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Math Help - Stokes Theorem

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    Stokes Theorem

    Verify Stokes Theorem ∬(∇xF).N dA where surface S is the paraboloid z = 0.5(x^2 + y^2) bound by the plane z=2, C is its boundary, and the vector field F = 3yi - xzj + yzk


    I had found (∇xF) = (z+x)i + (-z-3)k

    r = [u, v, 0.5(u^2 + v^2)]
    Therefore N= ru X rv = -ui -uj +k
    Therefore (∇xF).N = [(z+x), 0, (-z-3)].[-x, -y, 1]
    After that I substitute x = rcos(θ), y = rsin(θ), z = 0.5r^2
    Thus ∫(0-2)∫(0-2pi) (∇xF).N r dθdr

    But I cant seems to get the answer. Can anyone help? Help would greatly appreciated
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  2. #2
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    [QUOTE=HeheZz;510745]Verify Stokes Theorem ∬(∇xF).N dA where surface S is the paraboloid z = 0.5(x^2 + y^2) bound by the plane z=2, C is its boundary, and the vector field F = 3yi - xzj + yzk


    I had found (∇xF) = (z+x)i + (-z-3)k

    r = [u, v, 0.5(u^2 + v^2)]
    Therefore N= ru X rv = -ui -uj +k[/tex]
    First this is wrong because this N is not a unit vector.

    Therefore (∇xF).N = [(z+x), 0, (-z-3)].[-x, -y, 1]
    After that I substitute x = rcos(θ), y = rsin(θ), z = 0.5r^2
    Thus ∫(0-2)∫(0-2pi) (∇xF).N r dθdr

    But I cant seems to get the answer. Can anyone help? Help would greatly appreciated
    It seems to me simpler to use u= r \cos(\theta), v= r \sin(\theta) from the start: \rho(r, \theta)= r \cos(\theta)\vec{i}+ r\sin(\theta)\vec{j}+ (1/2)r^2\vec{j} so that \rho_r= \cos(\theta)\vec{i}+ \sin(\theta)\vec{j}+ r\vec{k} and \rho_\theta= -r \sin(\theta)\vec{i}+ r \cos(\theta)\vec{j}.

    Now, \rho_r\times\rho_\theta= -r^2 \cos(\theta)\vec{i}- r^2 \sin(\theta)\vec{j}+ r\vec{k}.

    There is no need to calculate "N" and "dS" separately since both would involve calculating the length of that vector which would then cancel. Instead \vec{N}dS= d\vec{S}= (-r^2 \cos(\theta)\vec{i}- r^2 \sin(\theta)\vec{j}+ r\vec{k})dr d\theta.

    \nabla F= (z+ x)\vec{i}+ (-z- 3)\vec{k}= ((1/2)r+ r \cos(\theta))\vec{i}- ((1/2)r- 3)\vec{j} so \nabla F\cdot d\vec{S}= (-r^2((1/2)\cos(\theta)+ \cos^2(\theta)- ((1/2)r- 3)r^2 \sin(\theta))dr d\theta.

    The parts involving only \sin(\theta) and \cos(\theta) are easy- the integral with respect to \theta is 0.

    That leaves only -\int_{r= 0}^2 \int_{\theta= 0}^{2\pi} r^2 \cos^2(\theta) d\theta dr.

    By the way, notice the negative sign in front of that. Doing a vector integration over a surface involves a choice of orientation. I chose to orient with "z component upward". Choosing z component downard will reverse the sign. Of course, in using Stoke's theorem, you also have to choose an orientation for the path integral around the boundary- which direction you will integrate around the boundary. An orientation for one implies a corresponding orientation for the other.
    Last edited by Jester; May 11th 2010 at 05:59 AM. Reason: fixed latex
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