A little nomenclature confusion here. Let us imagine we have a work function F=(x) = 3x + 1 and we integrate this work function as a function of x over the integral 1 to 5, the answer = 32, Before I get to my question. Allow me to clarify this first.
When I integrate the integral breaks up into two parts. The first part retains the dimensions of area ( width times length ) however the second has only one dimension. How do we reconcile this ? The units must always agree, no?
Now for the question I posted. A line integral is expressed as a dot product where each of the force components must be parameterized BUT work can also be defined WITHOUT an integral by a simple dot product assuming the force is constant. So what is it that I am integrating when I construct a line integral? This cannot be the same concept as work function but I should still end up with the same dimensions that are used by work in all three styles! Obviously I have missed the boat here on a few key points.
Perhaps the line integral represents the work done by an object as it meanders through some path. But still the units should agree and there is the rub for me.
You have a force $\vec{F}$, which is a vector field (for simplicity assume that $\vec{F}:\mathbb{R}^n\to \mathbb{R}^n$ smooth). If you consider the motion of a point particle describing a curve $\gamma:I\to \mathbb{R}^n $. Then the work of $\vec{F}$ along $\gamma$ is defined as the line integral $$W(\vec F,\gamma):=\int_\gamma \langle \vec{F},\cdot\rangle = \int_I \langle \vec F\circ \gamma, \dot{\gamma}\rangle dt$$ Here $\langle,\rangle:\mathbb{R}^n\times \mathbb{R}^n\to \mathbb{R}$ is the standard inner product of $\mathbb{R}$.
Now suppose that $\vec F$ is constant along a segment $\gamma:[0,1]\to \mathbb{R}^n$ $$\gamma(t) = x_0 + t \vec{v},$$ $$\vec F (\gamma(t))= \vec F_0$$ $x_0,v \in \mathbb{R}^n.$ It follows that $$W(\vec F,\gamma) = \int_I \langle \vec F\circ \gamma, \dot{\gamma}\rangle dt = \int_0^1 \langle \vec F_0, \vec v\rangle dt = \langle \vec F_0, \vec v\rangle$$