Gauss Law and surface integral

Question

Could somebody please explain to me the bottom line here. I don't understand how dS becomes r dtheta. I thought dS was supposed to be an outward pointing normal which is surely just r? I'm guessing theres something obvious im missing here.

Answer 1

$dS$ is an infinitesimal surface element. In this case it pertains to the surface of the cylinder which has been reduced to a circle. In particular it points in the $\hat{r}$ direction but its magnitude must represent infinitesimal surface area. In this case its magnitude is $rd\theta$ which is exactly the infinitesimal arc length of a circle with radius $r$.