Is the vector field $F(x,y)=(x+y,x^2)$ conservative?

Question

let's consider the vector field $F(x,y)=(x+y,x^2)$. It's defined over the whole $\mathbb{R}^2$. We can see immediately that $$\partial_x F_2 = \partial_y F_1$$

iff $x = \frac{1}{2}$

Since the domain (the plane) is simply connected, by Poincarè's lemma we know that if the form is closed, then it admits a potential. But in this case, it seems that it can be conservative only for $x=\frac{1}{2}$ and $y \in \mathbb{R}$. What does this mean? Actually, by integration of

$$\partial_x U(x,y)=x+y$$ $$\partial_y U(x,y)= x^2$$ I wasn't able to find a potential $U(x,y)$ as I found something whose gradient is not the vector field itself.

So I'm suspecting that this vector field is not conservative: but how can I deduce it from the fact that the differential form is closed only on the line $(\frac{1}{2},y)$?

Answer 1

You have to get $$2x = 2x$$.

Since you got: $$2x = 1$$ it is not conservative.

Here is a graph. Note that if you go up via the left or right and come back down the middle you wouldn't get a zero vector for a closed loop. desmos link