Fundamental Theorem of Line Integrals (We already did)
Let $C$ be a piecewise smooth curve that starts at $A$ and ends at $B$. If $f$ is continuously differentiable on an open region containing the curve $C$ (so $\vec F = \vec \nabla f$ is continuous), then we have $$ f(B)-f(A) = \int_C \vec \nabla f\cdot d\vec r = \int _C \vec F\cdot d\vec r $$
Green's Theorem
Let $C$ be a piece-smooth simple closed curve, oriented in a counter clockwise direction. Let $R$ be the region inside the curve $C$. Provided the vector field $F$ is continuously differentiable on an open region containing $R$, we then have $$\begin{align} \oint_C \vec F\cdot d\vec r &= \oint _C Mdx+Ndy = \iint_R N_x-M_y dA \text{ and } \\ \oint_C \vec F\cdot \hat n ds &=\oint _C Mdy-Ndx = \iint_R M_x+N_y dA. \end{align}$$
The Del Operator
Let $\vec \nabla = \left(\frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z}\right)$, and let $\vec F = (M,N,P)$. Then
- The gradient of $f$ is given by $$\vec \nabla f = \left(\frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z}\right)f = \left(\frac{\partial f}{\partial x},\frac{\partial f}{\partial y},\frac{\partial f}{\partial z}\right) .$$
- The divergence of a vector field is given by $$\begin{align}\vec \nabla \cdot \vec F &= \left(\frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z}\right)\cdot (M,N,P)\\ &= \frac{\partial M}{\partial x}+\frac{\partial N}{\partial y}+\frac{\partial P}{\partial z} = M_x+N_y+P_z\ .\end{align}$$
- The curl of a vector field is given by $$\begin{align}\vec \nabla \times \vec F &= \left(\frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z}\right)\times (M,N,P)\\ &= \left(\frac{\partial P}{\partial y}-\frac{\partial N}{\partial z},\frac{\partial M}{\partial z}-\frac{\partial P}{\partial x},\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y}\right)\\ &= \left(P_y-N_z,M_z-P_x,N_x-M_y\right) .\end{align}$$
Surface Integrals
Let $S$ be a surface, which is parametrized by $\vec r(u,v)$ for $a\leq u\leq b$ and $c(u)\leq v\leq d(u)$. A tiny bit of a surface area is given by $d\sigma = |\vec r_u\times\vec r_v|dudv$, which means the surface area is given by $$\sigma = \iint_Sd\sigma = \int_a^b\int_{c(u)}^{d(u)}|\vec r_u\times\vec r_v|dvdu.$$
Flux across a surface
Let $S$ be an orientable surface. Let $\hat n$ be a unit normal vector of $S$ that gives the orientation of $S$. Let $\vec F$ be a continuous vector field on $S$. Then the flux of $\vec F$ across $S$ is given by $$\Phi = \iint_S\vec F\cdot \hat n d\sigma$$. If $\vec r(u,v)$ is a smooth parametrization of $S$, and we let $\vec N= \vec r_u\times \vec r_v$, then this gives $\hat n = \pm\frac{ \vec N }{ |\vec N| }$, which means $$\Phi = \iint_S\vec F\cdot \hat n d\sigma = \iint_S\vec F\cdot \left(\pm \frac{\vec {N}}{|\vec N|}\right)|\vec N|dudv = \iint_S\vec F\cdot\left(\pm \vec {N}\right)dudv$$.
Divergence Theorem
Let $D$ be a solid domain in space, whose surface is given by the orientable piecewise smooth surface $S$. Let $\hat n$ be the unit normal vector of $S$ that points outwards from $D$. If $\vec F$ is continuously differentiable on an open region containing $D$, then we have $$ \iint _S \vec F\cdot \vec n d\sigma = \iiint_D M_x+N_y+P_z dV = \iiint_D \vec \nabla \cdot \vec F dV .$$
Stokes's Theorem
Let $S$ be an orientable surface in space, whose boundary is given by the piecewise smooth simple closed curve $C$. Let $\hat n$ be a unit normal vector of $S$ that gives the orientation of $S$, and let $C$ be given an orientation that agrees with $\vec n$ using the right-hand rule. If $\vec F$ is continuously differentiable on an open region containing $S$, then we have $$ \oint_C \vec F\cdot d\vec r = \oint _C Mdx+Ndy+Ndy =\iint _S \left(\vec \nabla\times \vec F\right)\cdot \vec n d\sigma .$$
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