I'm stuck on a question regarding work or perhaps better $\displaystyle \oint_{c} \vec{v} .d\vec{r}$

I know the vector $\displaystyle \vec{v} = -y\hat{i} + z\hat{j} + x\hat{k}$

I also know C is defined by $\displaystyle {z}^{2}+{x}^{2}=1$

Hence y = 0 and we're told the path starts at z = 0, x=1 then to x=0, z=1 back to the start (x=1, z=0)

I need to deduce the work or $\displaystyle \oint_{c} \vec{v} .d\vec{r}$ via stoke's law

- - - - - my attempt - - - -

Ok I keep coming out with zero for the answer, but im not certain it is correct as i'll explain in a minute:

$\displaystyle W = \iint_{s} (\nabla\times\vec{v}).d\vec{s}$

$\displaystyle d\vec{s} = (\vec{r}_x \times \vec{r}_y) dxdz$

However, for the matrix i keep obtaining zero.

i.e. here's the matrix code

\[ \left( \begin{array}{ccc}

\frac{\partial x}{\partial x} & \frac{\partial y}{\partial x} & \frac{\partial z}{\partial x} \\

\frac{\partial x}{\partial z} & \frac{\partial y}{\partial z} & \frac{\partial z}{\partial z} \end{array} \right)\]

Whereby there is no relationship between y with z and/or x. Resulting in N = 0 Hence W= 0

However my other method (which i need to know how to obtain the answer from stoke's rather than this) I obtained an answer of 2pi.

$\displaystyle W = \oint_{c} \vec{F}.d\vec{r} = \iint_{s} (\nabla \times \vec{v}).d\vec{s}$

Do dot product (as we're using f.dr here not stokes side) etc, notice y, dy = 0 xdz term left, use conversion of x = cos(u) z = sin(u)

Compute it all and i get w = pi.

- - - - - -- - -

I am not sure which one is correct, and why the other is wrong if one of them is correct, though my initial priority is to use stoke's theorem.

Any help is appreciated!