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Stoke It

  1. May 13, 2009 #1
    1. The problem statement, all variables and given/known data

    Use stokes theorem

    F = xyi + yzj + zxk on triangle 1,0,0,,,,,,,0,1,0,,,,,,0,0,1

    2. Relevant equations

    3. The attempt at a solution

    First i found the curl F

    curl F = -yi - zj - xk

    Then i found the equation of the plane for the triangle

    z = g(xy) = 1 - x - y

    Then i found the gradient

    f(xyz) = z - g(xy) = z - 1 + x + y, grad f = i + j + k

    so the integral will be dy from 0 to x-1, and dx from 0 to 1

    (-yi - zj - xk) dot (i + j + k) = -y - z - x

    integral (-y - z - x)dy = -y2/2 - zy - xy evaluated from 0 to x-1

    integral -(x-1)2/2 - z(x-1) - x(x-1) dx

    integral -x2/2 + x - 1/2 - zx + z - x2 + x dx evaluated from 0 to 2

    -1/2 + 1 - z/2 + z - 1/2 = 2 - z/2 should that z be in there
  2. jcsd
  3. May 14, 2009 #2

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    \oint\limits_C {{\rm \bar F \bullet }d{\rm \bar r} = \int\limits_S {\int {(\bar \nabla \times } } } {\rm \bar F}) \bullet {\rm \bar n }\;d{\rm A}

    Since the surface S can be any surface, any convenient surface can be chosen. This is best done here by choosing the three triangles formed by the three positive axes and the lines joining the three given points.

    Consider one such triangle -- for example, the one lying on the x-y plane, which is bounded by the x axis, y axis and the segment joining (1,0,0) and (0,1,0). The unit normal on this surface is constant and the double integral can be found very easily, taking the proper limits for x and y. The other two can be done similarly (or just by inspection using symmetry property) and all the three then added to get the result.
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