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We are given that $\vec{F}=k\left<y,x,0\right>$, and asked whether $\vec{F}$ is a conservative force. If yes, we are asked to find $U(x,y,z)$ and then find $\vec{F}$ back from $U$ and show it matches the original form.

Given $\vec{\nabla} \times\vec{F}=\vec{0}$, force is conservative.

Therefore, $U(x,y,z)=-\int_{r_0}^r \vec{F} \cdot d\vec{r}=-\int_0^x kydx-\int_0^y kxdy=-2xyk.$ (Note that the $z$ component is $0$).

Such a potential function yields $\vec{F}=-\vec{\nabla}U=2k\left<y,x,0\right>.$

Something must be wrong or I must be missing something because I get an extra factor of 2 and I do not understand why.

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Well you integrated it the wrong way.
$$\int ky \, dx + \int kx \, dy = \int k \,d(xy) $$
X and y are not independent.

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  • $\begingroup$ I am not quite following... . How are those two equivalent? $\endgroup$
    – Ptheguy
    Feb 7, 2017 at 4:48
  • $\begingroup$ Xdy +ydx= d(xy) .. Refer differential equations for better understanding $\endgroup$ Feb 7, 2017 at 5:42
  • $\begingroup$ So you're invoking the product rule? $\endgroup$
    – Ptheguy
    Feb 7, 2017 at 6:05
  • $\begingroup$ It'd be nice if you could elaborate a bit. Like why aren't $x$ and $y$ independent of one another? How do you combine the two integrals? etc. $\endgroup$
    – Ptheguy
    Feb 7, 2017 at 6:06
  • $\begingroup$ Your doubt here is primarily on differential equation and its solution, look it up. If you have a doubt on that, then ask $\endgroup$ Feb 7, 2017 at 7:10

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