What is the difference between an empty double integral and an empty surface integral?

Question

Consider the integration of $1$ over the surface of a sphere with radius $r$ which in this case would just be equal to $4\pi r^2$. The notation provided in my textbook looks like $$\iint_\Omega \,dA$$ where I am unsure of what $\Omega$ is supposed to denote. Given $S$ is the surface of the sphere, I am certain that the empty surface integral over $S$ is equal to $4\pi r^2$. I am confused as to why $$\iint_S \,dS = \iint_\Omega \,dA$$ for what I believe to be a problem with understanding what these notations mean.

EDIT: This is coming from a physics textbook which I have come to realize abuses notation quite a bit. Thus I think that they actually mean a surface integral when they write $$\iint_\Omega \,dA$$

Answer 1

When considering the integral $ \iint_{\Omega}dA $:

1. $\Omega$ is the symbol for the set of integration boundaries
2. $A$ is the symbol which denotes the set of variables to integrate.

In other words they are symbols or dummy variables and can be replaced by whichever letter or symbol you prefer, e.g., $S$.

This being said, in some books, there is a little distinction to be made between $dS$ and $dA$. Typically, $dS$ is an infinitesimal surface, while $dA$ is an infinitesimal variation of the parameters that describe the surface. As an example, consider your sphere $S$ with radius $r$; it can be parametrized as \begin{cases} x=r\cos\theta\sin\phi\\ y=r\sin\theta\sin\phi\quad\quad \theta,\phi\in [0,2\pi]\times [0,\pi]\\ z=r\cos\phi \end{cases} In this case, $dS\neq dA=d\theta d\phi$. The correct equality is $$dS=r^2\sin\phi dA=r^2\sin \phi d\theta d\phi $$

Answer 2

From what I understand, the general notation (for any problem) is $\iint_\Omega dA$ where A is the area of your shape and $\Omega$ stands for "The whole surface". In this case, the surface of the sphere is S and "The whole surface" is S as well. I do not see, however, how the notation would differ for a specific problem.