Example 1: Flux of $\vec{F}(\vec{r}) = \vec{r}/r^3$ Through a Sphere

If S is a sphere of radius R centered at the origin, what is the flux of $\vec{F} = \vec{r}/r^3$ out of this sphere?

At first glance, we might think to use the divergence theorem, since the surface is closed. We can easily calculate that $\mbox{div} \vec{F} = 0$so we might think that

$$\int_S \vec{F} \cdot \hat{n}dS = \int_W \mbox{div} \vec{F} dV = 0.$$

There's one big problem with this, though. At the origin (the center of this sphere,) $\vec{r} = \vec{0}$ and r = 0. So the vector field is not defined (it has a singularity) at one point inside W. That's enough to violate condition 5 above. Thus we cannot use the divergence theorem.

To correctly compute the flux, note that the outward normal to the surface is $\hat{n} = \vec{r}/r$. Thus, the vector field and surface are parallel everywhere. This tells us that

$$\vec{F}\cdot \hat{n}dS = \vert\vert\vec{F}\vert\vert\vert\vert\hat{n}\vert\vert dS = \frac{1}{R^2}dS.$$

Thus, the flux is easily calculated to be

$$\int_s \vec{F} \cdot \hat{n}dS = \int_s \frac{1}{R^2}dS = \f... ...c{1}{R^2}(\mbox{area of sphere}) = \frac{4\pi R^2}{R^2} = 4\pi.$$

Note that if the sphere were centered at the point (2,0,0) with a radius of 1, then the singularity of the vector field at the origin would lie outside the surface. The divergence theorem would then give us the flux as zero in one quick, easy step.