I am working on finding a surface integral solution for caculating an arbitrary point magnet field $\vec{B}$ distributed by a disc magnet via magnetic surface charge model. I used the assumption that surface charge is uniform. After I wrote out and simplified the surface integral with Weierstrass substitution, I obtained the following integrals that I need some hints to solve.
$ \int_{-\infty}^{\infty} \frac{\sqrt{1 + x^2}}{(a x^2 + b x + c)^{\frac{3}{2}}} dx$
$ \int_{-\infty}^{\infty} \frac{(1 + x^2)^{\frac{3}{2}}}{(a x^2 + b x + c)^{\frac{3}{2}}} dx $
where, $ b^2 -4 a c > 0 $, only when this arbitrary point is on the surface (which is always false), $ b^2 -4 a c = 0 $
$a, b, c $ can be treated as constants.
Hopefully, there are some integral tricks that I don't know. Please let me know. Thanks.
I tried to work towards elliptic integral as Claude Leibovici suggested. There are some updates on the integral.
It is a bit complicated. Let me state my variables.
- $\rho$ as the cylindrical coordinate distance of an arbitrary point, amplitude equals $\sqrt{x^2 + y^2}$, with unit direction vector $\vec{\rho}$.
- $\alpha$ as the angle between, x and $\rho$, equals $\arcsin{\frac{y}{\sqrt{x^2+y^2}}}$, with unit direction vector $\vec{\alpha}$.
- z axis remain the same in cylindrical coordinates.
- In the meanwhile, I used $\vec{x}, \vec{y}, \vec{z}$ for finding the $\vec{r_0} - \vec{r}$ vector.
- use $r, \theta$ and Cartesian coordinate to express the $\vec{r_0}$ vector on the disc magnet surface.
- $z_r:= z \pm h$, where $h$ is the half of the disc magnet thickness. For top surface, we use plus for $z_r$.
- $k:= \frac{\rho}{z_r}$ used later for simplification of the equation.
- $n^2:= \rho^2 + z_r^2 + R^2$, where $R$ is the radii of the disc magnet.
- $m^2 := (\rho - R)^2 + z_r^2 = n^2 - 2\rho R$
- $l^2:=(\rho + R)^2 + z_r^2 = n^2 + 2\rho R$
I used some of the approaches/tricks in this thread.
My approaches are:
- Firstly, write out the difference vector in Cartesian coordinates. Then, convert this vector to arbitrary point cylindrical coordinate used. This is for applying the lemma mentioned in above thread for periodical functions.
- Then, use symmetry to update the upper bound of the angle integral.
- Then, switch the order finish the radii integral first which is not difficult.
- Lastly, work towards the elliptic integral. However there are some annorying terms need to be taken care of.
I work on finding the magnetic field results separately, ignore the material constants leading the integral, we have:
$B_{\rho} \propto \int_0^R dr\int_0^\pi\frac{\rho - r\cos{\theta}}{(\rho^2 + z_r^2 + r^2 - 2\rho r\cos{\theta})^{3/2}} \vec{\rho}$
By solving the radii integral first I have the following results:
$B_{\rho} \propto \int_0^\pi d\theta[\frac{\rho R \sin^2{\theta} + z_r^2\cos{\theta}}{(z_r^2+\rho^2\sin^2{\theta})\sqrt{n^2 - 2\rho R\cos{\theta}}} - \frac{\cos{\theta}}{(1+k^2\sin^2{\theta})\sqrt{\rho^2 + z_r^2}}]\vec{\rho}$
rework the first term.
$B_{\rho} \propto \int_0^\pi d\theta[\frac{R}{\rho\sqrt{n^2 - 2\rho R\cos{\theta}}} + \frac{\rho \cos{\theta} - R}{\rho(1+k^2\sin^2{\theta})\sqrt{n^2 - 2\rho R\cos{\theta}}}- \frac{\cos{\theta}}{(1+k^2\sin^2{\theta})\sqrt{\rho^2 + z_r^2}}]\vec{\rho}$
Apply $\phi = 2 \theta$, we can find the first term as the elliptic integral of the first kind. And the third term is easy to find the integral. The difficulty for me, is the second term for integration.
I have tried two approaches. First one is to apply $\phi = 2 \theta$, however I fail to find the result in any kind of elliptic integral, because of the $\sin^2{2\phi}$ term. The other one is to use Weierstrass substitution instead.
The second terms is $\int_0^\infty dt\frac{[\rho + R +(R - \rho)t^2]\sqrt{1+t^2}}{[(1+t^2)^2+4k^2t^2]\sqrt{m^2 + l^2t^2}} \vec{\rho}$
Until now, this is my progression on failing to solve the integral on only $\vec{\rho}$ direction.
Update on Oct $3^{rd}$:
After revisit I found that I made a mistake without including the variable r in my double integrals. The original integral should be as follows:
$B_{\rho} \propto \int_0^R dr\int_0^\pi\frac{\rho r - r^2\cos{\theta}}{(\rho^2 + z_r^2 + r^2 - 2\rho r\cos{\theta})^{3/2}} \vec{\rho}$
After applying the same integration order, we can find the updated angle integrand in $\hat{\rho}$ direction again.
However the results are not simpler than before the mistaken version. I still need help with the integral.
