$\nabla \times \bf{u} \neq 0$ but $\oint_{c} \bf{u} \cdot \textit{d}r \textit{=0}$? 
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? 
 A: The vector field isn't conservative; if it were, then $\frac{\partial f}{\partial z} = xyz^2$ would give an $\frac{xyz^3}{3}$ term in $f$, which would give terms with $z$ in the $x$ and $y$ components of $\vec u$, but those clearly don't exist.
A: The vector field is not conservative. Being conservative means that FOR EVERY closed curve $C$, we must have $\oint_C \mathbf{u} \cdot \mathbf{dr} = 0$. However, you only checked that for one closed curve (namely that ellipse), the integral vanishes. You didn't check it for every closed curve.
Actually, the very fact that $\nabla \times \mathbf{u} \neq 0$ shows that the vector field is not conservative, because we have the theorem which says
\begin{align}
\text{ conservative} \implies \text{curl vanishes},
\end{align}
so, if we reason contrapositively, we get the equivalent statement
\begin{align}
\text{curl doesn't vanish $\implies$ not conservative.}
\end{align}

Anyway, if you want an explicit counterexample, try to calculate
\begin{align}
\int_{\gamma}\mathbf{u} \cdot \mathbf{dr}
\end{align}
where $\gamma:[0,2\pi] \to \Bbb{R}^3$ is the smooth closed curve $\gamma(t) = (\cos t, 1, \sin t)$. You'll find that the integral is strictly positive.
