Consider a moving particle m under the influence of gravity. What's the work done by gravity in moving m from A to B?
$$ \int_A^B \vec{F}\cdot \vec{dS}= \int_{rA}^{rB} \vec{F}\,dr $$
I din't understand this change of variables. Why the work can be calculated using dr instead of dS without considering the angle between dS and dr?
Thank you!
Imagine that the path connecting $A$ and $B$ is a curve $c$. At any point ${\bf r}$ the direction tangent to the curve $c$ can be written as the linear combination of the radial unitary vector $\hat{\bf r}$, and a perpendicular vector $\hat{\bf r}_\perp$ (in three dimensions, this is actually two vectors).
$$ d{\bf S} = dr\hat{\bf r} + dq \hat{\bf r}_\perp $$
Now, the gravitational force is radial so ${\bf F} = F{\bf \hat{r}}$, this means that
$$ {\bf F}\cdot d{\bf S} = (F{\bf \hat{r}})\cdot (dr\hat{\bf r} + dq \hat{\bf r}_\perp) = Fdr $$
In other words
$$ \int_A^B {\bf F}\cdot d{\bf S} = \int_{r_A}^{r^B}F dr $$