# Is the work in line integrals equivalent to the work as an area under the curve?

A little explanation is needed here. Let me use two dimensions. In a line integral a curve is given and usually one parametrizes yet another curve and then substitutes this into the original equation to eliminate on of the variables in the path.

I assume the path that a "particle" follows through some given curve is doing work as the first given path is a force acting on the particle. Or you could think of it in reverse but no matter the outcome is that a particle is doing work through a path given by an equation.

In single variable calculus we usually think of work as area under a curve using two dimensions. In a line integral we are thinking of summing all the infinitesimal pieces as the parametrized particle moves through a path.

Are these equivalent concepts of work? I assume force is always a vector quantity regardless of how work is used.

## 2 Answers

Force is indeed always a vector quantity.

In single variable calculus we usually think of work as area under a curve using two dimensions. In a line integral we are thinking of summing all the infinitesimal pieces as the parametrized particle moves through a path.

Are these equivalent concepts of work?

Generally, in classical mechanics the work done by a particle, say in a gravitational field, over some path is indeed the area under "the curve" i.e.

$$W = \Delta U = - \int \vec{F} \cdot \vec{r} dr$$

where $$U$$ is the potential energy, and where $$\vec{r}$$ points to the particle's position as it moves along the path relative to an appropriately chosen origin. This is a line integral (more generally these are called "path integrals" since the path does not always have to be a line through physical space). You can parameterize the force $$\vec{F}$$ and the displacement $$\vec{r}$$ with a parameter as you would with a purely mathematical path integral, but in introductory physics (especially basic mechanics) it's usually easy enough to not need such a parameterization. Consider for example a ball falling in free fall toward the surface of the Earth. $$\vec{F}$$ is always downward, and anti-parallel to $$\vec{r}$$, which simplifies the integral quite a bit without parameterizing it.

The above explanation is rather mathematical. Conceptually, the "area under the curve" and the "accumulation of infinitesimal pieces" conspire to provide the same physical information because the forces you are dealing with are conservative and so the work-kinetic energy (potential energy) theorem holds true.

Please let me know if this answer is unsatisfying to you!

Yes, you're right. Short answer: see the picture. If the movement is a 2D one (dark green curve on the floor plane).

The force is a vector, but $$|\vec{F}|\cdot cos(\theta)$$ is a scalar. For every point in space (x,y), there is a scalar $$|\vec{F}|\cdot cos(\theta)$$ associated, which you can write as z(x,y).

The path integral is like summing the value of the function, $$|\vec{F}|\cdot cos(\theta)$$, times $$dx$$, which is the definition of $$dW$$ $$dW=\vec{F}\cdot d\vec{x}=|\vec{F}| dx\ cos(\theta)$$

So the "area under the curve along the path" is work, yes. Anyways, you don't have to visualize it like that, just do an integral, it's a pure calculation without further physical meaning regarding the graph.

• Nice graph!!!!! – N. Steinle Oct 2 '18 at 15:20
• wow...you have both answered the question so I can understand ...kudos ....both are acceptable answers but I can only pick one as the answers ... – Sedumjoy Oct 2 '18 at 20:15