Find a closed path $C$ such that $\oint_C {\bf F} \cdot d{\bf r} \neq 0$, where $F = (y^2,x,0)$

Question

Consider the vector field $${\bf F}=(y^2,x,0).$$ Find a closed path $C$ such that $$\oint_C {\bf F} \cdot d{\bf r} \neq 0 .$$

My attempt:

I decided to try with the unit circle however the integral I get is very difficult. Would I be able to use Stoke's theorem to calculate this. If so, how would I do that?

I have calculated the $\operatorname{curl} {\bf F} = (0,0,1-2y).$

Answer 1

Hint You can use Stokes' Theorem, but since the third component of the vector field is $0$, we may as well restrict the vector field to the $xy$-plane and apply Green's Theorem: If we choose $C$ to be the appropriately oriented (say, piecewise $C^1$) boundary of a bounded region $D$ in the $xy$-plane, then $$\oint_C {\bf F} \cdot d{\bf r} = \iint_D \left(\frac{dQ}{dx} - \frac{dP}{dy}\right) dA .$$ Here $P$ and $Q$ are the $x$- and $y$- components of $\bf F$, and by definition the quantity in parentheses is the third component of the curl.

So for any region $D$ as above, the integral is $$\iint_D (1 - 2 y) \,dA .$$ We can see if, for example, we choose $D$ to be contained in the lower half-plane $\{y < 0\}$ and nonempty, then $$\oint_C {\bf F} \cdot d{\bf r} = \iint_D (1 - 2 y) \, dA \geq \iint_D dA > 0 .$$