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For the divergence in Cartesian coordinates, differentiate each component with respect to its own coordinate:The third component contributes zero because it is independent of $z$.

The three Cartesian contributions to the divergence are $2(y+z)$, $-2y$, and $-2z$. Hence

Take the Cartesian partial derivatives component by component. The divergence isThe constant third component has zero derivatives. The curl consequently has only two nonzero components:

Each Cartesian component of $a\mathbf{r}$ contributes $a$ to the divergence, while the constant vector contributes zero. Hence

The divergence is $y^2+x^2+z^2=r^2$. In spherical coordinates, the angular integral is $4\pi$ and the volume element contributes $r^{2}dr$. Thus

Direct calculation of the curl givesSince $\mathbf{a}\ne 0$, the first vector field cannot be a gradient field: the curl of a gradient is zero. The second is a smooth polynomial field on all of the simply connected space ${\mathbb{R}}^3$, so its vanishing curl makes it a conservative vector field. A potential is also constructed explicitly in the next part. Thus ${\mathbf{v}}_2$ is a conservative vector field and ${\mathbf{v}}_1$ is not. A vanishing divergence alone, as for ${\mathbf{v}}_1$, is a different property and does not establish conservativity.

Take the curl component by component:Its divergence is zero, as expected because the divergence of a curl is zero.

For a Cartesian vector field $\mathbf{F}=(F_x,F_y,F_z)$, its divergence is

Use the prism bounds read from the PDF diagram and the divergence from part (ii):Set $A=4-2x$ in the inner evaluation. Since $4x-9=-2A-1$, the inner integral is $A^2-(2A+1)A=-A^2-A$. Therefore

Each component of the cross product in part (a) is independent of the coordinate with respect to which the divergence differentiates it. Thus

The vector triple product rewrites the field as $\mathbf{b}(\mathbf{a}\cdot \mathbf{r})-\mathbf{r}(\mathbf{a}\cdot \mathbf{b})$. Its first term has divergence $\mathbf{b}\cdot \nabla (\mathbf{a}\cdot \mathbf{r})=\mathbf{a}\cdot \mathbf{b}$, while the second contributes $3\mathbf{a}\cdot \mathbf{b}$. Therefore

The fixed vectors have constant components, so $\mathbf{a}\cdot \mathbf{b}$ is constant, $\nabla \cdot \mathbf{a}=0$ and $\nabla \cdot \mathbf{r}=3$. Hence the divergence is

Take the outward orientation of the complete closed surface, including its flat base. The constant part of the vector field has zero divergence, and the radial part has divergence $3\mathbf{a}\cdot \mathbf{b}$. The divergence theorem gives its flux as that constant times the solid hemisphere's volume, $2\pi R^3/3$:The constant field may have nonzero flux through the curved part alone, but its curved and base contributions cancel on the closed surface.

The vector field has only a vertical component, namely $f(r)$. Its divergence is therefore its vertical partial derivative:

In the Cartesian coordinate system, the Laplacian of a scalar field is the sum of its three unmixed second partial derivatives. Each squared coordinate here contributes one, soEquivalently $\nabla \psi =(x,y,z)$ and its divergence is three.

The curl found in part (a) is constant, so its divergence isThis also follows from the identity divergence of a curl is zero, applicable here because the field is smooth everywhere.

The divergence isFor the tetrahedral domain, convenient integration bounds areConsequently its volume integral isPutting $h=2(1-x-y)$, the inner integral is $(2y+6)h+h^2=(16-4x)(1-x-y)$. ThereforeThusThere is an independent centroid check. The vertices are $(0,0,0),(1,0,0),(0,1,0),(0,0,2)$, giving volume $1/3$ and centroid $(1/4,1/4,1/2)$. An affine function averages to its value at the centroid, so $I=(1/3)(6+2/4+2/2)=5/2$. This is the integral of an affine function over a simplex.

Expanding the cross product givesTaking its divergence component by component,

Let $V$ be the enclosed region with piecewise smooth closed boundary $S$, oriented by its outward unit normal $\mathbf{n}$. For a continuously differentiable vector field on a neighbourhood of $V$, the divergence theorem statesThe outward flux integral equals the volume integral of the divergence.

The divergence of the position vector field is $\nabla \cdot \mathbf{r}=3$. At height $z$, the horizontal cross-section is a disk of area $\pi (1-z^2)$. By the divergence theorem, the outward flux integral is three times the enclosed volume:ThereforeA direct check separates the two pieces of the surface integral. On the unit sphere, the outward unit normal is $\mathbf{r}$, so $\mathbf{r}\cdot \mathbf{n}=1$ and the curved contribution is $2\pi (z_0+1)$. On the top disk, $\mathbf{n}={\mathbf{e}}_z$, giving $\pi z_0(1-z_0^2)$. Their sum is the same flux.