# Verification that vector field is not conservative

Reading through some notes on vector calculus, I came across the following exercise:

Let $F(x,y)= \left (\frac{-y}{\sqrt{x^2+y^2}},\frac{x}{\sqrt{x^2+y^2}}\right )$ be defined on $\mathbb{R}^2\backslash 0$. Show that $F$ is not conservative. I believe that I have a solution, I just want verification.

If $F$ were conservative then $\int_C F\cdot dr=0$ for every closed curve $C$. Therefore, we will try to compute the integral around the unit circle.

Now if we convert to polar coordinates then we have $x=\cos(\theta)$ and $y=\sin(\theta)$ and $r=\sqrt{x^2+y^2}$. Therefore the integral becomes $$\int_{0}^{2\pi} \frac{-\sin\theta}{r}(-\sin(\theta))\,\mathrm d\theta + \frac{\cos(\theta)}{r}\cos(\theta)\,\mathrm d\theta=\int_0^{2\pi}r^{-1}\,\mathrm d\theta=2\pi$$

Where the last $=$ is true because $r=1$. Therefore, we have found a simple closed curve which does not evaluate to zero and hence $F$ cannot be conservative.

Is my calculation and logic correct? Also, are there more slick ways to do this?

• Your work is correct. I can't think of a slicker way to do this. – kobe Dec 26 '16 at 20:00
• The only alternative is to prove the non existence of a potential function, but that's not going to be more elegant than what you did. – Paul Dec 26 '16 at 20:05
• How would one prove the non-existence of such a potential function? I'm asking because I would like to add this technique to my tool box.... for perhaps another problem. And thank you @kobe for verifying my work. – Clclstdnt Dec 26 '16 at 20:08
• @Clclstdnt the non-existence of a potential can be proven by calculating the curl of the vector field; if the curl if nonzero, then a potential cannot exist. – kobe Dec 26 '16 at 20:20
• Okay.... thanks for the input. – Clclstdnt Dec 26 '16 at 20:30

Your work is correct. I can't think of a slicker way to solve the problem, but another way to solve it is to show that $F$ has nonzero curl. The curl of $F$ will equal $\frac{1}{\sqrt{x^2 + y^2}}\bf{k}$, which is nonzero on $\Bbb R^2 \setminus \bf{0}$.
Note: I'm treating $F$ as the $3$-d vector field where the $z$-component is $0$.