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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:
    $\displaystyle F = ax\vec{i}$
    Where a is a constant and a sphere:
    $\displaystyle {x}^{2} + {y}^{2} + {z}^{2} \le {R}^{2}$
    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
    $\displaystyle \frac{4a\pi {R}^{3}}{3}$
    Though i haven't yet found a method to get near to an answer for the surface integral. I know:
    $\displaystyle dA = ds cos(\theta)$
    $\displaystyle cos(\theta) = \vec{n} . \vec{i or j or k}$
    $\displaystyle flux= \phi = \int \vec{F}.d\vec{S} = \iint \vec{F}.\vec{n} dA$

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

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

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

    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
    $\displaystyle 2a\pi{r}^{3} -( \frac{a\pi{r}^{3}}{6} sin(\theta)sin(2\theta))$

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

    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
    $\displaystyle \frac{\partial r}{\partial x}\frac{\partial \theta}{\partial y} - \frac{\partial r}{\partial y}\frac{\partial \theta}{\partial x}$ ?
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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 $\displaystyle \rho= R$, a constant: $\displaystyle x= R cos(\theta)sin(\phi)$, $\displaystyle y= R sin(\theta)sin(\phi)$, $\displaystyle z= R cos(\phi)$ so we can write the "position vector of any point on surface as
    $\displaystyle \vec{r}= R cos(\theta)sin(\phi)\vec{i}+ Rsin(\theta)sin(\phi)\vec{j}+ Rcos(\phi)\vec{k}$.

    The derivatives of that vector with respect to the two parameters,
    $\displaystyle \vec{r}_\theta= -R sin(\theta)sin(\phi)\vec{j}+ Rcos(\theta)sin(\phi)\vec{j}$ and
    $\displaystyle \vec{r}_\phi= Rcos(\theta)cos(\phi)\vec{i}+ Rsin(\theta)cos(\phi)\vec{j}- Rsin(\phi)\vec{k}$
    are tangent to the surface and their lengths contain information about the differentials of arc length. So their cross product
    $\displaystyle R^2 cos(\theta)sin^2(\phi)\vec{i}+ R^2sin(\theta)sin^2(\phi)\vec{j}+ R^2sin(\phi)cos(\phi)\vec{k}$
    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 $\displaystyle \vec{r}_\theta\times\vec{r}_\phi d\theta d\phi= d\vec{S}$.

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