I have seen answers to this question on this site already, though I still do not understand what line integrals and there results represent and would appreciate an oversimplified description. I have seen mention of work done in regards to line integrals over vector fields but I am a Mathematics student and have not studied Physics in over 4 years so I am somewhat clueless to how a lot of the maths I work with is used / interpreted in context to real life.
Any help would be greatly appreciated!
Relation to KE:
Work being done on an object = the change in the objects kinetic energy. This result can be derived from the definition of work
$$W= \int \vec{F} \cdot \vec{dr}$$ $$W= \int \vec{F} \cdot \vec{v}dt$$ $$W= \int m\frac{d\vec{v}}{dt} \cdot \vec{v}dt$$
$$W= \frac{1}{2}m|\vec{v}|^2$$
For a constant force, for an object moving in a straight line, Work is Force×displacement. $W = |\vec{F}||\vec{x}|$
This is only true for a constant force and an object moving in a straight line
Generally:
$$W= \int \vec{F} \cdot \vec{dr}$$
Situation:
Given I have some object that I know follows some specific PREDEFINED path, $\vec{r}(t)$. As it moves it is in the presence of some vector field $\vec{F}$
Differential line element: $\vec{r}(t) = x(t)\hat i +y(t)\hat j +z(t)\hat k$
$\frac{d\vec{r}(t)}{dt} = \frac{dx(t)}{dt}\hat i +\frac{dy(t)}{dt}\hat j +\frac{dz(t)}{dt}\hat k$
$d\vec{r}(t) = (\frac{dx(t)}{dt}\hat i +\frac{dy(t)}{dt}\hat j +\frac{dz(t)}{dt}\hat k) dt $
This represents an infinitesimal displacement that is traversed by the function $\vec{r}(t)$ in time dt
Vector field:
$\vec{F}$ is a vector field that has an assigned vector to every point in space usually in the form
$x(x,y,z)\hat i + y(x,y,z)\hat j + z(x,y,z)\hat k$
Dot product:
The quantity $\vec{F} \cdot \vec{dr}$ represents the infinitesimal amount of work being done on an object moving a distance of dr
The dot product between $\vec{F}$ and $\vec{dr}(t)$ is a measure of the alignment between these 2 vectors. If at a point in space a force vector points in the direction of the movement of the object, more work is done than if the force points perpendicular to the direction of motion.
The way I like to think about line integrals is that you are DEFINING a specific path that the object is taking. If at a point in time, the force is pointing perpendicular to the motion of an object, then no work can be done. Because if work WAS being done, then that force would cause a component of the displacement to be in the direction of the force( because it would be accelerating). So because we are defining a specific path, we are saying that IF an object WERE to be going on this path, then how much work would a force be doing on an object.
Another way to get a feel for why there's a dot product, is that that if a ball ORIGINALLY is moving in a straight line, and I exert a force perpendicular to the motion of the object, the force would cause an acceleration, changing the path of the ball to be curved, this dot product would "pick out" the displacement caused by the force(as it find the component in the direction of force), instead of adding up the original velocity of the ball.
Integral:
Because F and dr are changing as the ball moves we use an integral to sum up the contributions of all the $\vec{F} \cdot \vec{dr}$ elements as the line integral iterates in time to find the total amount of work done on the object.
ps:
When computing the line integral make sure that $\vec{F}$ is a function of the path $\vec{r}(t)$ in the form $ \vec{F}(\vec{r}(t))$ this is because we are evaluating the field at the position of the object at some time t
Line integral over a vector field doesn't differ a lot from a line integral over a scalar field. When you integrate over a scalar field you can visualize the integral:
$$\int_\mathcal{C} f(\mathbf{r}) ds$$
as the signed cross-sectional area between the curve and the graph of $f$. For line integrals over vector fields:
$$\int_\mathcal{C} f(\mathbf{r}) d\mathbf{r}$$
you can visualize the curve as the trajectory of the particle and the function $f$ as a force field. By integrating $f(\mathbf{r}) d\mathbf{r}$ you are summing the infinitesimal work $dw$. That is, you are calculating the work the force does on the particle.
You can imagine work as how much the force is pushing in the right direction (parallel to the displacement $d\mathbf{r}$ energy is added, work is positive, particle is accelarated). If the force is antiparallel to displacement the work is negative (energy is removed, particle is slowing down). In summary, you are just adding dot products between the vector field and the tangential to the curve vectors $d\mathbf{r}$.
Wikipedia also explain it very nice with some animated graphs. Have a look at the graph for vector fields where the "work" is translated as a typical integral under the curve.
