Hangup with vector integrals

tolove
Given,

$\sigma_{b} = \vec{P}\bullet\hat{n}$

Now, integrate both sides over a closed surface,

$\oint \sigma_{b} da = \oint (\vec{P}\bullet\hat{n}) da$

My math is fuzzy, and I don't really understand this next step.

$\oint \sigma_{b} da = \oint \vec{P} \bullet d\vec{a}$

What's going on here?

$d\vec{a}\equiv\vec{n}da$ is the ORIENTED area element, a vector in direction of the local normal vector, and with magnitude da.
It's simply a matter of notation. "da" is the "differential of area". "$$d\vec{a}$$" is defined as the unit normal vector times da. So $\vec{P}\cdot\vec{n}da= \vec{P}\cdot\ieft(\vec{n}da\right)= \vec{P}\cdot d\vec{a}$.