So, i have the following vector field $$\vec{v}=<(1+z^2)y,y-x,x^2y^2>$$and the path is the circle of radius 1 around the origin in the xy plane oriented clockwise, so $$\vec{r}=<sin(t),cos(t),0>$$ and I want to verify stokes theorem so, i want $$\int_{0}^{2\pi}\vec{v}\cdot\vec{r'}dt=\int\int_{disk}\vec{\nabla}\times\vec{v}\cdot\vec{n}dA$$ where $\vec{n}=\vec{k}$ and A is the region inside the disk inside the circle. The problem probably arises here, because I set to calculate the first term, and it is $$\int_{0}^{2\pi}\vec{v}\cdot\vec{r'}dt=\int_{0}^{2\pi}<(1+z^2)y,y-x,x^2y^2>\cdot<cos(t),-sin(t),0>dt$$, where i have to change the variables $x=sin(t), y=cos(t), z=0$ (this is the part i'm least sure about), which yields$$\int_{0}^{2\pi}\vec{v}\cdot\vec{r'}dt=\int_{0}^{2\pi} cos^2t+sin(t)cos(t)+sin^2(t)=0$$ because they are periodical functions with average zero being integrated in their own period. On the other hand, $$\int\int_{disk}\vec{\nabla}\times\vec{v}\cdot\vec{n}dA=\int\int_{disk}-2-z^2dA$$ which is not equal to zero.
Really having a hard time figuring out what i am getting wrong, i have a test tomorrow and with everything i have studied, i should know this already :(
$\vec{r} = <\cos t, \sin t, 0>$ so $\vec{r'} = <-\sin t, \cos t, 0>$. Also, who said that integrals of periodic functions over their period are always zero? $\cos^2 t \ge 0$ so its integral is non-zero. I did not check the curl calculation.