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Line Integral - Stokes theorem

  1. Nov 8, 2014 #1
    1. The problem statement, all variables and given/known data
    Hello
    I was given the vector field: [itex]\vec A (\vec r) =(−y(x^2+y^2),x(x^2+y^2),xyz)[/itex] and had to calculate the line integral [itex] \oint \vec A \cdot d \vec r [/itex] over a circle centered at the origin in the [itex]xy[/itex]-plane, with radius [itex] R [/itex], by using the theorem of Stokes.

    Another thing, when calculating [itex]∇× \vec A [/itex] I used Cartesian coordinates but when doing the integration I changed to polar coordinates, is that okay? Or do I have to do it all in Cartesian coordinates, because I did curl in Cartesian coordinates?

    2. Relevant equations
    Stokes equation:
    5f65e93751487f9350c194aa5f2bb8de.png

    3. The attempt at a solution
    My result is: [itex]2R^4π[/itex] Is that correct ? I evaluated both sides and got the same.

    Kind regards Alex
     
  2. jcsd
  3. Nov 8, 2014 #2

    vela

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    It's fine to change coordinates to make evaluating the integral easier whenever you want.

    I didn't actually do the integral on paper so I might have missed something, but the answer looks right to me.
     
  4. Nov 8, 2014 #3
    Okay, thanks for the reply :)
     
  5. Nov 9, 2014 #4

    rude man

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    I got 4πR4.

    del x A = 4R2 k
    dA = R2/2 k
    ∫2R2R2dθ from 0 to 2π = 4πR4.
     
    Last edited: Nov 9, 2014
  6. Nov 9, 2014 #5
    Thanks for doing the calculation but why is [itex] dA=\frac{R^2}{2} d \theta [/itex]. My curl vector is the same.
    I was using [itex]dxdy=rdrd \theta[/itex]
     
  7. Nov 9, 2014 #6

    rude man

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    I took the differential area as that of a triangle of area 1/2 bh = 1/2 R Rrdθ = 1/2 R2
    b = base
    h = height
    Proof: A = ∫dA = 1/2 R2∫dθ from 0 to 2π = πR2. :smile:

    You can do it your way too! Just requires integrating twice instead of once:
    dA = r dr dθ
    A = ∫∫r dr dθ= R2/2 (2θ) = πR2.
    So how about showing how you got 2πR4? One of us must have stumbled a bit in the dA department.
     
  8. Nov 9, 2014 #7

    vela

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    The curl is ##\nabla\times\vec{A} = 4r^2 \hat{k}##. It depends on variable ##r##, not constant ##R##, so you can't use that area element.
     
  9. Nov 9, 2014 #8
    Thanks for the replies guys. I'll write out my calculation then you can see where the mistake lies.
    [itex] \nabla \times \vec A = (xz - 0)i + (0 - yz)j + (3x^2+y^2 - (-x^2 - 3y^2))k [/itex]
    Chose a vektor normal to the circle:
    [itex] d \vec f = 0i+0j+k [/itex]
    The dot product is then:
    [itex] \nabla x \vec A \cdot d \vec f = 4x^2+4y^2 = 4r^2[/itex]
    Then I integrated over the circle, and changed to spherical coordinates where I used the area element: [itex]dA =rdrd \phi[/itex]:
    [itex] \int_{0}^{2\pi} \int_{0}^R 4r^2rdrd \phi = \int_{0}^{2Pi} \int_{0}^R 4r^3drd \phi = 2R^4\pi[/itex]
     
    Last edited: Nov 9, 2014
  10. Nov 9, 2014 #9

    rude man

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    I agree, I should not have made r constant in the curl expression.
     
  11. Nov 9, 2014 #10
    Is [itex] 2R^4\pi [/itex] then the correct result?
     
  12. Nov 9, 2014 #11

    vela

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    Yes.
     
  13. Nov 9, 2014 #12

    rude man

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    Yes, I say it is, and you were right to use r dr dθ as the elemental area.
     
  14. Nov 10, 2014 #13
    Okay, thanks for the help, will be back with some more problems :)
     
  15. Nov 10, 2014 #14

    rude man

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    Please do! I'll try not to mess up next time ... o:)
     
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