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Math Help - Computing the flux of the vector field F

  1. #1
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    Computing the flux of the vector field F

    Compute the flux of the vector field F = 5yi + 2j - 5xzk through the surface S, which is the surface y = x + z, with x + z ≤ 9, oriented in the positive y-direction.

    I understand that F = <5y, 2, -5xz>, but how do I find dA? and would the bounds for x and z be [0, 3]?

    please help...
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  2. #2
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    Re: Computing the flux of the vector field F

    I would change to polar coordinates in the xz-plane. That is, x= r cos(\theta), z= r sin(\theta), y= x^2+ z^2= r^2 with r\le 3
    Now the "position vector" for any point on that paraboid would be \vec{p}= x\vec{i}+ y\vec{j}+ z\vec{k}= r cos(\theta)\vec{i}+ r^2\vec{y}+ r sin(\theta)\vec{k}. The derivatives are \vec{p}_r= cos(\theta)\vec{i}+ 2r\vec{j}+ sin(\theta)\vec{k} and \vec{p}_\theta= -r sin(\theta)\vec{i}+ r cos(\theta)\vec{k}.

    The cross product of those vectors is
    \left|\begin{array}{ccc}\vec{i} & \vec{j} & \vec{k} \\ cos(\theta) & 2r & sin(\theta) \\ -r sin(\theta) & 0 & r cos(\theta)\end{array}\right|= 2r^2cos(\theta)\vec{i}- r\vec{j}+ 2r^2sin(\theta)\vec{k}
    so the "vector differential of surface area" is
    d\vec{S}= (2r^2cos(\theta)\vec{i}- r\vec{j}+ 2r^2sin(\theta))drd\theta

    But since we want "oriented in the positive y direction" we need to multiply throught by -1 to make the y-component positive:
    d\vec{S}= (-2r^2cos(\theta)\vec{i}+ r\vec{j}- 2r^2sin(\theta))drd\theta
    (That is the same as swapping the two derivative vectors in the cross product.)

    Since r\le 3, and there is circular symmetry on the xz-plane, the limits of integration are r= 0 to 3 and \theta= 0 to 2\pi.
    Last edited by HallsofIvy; April 13th 2013 at 10:46 AM.
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