# Work done by a constant vector field is 0?

We know that $$\oint \boldsymbol{F}\cdot d\boldsymbol{r}= \iint (\nabla \times \boldsymbol{F})\cdot d\boldsymbol{s}.$$ Now if $\boldsymbol{F}$ is a constant vector, then $\nabla \times \boldsymbol{F}=0$, this gives that $\oint \boldsymbol{F}\cdot d\boldsymbol{r}=0$. And $\oint \boldsymbol{F} \cdot d\boldsymbol{r}$ represents the work $$W=\int \boldsymbol{F} \cdot d\boldsymbol{r}$$ done by the vector field $\boldsymbol{F}$ along a curve. So this gives that work done by a constant vector field is $0$. How can it be possible?

This isn't surprising. Remember, $\oint$ means that you're integrating around a closed path. Without that requirement you can't use Stokes's Theorem to get a curl. All you've shown is that a constant vector does no work if you go in a big loop. This is the situation with, for instance, gravity close to Earth's surface. You throw a ball in the air, and when it returns to where it started it has the same amount of energy (it's just going the opposite direction).