The potential and the intensity of the gravitational field in the axis of a circular plate Calculate the potential and the intensity of the gravitational field at a distance $x> 0$ in the axis of thin homogeneous circular plate of radius $a$ and mass $M$.

Could anybody describe how to calculate this? Slowly and in detail. I'm helpless.
Answer is: potencial $\phi = - \frac {2 \kappa M}{a^2}(\sqrt{a^2+x^2}-x)$
and intesity $K = \frac {2 \kappa M}{a^2} \left( \frac{x}{\sqrt{a^2+x^2}}-1 \right)$
 A: First let us calculate the potential for a ring of radius $a$ at a distance $x$ from the center along the axis
Potential due to an infinitesimal mass element $dm$ will be $$\frac{-Gdm}{\sqrt{a^2+x^2}}$$
Potential due to the ring is then $$\int{\frac{-Gdm}{\sqrt{a^2+x^2}}}=\frac{-G}{\sqrt{a^2+x^2}}\int{dm}=\frac{-Gm}{\sqrt{a^2+x^2}}$$
Since $G, a, x$ are constant
Now let us break the disc into infinitesimal rings of mass $dm=2\pi rdr\frac{M}{\pi a^2} (=area * density)$
The potential due to a ring of radius $r$ and mass $dm$ as given above is $$\frac{-Gdm}{\sqrt{r^2+x^2}}=\frac{-2GMrdr}{a^2\sqrt{r^2+x^2}}$$
Integrating this from $0$ to $a$
$$\int{\frac{-2GMrdr}{a^2\sqrt{r^2+x^2}}}$$
$$=\frac{-GM}{a^2}\int{\frac{2rdr}{\sqrt{r^2+x^2}}}$$
putting $t^2=r^2+a^2$ and $2rdr=2tdt$
$$=\frac{-GM}{a^2}\int{\frac{2tdt}{\sqrt{t^2}}}$$
$$=\frac{-GM}{a^2}[2t]^{\sqrt{a^2+x^2}}_{x}$$
$$=\frac{-2GM}{a^2}({\sqrt{a^2+x^2}}-{x})$$
For intensity, it can be seen by symmetry that it is along the axis thus we work only with axial components
So, for ring
$$\int\frac{-Gdmcos\theta}{a^2+x^2}$$
where $\theta$ is half of the angle subtented by the point on the ring
$$cos\theta=\frac{x}{\sqrt{a^2+x^2}}$$
$$K=\int\frac{-Gxdm}{(a^2+x^2)^{3/2}}=\frac{-Gxm}{(a^2+x^2)^{3/2}}$$
For a disc, based on the same reasoning as in potential, it is
$$K=\int\frac{-Gxdm}{(r^2+x^2)^{3/2}}$$
$$=\int\frac{-2GMxrdr}{a^2(r^2+x^2)^{3/2}}$$
$$=\int\frac{-2GMxtdt}{a^2(t^2)^{3/2}}$$
$$=\int\frac{-2GMxdt}{a^2t^2}$$
$$=\frac{-2GMx}{a^2}[\frac{-1}{t}]^{\sqrt{a^2+x^2}}_x$$
$$K=\frac {2 G M}{a^2} \left( \frac{x}{\sqrt{a^2+x^2}}-1 \right)$$
A: Let's break the disc into small rings,
Here the mass in the disc is at a same distance $\sqrt{x^2+r^2}$ from the axis point A.
So, potential $$\mathrm dE=-\frac{G\,\mathrm dm}{\sqrt{x^2+r^2}}$$
Where $\mathrm dm$= mass of the ring. So, $$\mathrm dm=2\pi r\,\mathrm dr \times \frac M{\pi R^2}$$
Also $r=x\tan\phi$, so $\mathrm dr=x\sec^2\phi\,\mathrm d\phi$
So, the potential is $$\mathrm dP=-\frac{GM2\pi x\tan\phi x\sec^2\phi\,\mathrm  d\phi}{\left(x\sqrt{1+\tan^2\phi}\right)\times\pi R^2}=-\frac{2 GMx}{R^2}\times\tan\phi\sec\phi\,\mathrm d\phi$$
So, $$P=-\dfrac{2GMx}{R^2}\times\int\limits_0^{\tan^{-1}R/x} \tan\phi\sec\phi\,\mathrm d\phi$$
You can integrate it on your own.
You can proceed on similar basis for electric field. But before integrating you'll have to take components of field along axis and perpendicular because it's a vector and cannot be added directly.
You'll get field integrand as $$\mathrm dE_x=\frac{2GM}{R^2}\times \int\limits_0^{\tan^{-1}R/x}\frac{\tan\phi\sec\phi}{\sec^2\phi}\,\mathrm d\phi$$
Whereas along $Y$ you'll get no field by symmetric argument.
