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Stokes' Thm, intersection of sphere and plane

  1. Dec 19, 2012 #1

    CAF123

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    1. The problem statement, all variables and given/known data
    Use Stokes' Theorem to evaluate $$\int_{\gamma} y\,dx + z\,dy + x\,dz,$$ where ##\gamma## is the suitably oriented intersection of the surfaces ##x^2 + y^2 + z^2 = a^2## and ## x + y + z = 0##

    3. The attempt at a solution
    Stokes' says that this is equal to $$\iint_S (\underline{\nabla} \times \underline{F}) \cdot \underline{n}\,dA$$ So from the question, I can extract ##\underline{F} = \langle y,z,x \rangle## and compute ##\, \text{curl}\underline{F} ##. ##\,\,####\gamma## marks the boundary of the surface S, so in using Stokes' I can use any surface whose boundary is ##\gamma##.

    Let ##z = -x -y => x^2 + y^2 + (-x-y)^2 = a^2##, which when rearranged gives ##x^2 + y^2 +xy = a^2/2##. I put this into Wolfram Alpha and it is an ellipse, however I am unsure of how to show this. I.e I want to get what I have in the form ##\frac{x^2}{A^2} + \frac{y^2}{B^2} = 1.## I would complete the square but the## xy## term is annoying. Once I have it in the form of an ellipse, I am sure I can just say ##x = A\cos t, y = B\sin t## on the boundary, find normal, dot it with the curl and set the bounds. I am also a little bit unsure of what the bounds would be. If the result of curl F dotted with n resulted in 1, then I could just simply say that the surface integral is the area of the ellipse, which would greatly simplify things.
     
    Last edited: Dec 19, 2012
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  3. Dec 19, 2012 #2

    Dick

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    Try thinking about this problem geometrically using the surface integral. The sphere is centered at the origin. The plane passes through the origin. What does the intersection look like? It's not as complicated as a general ellipse. What is the curl? What is the normal? You can really do this whole problem in your head.
     
  4. Dec 19, 2012 #3

    CAF123

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    I would have thought that the intersection be a circle, but have I not shown that it is not a circle, ##x^2 + y^2 + xy = a^2/2## is not a circle?

    I computed the curl to be ##- \langle 1,1,1 \rangle## . Perhaps the normal would be in the direction into the first octant (or antiparallel to it).

    When you say you can do the whole problem in your head, do you mean because the computation is easy or is there some conceptual reason why the results are what they are?
     
  5. Dec 19, 2012 #4

    Dick

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    The conceptual reason is that the normal to the surface is constant and the curl is constant, so the dot product is constant. Integrating a constant over a surface whose area you know is pretty easy. ##x^2 + y^2 + xy = a^2/2## in the xy plane is probably an ellipse, it looks like the projection of the circle you want into the xy plane. But it's certainly not the boundary curve. Luckily, you don't need an equation for the boundary curve.
     
  6. Dec 19, 2012 #5

    CAF123

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    How should I compute the normal vector if I don't have a parametrisation?

    The projection of their intersection would be an ellipse in the xy plane?

    Is this the questions motivation for resorting to using Stokes' Thm?
     
  7. Dec 19, 2012 #6

    Dick

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    Yes, it's much easier to use Stokes' theorem than to do the path integral directly. Yes, if you take a circle in space and project it to the xy plane you generally get an ellipse. Finally, the normal to the intersection of the plane and the sphere is the same as the normal to the plane, isn't it?
     
  8. Dec 20, 2012 #7

    CAF123

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    Do you mean a circle inclined in space? E.g if I take a cylinder and chop it at some z, the projection onto xy would be a circle. But why is the intersection a circle anyway in this problem?

    This makes sense.

    So ## \hat{n} = \langle 1/\sqrt{3}, 1/\sqrt{3}, 1/\sqrt{3} \rangle## which gives ##
    \text{curl} \, \underline{F} \cdot \hat{n} dS = -\sqrt{3} dS. ##
     
  9. Dec 20, 2012 #8

    Dick

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    What kind of curves do you get when you cut a sphere with a plane? Just think about the geometry. Don't try to get the answer by eliminating z. That only gives you the xy part of the intersection.
     
  10. Dec 20, 2012 #9

    CAF123

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    In general, you get circles. Could you clarify what you mean by what I found was the 'xy' intersection?
     
  11. Dec 20, 2012 #10

    Dick

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    When you eliminate z you only have x and y in your equation. An equation that only has x and y in it describes a kind of vertical cylinder in space. What you got is an vertical elliptical cylinder which passes through the circle that the plane makes intersecting the sphere. It's just telling you the circle looks like an ellipse when viewed from an angle. It's not useful for doing a path integral.
     
  12. Dec 20, 2012 #11

    CAF123

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    Thanks!
     
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