differentialvolume
Having a spherical shell of internal radius of $r_0$ and thickness $\Delta r$, what is the volume element?
I started calculating and faced aome differential terms of degrees higher than 1. Should I ignore them? Would it be an approximation or this is the method to write elements?
The reason you need two integrals is because over the range $0$ to $4$ in $y$, the height of the shell is $2\sqrt y$ because it extends all the way across the $y$ axis. Over the range $4$ to $9$ in $y$, the height is $\sqrt y -(y-6)$. That changes the integrand, so split it into two integrals.
A sketch is below. The horizontal lines are what are revolved to make the shells. Note that the lower one extends from one side of the parabola to the other. The upper one extends from the line to the parabola. That is why the integral changes
differentials (in a way of standard analysis) are not very rigorous in mathematics
Standard or not has nothing to do with it.
Operations on differential notation are fully rigorous in standard mathematics, as a well-defined representation of properties of differentiable functions. If you want the differentials themselves to be mathematical objects that satisfy the rules of the differential notation, that can be done (and is done, in several different useful ways) in standard mathematics. Robinson NSA is only one option.
this approximation never really becomes equality, the smaller the angles the better it works, but still never equality!
That is true in all forms of infinitesimal reasoning except those with nilpotents (which can be defined by asserting that the order $k$ Taylor approximations are an equality, for a selected value of $k$). Calculations are always "up to higher order terms".
we actually have in mind that this quantity contains higher order terms, but they will vanish after we parametrise?
We have in mind that only a finite leading-order term is being calculated at the end, and the rest is an error term that is qualitatively smaller in the sense that it vanishes in the limit, or is infinitesimal or nilpotent or non-non-zero (which are some of the meanings of "qualitatively smaller" that are assigned in different formalizations of infinitesimal reasoning). For example, the product rule $d(FG) = (dF)G + F(dG)$ is true only up to second-order terms, which means ignoring $(dF)(dG)$ as something that will not contribute to the end result when calculating some quantity such as $d(FG)/dt$. If you do not want the limiting or infinitesimal answer, but a finite change in $FG$ over a finite interval of $t$, then the product rule is false and the higher order terms need to be accounted for. The calculus rules are only for handling the principal terms in asymptotic calculations that become "exact in the limit".
Best Answer
The volume of the shell is $4\pi r_0^2 \Delta r$. You can think at a solid with base surface the sphere and ''height'' $\Delta r$.
Or you can calculate the volume as: $$ \frac{4}{3}\pi (r_0+\Delta r)^3-\frac{4}{3}\pi r_0^3 $$
and ignore the terms that are of second or higher order in $\Delta r$.
Or you can use the volume element $dV=r^2\sin \varphi d\theta d \varphi dr$ and integrate with respect to $\varphi$ form $0$ to $\pi$ and with respect to $\theta$ from $0$ to $2\pi$.