The integral is path-independent if we can find a scalar function <em>f</em> such that grad(<em>f</em> ) = <em>A</em>. This requires
Take the first PDE and integrate both sides with respect to <em>x</em> to get
where <em>g</em> is assumed to be a function of <em>y</em> alone. Differentiating this with respect to <em>x</em> gives
which would mean <em>g</em> is *not* a function of only <em>y</em>, but also <em>x</em>, contradicting our assumption. So the integral is path-dependent.
Parameterize the unit circle (call it <em>C</em>) by the vector function,
with <em>t</em> between 0 and 2π.
Note that this parameterization takes <em>C</em> to have counter-clockwise orientation; if we compute the line integral of <em>A</em> over <em>C</em>, we can multiply the result by -1 to get the value of the integral in the opposite, clockwise direction.
Then
and the (counter-clockwise) integral over <em>C</em> is
and so the integral in the direction we want is -2π.
By the way, that the integral doesn't have a value of 0 is more evidence of the fact that the integral is path-dependent.