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Stokes theorem problem

  1. Feb 23, 2009 #1
    1. The problem statement, all variables and given/known data

    Please help me to check whether I did the right working for this problem. Thanks. The numerical answer is correct but I'm not very sure if the working is correct also.

    Find [tex]\int y dx + z dy + x dz[/tex] over the closed curve C which is the intersection of the surfaces whose equations are [tex]x + y = 2[/tex] and [tex]x^2 + y^2 + z^2 = 2(x+y) [/tex]

    3. The attempt at a solution

    First, I note that the integral required is the line integral for F . dr where F = (y, z, x). Since the curve is closed, we can apply Stokes' theorem. By Stokes' theorem [tex]\int F . dr = \iint (curl F) \cdot \b{n} dA[/tex].

    Curl F = (-1,-1,-1) after applying the cross product.

    Then I sketch the surfaces on the x-y axis and pick out the normal vector [tex]n = \frac{1}{\sqrt{2}}(\b{i} + \b{j})[/tex]. Also [tex]\iint dA = \pi r^2[/tex]. Then the answer for the line integral is [tex]-2\sqrt{2}\pi[/tex]
     
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  3. Feb 23, 2009 #2

    HallsofIvy

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    What surface, exactly, are you integrating over? You are told that the curve is the intersection of two surfaces. To use Stoke's theorem, you must integrate over the entire surface: both of the given surfaces.
     
  4. Feb 23, 2009 #3

    Dick

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    It looks like bigevil is integrating over the plane. You don't have to integrate over both surfaces. Either one should give you the same answer. I'm a little confused at you deduced the radius of the circle. But other than that, it's fine.
     
  5. Feb 23, 2009 #4
    Sorry Halls, the closed curve is the intersection of the two equations. That curve is closed and the surface it covers is a circle (of radius sqrt(2) I think) then I'm integrating over the plane.

    I don't know how to describe it, but I deduced by imagining that the line slices the sphere given. (I drew the whole thing in the x-y plane, ie looking from the 'top' down.) The plane (y + x = 2) runs through the centre of the sphere. I got radius sqrt 2 for the radius of the closed surface.

    Thanx Dick and Halls.
     
  6. Feb 23, 2009 #5

    Dick

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    Yes, exactly. The plane cuts through the center of a sphere with radius sqrt(2). You can use Stokes over that surface. As you did so well.
     
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