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Using Stokes law, calculate the work done along a curve

  1. Jan 4, 2014 #1
    1. The problem statement, all variables and given/known data
    Using Stokes law, calculate the work done along a curve ##\Gamma ## which is defined as edge of a spherical triangle in first octant of a sphere ##x^2+y^2+z^2=R^2##. Vector field is ##\vec{F}=(z^2,x^2,y^2)##.


    2. Relevant equations
    Stokes law: ##\int _{\partial \Sigma }\vec{F}d\vec{r}=\int \int _{\Sigma }\triangledown \times \vec{F}dA##


    3. The attempt at a solution

    firstly ##\triangledown \times \vec{F}=(2y,2z,2x)##.

    Now I should somehow parametrize that triangle but anything i try... comes out nasty. For example:
    ##x=t## for ##t\in \left [ 0,R \right ]## than also ##y=t## which would from ##x^2+y^2+z^2=R^2## mean that ##z^2=r^2-2t^2##

    Now i have parameterization as function of t only ##r(t)=(t,t,\sqrt{r^2-2t^2})## but... to calculate dS in stokes integral, I need parametrization of two paramateres... so... How do I continue?

    Is it ok if I just say that ##r(t,z)=(t,t,z)## where ##t\in \left [ 0,R \right ]## and ##z\in \left [ -\sqrt{r^2-2t^2},\sqrt{r^2-2t^2} \right ]##
     
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  3. Jan 4, 2014 #2

    LCKurtz

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    Do you have to use Stoke's theorem or does that just seem like the obvious choice to you? It is much easier to just work out the three line integrals. Lots of the variables are 0 on each curve of your spherical triangle.
     
  4. Jan 4, 2014 #3
    Yes, I HAVE to.
     
  5. Jan 4, 2014 #4

    LCKurtz

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    Then I would try spherical coordinates.$$
    \vec r(\theta,\phi)=\langle R\sin\phi\cos\theta,R\sin\phi\sin\theta,R\cos\phi\rangle$$Calculate ##\vec r_\theta## and ##\vec r_\phi##, and work out the integral$$
    \pm\int_0^{\frac \pi 2} \int_0^{\frac \pi 2}(\nabla \times \vec F) \cdot \vec r_\phi \times \vec r_\theta~
    d\phi d\theta$$where the ##\pm## sign is chosen to agree with the orientation which, by the way, you didn't specify.
     
    Last edited: Jan 4, 2014
  6. Jan 4, 2014 #5
    I think you meant ##\vec r(\theta,\phi)=\langle R\sin\phi\cos\theta,R\sin\phi\sin\theta,R\cos\phi\rangle##

    But you kinda lost me here already... Why would I want to integrate over a sphere (or one eight of it) if the problems says to integrate ##\vec{F}## along curve where this curve consists of three lines that together form a triangle?
     
  7. Jan 4, 2014 #6

    LCKurtz

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    Yes. I corrected it.

    That's why I asked you if you have to use Stoke's theorem. It lets you change the line integral around the boundary to a surface integral over the bounded surface. But that is no advantage in this problem. The line integrals are easy.

    And I assume you don't literally mean the boundary curve is a triangle. It is three circular arcs that form what you might call a spherical triangle.
     
  8. Jan 4, 2014 #7

    vela

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    How did you get y=t?
     
  9. Jan 5, 2014 #8
    Hmm, well ... The problems says to integrate along a positively oriented spherical triangle cut by ##x^2+y^2+z^2=R^2## .. To be honest with you, I never even thought about circular arcs.

    Yes, I was the most worried about that part. I simply imagined that since it is a sphere the triangle is symmetrical. I guess, that's not ok.
     
  10. Jan 5, 2014 #9
    Ok, I get it now. Now idea of how spherical triangle looks like was completely wrong! Thanks, LCKurtz!

    Now I agree with this:
    ##\pm\int_0^{\frac \pi 2} \int_0^{\frac \pi 2}(\nabla \times \vec F) \cdot \vec r_\phi \times \vec r_\theta~
    d\phi d\theta=\pm\int_0^{\frac \pi 2} \int_0^{\frac \pi 2}2R(\sin\theta \sin \phi,\cos \theta,\sin \theta \cos \phi )R^2(\sin^2\theta \cos \phi,\sin^2 \theta \sin \phi, \sin \theta cos \theta)##

    Which gives me some nasty integrals:

    ##\pm\int_0^{\frac \pi 2} \int_0^{\frac \pi 2}2R^3(\sin^3\theta \sin\phi \cos\phi +\sin^2\theta \sin\phi \cos\theta + \sin^2\theta \cos\theta \cos\phi)##

    Which of them is 0 and why?
     
  11. Jan 5, 2014 #10
    Is the result ##2R^3## ?
     
  12. Jan 5, 2014 #11

    vela

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    That's what I got. As a check, you could try calculating the line integrals and see if you get the same result.
     
  13. Jan 5, 2014 #12

    LCKurtz

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    That's what I get too, with a ##\pm## sign. You still have to give the orientation and choose the correct sign.
     
  14. Jan 6, 2014 #13
    I thought ##2R^3## including the sign.. But since you are asking about the sign, i am guessing it's -...
     
  15. Jan 6, 2014 #14

    LCKurtz

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    Guesses don't count. In problems like this, it matters which way you go around the curve. You haven't told us which way to go around the curve so the problem is incompletely stated. The direction around the curve determines the orientation of the corresponding bounded surface. You have to make sure your vector ##\vec r_\phi \times \vec r_\theta## agrees with that orientation to decide whether to use that vector or its negative in the formula. There is no way to know which is correct until you tell us which way around the curve to consider.
     
  16. Jan 6, 2014 #15
    There's nothing mentioned about the orientation of spherical triangle. So both signs should be ok.

    However, let's say that spherical triangle in positively oriented. Than the overall sign is positive (that's IF I understand this right).
     
  17. Jan 6, 2014 #16

    LCKurtz

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    To say the spherical triangle is "positively oriented" doesn't mean anything to me. The direction is usually specified with descriptions like "clockwise or "counterclockwise" as viewed from the origin" or more literally something like "moving from ##(R,0,0)## to ##(0,R,0)## along the curve in the xy plane then to ##(0,0,R)## in the yz plane, then etc... Then you use the right-hand rule to determine the orientation of the surface.
     
  18. Jan 6, 2014 #17
    Ok, our definition of positively oriented surface:

    If you stand on the edge of your surface (spherical triangle) and start walking, than the surface has positive orientation if it is on your left side and negative if it is on your right side.

    Does this make any sense?
     
  19. Jan 6, 2014 #18
    edit: opened another thread...
     
    Last edited: Jan 6, 2014
  20. Jan 6, 2014 #19

    LCKurtz

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    You started with a curve (spherical triangle). It is true that the direction you go around the curve determines the orientation of the surface with respect to that curve. The right hand rule will tell you which normal to use for positive orientation. But you have to know which way you are going around the curve to do that. Just like you can't say if the surface is on your right or left if you don't know which direction you are moving. If your book doesn't describe the direction are you sure it doesn't show the direction with a picture with little arrows on the curves to indicate which direction?
     
  21. Jan 6, 2014 #20
    Aaa, ok, thanks for this explanation. I'm 100% sure that there is no direction mentioned and believe me that there is no picture - otherwise I wouldn't have mistaken the curved lines for straight lines. :)
     
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