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The space ${\mathbb{R}}^3$ is simply connected, and the electric field has zero curl. Hence it is a conservative vector field. More explicitly, Stokes theorem gives zero circulation around any closed loop, since such a loop bounds a spanning surface in ${\mathbb{R}}^3$. Therefore the line integral between two fixed endpoints is independent of the chosen path.

Fix a reference point ${\mathbf{r}}_0$ and define the electric potentialTaking a short straight terminal segment in each coordinate direction shows ${\partial }_j\phi =-E_j$. ConsequentlyThe electric potential is determined up to an additive constant; the later condition at infinity fixes that constant for the finite charge distribution in part (b).

Let $S$ be a piecewise smooth oriented surface with piecewise smooth boundary $\partial S$, and let $\mathbf{F}$ be a continuously differentiable vector field on a neighbourhood of $S$. The Stokes theorem statesThe boundary orientation is induced by the chosen unit normal vector $\mathbf{n}$ through the right-hand rule: with the right thumb along $\mathbf{n}$, the curled fingers indicate positive boundary traversal. Equivalently, when walking along the boundary with one's head pointing along $\mathbf{n}$, the surface lies to the left. An upward oriented disk therefore has counterclockwise boundary traversal when viewed from above. Reversing the orientation reverses both the line integral and the surface integral.

On the unit circle, the positively oriented unit tangent vector is $\mathbf{u}=(-y,x,0)$, and $|\mathbf{u}|=1$. Hence $d\mathbf{r}=\mathbf{u}ds$ andSince the scalar field $\psi$ has no $z$ dependence, the curl product rule for a scalar field givesOn the unit upper hemisphere $S_1$, the outward unit normal vector is $(x,y,z)$, so its vertical component is $z=\cos \theta$. Applying the Stokes theorem to the vector field $\psi \mathbf{u}$ therefore provesThe smoothness needed here is that $\psi$ extend as a continuously differentiable scalar field to a neighbourhood of the projected unit disk. The left side is a scalar line integral with respect to arc length; the positive boundary orientation is used when identifying it with the vector line integral.

Choose the outward orientation on the upper half of the ellipsoid, giving counterclockwise boundary orientation on $C$ as viewed from $z>0$. Parametrize the boundary circle byThen $\mathbf{u}(\mathbf{c}(t))=(-\sin t,\cos t,0)={\mathbf{c}}^{\prime }(t)$, so the line integral isFor a direct surface integral, use the parametrized surfaceIts outward vector surface element is supplied by the cross productUsing the curl from a,The two integrals agree for every $\alpha >0$, verifying the Stokes theorem. Only the vertical component of the vector surface element enters, which explains the absence of $\alpha$ in the answer.

Take the boundary pieces in the orderOn the circular arc $C_1$, use $(x,y,z)=(2\cos \theta ,2\sin \theta ,0)$ for $0\le \theta \le \pi /2$. Then $\mathbf{F}=(0,0,1)$ and $dz=0$, so its line integral is zero.

On $C_2$, use $x=0$, $z=4-y^2$, with $y$ decreasing from two to zero. Since $dx=0$ and $F_y=0$, $\mathbf{F}\cdot d\mathbf{r}=dz$, giving four. On $C_3$, use $y=0$, $z=4-x^2$, with $x$ increasing from zero to two. Again $\mathbf{F}\cdot d\mathbf{r}=dz$, giving minus four. The total is $\boxed{0+4-4=0}$, agreeing with Stokes theorem.

The curl is $\nabla \times \mathbf{F}=(x,y,-2z)$. For the upward orientation of the paraboloid, the vector surface area element is $(2x,2y,1)dxdy$. Therefore Stokes theorem givesThe upward normal gives the prescribed anticlockwise boundary orientation viewed from above.

Choose the outward unit normal vector on the upper hemisphere. The curl of $\mathbf{F}=(4y,x,2z)$ isIn spherical polar coordinates, putHere $n_z=\cos \phi$ and $dS=a^2\sin \phi d\phi d\theta$. The surface integral is thereforeThe compatible boundary orientation follows increasing $t$ in $\mathbf{r}(t)=(a\cos t,a\sin t,0)$, $0\le t\le 2\pi$. On this circle,Thus the line integral iswhich agrees with the surface integral as required by the Stokes theorem.

Let $\Sigma$ be a compact, piecewise smooth oriented surface with piecewise smooth boundary, and let $\mathbf{F}$ be continuously differentiable on an open neighbourhood of $\Sigma$. The Stokes theorem statesThe boundary orientation must agree with the chosen unit normal vector by the right-hand rule. For an upward orientation on a horizontal disk, its boundary is traversed counterclockwise when viewed from above. Reversing the orientation reverses both the line integral and the surface integral.

Let $\Sigma$ be a piecewise smooth oriented surface, and let $\mathbf{F}$ be continuously differentiable on an open neighbourhood of it. Stokes theorem states thatThe boundary orientation is induced by the chosen unit normal vector through the right-hand rule. Smoothness on a neighbourhood makes the spatial partial derivatives in the curl well defined.

Choose the outward orientation of a surface on the lower half of the ellipsoid. At its lowest point the normal points down, so its boundary ellipse is traversed clockwise when viewed from above. UseThe line integral is

For an independent surface integral calculation, first compute $\nabla \times \mathbf{F}=(2y,-2x,1)$. Parametrize the lower half byThe outward vector area element isThusThe two sides agree. Choosing the inward unit normal vector and its corresponding counterclockwise boundary orientation changes both answers to $6\pi$.

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.

Let $a$ be an arbitrary constant vector. By the cyclic identity for the scalar triple product,Apply Stokes theorem to $\mathbf{F}\times a$. Since $a$ is constant, the curl product identity givesConsequentlyFor precision, the operator in the last line has componentsThe partial derivatives act on $\mathbf{F}$, with $\mathbf{n}$ multiplying their values; they do not differentiate $\mathbf{n}$. Since the dot products agree for every constant $a$, the cross-product form of Stokes theorem follows:Here $\wedge$ in the paper denotes the cross product, and the printed surface integral has its area element $dS$ understood.

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.

Take the outward orientation on the upper hemisphere; its boundary circle is then counterclockwise viewed from above. Parametrize that circle by $r(\varphi )=(r\cos \varphi ,r\sin \varphi ,0)$, $0\le \varphi \le 2\pi$. Then $G\cdot dr=r^2(3{\cos }^2\varphi -4{\sin }^2\varphi )d\varphi$, so Stokes theorem givesEquivalently $\nabla \times G=(0,0,-1)$, and the upward projected area of the hemisphere is $\pi r^2$. The opposite orientation gives the negative of this answer.

Orient the three faces outward from the tetrahedron. The two coordinate-plane faces have normals parallel to $-{\mathbf{e}}_x$ or $-{\mathbf{e}}_y$, so their flux of the curl is zero. On the inclined face $z=1-x-y$, its upward vector area is $(1,1,1)dxdy$, with $x,y\ge 0$, $x+y\le 1$. Thus the direct surface integral isThis surface induces the counterclockwise boundary $O\to X\to Y\to O$ viewed from above, where $X=(1,0,0)$ and $Y=(0,1,0)$. At $z=0$, $\mathbf{A}=(2x-y,0,0)$. On $OX$, take $(x,y)=(t,0)$, giving integral ${\int }_0^{1}2tdt=1$. On $XY$, take $(x,y)=(1-t,t)$, giving ${\int }_0^1(-2+3t)dt=-1/2$. On $YO$, $dx=0$ and the integral vanishes. ThereforeBoth sides have been calculated directly, verifying the Stokes theorem with consistent orientations.

For an oriented piecewise smooth surface $S$ with boundary $C=\partial S$ and a vector field $\mathbf{A}$ with continuous first derivatives near $S$, the Stokes theorem statesHere $\mathbf{n}$ is the chosen unit normal vector, $dS$ the positive area element, and $d\mathbf{r}$ the oriented tangent line element. The traversal of $C$ follows the right-hand rule relative to the surface orientation.

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 $\mathbf{n}$.

Orient the triangle by the path $A\to B\to C\to A$. The cross product $(B-A)\times (C-A)=(0,1,0)$ gives the positive normal vector. In the plane $y=1$, the triangular region is $x,z\ge 0$, $x+z\le 1$. The surface integral in Stokes theorem is thereforeFor the boundary line integral, $AB$ has $z=0$, so contributes zero. On $BC$, $x=0$ and only $dz$ is nonzero, so its contribution is also zero. Parametrize $CA$ by $(x,y,z)=(t,1,1-t)$, $0\le t\le 1$. Its contribution isThus both oriented integrals equal $1/15$. Reversing the orientation would reverse both signs.

Choose the upward orientation. The boundary is the circle $x^2+y^2=4$ at $z=1$, oriented counterclockwise from above. By Stokes theorem, the surface integral is ${\oint }_{\partial S}G\cdot dr$. On this boundary $dz=0$ and $z^2=1$, soThe first term integrates to zero around a closed curve. By Green's theorem, the second equals the disk integral ${\int }_{x^2+y^2\le 4}(-3y)dxdy=0$, by oddness in $y$. ThusHere reversing orientation leaves the zero value unchanged.

The shorter circular arc lies in the $x$-$y$ plane and can be written $\mathbf{r}(\theta )=(\cos \theta ,\sin \theta ,0)$, $0\le \theta \le \pi /2$. The first field is $(-\sin \theta ,\cos \theta ,-\cos \theta )$, whose dot product with ${\mathbf{r}}^{\prime }$ is one. HenceThe second field is radial while the circular tangent is perpendicular to the radius, so $\boxed{{\int }_{\mathrm{arc}}{\mathbf{F}}_2\cdot d\mathbf{r}=0}$. Thus the first field gives different values, $1$ and $\pi /2$, along the two paths, whereas the conservative second field gives the same endpoint difference.

As a check using Stokes theorem, the arc followed by the reversed segment encloses area $\pi /4-1/2$ in the plane. The constant curl has upward component two, giving circulation $2(\pi /4-1/2)=\pi /2-1$, exactly the difference above.