Results 1 to 3 of 3

Math Help - Flux across a surface

  1. #1
    Member
    Joined
    Aug 2010
    Posts
    77

    Flux across a surface

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

    I get:

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

    But i cannot find what is the right \mathbf{p}
    Last edited by FGT12; March 20th 2011 at 02:37 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,603
    Thanks
    1421
    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 \rho= a in spherical coordinates, x= a cos(\theta)sin(\phi), y= a sin(\theta)sin(\phi), and z= a cos(\phi).

    That is, we can write the "position vector" for any point on the sphere as
    \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:
    \vec{r}_\theta= -a sin(\theta)sin(\phi)\vec{i}+ a cos(\theta)sin(\phi)\vec{j}
    \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:
    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: \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 d\theta d\phi

    Your function is 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 \vec{i} and \vec{j} terms cancel and your integral is
    \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.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member
    Joined
    Mar 2010
    From
    Beijing, China
    Posts
    293
    Thanks
    23
    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
    \oint_{I+II+III+S} \langle \mathbf{F}, \mathbf{n} \rangle d\sigma = 0.
    So \int_S \langle \mathbf{F}, \mathbf{n} \rangle d\sigma
    = -\int_{I+II+III} \langle \mathbf{F}, \mathbf{n} \rangle d\sigma
    = \int_I y dydz - \int_{II}x dzdx + \int_{III}dxdy
    = \int_{III}dxdy
    = a^2\pi/4
    Last edited by xxp9; March 20th 2011 at 06:23 AM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Surface Flux Integral
    Posted in the Calculus Forum
    Replies: 6
    Last Post: May 3rd 2010, 09:46 AM
  2. another flux through surface question
    Posted in the Calculus Forum
    Replies: 19
    Last Post: April 26th 2009, 07:23 PM
  3. computing flux through a surface
    Posted in the Calculus Forum
    Replies: 4
    Last Post: April 21st 2009, 04:18 PM
  4. Surface Integrals of Flux
    Posted in the Calculus Forum
    Replies: 2
    Last Post: April 11th 2009, 01:12 PM
  5. Surface flux integral
    Posted in the Calculus Forum
    Replies: 2
    Last Post: May 23rd 2008, 07:27 PM

Search Tags


/mathhelpforum @mathhelpforum