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Thread: 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 $\displaystyle $F\left( {x,y,z} \right) = \left( {x^3 ,y^3 ,z^3 } \right)$$
    across the surface of the cone $\displaystyle $x^2 + y^2 = z^2$$ with 0 < z < H
    a) Directly
    b) Applying Gauss's theorem
    Last edited by mr fantastic; Jul 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: $\displaystyle x= r cos(\theta)$, $\displaystyle y= r sin(\theta)$, $\displaystyle z= r$ with $\displaystyle r$ from 0 to H and [/tex]\theta[/tex] from 0 to $\displaystyle 2\pi$.

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

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

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

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

    And what is "Gauss's theorem"?
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