Integrating an non-infinitesimal quantity - why does this work? I've derived something that appears to work, but I am not sure exactly how it does so as it relies on integrating a non-infinitesimal amount. I'll first outline the discrete version: I have a number of discrete sources, of finite size along the Z-axis. They are not infinitesimal in their extent, having radius $r_{c} = 0.5$. Each one of these sources emits spherically, with the concentration a distance $r$ away from the source given by
$C(r) = \left(r^2 + \frac{2r_{n}^3}{r} - 3r_{n}^2\right)$ 
where $r_{n}$ is the maximum extent the source can radiate to so that if $r>r_{n}$ then $C(r) = 0$. This is illustrated in the image below, in a situation where $r_{n} = 5$. We're interested in the total contribution from all visible sources ($r < r_{n}$) at some point along the x-axis $d$. 
 
Now let's take the example $d = 3$ (and noting picture above isn't to scale!). From a discrete point of view, we'd work out how many sources the point can "see" (in this case a,b,c and d) and manually sum them up, giving 
$C_{Total} = \Sigma C(r) = C(r_a) + C(r_b) + C(r_c) + C(r_d)= 31.986$
I was curious whether I could make this an integral expression: re-writing $r$ in terms of $d$ and $z$ I rewrote $C(d,z)$ as 
$C(d,z) = \left(d^2 + z^2 + \frac{2r_{n}^2}{\sqrt{d^2 + z^2}} - 3r_{n}^2 \right)$
And here I made a jump in logic I'm dubious about; I wrote
$\frac{dc}{dz} = \frac{1}{2r_{o}}C(d,z)$ 
to capture the fact that source emits from discrete spheres of radius $r_{o}$ whose centres are a distance of $2r_{o}$ from each other. I didn't expect this to work, but integrating along the z-axis between 0 and some maximum extent $Z_{m}$ yields
$\int \frac{dC}{dz} = \frac{1}{2r_{o}}\left( (d^2 - 3r_{n}^2)Z_{m} + Z_{m}^3/3 + 2r_{n}^3 log\left(\frac{\sqrt{d^2 + Z_{m}^2} + Z_{m}}{d}\right)\right) $
If we note that $Z_{m} = \sqrt{r_{n}^2 - d^2}$ then we can solve this at $d = 3$ and we get an answer of 31.986 , remarkably close to the answer found discretely. 
Is this sheer coincidence or does the method actually work, and if so, how? I always thought that integration needed infinitesimal quantities, unless my simple division was enough to approximate or capture that? 
 A: Here is a more clear description. Putting together a picture is too tedious right now.
Your concentrations $C_i(r)$ don't actually depend on $d$; they depend on the distance of their center from $d$. In your picture, you have stacked them on an axis. But your quantity $\sum C_i$ will be the same regardless of how they're arranged, as long as $C_1$ sits $r_1$ from $d$, $C_2$ sits $r_2$ from $d$, and so on. In other words, you can take that picture you drew, pivot each of those circles around $d$, and obtain the same result.
Next, you introduce a variable $z$, which goes up your axis. And you re-write each concentration in terms of that variable $z$. However, look at the path $z$ takes -- it simply goes through the diameter of each $C$. This is because for any fixed $r_i$, $C_i$ is a function of its radius $r$, which is essentially a function of its diameter.
So if you re-arrange those circles, and draw new diameters through them, and "integrate" along those lines, you'll get the same result. It doesn't matter how they're placed. You've just happened to arrange them in such a way, and picked the diameters in such a way, that the diameters of each circle meet at their endpoints and form a straight line.
Now, finally, the integration trick works because you've first differentiated. Essentially, all you're doing is $\int_a^b \frac{dF}{dx}\ dx = F(b) - F(a)$ -- the fundamental theorem of calculus.
