Surface integral of vector field **F** over a unit ball E I'm working on a problem of a vector field $F=\langle x,y,z\rangle $ over a region $E$ i.e. the unit ball $ x^2+y^2+z^2 \le 1 $.
This is an easy surface integral to calculate using the Divergence Theorem:
$$ \iiint_E {\rm div}(F)\ dV = \iint_{S=\partial E} \vec{F}\cdot d{\bf S}$$
However, to confirm the divergence theorem by the direct calculation of the surface integral, how should the bounds on the double integral for a unit ball be chosen?
Since, $div(\vec{F}) = 0$ in this case, hence, it's just a triple integral over volume $\iiint_E 1\,\, dV$. Further, E is a solid ball and $\iiint_E $ results in a volume calculation, we get $\frac{4\pi r^3}{3} = \frac{4\pi 1^3}{3} = \frac{4\pi }{3}$. So this is the final numerical answer I'm looking for, in order to verify the divergence theorem.
 A: After reading a few examples on the web and in my textbook I think I may be have figured this out. It seems that the hardest (or most tedious) part of this problem is actually performing the calculus. Indeed, I needed to calculate the surface integral of a unit sphere floating around in the given vector field, because a unit sphere is the surface that lines a unit ball that was given in the problem.
A unit sphere can be parametrized using spherical coordinates:
$$ r(\phi,\theta) = \sin(\phi)\cos(\theta)\hat{i} + \sin(\phi)\sin(\theta)\hat{j}+\cos{\phi}\hat{k}$$
By definition, a surface integral:
$$ \iint_S \vec{F} \cdot d\vec{S}= \iint_D\vec{F}(\vec{r}(\phi,\theta))\cdot(\vec{r}_u \times \vec{r}_v) dA$$
If you have some time and decide to crank this out and set up the integral:
$$\vec{F}(\vec{r}(\phi,\theta)) = \cos(\phi)\hat{i}+\sin(\phi)\sin(\theta)\hat{j}+\sin
(\phi)\cos(\theta)\hat{k} $$
$$ \vec{r}_\phi \times \vec{r}_\theta = \sin^2(\phi)\cos(\theta)\hat{i}+\sin^2(\phi)\sin(\theta)\hat{j}+\sin(\phi)cos(\phi)\hat{k}$$
Then:
$$\vec{F}\cdot(\vec{r}_u \times \vec{r}_v) = \cos(\phi)sin^2(\phi)\cos(\theta)+\sin^2(\phi)sin^2(\theta)+\sin^2(\phi)\cos(\phi)\cos(\theta)$$
The integral is now ready to put in the oven:
$$\iint_D [\cos(\phi)sin^2(\phi)\cos(\theta)+\sin^2(\phi)sin^2(\theta)+\sin^2(\phi)\cos(\phi)\cos(\theta)] dA$$
$$= \int_0^{2\pi}\int_0^\pi [\cos(\phi)sin^2(\phi)\cos(\theta)+\sin^2(\phi)sin^2(\theta)+\sin^2(\phi)\cos(\phi)\cos(\theta)] d\phi d\theta$$
$$= 2 \int_0^\pi\sin^2(\phi)\cos(\phi) d\phi \int_0^{2\pi}\cos(\theta)d\theta\int_0^\pi\sin^3(\phi)d\phi\int_0^{2\pi}\sin^2\theta d\theta$$
$$= 0 + \int_0^\pi \sin^3(\phi)d\phi \int_0^{2\pi} \sin^2(\theta)d\theta $$
$$ = \frac{4\pi}{3} $$
