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I'm trying to find the potential function for $F(x,y)=<x^2-y;x-y>$

So I started by integrating the first component: $$\int (x^2-y) dx = \frac {x^3}{3}-yx+h(y)$$ From that, I took a partial derivative with respect to $y$ and got: $$f_y=-x+h'(y)$$

So I'm now trying to find what $h(y)$ is, so I set the previous derivative equal to the y-derivative I got from $F(x,y)$: $$-x+h'(y) = x-y$$ But, here's the problem: solving for $h'(y)$ gives me $h'(y)=2x-y$ and if I integrate this I get an extra $x$ that ruins my whole thing: $$h(y)=\int (2x-y)dy = 2xy- \frac 1 2 y^2$$ If I try to plug this into my potential function I'd get $\frac {x^3}{3}-yx+2xy- \frac 1 2 y^2$ which doesn't work. However, if I could just get rid of the minus sign in the $f_y$ I found, everything would be better. But on what basis can I get rid of it? Is it possible? Or I'm just missing something in my whole process?

Floella
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    It looks like your $F$ does not have a potential function. Did you copy down the wrong problem? – KittyL Dec 15 '16 at 20:51
  • No, I just checked and it's that $F$. But I didn't know that "no potential function" was possible. Is this process I followed enough to state that my $F$ doesn't have a potential function? Thanks :) – Floella Dec 15 '16 at 20:53
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    @Anarelle Yes, it is possible for a vector field to not be a potential field.

    The short way to justify your answer is to say that $F$ has nonzero curl. (A potential vector field is irrotational: https://en.wikipedia.org/wiki/Potential)

     – avs Dec 15 '16 at 20:53
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    $F$ has a potential function when it is conservative, i.e., it satisfies $\partial F_2/\partial x=\partial F_1/\partial y$. – KittyL Dec 15 '16 at 20:54
  • Sorry, I'm not familiar with this notation yet (just starting to learn about conservative fields). What does $\partial F_2$ stand for? – Floella Dec 15 '16 at 21:01
  • @Anarelle, partial derivative. – Martín-Blas Pérez Pinilla Dec 15 '16 at 21:05
  • Yes, sorry, I meant the $F_2$ and $F_1$ parts (never seen those subscript 1 and 2 before). Thank you. – Floella Dec 15 '16 at 21:06

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