Q: calculate work done by force $F(x,y) = xy $ ${\bf i} + (y-x)\bf j$ over $c$ where $c$ is the unit circle.
So this is what I did: since the curve is the unit circle then $x=\cos t$ and $y=\sin t$ and $t \in [0,2\pi]$ Then $$dx=-\sin t\,dt \text{ and } dy=\cos t\,dt$$
But when I solve the line integral I don't get $0$ as an answer: $$\int _0^{2\pi} \cos t \sin t (-\sin t)+(\sin t- \cos t) \cos t\,dt$$ This integral is not zero. What am I doing wrong?? It's a closed path.....
If $F=M i +N j$ is a vector field over a simply connected and open set $D$, it is a conservative field if the first partial derivatives of $M,N$ are continuous in $D$ and $\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$. This is not true in your case. That means your vector field is not conservative. Therefore, it's not necessary that the integral on a closed path is zero.
Edit if $F=P i +Q j +R k,$ then checking whether the curl $ \Delta \times F =0$ or not is a faster way to verify whether the given field is conservative or not.