What does the line integral of $\int\vec{s}\cdot d\vec{F}$ represent, where $s$ is the total displacement, and $dF$ is the force differential?

Asked Jul 18 '21 at 15:58

Active Jul 18 '21 at 16:46

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I recently started reading Feynman Lectures on Physics (Vol 1). I came across the formula for work done as,

Work done is $\int\vec{F}\cdot d\vec{s}$

My question is:

What does $\int\vec{s}\cdot d\vec{F}$ represent?

Does it even make sense to calculate the product of total displacement and small amount of force?

or does $\int\vec{F}\cdot d\vec{s}$ have the same meaning as $\int\vec{s}\cdot d\vec{F}$, they both represent the work done?

edited Jul 18 '21 at 16:46

Qmechanic

asked Jul 18 '21 at 15:58

cart3ch

The force is in principle a function $F(x,\dot{x},t)$, so you can integrate it along paths. I don't understand what sort of mathematical object you think $S$ is so that you could integrate it. – ACuriousMind Jul 18 '21 at 16:03
Since $y=x^3\implies ydx=x^3dx,,xdy=3x^3dx$, you can't just commute quantities either side of $d$. – J.G. Jul 18 '21 at 16:19
and $\vec s$ is not a vector, but the tanget $\vec{ds}$ is. – JEB Jul 18 '21 at 16:21
Possible duplicate: https://physics.stackexchange.com/q/382726/2451 – Qmechanic Jul 18 '21 at 16:48

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