# 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).
It is possible for conservative forces. Closed integral of a conservative force (say Electric force) is zero. 