Feynman lectures, Volume I, chapter 13-4 While reading Feynman lectures on Physics, volume I, Chapter 13-4, I found following assumption, which I don't understand:

Then, since $r^2 = \rho^2 + a^2$, $\rho\,d\rho = r\,dr$. Therefore ... 

leading to equation 13.17.
Could somebody please explain why is it correct? However I think that it relates to derivation, but I don´t know how.
 A: $$r^2=\rho^2+a^2 \tag{1}$$
We know that $r$ and $\rho$ are variables while $a$ is a constant.
Differentiate both sides of $(1)$ wrt some arbitrary variable $u$ (because we knew that both $r$ and $\rho$ are variables,their change will depend on some other variables. Thus we can treat them as $r(u)$ and $\rho(u)$) and use the chain rule
$$2r\frac{\rm dr}{\rm du}=2\rho\frac{\rm d\rho}{\rm du}+0$$
$$r\;\frac{\rm dr}{\rm du}=\rho\;\frac{\rm d\rho}{\rm du}\tag{2}$$
Now recall the differential of a function $f(x,y)$ is defined as
$$\rm df=\frac{\partial f}{\partial x}\;\rm dx+\frac{\partial f}{\partial y}\;\rm dy$$
Therefore for $r$ its differential is
$$\rm dr=\frac{\rm dr}{\rm du}\;\rm du$$
Now multiply both sides by $r$ to get
$$r\;\rm dr=r\frac{\rm dr}{\rm du}\;\rm du$$
Now substitute $(2)$ to obtain
$$r\;\rm dr=\rho\frac{\rm d\rho}{du}\;\rm du$$
Now use the differential definition again to collapse the above into the required result
$$r\rm \;dr=\rho \rm\; d\rho$$
The chain rule is the reason why in physics we can use the short-cut in $(2)$ by formally "cancelling the $\rm du$ both sides to obtain the result directly"
A: Maybe a change of notation can help: if $f^2(x)=x^2+c$, then (differentiating both sides) $f(x)f'(x)=x$.
A: This is a mathematical question but still physically relevant.
The point here is that you need to take the differential on both sides with respect to the different variables.
You have the equation $r^2=\rho^2+a^2$ where $r$ and $\rho$ are both variables but $a$ is a constant (Assuming this from the result, since you have not provided enough context.)
You then take the differential of $r$ on the left side and the differential of $\rho$ on the right side and get the required result.
$$2r\rm dr=2\rho \rm d\rho \Rightarrow r\: \rm dr=\rho \: \rm d\rho$$
A: The result that the gravity emerging from a infinitly extended sheet of matter is independent of the distance ( it´s a uniform gravityfield) from the sheet can be easily seen if we consider the lines of gravity. They are all parallel, so there is no de- or increase with distance from the sheet. 
A problem arises if we consider (in the general relativity theory) the pace of time.
