The integrals:
$$
\oint \frac{r\,dr\,d\phi}{\left(L^2+r^2+h^2+2Lr\cos\phi\right)^{3/2}}\\
\oint \frac{dx\,dy}{\left((L+x)^2+y^2+h^2\right)^{3/2}}
$$
If we have a point charge at the origin and we want to find the flux through a disk of radius $R$ which is located at $(x,0,z)$ and lies parallel to the $x-y$ plane we will have to do a tricky integral. We can write down the exact $E.dA$ or we can try and find the solid angle (turns out it is like finding the flux through an ellipse at $(0,0,z)$).
What I'm looking for is the answer to that integral or the solid angle. I thought the integral wasn't doable. But recently I came across a physical way of finding it using greens reciprocity theorem. If you can find it from that, then the integral should be doable as well!
To be clear: this is entirely a math problem. We have an inverse square vector field. We want its flux through a disk.
for example take the (x=0) situation.we can simply use gauss law and just find the solid angle to obtain the flux.ot write out the integral by using r and theta as variables.but even at this situation doing it with dxdy integral is tough.
Best Answer
The solid angle can be expressed in terms of complete elliptic integrals of first and third kind.
I believe this is first derived by F. Paxton in 1959.
A copy of the article can be found here.
Below is my attempt to derive the same result in an alternate manner.
The integral is indeed trickier than I thought.
To avoid confusion with the usage of $x,y,z$ as coordinates, let us change the problem and assume the circular disk $C$ we are dealing with lies in the plane $z = b$, centered at $(a,0,b)$ with radius $s$. We will assume $a, b > 0$ and $a \ne s$.
In terms of ordinary spherical polar coordinates $(r,\theta,\phi)$, points on the plane $z = b$ can be parametrized as:
$$(x,y,z) = (b\tan\theta \cos\phi,\,b\tan\theta\sin\phi,\, b)$$ Let $\rho = \sqrt{x^2+y^2} = b\tan\theta$, the "element" for integrating the solid angle is given by:
$$d\Omega = \sin\theta d\theta \wedge d\phi =\frac{\tan\theta d\tan\theta}{\sqrt{1 + \tan\theta^2}^3} \wedge d\phi = \frac{b \rho d\rho \wedge d\phi}{\sqrt{\rho^2+b^2}^3} = \frac{b\;dx \wedge dy}{\sqrt{\rho^2+b^2}^3} $$ Introduce complex coorindates $\eta = x + iy$ and $\bar{\eta} = x - iy$, the solid angle extended by $C$ can be rewritten as:
$$\begin{align} \Omega_{C} = & \int_{C} d\Omega = \frac{b}{2i}\int_{C} \frac{d\bar{\eta} \wedge d\eta}{\sqrt{\eta\bar{\eta}+b^2}^3} =\color{blue}{^{[1]}} \frac{b}{i} \int_C d\left(\frac{1}{\eta}\left( \frac{1}{b} - \frac{1}{\sqrt{\eta\bar{\eta}+b^2}}\right)\right) \wedge d\eta\\ = & \frac{b}{i} \int_{\partial C} \left(\frac{1}{b} - \frac{1}{\sqrt{\eta\bar{\eta} + b^2}}\right) \frac{d\eta}{\eta} = 2\pi \delta_{C} + ib \int_{\partial C} \frac{d\eta}{\eta \sqrt{\eta\bar{\eta} + b^2}} \end{align}$$ where $\delta_{C} = 1 \text{ or } 0$ depends on whether $s > a$ or $< a$. In other words, whether $\partial C$ contains $0$ or not.
On $\partial C$, we can parametrize $\eta$ as $s e^{it} + a$ and $\bar{\eta}$ as $s e^{-it} + a$. Substitute this into above expression, we obtain:
$$\begin{align} &\Omega_C - 2\pi \delta_{C}\\ = &ib \int_{-\pi}^{\pi} \frac{ dt }{\sqrt{ s^2 + a^2 + b^2 + 2sa\cos t}}\frac{ is e^{it}}{s e^{it} + a }\\ = & -b \int_{0}^{\pi} \frac{ dt }{\sqrt{ s^2 + a^2 + b^2 + 2sa\cos t}} \left( \frac{s e^{it}}{s e^{it} + a } + \frac{s e^{-it}}{s e^{-it} + a } \right)\\ = & -b \int_{0}^{\pi} \frac{ dt }{\sqrt{ s^2 + a^2 + b^2 + 2sa\cos t}} \frac{2s(s + a\cos t)}{s^2 + a^2 + 2sa\cos t}\\ = & -b \int_{0}^{\pi} \frac{ dt }{\sqrt{ s^2 + a^2 + b^2 + 2sa\cos t}} \left( 1 + \frac{s^2 - a^2}{s^2 + a^2 + 2sa\cos t} \right)\\ = & -2b \int_{0}^{\pi/2} \frac{ dt }{\sqrt{ s^2 + a^2 + b^2 + 2sa\cos(2t)}} \left( 1 + \frac{s^2 - a^2}{s^2 + a^2 + 2sa\cos(2t)} \right)\\ = & -2b \int_{0}^{\pi/2} \frac{ dt }{\sqrt{ (s+a)^2 + b^2 - 4sa\sin^2(t)}} \left( 1 + \frac{s^2 - a^2}{(s+a)^2 - 4sa\sin^2(t)} \right) \end{align}$$ The last integral can be expressed in terms of the complete elliptic integral of first and third kind: $$ K(k) = \int_{0}^{\frac{\pi}{2}} \frac{dt}{\sqrt{1-k^2\sin^{2}t}} \quad\text{ and }\quad \Pi(n,k) = \int_{0}^{\frac{\pi}{2}} \frac{dt}{(1-n\sin^{2}t)\sqrt{1-k^2\sin^{2}t}} $$
The final results are: $$\Omega_{C} = 2\pi\delta_C -\frac{2b}{\sqrt{(s+a)^2+b^2}} \left(\;K(k) + \left(\frac{s-a}{s+a}\right) \Pi(n,k)\;\right) $$ where $\displaystyle \;\;k = \sqrt{\frac{4sa}{(s+a)^2+b^2}} \quad\text{ and }\quad n = \frac{4sa}{(s+a)^2} $.
Notes
$\color{blue}{[1]}$ When $C$ contains $(0,0,b)$, we need the $\frac{1}{b}$ term. It regularize the 1-form at $\eta = 0$ and make Stroke's theorem continue to work.