= Solution
Construct a global <vector-field potential> rather than relying only on vanishing <curl>:
$$
\phi(x,y,z)=x^2z+\frac{y^2}{2},\qquad \nabla\phi=(2xz,y,x^2)=\mathbf F.
$$
Thus $\mathbf F$ is a <conservative vector field>. The endpoints of the given <parametric curve> are $\mathbf r(0)=(3,0,0)$ and $\mathbf r(\pi/2)=(0,2,1)$. The <fundamental theorem for line integrals> gives
$$
\boxed{\int_C\mathbf F\cdot d\mathbf r=\phi(0,2,1)-\phi(3,0,0)=2.}
$$
The intervening turns of the curve have no effect because this <line integral> is <path independent>.
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