Post History

What is $\mathcal{O}$ in integration of potential?

I saw following equation in Griffiths EM 
$$V(r)=-\int_\mathcal{O}^r \vec E \cdot d\vec l$$

While the surface was closed but not symmetrical circular. At first I was thinking $\mathcal O$ was representing closed surface and $r$ was representing radius of that circle. But when I noticed the circle isn't symmetrical I claimed that my thought was wrong. The circle was started at point A and end at point B, After looking at that non-symmetrical circle I thought Griffiths will write an integrand like this : $$V(r)=-\int_A^B \vec E \cdot d\vec l$$ but his method confused me.