1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Verification of Stoke's Theorem for a Cylinder

  1. Oct 30, 2014 #1
    1. The problem statement, all variables and given/known data

    3sPl6GJ.jpg
    2. Relevant equations
    Stoke's Theorem:
    5f65e93751487f9350c194aa5f2bb8de.png
    3. The attempt at a solution
    ∇×A = (3x,-y,-2(z+y))

    I have parametric equation for wall and bottom:
    Wall: x(θ,z) = acosθ ; y(θ,z) = asinθ ; z(θ,z) = z [0≤θ≤2π],[0≤z≤h]
    Bottom: x(θ,r) = rcosθ ; y(θ,r) = rsinθ ; z(θ,r) = 0 [0≤θ≤2π],[0≤r≤a]

    And then I get lost, other examples I have tried to understand the method from are going straight over my head, any help much appreciated.
     
  2. jcsd
  3. Oct 30, 2014 #2

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    Take a stab at writing down the integral on the lefthand side. Surely you can at least write down ##\nabla\times\vec{A}## in terms of the cylindrical coordinates now.
     
  4. Oct 30, 2014 #3
    ∇×A in cylindrical coordinates is (3acosθ,-asinθ,-2(z+asinθ)) and then I integrate w.r.t. θ and z?
     
  5. Oct 30, 2014 #4

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    It depends on which surface you're integrating over. The integral over ##\Sigma## will be the sum of the integral over the bottom and the integral over the wall.
     
  6. Oct 30, 2014 #5

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    For the lateral side you have ##\vec R(\theta,z) = \langle a\cos\theta,a\sin\theta,z\rangle##. Now you need to calculate$$
    d\Sigma=\pm \vec R_z\times \vec R_\theta$$with the sign chosen to agree with the orientation. Then you will be ready to integrate$$
    \iint_\Sigma \nabla \times \vec F\cdot d\Sigma$$
     
  7. Oct 31, 2014 #6
    Rz = (0,0,1) and Rθ = (-asinθ,acosθ,0)
    use -Rz because it the bottom face

    giving dΣ = (-acosθ,asinθ,0)

    therefore the integral is ∫dz (3x,-y,-2(z+y))⋅(-acosθ,asinθ,0)
     
  8. Oct 31, 2014 #7

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    You have two surfaces, the lateral surface and the bottom face. Don't mix them up or combine them.

    I assume you are doing the lateral surface here. Your ##d\Sigma## has a sign error and it needs ##d\theta dz##. Also your integral must be in terms of those variables.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Verification of Stoke's Theorem for a Cylinder
Loading...