• anonymous
  • Stacey Warren - Expert
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  • chestercat
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  • anonymous
Isn't Green's Theorem just a special case of Stokes' Theorem?? from Green's Theorem we have \[\oint_\limits{C}(\phi dx+\psi dy)=\iint_\limits{R}(\frac{\partial \psi}{\partial x}-\frac{\partial \phi}{\partial y})dxdy\] Now if we let \[\vec F=\phi \hat i+\psi \hat j+0 \hat k\] and \[d \vec r=dx \hat i+dy \hat j+dz \hat k\] therefore we can write \[\phi dx+\psi dy=\vec F. d \vec r\] Then we can write \[\frac{\partial \psi}{\partial x}-\frac{\partial \phi}{\partial y}\] as \[\hat k.(\vec \nabla \times \vec F)\] To confirm For a scalar triple product we have \[\vec a.(\vec b \times \vec c) =\begin{vmatrix}a_{x} & a_{y} & a_{z} \\ b_{x} & b_{y} & b_{z} \\ c_{x} & c_{y} & c_{z} \end{vmatrix}\] So, if we expand we get \[\det=\begin{vmatrix}0 & 0 & 1 \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial }{\partial z} \\ \phi & \psi & 0 \end{vmatrix}\] \[\det=\begin{vmatrix}\frac{\partial}{\partial x} & \frac{\partial}{\partial y} \\ \phi & \psi\end{vmatrix}=\frac{\partial \psi}{\partial x}-\frac{\partial \phi}{\partial y}\] So we have \[\oint_\limits{C} \vec F.d \vec r=\iint_\limits{R}[(\vec \nabla \times \vec F). \hat k]dxdy\] From Stokes theorem we have \[\oint_\limits{C} \vec F . d \vec r=\iint_\limits{S}[(\vec \nabla \times \vec F).\hat n]ds\] Now we can write \[ds=\frac{dxdy}{|(\pm \hat k).\hat n|}=\frac{dxdy}{\hat k . \hat n}\] and when \[\hat n=\hat k\] we have \[\oint \vec F . d \vec r=\iint_\limits{R}[(\vec \nabla \times \vec F).\hat k]\frac{dxdy}{\hat k . \hat k}=\iint_\limits{R}[(\vec \nabla \times \vec F).\hat k]dxdy\] So it's a special case when we consider the projection of S on xy-plane(R) and of course for a surface in xy plane, it's normal will be along z or -z axis. It is essentially a transformation of Stokes Theorem from 3 dimensional surface to a 2 dimensional surface

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