[Mathematics][MIT 18.02]Detailed discussions about 2-D and 3-D integrals and their connections

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

General:

• Line integral(Work and in the plane)

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

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

• Gradient fields & path-independence

Condition:

　　$curl(\vec{F}) = 0$ and $\vec{F}$ is defined in a simple-connected region,

　　　　　　　　in which $\displaystyle curl(\vec{F}) = N_{x} - M_{y}$ if $\vec{F} = $ 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} = $

　　　　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. (...)

• Flux in plane & space

in the plane:

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

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

in the space(or specifically, surface):

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

　　Case 2. $\displaystyle \iint_{S}\vec{F}\cdot\hat{n}\mathrm{d}S=\iint_{S}\vec{F}\cdot(\pm\frac{\vec{N}}{\vec{N}\cdot\hat{k}}\mathrm{d}x\mathrm{d}y)$, if we are given the normal vector of the surface,or specifically, $g(x,y,z) = 0$

Addition(general case of the second): let's say $x = x(u,v)$ and $y = y(u,v)$ and $z = z(u,v)$ describe a surface, then we can get the $\displaystyle \hat{n}\mathrm{d}S$ by changing $u$ and $v$ a little bit. Specifically, we begin at $\displaystyle (x(u,v),y(u,v),z(u,v))$. By changing $u$ a little bit($\Delta u$), then we arrive at $\displaystyle (x(u+\Delta u,v),y(u+\Delta u,v),z(u+\Delta u,v))$.

Using linear approximation, we get $\displaystyle (x(u,v) + x_{u}\Delta u,y(u,v) + y_{u}\Delta u,z(u,v) + z_{u}\Delta u)$, so the difference is $\displaystyle (x_{u}\Delta u, y_{u}\Delta u, z_{u}\Delta u) = \Delta u(x_{u},y_{u},z_{u})$, let's set it to be $\vec{r_{1}}$.

In the same way, we can get $\displaystyle \vec{r_{2}} = \Delta v(x_{v},y_{v},z_{v})$ by changing $v$ a little bit. Thus we take the limits (meaning replace $\Delta$ with $\mathrm{d}$) and we can derive the corresponding $\displaystyle \hat{n}\mathrm{d}S$ from it.

So $\displaystyle \hat{n}\mathrm{d}S = \vec{r_{1}}\times\vec{r_{2}} = \times\mathrm{d}u\mathrm{d}v$.

(of course we can use position vector $\displaystyle \vec{r} = \vec{r}(u,v) = $ to simplify the problem, namely $\displaystyle \frac{\partial \vec{r}}{\partial u}\mathrm{d}u$ is the $\vec{r_{1}}$)

Association:

Work(line integral):

• 2-D(Green's Theorem):

　　$\displaystyle \oint_{C}\vec{F}\cdot\mathrm{d}\vec{r} = \iint_{R}curl(\vec{F})\mathrm{d}A$

• 3-D(Stoke's Theorem):

　　$\displaystyle \oint_{C}\vec{F}\cdot\mathrm{d}\vec{r} = \iint_{S}curl(\vec{F})\hat{n}\mathrm{d}S$,in which $S$ means any surface bounded by this curve and $curl(\vec{F})=\nabla\times\vec{F}$.

Flux:

• 2-D(Green's Theorem):

　　$\displaystyle \oint_{C}\vec{F}\cdot\hat{n}\mathrm{d}s = \iint_{R}div(\vec{F})\mathrm{d}A$,in which $\vec{F} = $ and $div(\vec{F}) = P_{x} + Q_{y}$.

• 3-D(Divergence Theorem):

　　$\displaystyle\oiint_{S}\vec{F}\cdot\hat{n}\mathrm{d}S = \iiint_{R}div(\vec{F})\mathrm{d}V$, in which $\vec{F} = $ and $div(\vec{F}) = P_{x} + Q_{y} + R_{z}$, in which the direction of $\hat{n}$ is determined by the right-hand rule.