Once you have determined that the field is conservative (that the differential is exact), for example, by observing that \(\displaystyle \frac{\partial(2xy^2z)}{\partial y}= \frac{(\partial2x^2yz)}{\partial x}\), \(\displaystyle \frac{\partial(2xy^2z)}{\partial z}= \frac{\partial(x^2y^2)}{\partial x}\), and that \(\displaystyle \frac{\partial(2x^2yz)}{\partial z}= \frac{\partial(x^2y^2)}{\partial y}\), you don't really have to find an "anti-derivative"- the integral around any closed path is 0.

ban26ana, if the differential were not exact, any one dimensional path can be written in terms of a single parameter. Here, since the path is the circle with center at (0, 0), in the z= 1 plane, with radius 1, so we can use the "standard" parameterization for the unit circle, \(\displaystyle x= cos(\theta)\), \(\displaystyle y= sin(\theta)\) with \(\displaystyle z= 1\). Then \(\displaystyle dx= -sin(\theta)d\theta\), \(\displaystyle dy= cos(\theta)d\theta\) and \(\displaystyle dz= 0\). \(\displaystyle d\vec{s}= (-sin(\theta)\vec{i}+ cos(\theta)\vec{j})d\theta\). Integrating around the path in the "positive orientation" would be integrating from 0 to \(\displaystyle 2\pi\) and in the "negative orientation", from \(\displaystyle 2\pi\) to 0 (or, equivalently, from 0 to \(\displaystyle -2\pi\)).

Here, \(\displaystyle \vec{F}= 2xy^2z\vec{i}+ 2x^2yz\vec{j}+ x^2y^2\vec{k}\)\(\displaystyle = 2cos(\theta)sin^2(\theta)\vec{i}+ 2cos^2(\theta)sin(\theta)\vec{j}+ cos^2(\theta)sin^2(\theta)\vec{k}\) so \(\displaystyle \vec{F}\cdot d\vec{s}= (-2cos^2(\theta)sin^2(\theta)+ 2cos^2(\theta)sin^2(\theta))d\theta= 0\).

Since it is true, for any function G(x,y,z) with continuous second derivatives, that \(\displaystyle \nabla\times \nabla G= \vec{0}\), Chris L T521's observation that this \(\displaystyle \vec{F}(x,y,z)= \nabla G\) for a specific G, is the same as saying \(\displaystyle \nabla\times\vec{F}(x,y,z)= 0\) and so, by Stoke's theorem, \(\displaystyle \oint\vec{F}\cdot d\vec{s}= \int\int \nabla\times\vec{F}\cdot d\vec{S}= \int\int \vec{0}\cdot d\vec{S}= 0\).