The divergence of F:

$\nabla ·\mathbf{F}\={\partial }_x\left(x y^3\right)\+{\partial }_y\left(y z\right)\+{\partial }_z\left(x^{2}z\right)\=y^3\+z\+x^2$ 

Implement the integral of $\nabla ·\mathbf{F}$ over the interior of $R$:

${\int }_{-1}^1{\int }_{-\sqrt{1-x^2}}^{{\sqrt{1-x^2}}_{}}{\int }_{x^2\+y^2}^{2-x^2-y^2}\left(y^3\+z\+x^2\right) \mathit{\ⅆ}z \mathit{\ⅆ}y \mathit{\ⅆ}x$ = $\frac{7}{6} \mathrm{\π}$ 

To compute the flux through $R$, note that there are two boundaries, the lower paraboloid $z\=x^2\+y^2$, and the upper paraboloid $z\=2-x^2-y^2$. The intersection of these two bounding surfaces is a circle of radius 1 lying in the plane $z\=1$.

To compute the flux through the lower paraboloid, note that on it

If this be integrated over the unit disk, the result is

${\int }_{-1}^1{\int }_{-\sqrt{1-x^2}}^{{\sqrt{1-x^2}}_{}}\left(2 x^2{y}^3\+\left(2 y^2-x^2\right) \left(x^2\+y^2\right)\right) \mathrm{dy} \mathrm{dx}\=\frac{\mathrm{\π}}{6}$ 

On the upper paraboloid,

Its integral over the unit disk is

${\int }_{-1}^1{\int }_{-\sqrt{1-x^2}}^{{\sqrt{1-x^2}}_{}}\left(2 x^2{y}^3\+\left(2 y^2-x^2\right) \left(2- x^2- y^2\right)\right) \mathrm{dy} \mathrm{dx}\=\mathrm{\pi }$ 

The total flux is then $\left(\frac{1}{6}\+1\right) \mathrm{\π}\=\frac{7}{6} \mathrm{\π}$, the same value obtained for the volume integral of the divergence, as predicted by the Divergence theorem.

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.

There are two parts to the boundary of $R$, the lower paraboloid $z\=x^2\+y^2$ and the upper paraboloid $z\=2-x^2-y^2$. The intersection of these two bounding surfaces is a circle of radius 1 lying in the plane $z\=1$. For the flux through the lower paraboloid, use a task template.

For the "open" surface $z\=x^2\+y^2$, Maple uses the normal ${\mathbf{R}}_x\times {\mathbf{R}}_y$, where $\mathbf{R}\=x \mathbf{i}\+y \mathbf{j}\+\left(x^2\+y^2\right) \mathbf{k}$ is a position-vector description of this paraboloid. This gives a normal that is upward, and hence inward for the closed region $R$. Because $R$ is a closed region, the normal should be outward, and hence downward; the flux through the lower paraboloid is actually $\mathrm{\π}\/6$.$$

For the flux through the upper paraboloid can be obtained with the same task template.

The total flux is then $\left(\frac{1}{6}\+1\right) \mathrm{\π}\=\frac{7}{6} \mathrm{\π}$, the same value obtained for the volume integral of the divergence, as predicted by the Divergence theorem.

For the "open" surface $z\=x^2\+y^2$, Maple uses the normal ${\mathbf{R}}_x\times {\mathbf{R}}_y$, where $\mathbf{R}\=x \mathbf{i}\+y \mathbf{j}\+\left(x^2\+y^2\right) \mathbf{k}$ is a position-vector description of the paraboloid. This gives a normal that is upward, and hence inward for the closed region $R$. Because $R$ is a closed region, the normal should be outward, and hence downward; the flux through the paraboloid is actually $\mathrm{\π}\/6$.$$

The flux on the upper paraboloid can also be obtained with the Flux command.

The total flux is then $\left(\frac{1}{6}\+1\right) \mathrm{\π}\=\frac{7}{6} \mathrm{\π}$, the same value obtained for the volume integral of the divergence, as predicted by the Divergence theorem.