I need a check on the following exercise:
Let $S$ be the surface $$S= \{(x,y,z):y=x^2+z^2, y \in [0,1] \}$$ and $F(x,y,z)=(yx,x+z,yz)$ a vector field. Compute $$\oint_{\partial S} F \cdot dr$$ Is $F$ conservative?
The boundary $\partial S$ is parametrized by $\gamma(t)=(\cos(t),1,\sin(t))$, so $$\int_0^{2 \pi} F(\gamma(t)) \cdot \gamma'(t)dt =\int_0^{2 \pi} (\cos(t),\cos(t)+\sin(t),\sin(t)) \cdot (-\sin(t),0,\cos(t)) dt = 0$$
Unfortunately, this is not enough to say that it's conservative, because we need $\oint_{\gamma} F =0$ for every closed curve. I just noticed that since $\mathbb{R}^2$ is simply connected the fact that $\nabla \times F=\vec{0}$ is equivalent to the fact that $F$ is conservative. However,here $\nabla \times F = [z-1,0,1+x]$ is not zero, except for $z=1$ and $x=-1$. Can I say it's not conservative?
$curl F = \nabla \times F = \begin{vmatrix} i & j & k \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ yx & x+z & yz\end {vmatrix} = \, (z-1,0,1-x) \ne 0$
So yes it is not a conservative vector field.
Your working is correct and the way the vector field is set up, $x$ and $z$ components will cancel out over any circle parallel to $XZ$ plane. But that may not be the case if the circle was not parallel to $XZ$ plane or if it was another closed curve, say ellipse.
Also be careful in orientation of the curve to make sure it matches the orientation of the surface. In this case it does not matter as the integral is zero and the question does not state orientation for the surface explicitly.