Volume of spherical shell with $dr$ thickness Let's consider two spheres in the $(x,y,z)$ 3D-space, both centered in the origin: the inner with radius $r$ and the outer with radius $r + dr$. 
To compute the volume of the spherical shell between their two surfaces, one should simply proceed as follows:
$$dV = \frac{4}{3} \pi (r + dr)^3 - \frac{4}{3} \pi r^3 = \frac{4}{3} \pi (r^3 + 3 r^2 dr + 3 r dr^2 + dr^3 - r^3)$$
$$dV = \frac{4}{3} \pi (3 r^2 dr + 3 r dr^2 + dr^3)$$
but usually the last two terms between the brackets are neglected. I know $dr$ is infinitesimal, but can infinitesimals (raised to $n$-th power, $n>1$) be neglected and still obtain an exact result? How can this be justified?
 A: Let’s forget about “infinitesimal” quantities and just look at the difference that you’ve computed: $$V(r+h) = V(r)+4\pi r^2h+4\pi rh^2+\frac43\pi h^3.\tag{1}$$ Now, one way to define the differential of a function is as the best linear approximation to the way its value changes near a given point, i.e., $dV_r$ is a linear function such that $$V(r+h)=V(r)+dV_r[h]+\phi(h)$$ where the error $\phi(h)$ “goes to zero” faster than $h$. More formally, $\lim_{h\to0}{\phi(h)\over h}=0$, or, $\phi(h)=o(h)$. You can check for yourself that this is consistent with the usual limit-of-a-quotient definition of a derivative of a single-variable function. This condition ensures that the total accumulated error that you get by adding up these approximations to the shell volumes can be made arbitrarily small by making the shells thin enough.  
In practical terms, what the above means is that when computing a differential we can collect all terms that are higher than linear in the displacement $h$ into the error term $\phi(h)$. Equation (1) then becomes $$V(r+h)=V(r)+4\pi r^2h+\phi(h)$$ and so $$dV_r[h]=4\pi r^2h,$$ but $h$ is just $dr$ in disguise.  
The key idea to take away from this is that $dV$ is a linear approximation to the change in volume relative to a change in radius, so we can throw out the higher-order terms.
A: You seem to think neglecting something is inaccuracy.
OK, if you want exact result do not go in for $ dr$ at all, but use full $ \Delta V= 4 \pi (R^3-r^3)/3$ instead. 
When $R \approx r $ then the radial difference is $ dr$, higher order terms you gave in last line can be neglected with $ dr$ powers.
Using differentiation of volume $ V = 4 \pi r^3/3 $ w.r.t. radius,we get area (rate of change of volume is area)
$$ \frac {dV} {dr}= 4 \pi r^{2}, {dV}= 4 \pi r^2 \, dr  $$
So the procedure has automatically neglected higher powers without such calculation.
A: If $dr$ is defined as infinitessimal at the outset then presumably it is not the thickness  of a single shell that you are interested in but something else such as the volume of a sphere made up of an infinitely large number of infinitessimally thick shells.
One way to mentally rationalize the neglect of the minor terms in such a case is to examine the magnitude of their total summed over all the shells in the sphere. Then observe how the relative magnitude of that sum diminishes (relative to the magnitude of the sum of the major term) as you increase the number (and decrease the thickness) of the shells.
For example, assuming the volume of a sphere is given by $\frac{4\pi }{3}R^3$, we can derive an exact formula for the volume of any spherical shell as 
$$Vshell = \frac{4\pi }{3}(3r^2 h + \frac{h^3}{4})$$
where $h$ is shell thickness and $r$ is the radius to the middle of the shell.  This expression can be used to calculate the exact volume of a sphere composed of a small number of shells with finite thickness $h$.  
$$Vsphere = \frac{4\pi }{3}\sum_{r=1} ^{r=n}  (3 r^2 h + \frac{h^3}{4} )$$
The minor term $\frac{4\pi }{3}\frac{h^3}{4}$ cannot be neglected in this case. 
For example in a sphere of radius $R$ and $n=4$ shells, each of thickness $h=R/4$, the total sum of the minor term represents 1/64th of the shell volume. (In this example the shell radii are $\frac{h}{2}$, $\frac{3h}{2}$, $\frac{5h}{2}$ and $\frac{7h}{2}$).
In all cases, the total of the minor term summed over all the shells is $\frac{4\pi }{3}\frac{nh^3}{4}$ which (using $n=\frac{R}{h})$ reduces to $\frac{\pi}{3}Rh^2$.  It is clear that this minor total will become progressively smaller as the shell thickness h is reduced, tending to zero as $h$ tends to $\frac{1}{\infty}$.  Concurrently (as $h$ is reduced) the major term total will tend towards $\frac{4\pi }{3}R^3$ with the decrease in $h$ being compensated completely by the increase in the number of shells ($n$).
