earlier potential integral is therefore interpreted up to a radius-independent divergent constant for this example; the force calculation remains valid. Truncating the <Mestel disk> gives a more physical finite system but changes the exact flat curve near its edges.
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