This problem has myself and quite a few others stumped. Someone on another forum recommended this place, so I thought I'd give it a try. The question is this:
Flux of a vector $v$ through a surface $S$ is given by the integral...
$$\iint_S(v \cdot n) \, ds$$
...where $n$ is a unit vector on $S$ directed outward, and $ds$ is the area element on $S$. Find analytically the flux of the vector field $v = (\cos{2 \pi x}, y^3, z^3)$ through the unit sphere with the center at the origin.
When finding flux through a closed surface (such as a sphere) you can use the divergence theorem. This means: $$\iint_S (v \cdot n)ds = \iiint_E \nabla \cdot v \,dV$$ where $E$ is the (solid) unit ball.
$\nabla \cdot v = -2 \pi \sin(2 \pi x)+3y^2+3z^2$. So you simply need to calculate: $$\iiint_{x^2+y^2+z^2 \leq 1} -2\pi\sin(2\pi x)+3(y^2+z^2)\,dV$$
Now sine is an odd function and the region is symmetric with respect to $x$ so this term yields 0 (toss it out): $$\iiint_{x^2+y^2+z^2 \leq 1} 3(y^2+z^2)\,dV$$
In cylindrical coordinates [$x=x$ ,$y=r\cos(\theta)$, $z=r\sin(\theta)$]: $$\int_0^{2\pi}\int_0^1 \int_{-\sqrt{1-r^2}}^{\sqrt{1-r^2}} 3r^2 \cdot r \,dx\,dr\,d\theta = 2\pi \int_0^1 6r^3\sqrt{1-r^2}\,dr = \frac{8}{5}\pi$$