Using infinitesimals to find the volume of a sphere/surface area of a sphere I've always thought of $dx$ at the end of an integral as a "full stop" or something to tell me what variable I'm integrating with respect to.
I looked up the derivation of the formula for volume of a sphere, and here, $dx$ is taken as an infinitesimally small change which is multiplied by the area of a disc($\pi r^2$) giving $\displaystyle V = 2\pi \int_0^r x^2 dy$ which is the sum of these infinitesimals. 

Now I'm really confused. Is it correct to think of it this way? Is there any other way to prove this result without using infinitesimals? 
Part two to my question:
Using this same logic of using infinitesimals, I tried to find the surface area of a sphere and looked at it as the sum of infinite rings.
$\displaystyle A =2 \int_0^r 2\pi x dy$
$=> \displaystyle A = 2 \int_0^r 2\pi \sqrt{r^2-y^2} dy$
But this is wrong. Why?

You have to think in a geometrical way. So the line integral is against a line element $dx$. Then the area integral is against an area element $da$, the volume integral - against a volume element $dv$ and so on.
The area element is an outer product of two vectors (in the sense of geometric algebra) $da =dx \wedge dy $, the volume -- of area elemnt and a line element
$dv = da  \wedge dz = dx \wedge dy \wedge dz $.
So the coordinate transformations will produce isometric elements  - i.e. scaled by the correct Jacobian of the transformation.
So in this case in polar coordinates  $|da | = |dx \wedge dy | = \rho \, d \rho \, d \phi  $.
So that 
$$
A = \int \int |da | = \int_{0}^{2 \pi} \, d \phi \int_{0}^{r} \rho \, d \rho = \pi r^2 
$$
