Doubt in area of an infinitesimally thin ring I want to find out area of a ring of an infinitesimal width in a derivation of electrostatics. So here's how my teacher explained it. Let the inner radius of thin ring be $r$ and outer be $r + dr$.
Area of thin ring
$$dA = π(r + dr)² - πr²
        = π(r² + (dr)² + 2r dr) - πr²
        = π(dr)² +π(2r dr)
        = π(2r dr).$$
My doubt is that why do we ignore the term with $(dr)²$. What determines to ignore a term or contain the term in the expression like $π(2r dr)$ is contained? If we say that $(dr)²$ is infinitesimally small then can't we say that the term with $dr$ is also infinitesimally small to be ignored. 
 A: Notice, $dr$ is infinitesimal small which tends to zero i.e. $dr\to 0$ but $dr\ne0$. Now the square of infinitesimal small length i.e. $(dr)^2$ is even much much smaller. 
We can say that $2\pi rdr$ is much larger than $(dr)^2$. Thus adding $(dr)^2$ to $2\pi rdr$ doesn't make any valuable difference. Therefore $(dr)^2$ is neglected while adding to $2\pi rdr$ although $2\pi rdr$ is very small.
$$2\pi rdr+(dr)^2\approx 2\pi rdr\quad \quad (\because \  \ \ (dr)^2\ll2\pi r dr)$$
Take a simple example of very small number say $10^{-15}$ if we add it to its square $(10^{-15})^2$ it makes no difference we can say that sum is approximately $10^{-15}$ i.e. $10^{-15}+(10^{-15})^2\approx10^{-15}$
A: Consider a dx which is small but finite.  If dx = 1E-6 m, then (dx)^2 = 1E-12 m.  Then add them.
A: We have π()²+π(2) = π(dr + 2). We need to keep dr somewhere, but as dr -> 0 it becomes more and more negligible compared to 2r so (dr + 2) can be replaced by 2r.
More generally, consider quantities x and x². If x = 1, x² = 1, if x = .5, x² = .25, if x = .25, x² = .06 etc.
As x gets smaller, the ratio of x² to x gets smaller (as it equals x). So in the limit x -> 0, we can ignore x² terms compared to x.
