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Differentiating the assumed primitive identity and using Cartan's magic formula givesApply the pullback of a differential form by ${\rho }_t^{-1}$ and rearrange:This smooth family of Hamiltonian functions proves that ${\rho }_t$ is a Hamiltonian isotopy in the convention of part (b).

To connect this with the preceding assertion that every ${\rho }_t^{\ast }\lambda -\lambda$ is an exact differential form, the smooth choice of primitives causes no additional obstruction. Fix $p_0\in M$ and normalize $F_t(p_0)=0$. Define $F_t(p)$ by the line integral of ${\rho }_t^{\ast }\lambda -\lambda$ from $p_0$ to $p$. The fundamental theorem for line integrals makes this independent of the path. Near any $p$, use one fixed path to a nearby chart center followed by coordinate line segments; this expression is smooth jointly in $t,p$. Thus these normalized primitives form the required smooth family.

Parametrize the straight segment by $x=t$, $y=1+t$, $0\le t\le 1$. The line integral becomes

Because $u^0=c,dt/d\tau =\gamma c$ is constant, $dt/d\tau =\gamma$. The coordinate speed along the circle is consequentlysoFor the convention in the question, take $qB>0$, so $\omega >0$ and $\Delta \tau =2\pi /\omega$.

On the circular solution, the free part of the relativistic charged-particle action isIn the gauge $A_y=Bx$, the interaction is the closed line integralThe orbit is clockwise for $qB>0$, so $\oint xdy=-\pi R^2$. ThusCombining the two terms and using ${\gamma }^2{v}^2=c^2({\gamma }^2-1)$ gives the charged-particle action over one relativistic cyclotron orbitAlthough the intermediate magnetic line integral depends on the chosen magnetic vector potential, the closed-orbit result is gauge invariant.

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).

Here a potential of a conservative vector field means a single-valued scalar field $\psi$ with $\nabla \psi =\mathbf{F}$. By the fundamental theorem for line integrals, its line integral around every closed curve must vanish. Here the angular circulation obstruction to a cylindrical potential lets us test this using a circle about the axis.

On $V_1$, the circle from i lies entirely in the region and has line integral $4\sqrt{2}$. Therefore there is no potential of a conservative vector field on $V_1$.

On $V_2$, choose the continuous angular branch $-\pi <\theta <\pi$. By a, the equation for a potential of a conservative vector field reduces to ${\psi }_r={\psi }_z=0$ and ${\psi }_{\theta }=\sqrt{2}\cos (\theta /2)$. ThusThe denominator is positive throughout $V_2$, including when $y=0$ and $x>0$, so this is a well-defined differentiable potential of a conservative vector field on the entire region.

For $V_3$, choose the particular circle $x=2\cos \theta$, $y=2\sin \theta$, $z=0$. Its only point on the removed half-plane's supporting cut is $(-2,0,0)$, where $(x+2{)}^2+z^2=0<1$; this point is retained by $V_3$. Thus the whole circle lies in $V_3$ and again has line integral $4\sqrt{2}$. Consequently there is no potential of a conservative vector field on $V_3$. The opening in the cut allows a closed curve to go around the excluded axis.

Finally $V_4\subset V_2$. Restricting the potential of a conservative vector field already constructed on $V_2$ givesThe extra excluded cylinder does not invalidate an existing potential of a conservative vector field. Thus being simply connected is a sufficient condition in the usual curl-free criterion, not a necessary condition for a particular vector field to have a potential of a conservative vector field.

In cylindrical coordinates, $(-y,x,0)=r{\mathbf{e}}_{\theta }$ and $x+r=r(1+\cos \theta )$. Hence the vector field simplifies toTake the orientation of $C$ to be increasing $\theta$, as in its given parametrization. On the circle, $d\mathbf{r}=a{\mathbf{e}}_{\theta }d\theta$, so its line integral isWe used $\sqrt{1+\cos \theta }=\sqrt{2}|\cos (\theta /2)|$; the cosine is nonnegative on this interval. In particular, the answer does not depend on $a$.

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.

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.

Choose the scalar fieldThe allowed vector-valued Stokes identity givesEvery boundary point lies on the unit circle in the $xy$-plane, so $f=1/2$ throughout the boundary. Since the line integral of $d\mathbf{r}$ around a closed curve is the zero displacement,Only the boundary value of this scalar field mattered: the interior of the surface need not be part of a sphere. Either consistent choice of orientation gives the same zero answer.

Use the outward orientation on the unit hemisphere and counterclockwise boundary orientation on the unit circle. With $\mathbf{r}(t)=(\cos t,\sin t,0)$ and $f=x=\cos t$, the vector line integral isFor the surface integral, $\nabla f=\mathbf{i}$ and the outward unit normal vector is $\mathbf{n}=(x,y,z)$, since the radius is one. The cross product isThe integral of $y$ is zero by reflection symmetry. Using the same spherical polar coordinates as in ii, with $a=1$, the other component isHenceverifying the vector-valued Stokes identity directly, component by component. The wedge symbol in the paper denotes the cross product here.

A scalar potential is $U(x,y)=x^{3}y+y^4$, since $\nabla U=(3x^{2}y,x^3+4y^3)$. The fundamental theorem for line integrals therefore reduces the line integral to its endpoints. The curve starts at $(0,1)$ and ends at $(2,1)$, so

In polar coordinates, put $x=r\cos \theta$ and $y=r\sin \theta$. Differentiating along the curve givesThe cardioid has $r\ge 0$ and is traversed once counterclockwise as $\theta$ increases from $0$ to $2\pi$. Its cusp at $\theta =\pi$ causes no difficulty: split the line integral there, or exclude a small arc and take its limit. Thus signed area as a boundary integral gives

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 ${\partial }_x{F}_y=0$.

Compute the line integral on each straight segment. On the first, $y=z=0$ and only $dx$ contributes; on the second, $x=1,z=0$ and only $dy$ contributes; on the third, $x=y=1$ and only $dz$ contributes. ThereforeAlthough $F_x=9$ on the second segment, that term does no work there because $dx=0$.