I was looking at Chow and Knopf's book, The Ricci Flow: An Introduction, and noticed that the normalized Ricci flow is defined as
$$ \frac{\partial}{\partial t}g = -2\text{Ric} + \frac{2}{n}rg \tag{1} $$ where $$ r = \frac{\int_M S \, d\mu}{\int_M d\mu}, $$ and $S$ is the scalar curvature.
Later on, the normalized Ricci flow for a 2-dimensional, closed Riemannian manifold is given as $$ \frac{\partial}{\partial t}g = (r-S)g \tag{2}\\ $$
I am now wondering how one goes from (1) to (2). Given the form of the traceless Ricci tensor $$ \overset{\circ}{\text{Ric}} = \text{Ric}-\frac{1}{n}Sg, $$ I am led to wonder if the manifold being closed somehow results in $\overset{\circ}{\text{Ric}}=0$ so that for the $n=2$ case, $\text{Ric}=\frac{1}{2}Sg$, which in turn gives (2).
Is this how (2) is obtained? If not, could someone please point me in the right direction? Thanks!