The divergence of F:

$\nabla ·\mathbf{F}\={\partial }_x\left(x\right)\+{\partial }_y\left(y\right)\+{\partial }_z\left(z\right)\=1\+1\+1\=3$ 

Using spherical coordinates (volume element: ${\mathrm{\ρ}}^2\sin \left(\mathrm{\φ}\right)$), the integral of $\nabla ·\mathbf{F}$ over the unit sphere:

${\int }_0^{2 \mathrm{\π}}{\int }_0^{\mathrm{\π}}{\int }_0^{1}3 {\mathrm{\ρ}}^2\sin \left(\mathrm{\φ}\right) \ⅆ\mathrm{\ρ} \ⅆ\mathrm{\φ} \ⅆ\mathrm{\θ}$ = $4 \mathrm{\π}$ 

The surface integral of $\mathbf{F}·\mathbf{N} \mathrm{d\σ}$ is the flux of F through the surface $S$, with N being a unit normal on $S$ and $\mathrm{d\σ}$ being the surface-area element $\mathrm{dA}\/\left|z\right|$.

Describe $S$ in Cartesian coordinates by $f\equiv x^2\+y^2\+z^2\=1$, so that $\mathbf{N}\=x \mathbf{i}\+y \mathbf{j}\+z \mathbf{k}$ by normalizing $\nabla f$. Since $\mathbf{F}·\mathbf{N}\=x^2\+y^2\+z^2\=1$ on $S$, the integral of $\mathbf{F}·\mathbf{N} \mathrm{d\σ}$ on $S$ is just the integral of $1\cdot \mathrm{d\σ}$ over the surface of the unit sphere. This is the surface area of the sphere, and that is known to be $4 \mathrm{\π}$.

The purist who wants to evaluate the surface integral in more detail would have to obtain

$\mathrm{d\σ}\=\genfrac{}{}{0ex}{}{\frac{∥\nabla f∥}{\|f_z\|}}{\phantom{x\=a}}|\genfrac{}{}{0ex}{}{\phantom{\mathrm{f(x)}}}{S}$ $\mathrm{dA}$ = $\frac{2}{2\left|z\right|} \mathrm{dA}\=\frac{\mathrm{dA}}{\|z\|}$ $$

and then write the integral over the upper hemisphere in polar coordinates as

${\int }_0^{2 \mathrm{\π}}{\int }_0^1\frac{r}{\sqrt{1-r^2}} \mathrm{dr} d\mathrm{\θ}$ = $2 \mathrm{\π}$ 

On the lower hemisphere where $z$ is negative, $\mathrm{d\σ}\=\mathrm{dA}\/\left|z\right|$ gives the same integral as evaluated on the upper hemisphere because of the absolute value on $z$. Hence, the contribution to the flux through the lower hemisphere is the same $2 \mathrm{\π}$, so that the total flux through the sphere is $4 \mathrm{\π}$.$$

The Student VectorCalculus package is needed for calculating the divergence, but it then conflicts with any multidimensional integral set from the Calculus palette. Hence, the Student MultivariateCalculus package is installed to gain Context Panel access to the MultiInt command.

In the Divergence theorem, the volume integral on the left and the surface flux integral on the right both have the value $4 \mathrm{\π}$.$$

In the Divergence theorem, the volume integral on the left and the surface flux integral on the right both have the value $4 \mathrm{\π}$.$$