Physical meaning behind the line integral of a vector field

Question

For the line integral of non-vector field functions, I know that you can kind of consider the line integral to be the area of a fence, with the base of the fence slinking along the curve of interest and the roof of the fence bordering on a function f(x,y). But for the line integral of a vector field, I'm not sure how I should visualize or conceptualize it. Any help would be awesome!

Answer 1

I think the easiest intuition comes from physics: if you have a particle in a force field $F$ and it moves along the path $C$, then $\int_C F \cdot dr$ is the change in the kinetic energy of the particle as it traverses the path. In particular this is big and positive if the particle's motion is aligned with the force field and it is big and negative if the particle's motion is opposed to the force field.

Answer 2

Your question is quite general and this is not the most proper place for general explanations. So just some points to pin up. Given a function $f(x,y)$ and a curve expressed in terms of a parameter $s$, then you can imagine as you do to take a vertical $z$ to the $x$-$y$ plane $$ z = f(x,y)\quad C:\left\{ \begin{gathered} x(s) \hfill \\ y(s) \hfill \\ \end{gathered} \right. $$ and figure out that the integral $$ \int_C {f(x,y)\,ds} = \int_C {z(x(s),y(s))\,ds} = \int_C {z(s)\,ds} $$ be the area of the "fence" raised along the curve and having heigth $z$. But it shall be made clear that the base-line along $C$ is measured according to the $s$ values, so it would be actually an area if $s$ corresponds to the euclidean length taken along the curve.

Now consider a vector $$ \mathbf{v} = \mathbf{v}(x,y) = \left( {\begin{array}{*{20}c} {f(x,y)} \\ {g(x,y)} \\ \end{array} } \right) $$ 1) The integral $$ \int_C {\mathbf{v}(x,y)\,ds} = \text{vector} = \int_C {\left( {\begin{array}{*{20}c} {f(x,y)} \\ {g(x,y)} \\ \end{array} } \right)\,ds} = \left( {\begin{array}{*{20}c} {\int_C {f(s)\,ds} } \\ {\int_C {g(s)\,ds} } \\ \end{array} } \right) $$ is a vector itself, and you can imagine it as grouping the areas of 2 fences.
You find this type of integral, e.g. when $\mathbf{v} $ is a pressure (force/meter) field and you want to find the total force exerted on a slender beam shaped as $C$.
2) You may also encounter a integral such as $$ \begin{gathered} \int_C {\mathbf{v}(x,y)\, \cdot d\mathbf{s}} = \text{scalar} = \int_C {\mathbf{v}(x,y)\, \cdot \mathbf{s}(x,y)ds} = \int_C {\mathbf{v}(s)\, \cdot \mathbf{s}(s)ds} = \hfill \\ = \int_C {f(x,y)ds_x + g(x,y)ds_y \,} = \int_C {v_{//} (x,y)\,ds} = \cdots \hfill \\ \end{gathered} $$ where $d\mathbf{s}$ is the vector tangential to the curve and of length $ds$, so also equal to $\mathbf{s} (x,y)=\mathbf{s}(s)$, unitary tangent vector, multiplied by $ds$. Since the dot product results in a scalar, depending on $(x,y)$, i.e. on $s$, then whole integral actually boils down to a integral of a scalar function, thus a scalar.
E.g., when $\mathbf{v} $ is a force field, the integral will give you the work made on a particle moving along the curve $C$.
Your fence will have an heigth which is the component of $\mathbf{v} $ tangential to the curve.
3) Another common integral is $$ \begin{gathered} \int_C {\mathbf{v}(x,y)\, \cdot d\mathbf{n}} = \text{scalar} = \int_C {f(x,y)dn_x + g(x,y)dn_y \,} = \hfill \\ = \int_C {f(x,y)ds_y - g(x,y)ds_x \,} = \int_C {v_ \bot \,(x,y)\,ds} = \int_C {v_ \bot \,(s)\,ds} = \cdots \hfill \\ \end{gathered} $$ where, this time, $d\mathbf{n}$ is a vector normal (outwards) to the curve. Apart from that, it is mathematically treated same as the precedent. The physical meaning is that, when $\mathbf{v} $ is e.g. the 2D velocity field of the particles in a fluid, the integral will give the (outwards) flux (= volume(area)/time) of the fluid across $C$.

4) Finally, another common integral is $$ \int_C {\mathbf{v}(x,y)\, \times d\mathbf{s}} = \text{vector} = \int_C {\mathbf{v}(x,y)\, \times \mathbf{s}(x,y)\,ds} = \int_C {\mathbf{v}(s)\, \times \mathbf{s}(s)\,ds} = \cdots $$ as shown, it reduces to the integral of a vector, so same as in case 1).
Physically, if e.g. $\mathbf{v}$ is a magnetic field, then the integral will give the total force exerted on a wire shaped as $C$, in which a unitary electric current is maintained.