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Math Help - Surface integral solved directly and using Gauss' Theorem.

  1. #1
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    Surface integral solved directly and using Gauss' Theorem.

    To find the flow of the field $F\left( {x,y,z} \right) = \left( {x^3 ,y^3 ,z^3 } \right)$
    across the surface of the cone $x^2  + y^2  = z^2$ with 0 < z < H
    a) Directly
    b) Applying Gauss's theorem
    Last edited by mr fantastic; July 19th 2010 at 12:02 AM. Reason: Re-titled.
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  2. #2
    MHF Contributor

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    What have you tried?

    I will give you this much: we can write the cone in parametric coordinates using polar coordinates: x= r cos(\theta), y= r sin(\theta), z= r with r from 0 to H and [/tex]\theta[/tex] from 0 to 2\pi.

    With that you can write the "position vector" of a point on the surface of the cone as \vec{p}(r, \theta)= r cos(\theta)\vec{i}+ r sin(\theta)\vec{j}+ r \vec{k}.

    The deriviatives of that with respect to r and \theta:
    \vec{p}_r= cos(\theta)\vec{i}+ sin(\theta)\vec{j}+ \vec{k} and
    \vec{p}_\theta= -r sin(\theta)\vec{i}+ r cos(\theta)\vec{j}
    are in the tangent plane to the cone and their cross product is
    -r cos(\theta)\vec{i}- r sin(\theta)\vec{j}+ r \vec{k}.

    The (upward oriented) "vector differential of area" is (-r cos(\theta)\vec{i}- r sin(\theta)\vec{j}+ r \vec{k})drd\theta

    Take the dot product of your field with that and integrate.

    And what is "Gauss's theorem"?
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