Line Integral Harmonization

Question

Is there a connection between line integrals over scalar fields and line integrals over vector fields? For example, do the pair $f(x, y)$ and $F(x, y)$ which stand in a potential function and gradient relationship have the same line integral for any specified curve, when $F(x, y)$ is drawn in the $xy$-plane of the $3$-space where $f(x, y)$ lives? I feel like I'm missing some geometric connection that justifies calling both by the name "line integral."

Answer 1

Line integrals over scalar and vector fields have different physical interpretations. A line integral over a scalar field is like finding the area between the graph of the curve and the xy plane. A line integral over a vector field is interpreted as work.

Answer 2

The scalar field $f(x,y)$ is a function $f:\mathbb{R}^2 \to \mathbb{R}$ and can be thinked as the ''height'' of a surface over the $xy$ plane. Given a line $\gamma(t)=(x(t),y(t))$ parametrized by $t\in[a,b]$, the line integral of $f$ on this line is defined as $$ \int_\gamma f(x,y)ds=\int_a^b f(x(t),y(t)\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2} dt $$ (note that here the $ds$ is a length along the path $\gamma$)

This is essentially an extension to a ''curved coordinate'' of the usual integral of a real function of a real variable and represents the area between the surface $z=f(x,y)$ and the line $\gamma$, as you can see intuitively in the nice animated figure in the wikipedia.

The integral of a vector field $\vec F(x,y)$ over $\gamma$ is an entirely different thing. In this case $\vec F(x,y)$ cannot be wieved as a surface, and the intuitive meaning of the integral is not an area but the work done by the force $\vec F$ on the path $\gamma$.

In this case the integral is defined as $$ \int_\gamma \vec F\cdot d\vec r = \int_\gamma \left(F_x dx+F_ydy\right)=\int_a^b \left(F_x(x(t),y(t))\frac{dx}{dt}+F_y(x(t),y(t))\frac{dy}{dt} \right)dt $$ (note that here $d\vec r$ is a vector, and $F_x dx+F_ydy$ is a differential 1-form)

If $\vec F(x,y)$ is the gradient of the scalar field $f(x,y)$ than this means that $F_x dx+F_ydy$ is an exact form and we have:

$$ \int_\gamma \vec F\cdot d\vec r=f(b)-f(a) $$ that is essentially a generalization of the fundamental theorem of calculus.