Consider the vector field $\vec{u}=(xy^2,x^2y,xyz^2)$
The curl of the vector field is $$\nabla \times\vec{u}=(xz^2,-yz^2,0)$$ Consider the line integral of $\vec{u}$ around the ellipse $C$ $x^2+4y^2=1, z=-1$.
With $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, a=1, b=\frac{1}{2}$, the parameterisation gives $$\vec{r}=(x,y,z)=(cos\theta,\frac{1}{2}sin\theta,-1)$$ $$d\vec{r}=(-sin\theta,\frac{1}{2}cos\theta,0))$$
$$\vec{u}=(\frac{1}{4}cos\theta sin^2\theta,\frac{1}{2}cos^2\theta sin\theta,\frac{1}{2}sin\theta cos\theta) $$ $$\oint_{C} \vec{u} \cdot d\vec{r}=\frac{1}{4}\int^{2\pi}_0sin\theta cos\theta(cos^2\theta -sin^2\theta)d\theta=0$$ (I got the same result by working in cartesian cooridnates without parameterizing the curve)
But this does not make sense because $$\nabla\times \vec{u}=\lim_{\delta S \to 0}\frac{1}{\delta S}\oint_{\delta C}\vec{u} \cdot d\vec{r}$$ so if a vector field is conservative, its curl should be zero.
Can someone please explain where my conceptual errors lie?