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Thread: Surface integration

  1. #1
    Junior Member
    Mar 2011

    Surface integration

    I think my method is correct here, but i must of gone wrong somewhere (note i cannot use divergence theorem)

    I have a vector field:
    F = ax\vec{i}
    Where a is a constant and a sphere:
    {x}^{2} + {y}^{2} + {z}^{2} \le {R}^{2}
    R is the radius.

    Now i have to do a surface integral i.e. not gauss's divergence theorem of a volume integral. Though i did this in order to match my answers and i got
    \frac{4a\pi {R}^{3}}{3}
    Though i haven't yet found a method to get near to an answer for the surface integral. I know:
    dA = ds cos(\theta)
    cos(\theta) = \vec{n} . \vec{i or j or k}
    flux= \phi = \int \vec{F}.d\vec{S} = \iint \vec{F}.\vec{n} dA

    Though in this situation dx, dy and dz are all present:
    \phi = \iint \vec{F}.\vec{n} \frac{dxdy}{\vec{n}.\vec{k}} + \iint \vec{F}.\vec{n} \frac{dxdz}{\vec{n}.\vec{j}} + \iint \vec{F}.\vec{n} \frac{dydz}{\vec{n}.\vec{i}}
    \vec{N} = \nabla g(x,y,z) = \nabla ({x}^{2} + {y}^{2} + {z}^{2} - {R}^{2}) = 2x\vec{i} + 2y\vec{j} + 2z\vec{k}
    \vec{n} = \frac{\vec{N}}{|N|}

    then i sub in n into the flux do the dot products, cancel the 1/|N| terms and get:
    \iint \frac{a{x}^{2}}{z} dxdy + \iint ax dydz + \iint \frac{a{x}^{2}}{y} dzdx

    Now i introduce spherical coordinates whereby:
    dx dy = |J1| dr d\theta = 0
    dy dz = |J2| d\theta d\phi = {r}^{2}{sin}^{2}(\theta) cos(\phi) d\theta d\phi
    dxdz = |J3| dr d\phi = \frac{r}{2} sin(2\theta)sin(\phi)

    Now it looks a little messy but is still all doable - albeit a little long winded but im not allowed to use the divergence theorem.

    To cut it short i get
    2a\pi{r}^{3} -( \frac{a\pi{r}^{3}}{6} sin(\theta)sin(2\theta))

    Which is not the obvious and the one i got from divergence theorem:
    \frac{4}{3} a\pi {r}^{3}

    Any help is much appreciated and thanks for reading
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  2. #2
    Junior Member
    Mar 2011

    Re: Surface integration

    If i may add, one point I was surprised about was the fact that the first jacobian came out to be zero, though i beleive i have done it correctly
    \frac{\partial r}{\partial x}\frac{\partial \theta}{\partial y} - \frac{\partial r}{\partial y}\frac{\partial \theta}{\partial x} ?
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  3. #3
    MHF Contributor

    Apr 2005

    Re: Surface integration

    I really dislike that way of doing a surface integral. Instead, I would do this:
    Start by using spherical coordinates, with \rho= R, a constant: x= R cos(\theta)sin(\phi), y= R sin(\theta)sin(\phi), z= R cos(\phi) so we can write the "position vector of any point on surface as
    \vec{r}= R cos(\theta)sin(\phi)\vec{i}+ Rsin(\theta)sin(\phi)\vec{j}+ Rcos(\phi)\vec{k}.

    The derivatives of that vector with respect to the two parameters,
    \vec{r}_\theta= -R sin(\theta)sin(\phi)\vec{j}+ Rcos(\theta)sin(\phi)\vec{j} and
    \vec{r}_\phi= Rcos(\theta)cos(\phi)\vec{i}+ Rsin(\theta)cos(\phi)\vec{j}- Rsin(\phi)\vec{k}
    are tangent to the surface and their lengths contain information about the differentials of arc length. So their cross product
    R^2 cos(\theta)sin^2(\phi)\vec{i}+ R^2sin(\theta)sin^2(\phi)\vec{j}+ R^2sin(\phi)cos(\phi)\vec{k}
    is normal to the surface and contains all differential information.
    (I chose the order of multiplication to give the outward normal)

    In other words, that is \vec{r}_\theta\times\vec{r}_\phi d\theta d\phi= d\vec{S}.

    Now, \vec{F}= ax\vec{i}= aRcos(\theta)sin(\phi)\vec{i} so that
    \vec{F}\cdot d\vec{S}k= aR^3cos^2(\theta)sin^3(\phi)d\theta d\phi
    and your integral is
    aR^3\int_{\theta= 0}^{2\pi}cos^2(\theta)d\theta\int_{\phi= 0}^\pi sin^3(\phi)d\phi
    which does, in fact, give \frac{4}{3}a\pi R^3.
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