If a line integral is 0, does the function have to be conservative?
Take the line integral $\oint_{C}^{}y^4dx+2xy^3dy$, which is equal to 0 over the bounded region C: $x^2+2y^2=2$. However, I was under the impression that if a line integral equaled 0, $\frac{\partial P}{\partial y}=\frac{\partial Q}{\partial x}$ would be true based on Clairaut's theorem. However, P and Q in this line integral, $\frac{\partial P}{\partial y}=4y^3$ and $\frac{\partial Q}{\partial x}=2y^3$, which aren't equal but are instead scalar multiples of each other.
Is this function $\vec F(x,y)=P\hat{\imath}+Q\hat{\jmath}$ conservative, or am I missing something?
If a vector field is conservative in a given domain then its line integral over every closed curve in that domain is zero. Just because the line integral is zero for the given curve does not necessarily mean that it is zero for every closed curve. For example, is it zero over a unit square with vertices at $(0, 0), (1, 0), (1, 1)$ and $(0, 1)$?
In general note that the integral of an odd function is zero if the region has symmetry about $x$ or $y$ axis. In this specific case, assuming $C$ is $x^2 + 2y^2 = 1$ then parametrizing as $r(t) = (\cos t, \frac{1}{\sqrt2} \sin t), 0 \leq t \leq 2\pi$ and $r'(t) = (- \sin t, \frac{1}{\sqrt2} \cos t)$.
So, $ \vec F \cdot r'(t) = -\dfrac{\sin^5t}{4} + \dfrac{\cos^2 t \sin^3 t}{2} = \dfrac{\sin^3 t}{2} - \dfrac{3 \sin^5 t}{4}$.
Integral of odd power of $\sin$ function over $(0, 2\pi)$ is always zero.