A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing


  • one year ago


  • This Question is Closed
  1. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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

  2. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...


  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.