We had this example in class the other day, and the professor didn't not walk through how he obtained it.
Compute $\int_C \vec f \cdot d\vec r = \langle 4x^3y^2 - 2xy^3, 2x^4y - 3x^2y^2 + 4y^3\rangle$
The answer was written as $F(x,y) = x^4y^2 - x^2y^3 + y^4$
Maybe someone can confirm I did it the right way, here's my work:
$$P_y = 8x^3y - 6xy^2$$
$$Q_x = 8x^3y - 6xy^2$$
So it might be the gradient field, since they agree.
$$F_x = 4x^3y^2 - 2xy^3$$
$$\int F_x \ dx = x^4y^2 - x^2y^3 + g'(y)$$
$$F_y = 2x^4y - 3x^2y^2 + 4y^3$$
$$\int F_y \ dy = x^4y^2 - x^2y^3 + y^4$$
So $g'(y)$ is supposed to be $y^4$ since I guess $\int F_x \,dx$ and $\int F_y \,dy$ are supposed to be equal?
Then $F(x,y) = x^4y^2 - x^2y^3 + y^4$