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Thread: Flux across a surface

  1. #1
    Aug 2010

    Flux across a surface

    Find the flux of the field $\displaystyle \mathbf{F}$ across the portion of the sphere $\displaystyle x^2+y^2+z^2=a^2$ in the first octant in the direction away from the origin, where $\displaystyle \mathbf{F} = y\mathbf{i} -x\mathbf{j} + \mathbf{k}$

    I get:

    Flux = $\displaystyle \displaystyle \int_S \mathbf{F} . \mathbf{n} d\sigma = \int_R \mathbf{F}. \frac{\pm \bigtriangledown g}{\bigtriangledown . \mathbf{p}} dA$
    $\displaystyle \bigtriangledown g = 2x\mathbf{i}+2y\mathbf{j}+2z\mathbf{k}$

    But i cannot find what is the right $\displaystyle \mathbf{p}$
    Last edited by FGT12; Mar 20th 2011 at 02:37 AM.
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  2. #2
    MHF Contributor

    Apr 2005
    Well, where ever you got that formula should tell you what "p" is- but I don't recognize that formula.

    Here is how I would do that problem. First, because of the spherical symmetry, I would use spherical coordinates to get parametric equations for the sphere. Taking $\displaystyle \rho= a$ in spherical coordinates, $\displaystyle x= a cos(\theta)sin(\phi)$, $\displaystyle y= a sin(\theta)sin(\phi)$, and $\displaystyle z= a cos(\phi)$.

    That is, we can write the "position vector" for any point on the sphere as
    $\displaystyle \vec{r}(\theta, \phi)= x\vec{i}+ y\vec{j}+ z\vec{k}= a cos(\theta)sin(\phi)\vec{i}+ a sin(\theta)sin(\phi)\vec{j}+ a cos(\phi)\vec{k}$

    The derivatives with respect to the parameter are vectors in the tangent plane at each point:
    $\displaystyle \vec{r}_\theta= -a sin(\theta)sin(\phi)\vec{i}+ a cos(\theta)sin(\phi)\vec{j}$
    $\displaystyle \vec{r}_\phi= a cos(\theta)cos(\phi)\vec{i}+ a sin(\theta)cos(\phi)\vec{j}- a sin(\phi)\vec{k}$.

    The cross product of those two vectors, the "fundamental vector product" for the surface, is normal to the surface at each point and, in fact, gives the vector "differential of suface area:
    $\displaystyle a^2cos(\theta)sin(\phi)cos(\phi)\vec{i}+ a^2sin(\theta)sin(\phi)cos(\phi)\vec{j}+ a^2sin(\phi)cos(\phi)\vec{k}$
    (Of course, changing the order in which you take the cross product changes the sign: $\displaystyle \vec{r}_\theta\times\vec{r}_\phi= -\vec{r}_\phi\times\vec{r}_\theta$. We know each component is "+" because the problem specifically says "in the direction away from the origin")

    The flux of any vector function across that sphere is the integral of the dot product of the vector function with that vector $\displaystyle d\theta d\phi$

    Your function is $\displaystyle F(x,y,z)= y\vec{i}- x\vec{j}+ \vec{k}= a sin(\theta)sin(\phi)\vec{i}- a cos(\theta)\vec{j}+ \vec{k}$ so you should find that the $\displaystyle \vec{i}$ and $\displaystyle \vec{j}$ terms cancel and your integral is
    $\displaystyle \int_{\theta= 0}^{2\pi}\int_{\phi= 0}^\pi a^2 sin(\phi)cos(\phi)d\phi d\theta$
    which is very easy.

    Because of the symmetry here, the answer should not be too surprizing.
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  3. #3
    Senior Member
    Mar 2010
    Beijing, China
    an alternative way to solve, using the Stokes formula.

    Notice that divF=0 so the net flux for a closed surface is 0. Choose the 3 plane parts: I the zoy plane, II the zox plane, III the xoy plane. These 3 parts together with the sphere surface part S form a closed surface and we have
    $\displaystyle \oint_{I+II+III+S} \langle \mathbf{F}, \mathbf{n} \rangle d\sigma = 0$.
    So $\displaystyle \int_S \langle \mathbf{F}, \mathbf{n} \rangle d\sigma $
    = $\displaystyle -\int_{I+II+III} \langle \mathbf{F}, \mathbf{n} \rangle d\sigma$
    = $\displaystyle \int_I y dydz - \int_{II}x dzdx + \int_{III}dxdy $
    = $\displaystyle \int_{III}dxdy $
    = $\displaystyle a^2\pi/4$
    Last edited by xxp9; Mar 20th 2011 at 06:23 AM.
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