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Parametrize the surface of revolution byAn oriented vector surface area element isUse the vector field $\mathbf{F}=(x/2,y/2,0)$, whose divergence is one. If this parametrization has the outward orientation on an enclosing surface, the divergence theorem givesThe second equality is integration by parts, using the endpoint values. For example this outward convention applies when the positive-$t$ branch lies above the negative-$t$ branch, $f(t)>f(-t)$ for $0<t<\pi /2$.

The stated $C^1$ regularity and endpoint conditions alone do not guarantee an embedded enclosing surface: branches can cross or coincide. Without an appropriate orientation assumption the flux formula is an algebraic signed volume. If “inside” is defined as the region between the two graph branches for each radius, its nonnegative geometric volume is insteadThis explicitly handles a reversal or change in the order of the two branches.

For each of the three constant Cartesian unit vectors ${\mathbf{e}}_j$, apply the divergence theorem to the vector field $\varphi {\mathbf{e}}_j$. Its divergence is ${\partial }_j\varphi$, soThese are precisely the three components ofThus the volume integral of the gradient is the boundary surface integral weighted by the vector surface element $d\mathbf{S}=\mathbf{n}dS$. The regularity needed here is that $\varphi$ be continuously differentiable on a neighbourhood of the closure of $R$.

Use cylindrical coordinates $(\rho ,\theta ,z)$. The right circular cone is described by $0\le z\le 1$ and $0\le \rho \le a(1-z)$, and the divergence of $\mathbf{F}=(0,0,z^2)$ is $2z$. Direct integration gives

To check the boundary flux integral independently, first consider the base $z=0$. Its outward unit normal vector is $-\mathbf{k}$, but $\mathbf{F}=0$ there, so its flux is zero. Parametrize the lateral surface byThe cross productpoints outward: its horizontal component points away from the axis of the cone. This gives the vector surface element without separately normalizing it. Consequently the lateral flux integral isAdding the zero base contribution agrees with the volume integral, verifying the divergence theorem. The apex has zero surface area; the vector field is smooth there, so no additional flux is hidden at that point.

The displayed vector field is defined on the punctured plane $U={\mathbb{R}}^2\setminus \{(0,0)\}$; it is undefined at the origin. Take the counterclockwise unit circle $\mathbf{r}(t)=(\cos t,\sin t)$, $0\le t\le 2\pi$. Along it,By the fundamental theorem for line integrals, it has no single-valued potential of a conservative vector field on $U$. In particular it cannot be the restriction of the gradient of a smooth function on all of ${\mathbb{R}}^2$.

For $(x,y)\ne (0,0)$, direct differentiation givesThere is no contradiction with Green's theorem: $\mathbf{F}$ is not defined, let alone continuously differentiable, throughout the disk enclosed by $C$. On an annulus the theorem does apply; the counterclockwise outer integral is $2\pi$, and the clockwise inner integral is $-2\pi$, so their sum is zero. The hole prevents this curl-free vector field from being a conservative vector field on its entire domain.

For an oriented piecewise smooth parametric curve $\mathbf{r}:[a,b]\to {\mathbb{R}}^3$ and a continuous vector field $\mathbf{F}$ along it, the vector-field line integral isFor a piecewise smooth curve, sum these integrals over its smooth pieces. The dot product selects the component of the vector field in the direction of motion.

If $\mathbf{F}=\nabla \varphi$ for a smooth function $\varphi$, the multivariable chain rule givesThe fundamental theorem of calculus, with cancellation at intermediate endpoints, therefore proves the fundamental theorem for line integrals:On a closed curve $\mathbf{r}(a)=\mathbf{r}(b)$, so $\boxed{{\oint }_C\nabla \varphi \cdot d\mathbf{r}=0}$.

For a continuous vector field along an oriented piecewise continuously differentiable parametric curve $\gamma :[a,b]\to {\mathbb{R}}^3$, define its line integral byadding the integrals over the smooth pieces when necessary. The value refers to the specified traversal: going around a curve twice counts twice.

Parametrize the straight segment by $\mathbf{r}(t)=(t,t,t)$, $0\le t\le 1$. Then the line integral isThe three different answers are possible because this vector field is not a conservative vector field; for example ${\partial }_y{F}_x=6$ whereas ${\partial }_x{F}_y=0$.

For the parametric curve $\mathbf{r}(t)=(t,t^2,t^3)$, $0\le t\le 1$, the vector field and the tangent vector areConsequently the line integral is

The sufficiency claim requires a domain qualification. For a continuously differentiable vector field on all of ${\mathbb{R}}^3$, or more generally on a simply connected open domain, the equivalence is valid. On an arbitrary domain, vanishing curl is necessary but may fail to be sufficient.

Necessity. Suppose every closed-curve line integral vanishes. Fix a point $p$ in the open domain and a unit vector $a$. A sufficiently small planar disk $D_{\epsilon }$ of radius $\epsilon$, centred at $p$ and normal to $a$, lies in the domain. Stokes theorem givesDivide by $\pi {\epsilon }^2$ and let $\epsilon \downarrow 0$. By continuity, the average tends to $(\nabla \times \mathbf{F})(p)\cdot a$. Taking the three coordinate directions proves $\boxed{\nabla \times \mathbf{F}=0}$ everywhere.

Sufficiency on ${\mathbb{R}}^3$. A direct proof constructs a potential of a conservative vector field. PutWrite $F_j$ for the components of $\mathbf{F}$. Vanishing curl says ${\partial }_i{F}_j={\partial }_j{F}_i$. Differentiation under the integral, followed by the chain rule, givesThis is the radial integral potential for a curl-free field. The fundamental theorem for line integrals now gives zero on every closed curve. On a simply connected open domain the analogous argument uses path integrals: every loop can be contracted within the domain, and applying Stokes theorem to a piecewise smooth contraction makes its circulation zero. Equivalently, whenever a particular loop bounds an oriented spanning surface entirely in the domain, Stokes theorem immediately gives zero for that loop.

Why the domain matters. On $U={\mathbb{R}}^3\setminus \{(0,0,z):z\in \mathbb{R}\}$, the smooth vector fieldhas zero curl, but its line integral around the unit circle in $z=0$ is $2\pi$, as in ii. No spanning surface for this loop can lie entirely in $U$. The literal claim without a condition on the domain therefore has this counterexample.

The middle component suggests a scalar potential with $-{\varphi }_y=2y{\cosh }^2(xz)$, so $\varphi =-y^2{\cosh }^2(xz)+h(x,z)$. The chain rule and $2\sinh t\cosh t=\sinh (2t)$ giveMatching the other two components requires $h_x=h_z=0$. ThereforeIts negative gradient is the given vector field in all three components; the additive constant has no effect on that gradient.

Evaluate the line integral separately on the three oriented line segments. Their parametrizations and contributions areThusThe different value in i demonstrates path dependence. The vector field in this part is different from the conservative vector field in part (a).

For the curved path $r(t)=(t,t^2,t^3)$, $r^{\prime }(t)=(1,2t,3t^2)$. The vector field along this parametric curve isIts dot product with the tangent is $(t^2+2t)e^t+13t^{12}$. Therefore the direct line integral is

Parametrize the straight line by $r(t)=(t,t,t)$, $0\le t\le 1$, with $r^{\prime }(t)=(1,1,1)$. Substituting into the vector field and taking the dot product givesHence the direct line integral is

Compute the curl component by component from the given vector field:For example, $B_x={\partial }_y{A}_z-{\partial }_z{A}_y=0-(-2xz)$ and $B_z={\partial }_x{A}_y-{\partial }_y{A}_x=(1-z^2)-(z^2-1)$. Its divergence isas expected because the divergence of a curl is zero.

Take $C$ counterclockwise when viewed from positive $z$, the disc $D$ upward oriented, and the hemisphere $H$ outward oriented. These choices give both surfaces the same positive boundary orientation. Parametrize $C$ by $r(t)=(\cos t,\sin t,0)$, $0\le t\le 2\pi$. On it, $A=(-\sin t,\cos t,0)$ and $dr=(-\sin t,\cos t,0)dt$, so the line integral isOn $D$, use polar coordinates $(x,y)=(r\cos \varphi ,r\sin \varphi )$ and $d\mathbf{S}=(0,0,1)rdrd\varphi$. Since $B_z=2$ there, the surface flux isOn the unit hemisphere, the outward unit normal vector is $n=(x,y,z)$. Since $x^2+y^2=1-z^2$, we have $B\cdot n=4z(1-z^2)$. In spherical polar coordinates, $z=\cos \theta$ and $dS=\sin \theta d\theta d\varphi$, giving the explicit surface integralIndeed the inner integral is $[{\sin }^4\theta {]}_0^{\pi /2}=1$.

These equalities illustrate Stokes theorem twice: $B=\nabla \times A$ and either upward $D$ or outward $H$ spans the oriented circle $C$, so each surface flux equals the same line integral. For the divergence theorem, close the upper half-ball by adjoining the downward oriented base, namely $-D$. Its outward surface flux isThe minus sign on the base is required because its outward normal vector points down. Reversing a specified orientation reverses that integral's sign.

For a bounded volume $V$ with piecewise smooth closed boundary $\partial V$, oriented by the outward unit normal vector, and a continuously differentiable vector field on a neighborhood of $\overline{V}$, the divergence theorem statesFor an oriented piecewise smooth surface $S$ with piecewise smooth boundary $C=\partial S$, and a continuously differentiable vector field on a neighborhood of $S$, Stokes theorem statesThe boundary orientation follows the right-hand rule relative to the chosen normal vector. Reversing that orientation reverses both sides of the latter identity.

Let $\Omega$ be the interior of the tetrahedron, and take any constant vector field $\mathbf{q}$. Its divergence is zero, so the divergence theorem givesChoosing $\mathbf{q}$ to be each Cartesian basis vector proves ${\sum }_i{\mathbf{V}}_i=0$. This is the general fact that vector areas of a closed polyhedron sum to zero, specialized to four faces. The outward orientations are essential: the scalar face areas are all positive and do not sum to zero.

For a sufficiently smooth vector field on an oriented piecewise-smooth surface $S$ with boundary $\partial S$, Stokes theorem statesHere $\mathbf{n}$ is the chosen unit normal, $dS$ is the scalar surface-area element, and $d\mathbf{r}$ follows the positively oriented boundary. The boundary orientation is fixed by the right-hand rule: the fingers curl along $\partial S$ when the thumb points along $\mathbf{n}$.

Integrating the first component with respect to $x$ suggests $\varphi =xy+g(y,z)$. Matching the other components gives $g=\sin (yz)+C$. Direct differentiation verifiesThus the vector field is conservative, with the displayed scalar potential.

For positive semiaxes $a,b$, the cross product of the two parametric tangent vectors gives the outward oriented surface element of an ellipsoid of revolution:Substituting the parametrization into the vector field and taking its dot product with this element givesThe middle term has zero integral over $0\le \varphi \le 2\pi$. In the other terms, ${\int }_0^{2\pi }{\cos }^2\varphi d\varphi =\pi$, ${\int }_0^{\pi }{\sin }^3\theta {\cos }^2\theta d\theta =4/15$, and ${\int }_0^{\pi }\sin \theta {\cos }^4\theta d\theta =2/5$. The directly evaluated flux is therefore

For a bounded region $V$ with piecewise smooth closed boundary $S$, outward unit normal vector $\mathbf{n}$ and continuously differentiable vector field $\mathbf{F}$, the divergence theorem states ${\int }_S\mathbf{F}\cdot \mathbf{n}dS={\int }_V\nabla \cdot \mathbf{F}dV$. Here $dS$ is scalar surface area, $dV$ is volume, and $d\mathbf{S}=\mathbf{n}dS$ is the oriented area element.

For $\mathbf{F}=\mathbf{r}$, its divergence is three. The ellipsoid has semiaxes $a,a,b$, so its volume is $4\pi a^{2}b/3$, obtained by scaling the unit ball. Consequently