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Since $\mathbf{a}$ is constant, componentwise differentiation gives $\nabla \cdot (\mathbf{a}\varphi )={\sum }_i{a}_i{\partial }_i\varphi =\mathbf{a}\cdot \nabla \varphi$. Applying the divergence theorem to this vector field givesThis holds for every constant vector $\mathbf{a}$, so the vector integrals agree. This is the vector gradient form of the divergence theorem. With $\varphi =z$, the gradient is the constant unit vector ${\mathbf{e}}_z$. Therefore

Parametrize the directed segment by $\mathbf{r}(t)=(t,t,0)$, $0\le t\le 1$. Its tangent is $(1,1,0)$, while the restricted vector field is $(t^3,t^3,0)$. The line integral is thereforeAlternatively, $\mathbf{G}=\nabla (x^2{y}^2/2)$, so the fundamental theorem for line integrals gives the same endpoint difference.

Only the $x$ component of the vector field is nonzero, so $\mathbf{F}\cdot d\mathbf{r}=(x^2+y^2)dx$. On the vertical segment, $dx=0$ throughout. Hence

For a sufficiently smooth vector field on a volume $V$ and its closed boundary $S$, the divergence theorem stateswith the unit normal directed outwards. HereHence the net volume flux through every closed surface in the liquid is zero. This verifies incompressible flow directly. It does not require the velocity to vanish pointwise on an arbitrary closed surface; inward and outward contributions can cancel.

The curl of the first vector field iswhich is not identically zero. Thus $\mathbf{F}$ is not conservative on ${\mathbb{R}}^3$ and cannot equal a negative gradient there.

For the second field, direct differentiation showsTherefore $\mathbf{G}$ is conservative, and the requested negative-gradient potential isIts curl vanishes, consistently with the curl of a gradient identity.

Write $\mathbf{a}=(a_x,a_y,a_z)$. The cross product gives ${\mathbf{v}}_1=(a_{y}z-a_{z}y,a_{z}x-a_{x}z,a_{x}y-a_{y}x)$, whose three diagonal partial derivatives vanish. For the second vector field, direct differentiation gives ${\partial }_x(2xy)=2y$, ${\partial }_y(x^2+3y^2+z^2)=6y$, and ${\partial }_z(2yz)=2y$. 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.

On a connected domain, a vector field is a conservative vector field if and only if it is the gradient of a single-valued scalar potential, $\boxed{\mathbf{H}=\nabla \varphi }$. Equivalently, its line integral is path independent, or its integral around every closed curve is zero. For a continuously differentiable field on a simply connected domain, this is equivalent to $\nabla \times \mathbf{H}=0$. Without the domain hypothesis, vanishing curl alone is only a local test and is not the necessary-and-sufficient global statement.

For a smooth oriented surface $S$ with piecewise smooth boundary $C=\partial S$ and a continuously differentiable vector field $\mathbf{G}$, the Stokes theorem statesThe boundary orientation is fixed by the right-hand rule: curling the fingers along the positive boundary direction makes the thumb point along the surface normal. For an outward upper cap, its boundary ellipse is traversed counterclockwise when viewed from the positive $z$ direction. The figure shows two possible caps with the same oriented boundary, anticipating the last part.

The homogeneous equation is the dot product condition $(2y,-x)\cdot \nabla u=0$. A suitable vector field is thereforeWhere nonzero, it is tangent to the level curves of $u$ and perpendicular to its gradient. At the origin this vector vanishes, so no direction is defined there.

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

The curl of the constant vector field $\mathbf{a}\times \mathbf{b}$ vanishes. The curl of the position field $\mathbf{r}=(x,y,z)$ also vanishes, since all its cross-coordinate partial derivatives are zero. Thus

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:

The vector field is radial, while the differential displacement along a centred circle is tangential. These directions are orthogonal, so the dot product vanishes at every point. Hence

Write $q(x)=1/[x(1+(\ln x{)}^2)]$. The vector field is undefined at $x=0$ because $q$ diverges there, but this component is tangential to that face. The face fluxes have finite improper integral limits. In particular, the substitution $s=\ln x$ givesUsing the outward unit normal on each face, the flux contributions areHenceFor a precise justification at the singular face, first use the truncated cube $\epsilon \le x\le 1$. Its opposite $y$-face integrals cancel, the $x$-faces contribute $e^{-1}-e^{-\epsilon }$, and the upper $z$-face contributes $1-\epsilon$. The limit as $\epsilon \downarrow 0$ is $e^{-1}$. On each truncated cube the divergence theorem gives the same answer, since $\nabla \cdot \mathbf{F}=1-e^{-x}$. Thus the result is a flux of a vector field with an integrable tangential singularity, rather than an application of the classical smooth-field theorem directly at $x=0$.

Compute the flux directly using the outward vector area element, as required. Restrict the vector field to the surface parametric equation and take the dot product:The terms quadratic in $r$ cancel. Therefore the surface integral isThis is a direct evaluation of the flux of the position vector through a cone.

The new cone can be parametrized byThe outward vector area element is $r(\frac{b}{a}\cos \theta ,1,\frac{b}{a}\sin \theta )drd\theta$. This is ${\mathbf{R}}_{\theta }^{\prime }\times {\mathbf{R}}_r^{\prime }drd\theta$; reversing the order is needed to retain the outward orientation after the exchange of coordinates. Its dot product with $\mathbf{F}({\mathbf{R}}^{\prime })$ is again $brdrd\theta$. HenceThis equality also follows because a rigid rotation takes the first cone into the second and leaves the radial vector field $\mathbf{F}(\mathbf{x})=\mathbf{x}$ invariant.

Parametrize the segment in its stated direction by $\mathbf{x}(t)=(1-t,t,t)$, $0\le t\le 1$. Then $d\mathbf{x}=(-1,1,1)dt$, and substitution into the two vector fields givesTaking each dot product with $d\mathbf{x}$ evaluates the line integrals:

The proposed parametric equation traces precisely the same oriented segment, with a different speed. Set $u=t^n$, which increases from zero to one, and $du=n{t}^{n-1}dt$. Direct substitution gives the line integralsThus both values are unchanged. The derivative of $u$ vanishes at the initial endpoint for $n>1$, but the monotone integration by substitution is still valid. This is invariance under reparametrization of one curve, which applies even to a nonconservative vector field; it is not evidence of general path independence.