1. ## 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:
${x}^{2} + {y}^{2} + {z}^{2} \le {R}^{2}$

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

2. ## 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
$\frac{\partial r}{\partial x}\frac{\partial \theta}{\partial y} - \frac{\partial r}{\partial y}\frac{\partial \theta}{\partial x}$ ?

3. ## 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
$\vec{F}\cdot d\vec{S}k= aR^3cos^2(\theta)sin^3(\phi)d\theta d\phi$
$aR^3\int_{\theta= 0}^{2\pi}cos^2(\theta)d\theta\int_{\phi= 0}^\pi sin^3(\phi)d\phi$
which does, in fact, give $\frac{4}{3}a\pi R^3$.