I want to extend the area formula of the closed curve as follows :
First for some curve $C(t) = (x(t), y(t))$. I know by Green's theorem, then the area have the following form :
\begin{align}A = \iint_A dx \, dy = \frac{1}{2} \oint_C (-y\, dx + x\, dy) = \frac{1}{2} \oint_C \vec{F}\cdot d\vec{r} \end{align}
where $\vec{F}=(-y,x,0)$ and $d\vec{r} = (dx,dy,dz)$. Note one can treat $\vec{F} = \hat{n} \times \vec{r}$ where $\hat{n}$ is the unit normal vector and $\vec{r}=(x,y,z)$. [Since we are considering $xy$ plane $\hat{n} = (0,0,1)$]
I want to express this with $\gamma$ and Fernet-Serret formulas, i.e., $T,B,N$
From somewhere else, I found \begin{align} A = \frac{1}{2} \oint_{\gamma} \gamma \times T \cdot B ds \end{align} where $T$ is tangent vector and $B$ is binormal vector. Is this formula right? (For simple case as circle $(r\sin(t), r\cos(t))$ this formula seems work) And How one proves this?