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Math Help - using stokes' theorem to evaluate

  1. #1
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    using stokes' theorem to evaluate

    I have to evaluate ∬ curlF.dS where F (x,y,z)= (xyz, xy, yzx^2)
    s
    S consists of the top and the four sides( but not the bottom) of the cube with vertices(1, 1, 1) oriented onward.

    I calculated curl F but I wasnt sure how to go from there.

    please help!
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  2. #2
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    If you have computed \mathop{curl}(\vec{F}), then the surface integral on, for instance, the face of the cube with x=1 is: \int_{-1}^1\int_{-1}^1 \mathop{curl}(\vec{F})(1,y,z)\cdot \vec{e}_x dy\,dz, where \vec{e}_x is a unit vector in direction x (hence orthogonal to the face), meaning that you integrate the x-component of \mathop{curl}(\vec{F}) over the face. And you have to do this for each of the 5 faces. Good luck!

    Or you remember the title "using Stokes' thm to evaluate..." and write your surface integral (over \Sigma: the cube minus the bottom face) as the integral \int_{\partial\Sigma}\vec{F}\cdot d\vec{r} over the boundary of \Sigma, which is a square. This reduces the problem to the computation of four line integrals (and you don't need the curl). For instance, one of them is: \int_{-1}^1 \vec{F}(1,y,-1)\cdot \vec{e}_y dy.

    You also have to care about the orientation of the boundary: it is defined by the orientation of the surface.

    Laurent.
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  3. #3
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    would you have:

    curl F = (x^2z) \ \vec{i} -(2xyz-xy) \ \vec{j} + (y-xz) \ \vec{k}

    where P=(x^2z), \ Q=-(2xyz-xy), \ R=(y-xz)

    so since we want all sides but the bottom then would it be:

    top: z=g(x,y)=1
    left side: y=g(x,z)=-1, right side: y=g(x,z)=1
    front: x=g(y,z)=1, back: x=g(y,z)=-1

    now since \iint_S F \cdot \ dS = \iint_D \left(-P \frac{\partial g}{\partial x} -Q \frac{\partial g}{\partial y} +R \right) \ dA

    so for the top I would have:

    \int_{-1}^1 \int_{-1}^1 (y-xz) \ dx \ dy = \int_{-1}^1 \int_{-1}^1 (y-x) \ dx \ dy

    now for the left side:

    \int_{-1}^1 \int_{-1}^1 -(-(2xyz-xy)) \ dx \ dz = \int_{-1}^1 \int_{-1}^1 (2x(-1)z+x)) \ dx \ dz

    and for the front:

    \int_{-1}^1 \int_{-1}^1 -(x^2z) \ dz \ dy = \int_{-1}^1 \int_{-1}^1 -(z) \ dz \ dy

    where the right and back would simply be replaced by 1 and -1.

    is this correct?
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  4. #4
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    Quote Originally Posted by lllll View Post
    is this correct?
    Sorry for the delay... This seems correct.

    Besides, I've just noticed this can be shortened: if you integrate \mathop{curl}(\vec{F}) over the surface of the whole cube (bottom included), this gives 0. This is a consequence of the divergence theorem and \mathop{div}(\mathop{curl})=0 (or a consequence of Stokes' theorem, with an empty boundary). Therefore, it suffices to find the integral of \mathop{curl}(\vec{F}) of the bottom, and take its opposite. This finally gives -\int_{-1}^1\int_{-1}^1 -(y-x(-1)) dx dy = 0, disappointedly...
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