Intuition on why a field is not conservative

Question

I know the definition of a conservative vector field is that the integral $\int_{C} F \cdot dr$ is path independent, or it is zero if $C$ is a closed loop, or the curl of $F$ is zero (in a simply connected region), or there exists a scalar field $\phi$ such that $\nabla \phi = F$. But what is the intuition behind this example?
$F = (x^2 - xy)i + (y^2 - xy)j$. The curl is obviously non-zero, but what is the intuition, if any, for the result that this vector field is not conservative? Perhaps the vector field isn't """"evenly distributed""" (as in, the force on each point in the vector field has no symmetry to cancel out (in the case of a closed loop)?

Answer 1

To better understand such things you probably benefit from plotting the vector fields. You should be able to sketch them by hand, if you shorten the vectors down in scale. Wolfram Alpha will do that for you, but to learn from it you should manually/mentally verify its output - particularly the changes in the direction of the flow as well as the (relative) magnitude (WA scales the lengths way down). Here's what this field looks like.

As pointed out by Semiclassical, the part $x^2\vec{i}+y^2\vec{j}$ is actually conservative, so we may want to examine the non-conservative $-xy\vec{i}-xy\vec{j}$ more closely. Here it is.

Now is a good time to give that CAS-output a mental check. The vector $\vec{i}+\vec{j}$ points at Northeast. The multiplier $-xy$ changes signs from one quadrant to another. It is negative in quadrants one and three, and we see that the field points at Southwest there. It is positive in quadrants two and four, and the field points at Northeast as it should. I want to also observe that $xy=0$ on the coordinate axes, and we do see that the field vanishes there. This is largely coincidental, but will come to the fore in what follows.

To show that a vector field is not conservative it suffices to exhibit a closed path such that the corresponding path integral does not vanish. Here I take advantage of the fact that the second field vanishes on the axes. So I use a triangular path that starts from the origin, first goes straight to the point $(-1,0)$, then diagonally (with the flow!!) to the point $(0,1)$, and then returns to the origin. Behold.

Here it is totally clear that the horizontal and vertical line segments won't contribute anything to the path integral. From the picture it is equally clear that the diagonal line segment contributes something positive. Thus the path integral over the closed triangle is positive, and this field cannot be conservative.

If we look at the same path with the original field it looks like the following:

The details are a bit different, but the conclusion remains the same. From the picture you should get the feeling that this time the contributions of the vertical and the horizontal segments exactly cancel each other. This is hardly a surprise because those come only from the radially symmetric field $x^2\vec{i}+y^2\vec{j}$. This time the field is not perfectly aligned with the path along the diagonal line segment, but we still see that the path and the field are heading in the same general direction (Northeast) and their inner product is positive along this line segment all the time. As @epimorphic pointed out the above is not correct, because $f(x,y)=x^2\vec{i}+y^2\vec{j}$, while conservative with potential $\phi(x,y)=(x^3+y^3)/3$, is not radially symmetric. In fact, the first and the last line segment now both go against the flow. But, because $f$ is conservative, those two contributions are cancelled against the path integral of $f$ along the diagonal. If you squint real hard, you can actually see this in the picture. I was too optimistic/simple-minded.

Leaving the actual calculation of the path integrals related to this example to you. Please do a few of them using these images and observations as tests for correctness.