Closed loops on nonconservative vector field

Question

Consider the following (very simple) nonconservative vector field:

$$V = (y, -x)$$

sketched in the following figure:

Roughly speaking, the integral along the central red circle is obviously $\neq 0$.

But what about the lower red circle? Intuitively I'd say it has too a nonvanishing integral...some sources say it's so, but some other say it vanishes...who's right?

Answer 1

With

$V = (y, -x) = (V_x, V_y), \tag 1$

we have

$\nabla \times V = \dfrac{\partial V_y}{\partial x} - \dfrac{\partial V_x}{\partial y} = -1 - 1 = -2; \tag 2$

by Stokes' theorem, the integral of $V$ over any open region $\Omega$ with $\partial \Omega = C(t)$, a simple closed curve, is given by

$\displaystyle \int_C V \cdot \vec{ds} = \int_\Omega (\nabla \times V) \; dA = \int_\Omega (-2)\; dA = -2 \cdot \text{area}(\Omega) \ne 0; \tag 3$

since the lower red circle in the figure obviously bounds a region of non-zero area, we see that

$\displaystyle \int_C V \cdot \vec{ds} \ne 0 \tag 4$

where $C$ is taken to be this circle.