The Flux of vector field F across unit sphere centered on the origin Vector Field F is given by
$F= 3z^2i + 2yj + xk$
Calculate the the flux of $F$ across the sphere of radius 1 centered on the origin
Hint is to use the Divergence theorem but I am having trouble applying it 
Any help is appreciated
 A: The divergence theorem says that
$\int_{\partial M} \vec X \cdot \vec n dS = \int_M \nabla \cdot \vec X dV, \tag{1}$
where $M$ is a closed,  bounded region in $\Bbb R^3$ with smooth boundary $\partial M$ which is an orientable surface in $\Bbb R^3$ with outward pointing normal field $\vec n$; here $dV$ is the volume element in $M$ and $dS$ is the area element on $\partial M$.  The flux of $\vec X$ through $\partial M$ is, by definition, the left-hand integral in (1).  In the present case, it is particulary easy to evaluate the volume integral on the right-hand side of (1); taking $\vec X = F = 3z^2 i + 2y j +  xk$, we see that
$\nabla \cdot F = \dfrac{\partial(3z^2)}{\partial x} + \dfrac{\partial(2y)}{\partial y} + \dfrac{\partial(x)}{\partial z} = 2; \tag{2}$
the flux of $F$ through the unit sphere $S^2$ is then,  by (1),
$\int_{S^2} F \cdot \vec n dS = \int_B 2 dV = 2 \int_B dV, \tag{3}$
where $B = \{(x, y, z) \in \Bbb R^3 \mid x^2 + y^2 +z^2 \le 1 \}$ is the unit ball.  We have
$\int_B dV = \dfrac{4 \pi}{3}, \tag{4}$
so that
$\int_{S^2} F \cdot \vec n dS = \dfrac{8 \pi}{3}. \tag{5}$
Hope this helps.  Cheerio,
and as always,
Fiat Lux!!!
