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Math Help - Flux

  1. #1
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    Flux

    Find the flux of F across the surface sigma by expressing sigma parametrically.

    F(x,y,z)=xi+yj+zk; sigma is the portion of the cylinder x^2+z^2=1 between the planes y=1 and y=-2, oriented by outward unit normals.
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  2. #2
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    Find the flux of F across the surface sigma by expressing sigma parametrically.

    F(x,y,z)=xi+yj+zk; sigma is the portion of the cylinder x^2+z^2=1 between the planes y=1 and y=-2, oriented by outward unit normals.

    Parametrize the given surface as x = 1 cos \theta , y = t and z = 1 sin \theta. So we have the limits as 0 \leq \theta \leq 2\pi and 1 \leq t \leq -2.

    The surface is now \Phi(\theta,t) = (cos\theta, t ,sin\theta).
    To calculate the normal to the surface, we have, \left(\frac{d\Phi}{d\theta}(\theta,t) \times \frac{d\Phi}{dt}(\theta,t)\right).
    \frac{d\Phi}{d\theta}(\theta,t) = (-sin\theta, 0 , cos\theta).

    \frac{d\Phi}{dt}(\theta,t) = (0, 1, 0)
    \frac{d\Phi}{d\theta}(\theta,t) \times \frac{d\Phi}{dt}(\theta,t) = \begin{array}{|l c r|}i&j&k\\-sin\theta&0&cos\theta\\0&1&0\end{array} = (-cos\theta, 0 , -sin\theta)

    The flux is now calculated as \int_1^{-2}\int_0^{2\pi} F(\Phi(\theta,t)) . \left(\frac{d\Phi}{d\theta}(\theta,t) \times \frac{d\Phi}{dt}(\theta,t)\right) d\theta \ dt

    Could you substitute and complete this?
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  3. #3
    MHF Contributor Calculus26's Avatar
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    The term

    has to be in the same dirction os the orientation vector n . since n is outward we would use [cos(theta),0, sin(theta)]

    Also the outer limit of integration shoud be -2 to 1

    Of course these cancel and you would have the correct answer --if thats what you had in mind I apologize for my remarks.
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