The surface area of a ring: $\pi[(r+dr)^2 - r^2]$ or $2\pi r\,dr$?

Question

I know this may be really simple but here it is nonetheless. Let's say that I have a ring with a radius of $r$ and width of $dr$. I'm trying to find the surface $dS$ of the ring. Isn't it $dS = \pi[(r+dr)^2 - r^2]$? Because in my physics book it says that $dS = 2\pi rdr$. I've done the calculations and I find that $dS = \pi dr(2r+dr)$.

Answer 1

The formula $$dS = \pi [(r+dr)^2 - r^2] = 2 \pi r \,dr + \pi (dr)^2 \tag1$$ is correct. But your book takes one step further: when $dr$ is small, the term $\pi (dr^2)$ is much smaller than $2 \pi r \,dr$, which makes it dispensable in certain contexts. For example, if you are setting up an integral over the region by slicing it into circular rings of size $dr$, then the total contribution of the terms with $(dr)^2$ will tend to zero as $dr\to 0$.

It is also correct to write $$dS = 2 \pi r \,dr \tag2$$ except that now the meaning of $dS$ and $dr$ is not the same as in (1); they no longer refer to concrete geometric shapes, but rather describe the limiting behavior of a family of such shapes.