Since it is just a sort of discussion, I will just give the formula and condition without proving them or leaving examples.
$\displaystyle \int_{C}\vec{F}\cdot \mathrm{d}\vec{r} = \int_{C}M\mathrm{d}x+N\mathrm{d}y$, in which $\vec{F} = <M,N>$
Method: Express $x$ and $y$ in a single variable (OR means parameterization).
$curl(\vec{F}) = 0$ and $\vec{F}$ is defined in a simple-connected region,
in which $\displaystyle curl(\vec{F}) = N_{x} - M_{y}$ if $\vec{F} = <M,N>$ AND $\displaystyle curl(\vec{F}) = \nabla\times\vec{F}$(namely$\displaystyle \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 $\vec{F} = <P,Q,R>$
then $\vec{F} = \nabla f$, or $\vec{F}$ is the partial derivative vector of some vector field.
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. (...)
$\hat{n} = \hat{T}$ rotated 90 degrees clockwise $=<\mathrm{d}y,-\mathrm{d}x>$
$\displaystyle \int_{C}\vec{F}\cdot\hat{n}\mathrm{d}s = \int_{C}P\mathrm{d}y-Q\mathrm{d}x$, in which $\vec{F} = <P,Q>$
$\displaystyle \iint_{S}\vec{F}\cdot\hat{n}\mathrm{d}S = \iint_{S}\vec{F}\cdot(<-f_{x},-f_{y},1>\mathrm{d}x\mathrm{d}y)$, if we use $z = f(x,y)$ to describe the surface.
$\displaystyle =\iint_{S}\vec{F}\cdot(\pm\frac{\vec{N}}{\vec{N}\cdot\hat{k}}\mathrm{d}x\mathrm{d}y)$, if we are given the normal vector of the surface,or specifically, $g(x,y,z) = 0$
$\displaystyle \oint_{C}\vec{F}\cdot\mathrm{d}\vec{r} = \iint_{R}curl(\vec{F})\mathrm{d}A$
$\displaystyle \oint_{C}\vec{F}\cdot\mathrm{d}\vec{r} = \iint_{S}curl(\vec{F})\hat{n}\mathrm{d}S$,in which $S$ means any surface bounded by this curve and $curl(\vec{F})=\nabla\times\vec{F}$.
$\displaystyle \oint_{C}\vec{F}\cdot\hat{n}\mathrm{d}s = \iint_{R}div(\vec{F})\mathrm{d}A$,in which $\vec{F} = <P,Q>$ and $div(\vec{F}) = P_{x} + Q_{y}$.
$\displaystyle\oiint_{S}\vec{F}\cdot\hat{n}\mathrm{d}S = \iiint_{R}div(\vec{F})\mathrm{d}V$, in which $\vec{F} = <P,Q,R>$ and $div(\vec{F}) = P_{x} + Q_{y} + R_{z}$.
[Mathematics][MIT 18.02]Detailed discussions about 2-D and 3-D integral and their connections
原文:https://www.cnblogs.com/raymondjiang/p/11715381.html