What does it mean for an integral to be independent of a path? Take the path $x^4-6xy^3=4y^2$ for
$$
\int_{(0, 0)}^{(2,1)} \left(10x^4-2xy^3\right)\,{dx}-3x^2y^2\, dy
$$
My textbook says the integral is independent of the path since
$$
\frac{\partial}{\partial y}\left(10x^4-2xy^3\right)
= \frac{\partial}{\partial x}\left(3x^2y^2\right)\,{dy}.
$$
But I've no idea what these partial derivatives (from Green's theorem) being equal has to do with it.
Best Answer
An integral is path independent if it only depends on the starting and finishing points.
Conservative fields $F$ are path independent, as they can be written as a gradient of a function $f$: $$ F=\nabla f $$ Consequently, on any curve $C=\{ r(t)\;|\; t\in [a,b]\}$, by the fundamental theorem of calculus $$ \int_C F\, dr = \int_C \nabla f \, dr = f(r(b))-f(r(a)), $$ in other words the integral only depends on $r(b)$ and $r(a)$: it is path independent.
Now, to link this with your question, when the partial derivatives from Green theorem are equal, the field is conservative (provided it is defined on a simply connected domain). Therefore, in your case, the path integral is path independent indeed.