# Integrating along a line of points; two different approaches, two different problems

I have an integration question, and I've taken two different approaches which both have serious flaws and don't agree; I'll outline them here and would be grateful for any guidance. I have an infinite line of little point sources. Each of these sources radiates a concentration outwards spherically, which is slowly consumed at a rate $a$ - it can diffuse out to a maximum distance of $r_{c}$ - I've solved this for ONE point source, and it works well. In this case, the concentration $p$ at a distance $r$ is simply

$p = F(r^2 + (2r_c^3)/r - 3r_c^2)$

where $F$ is simply a known constant.

Approach one - Integrating along line attempt

Now, Imagine a perpendicular distance $d$ away from the line, as in the diagram*. I can rewrite $p$ in terms of $z$ and $d$ like this;

$p = F(d^2 + z^2 + \frac{ (2r_c^3)}{\sqrt{d^2 + z^2}} - 3r_c^2)$

A point $d$ away can get contributions from only elements of $z$ between $+/- \sqrt{r_{c}^2 - d^2}$ - now, we want the TOTAL concentration due to all the line elements between these limits; if I (very naively) do the following, the definite integral between these limits is;

$P= \int p dz = 2F(\frac{\sqrt{r_c^2 - d^2}^3}{3} + 2r_c^3[log(\frac{r_c + \sqrt{r_c^2 - d^2}}{d})] + (d^2 - 3r_c^2 )(\sqrt{r_c^2 - d^2 }) )$

For comparison, I made a numerical solution and compared them; This solution gives the same form as the numerical version, but it is not enough; firstly, the integration introduces an extra dimension of metres (m) and the solution needs normalisation; to yield correct solution, $P$ has to be divided by something of the same order of magnitude as $r_{c}$ or $d$ but I cannot think of a physical reason why this should be, or a justification for doing so. Does anybody know where and how I have screwed up, or a better approach to take?

Approach Two - An angular attempt

In order to try and circumvent the unit / normalisation problem, I tried to recast the question in terms of angular contribution and integrate it that way; using same geometry as per the diagram*, and in this case, $p$ can be recast as

$p(\theta) = F(d^2\sec^2{\theta} + \frac{2r_c^3}{d}\cos{\theta} -3r_c^3$)

The maximum angle that can be subtended for any value of $d$ is $\theta_{max} = \arccos(d/r_c)$. Integrating this with respect to $\theta$ between the limits of $+/- \theta_{max}$ yields

$P(\theta) = 2F(d^2\tan(\theta_{max}) + \frac{2r_c^3}{d}\sin{\theta_{max}} -3r_c^3\theta_{max})$

This avoids the dimensional inconsistency, but does NOT yield a solution of the same form of the numerical simulation, nor the previous attempt - this leads me to conclude I'm not solving what I think I'm solving, but I'm unclear as to where I've messed up.

Anyone have any ideas ? I've also tried a partial differential approach but had no joy with that either....

*diagram couldn't be inserted into body as I'm a new user; hopefully this link will work -

http://i.stack.imgur.com/baLjx.png Also, if anyone is curious - for simulation purposes, $F = 1.081 x 10^8$ and $r_c = 1 x 10^{-4}$. The values of $d$ can be anywhere between $5x10^{-6}$ to $r_{c}$.

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Maybe it's only me, but terms like "point sources" (for a line), "radiates a concentration", "consumed" (the radiation, apparently), etc. are not usual mathematical terms and thus this question perhaps doesn't belong here. Nevertheless I won't vote to close it now until I hear other opinions. –  DonAntonio May 24 '13 at 13:07
You're correct - they're not mathematical terms, and stem from a very real physics problem - however, I'm pretty sure the problem is mathematical rather than physical and would welcome the advice of mathematicians in finding out where I screwed up - particularly if anyone can see why the two approaches yield such different results.... –  DRG May 24 '13 at 13:23