Since it is just a sort of discussion, I will just give the formula and condition without proving them or leaving examples.

General:

　　　　$\int_{C}\vec{F}\cdot \mathrm{d}\vec{r} = \int_{C}M\mathrm{d}x+N\mathrm{d}y$, in which $\vec{F} = $

　　　　　　Method: Express $x$ and $y$ in a single variable (OR means parameterization).

　　　　Condition:

　　　　　　$curl(\vec{F}) = 0$ and $\vec{F}$ is defined in a simple-connected region, in which $curl(\vec{F}) = N_{x} - M_{y}$ if $\vec{F} = $ and $curl(\vec{F}) = \nabla\times\vec{F}$(namely

\begin{vmatrix}

\hat{i} & \hat{j} & \hat{k} \\

\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\

P & Q & R

\end{vmatrix}

) if $\vec{F} = $

　　　　　　then $vec{F} = \nabla f$, or $vec{F}$ is the partial derivative vector of some vector field.

　　　　The method of finding the potential:

　　　　　　Method 1. Do line integral. Integral along the x-axis and y-axis and z-axis, if they exist. (Using path-independence)

　　　　　　Method 2. Integral one component of $\vec{F}$ and then differential it over another variable and compare. (…)

　　　　in the plane:

　　　　　　$\hat{n} = \hat{T}$ rotated 90 degrees clockwise $=<\mathrm{d}y,-\mathrm{d}x>$

　　　　　　$\int_{C}\vec{F}\cdot\hat{n}\mathrm{d}s = \int_{C}P\mathrm{d}y-Q\mathrm{d}x$, in which $\vec{F} = $

　　　　in the space(or specifically, surface):

　　　　　　$\iint_{S}\vec{F}\cdot\hat{n}\mathrm{d}S = \iint_{S}\vec{F}\cdot(<-f_{x},-f_{y},1>\mathrm{d}x\mathrm{d}y)$, if we use $z = f(x,y)$ to describe the surface.

　　　　　　　　　　　　　　　　　　　　　　　　$=\iint_{S}\vec{F}\cdot(\pm\frac{\vec{N}}{\vec{N}\cdot\hat{k}}\mathrm{d}x\mathrm{d}y)$, if we are given the normal vector of the surface,or specifically, $g(x,y,z) = 0$

Association: