Alright, I was trying to find the gravitational field due to a solid hemisphere (of mass $M$ and radius $R$) on an axial point, IN A NON-CONVENTIONAL MANNER.
Consider an elementary disk of thickness $dt=R d\theta$ (sorry! I couldn't include that in the diagram) at a distance $OB=p$ from the centre of the base of hemisphere. Let its radius be $PB=b$. As the density of the hemisphere is $\rho= \frac {3M}{2\pi R^3}$, the surface density of the disk is, $\sigma = \rho dt$.
Hence field due to this disk at an axial point $A$ $(OA=z)$, is,
$$
dg = -2\pi G \sigma \left[ 1- \frac {p+z}{\sqrt{b^2 + (p+z)^2}} \right]
= - \frac {3GM}{R^2} \left[ 1- \frac {R \sin \theta +z}{\sqrt{R^2 + z^2 +2Rz\sin \theta}} \right] d\theta
$$
And now this HORRENDOUS INTEGRAL:
$$
g(z)=\int dg= - \frac {3GM}{R^2} \int^{\pi/2}_0 \left[ 1- \frac {R \sin \theta +z}{\sqrt{R^2 + z^2 +2Rz\sin \theta}} \right] d\theta
$$
At least Maxima has been unable to solve it. So I guessed it would be a theorem or a special function. Any help or suggestion please. Thanks!!!
An attempt
Consider the general form of the integral: $$ \int_0^{\pi/2} \frac {A\sin \theta + B}{\sqrt{a\sin \theta +b}} d\theta $$ By substituting $t^2=a\sin \theta +b$ we get: $$ \frac {2}{a}\int_{\sqrt{b}}^{\sqrt{a+b}} \frac {At^2+(aB-Ab)}{\sqrt{(a^2-b^2)+2bt^2-t^4}} dt $$ Is there any way to get rid of the square root?
The first part of the integrand is trivial.
The second part is of the form $$\frac{a+b\sin\theta}{\sqrt{c+d\sin\theta}}.$$ Use the substitution $\sin\theta=\frac cd\tan^2\psi.$