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Math Help - Stokes theorem - Calculating circulation of vector field in a geometric problem

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    Stokes theorem - Calculating circulation of vector field in a geometric problem

    Hey all!
    I've been staring myself blind at this problem for the last few hours and my hair is turning grey.

    The Question:
    Calculate, using Stokes theorem the circulation of the vector field

    A = (yz + y -z, xz + 5x, xy + 2y)

    along the line L between the surfaces

    x^2+y^2+z^2 = 1 and x+y = 1

    L is oriented so that the positive direction in (1,0,0) is given by vector (0,0,1).

    Attempts at solution:

    First I draw the surfaces. I see that we have a sphere that is cut in the x-y plane giving us a sphere segment.

    Using the stokes theorem I get

    curl(A) = (2,-1,4) (checked like 50 times :P )

    now I need to find the normal vector.

    I assume that what I have is a level surface. This is because the definition of a level surface is f(x,y,z) = constant.

    This I take x^2+y^2+z^2=x+y (this is where they meet)
    so f(x,y,z) = x^2+x + y^2+y + z^2

    n = grad(f) = (2x-1,2y-1,2z) / |grad(f)|

    dS = n dS , where I assume dS = Area of the domain D (shadow of the sphere segment) as this is the line in the original question.
    so,
    dS = |grad f| dxdy

    [double integral] curl(A) * n dS = (2,-1,4) * (2x-1,2y-1,2z) / |grad(f)| *
    |grad(f)| dx dy

    which leaves me with
    [double integral] 4x-2 -2y+1 + 8z dx dy = [D.I] 4x-2y+8z -1 dx dy

    As this is a area of a shadow with height 0 we get z = 0

    Using this and integrating over the area simply does not work.
    The thing is I really want to use spherical coordinates to solve this problem but I have tried over and over again and I always fail.

    What I need help with
    Is there any way possible to solve this problem using cylindrical coords? If so, what normal vector should I use aswell as what dS should I take?

    While writing this I got an Idea.. I'll update if It worked out.

    EDIT: Problem solved! I used f(x,y) = x + y - 1 and integration over the area of the shade of the segment worked! Yes! I feel like I actually learned something
    Last edited by Hanga; March 6th 2011 at 06:41 AM.
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