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Math Help - Stokes' Theorem

  1. #1
    Super Member angel.white's Avatar
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    Stokes' Theorem

    Use Stokes' Theorem to evaluate \int_C \vec F \cdot d\vec r. C is oriented counterclockwise as viewed from above.

    \vec F (x,y,z) = <x+y^2, y+z^2, z+x^2>, C is the triangle with vertices (1,0,0), (0,1,0), and (0,0,1)

    -----

    I've computed the curl as <-2z, -2x, -2y> and the equation of the plane as z=1-x-y

    Stokes' theorem says \int_C \vec F \cdot d\vec r = \int \int_S curl \vec F \cdot d\vec s

    But I can't figure out how to evaluate the second part, since the curl is a vector, and the surface has three variables, I don't know how to get it into a form I can integrate.
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    Quote Originally Posted by angel.white View Post
    Use Stokes' Theorem to evaluate \int_C \vec F \cdot d\vec r. C is oriented counterclockwise as viewed from above.

    \vec F (x,y,z) = <x+y^2, y+z^2, z+x^2>, C is the triangle with vertices (1,0,0), (0,1,0), and (0,0,1)

    -----

    I've computed the curl as <-2z, -2x, -2y> and the equation of the plane as z=1-x-y

    Stokes' theorem says \int_C \vec F \cdot d\vec r = \int \int_S curl \vec F \cdot d\vec s

    But I can't figure out how to evaluate the second part, since the curl is a vector, and the surface has three variables, I don't know how to get it into a form I can integrate.
    Does this thread help: http://www.mathhelpforum.com/math-he...s-theorem.html

    By the way, your curl and equation for the plane look OK.
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  3. #3
    Super Member angel.white's Avatar
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    Quote Originally Posted by mr fantastic View Post
    Does this thread help: http://www.mathhelpforum.com/math-he...s-theorem.html

    By the way, your curl and equation for the plane look OK.
    It is still unclear to me, perhaps I do not understand what is meant by d\vec S.
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    Quote Originally Posted by angel.white View Post
    It is still unclear to me, perhaps I do not understand what is meant by d\vec S.
    \vec{dS} is an infinitesimal element of directed surface.

    \vec{dS} = \hat{n} dS where \hat{n} is a unit normal vector to S.

    The formulae I quote in the other thread follow from this.

    For your surface (plane), \vec{n} = \frac{\partial z}{\partial x} \, i + \frac{\partial z}{\partial y} \, j - k = -i - j - k. Then:

    \nabla F \cdot \vec{dS} = (-2z \, i - 2x \, j - 2y \, k) \cdot (-i - j - k) \, dx \, dy = 2 (z + x + y) \, dx \, dy = 2 \, dx \, dy

    where I've substituted z = 1 - x - y.

    The region of integration in the xy-plane is clearly the right-triangle with vertices at (0, 0), (1, 0) and (0, 1).

    So your line integral is simply twice the area of this triangle: 1.
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