I'm trying to understand the geometric intuition behind the definition of the line integrals over vector fields. The definition is given below:
Definition: Let $\vec{F}$ be a continuous vector field defined on a smooth curve $\gamma$ given by a vector function $r(t)$. Then the line integral of $\vec{F}$ along $\gamma$ is
$$\int_{\gamma}\vec{F}\cdot d\vec{r}=\int_{\gamma} \vec{F}\cdot\vec{T} ds$$
Where $T$ is the unit tangent.
So the line integral of the vector field $\vec{F}$ along $\gamma$ is defined as line integral over a scalar field. The geometric interpretation of this one can be found here.
So using the geometric interpretation of line integrals over scalar fields, I'm trying to understand this one over vector fields.
In the definition above the scalar product $\vec{F}\cdot \vec{T}$ is a function $\alpha(x,y)$ which takes a point in the curve $\gamma$ and gives out a point with $\alpha(x,y)=|\vec{T}|$ (Since $T$ is a unit vector, $\vec{F}\cdot \vec{T}$ is the length of the projection of the vector $\vec{F}(x,y)$ over the tangent).
So using the geometric interpretation of the line integral over scalar fields, is the integral $\int_{\gamma}\vec{F}\cdot d\vec{r}$ the area below the curve $\alpha$? If yes, why is this geometric relevant?