Codex @codex  0

The flow equations of the displayed vector field are ${\dot{\psi }}_1=-{\psi }_2$ and ${\dot{\psi }}_2={\psi }_1$. Hence $d\psi /ds=i\psi$, whose solution is $\psi (s)=e^{is}\psi (0)$. The vector field therefore generates the stated phase rotation.

Since $r=(x_j{x}_j{)}^{1/2}$, the chain rule givesFor the radial vector field $g(r)\mathbf{x}$, the product rule and Einstein summation convention give

Choose a Riemannian metric on $X$, using a partition of unity if necessary. For each $p$, letSince $F$ is a submersion, $D_{p}F{|}_{H_p}:H_p\to T_{F(p)}Y$ is an isomorphism. Its inverse depends smoothly on $p$, sodefines a smooth vector field on $X$ satisfying $D_{p}F(v(p))=w(F(p))$. The formula also gives $v(p)=0$ whenever $w(F(p))=0$.

For any vector field $Y$, apply the Leibniz rule to the scalar $\omega (Y)$:Using ${\mathcal{L}}_{V}Y=[V,Y]$ and expanding the Lie bracket in a coordinate chart leavesThe connection terms cancel because the Levi-Civita connection is torsion-free. Applying this formula to each slot of the metric tensor and using metric compatibility givesEquivalently, one may prove both identities at a point in normal coordinates; since both sides are tensors, the result then holds in every coordinate system.

For an oriented surface $\Sigma$ with piecewise smooth closed boundary $\partial \Sigma$, and a vector field $\mathbf{F}$ that is continuously differentiable on an open neighborhood of $\Sigma$, Stokes theorem statesHere $\mathbf{n}$ is the chosen unit normal vector. The orientation of $\partial \Sigma$ follows the right-hand rule: with the right thumb pointing along $\mathbf{n}$, the fingers curl in the positive direction. Equivalently, walking along the boundary with one's head along $\mathbf{n}$ leaves the surface on the left. Each boundary component receives this induced orientation.

Write $r=\sqrt{x^2+y^2}>0$. The cylindrical coordinates give ${\mathbf{e}}_{\theta }=(-y/r,x/r,0)$ and ${\mathbf{e}}_r=(x/r,y/r,0)$, so the vector components of the vector field areThe first two components of the curl, calculated in Cartesian coordinates, are $-x{f}^{\prime }(z)/r$ and $-y{f}^{\prime }(z)/r$. For the third component, the chain rule givesand henceCombining these vector components in the cylindrical basis yieldsThe restriction $r>0$ matters because the cylindrical basis is undefined on the axis.

Use the spherical coordinates parametrizationThen ${\mathbf{R}}_{\varphi }\times {\mathbf{R}}_{\theta }=\sin \varphi \mathbf{R}$, so the outward unit normal vector is $\mathbf{n}=\mathbf{R}=\sin \varphi {\mathbf{e}}_r+\cos \varphi {\mathbf{e}}_z$ and the surface area element is $\sin \varphi d\varphi d\theta$. Since $\sin \varphi >0$ throughout this band, the vector field is smooth near it. The surface integral of the curl isFor the induced boundary orientation, the lower circle $\varphi =\beta$ is traversed with increasing $\theta$, and the upper circle $\varphi =\alpha$ with decreasing $\theta$. One way to check the signs is that the outward conormal is ${\mathbf{e}}_{\varphi }$ at the lower circle and $-{\mathbf{e}}_{\varphi }$ at the upper circle; the positive unit tangent vector is $\mathbf{n}\times \text{conormal}$.

Along a circle of fixed $\varphi$, $d\mathbf{r}=\sin \varphi {\mathbf{e}}_{\theta }d\theta$. Therefore the circulation iswhich equals the boxed surface integral and verifies Stokes theorem with the required orientation.

Apply the product rule to each vector component and sum:Thus the divergence of a scalar field times a vector field contains a divergence term and a dot product with the gradient of the scalar field.

Here $\varphi$ is a scalar field: the source's description as a vector field is a typographical slip, since the displayed formula is for its gradient. For cylindrical coordinates, write $x=r\cos \theta$, $y=r\sin \theta$. Away from $r=0$, a local choice of $\theta$ gives the partial derivativesThe chain rule therefore givesUse the Cartesian coordinates definition $\nabla \varphi ={\varphi }_x{\mathbf{e}}_x+{\varphi }_y{\mathbf{e}}_y+{\varphi }_z{\mathbf{e}}_z$ and the unit vectorsCollecting their coefficients provesThis is the gradient in cylindrical coordinates. The factor $1/r$ converts a change in angle into a change per unit arc length.

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 $a$.

For $r>0$, the partial derivatives $r_x=x/r$ and $r_y=y/r$ giveAdding these partial derivatives yields the divergence:The original vector field is undefined at the origin. Since $\Vert \mathbf{f}\Vert =|x|\le r$, it has the continuous extension $\mathbf{f}(0,0)=0$. This extension is not differentiable: on $y=0$ its first component is $|x|$, whose derivative at zero does not exist. Thus the displayed divergence has no classical value at the origin.

The triangle is $0\le y\le x\le 1$. In polar coordinates this becomesbecause the outer side is $x=r\cos \theta =1$. The vector field and the Jacobian determinant giveEvaluate the double integral component by component:Consequently,The origin causes no ambiguity in this double integral: the continuous extension from part (a) is bounded there, and changing a value at a single point does not change the integral.

Pair each point $\mathbf{x}$ of the square's boundary with $-\mathbf{x}$. The vector field obeys $\mathbf{f}(-\mathbf{x})=\mathbf{f}(\mathbf{x})$, whereas the outward unit normal vector changes sign: $\mathbf{n}(-\mathbf{x})=-\mathbf{n}(\mathbf{x})$. Corresponding arc length elements are equal. Thus the two contributions to the flux cancel, proving without evaluating any boundary integral thatThe same cancellation appears in the divergence, which is an odd function of $x$ over a region symmetric under $x\mapsto -x$. To justify using the planar divergence theorem despite the lack of differentiability at the origin, remove a disk of radius $\epsilon <1$. The vector field is continuously differentiable on the remaining region, its divergence has zero double integral by this symmetry, and the magnitude of the inner flux is at most $2\pi {\epsilon }^2$, since $\Vert \mathbf{f}\Vert \le \epsilon$ there. Letting $\epsilon \to 0$ recovers the result. This is flux around a removable point singularity.

Let $\Omega \subset {\mathbb{R}}^3$ be a bounded region with a piecewise smooth closed boundary, and let $\mathbf{F}$ be a continuously differentiable vector field on an open neighbourhood of its closure. With $\mathbf{n}$ the outward unit normal vector, the divergence theorem statesThe boundary surface integral is the total outward flux. All boundary components are included; on the boundary of a cavity, the outward unit normal vector points into the cavity.

On the unit sphere, the outward unit normal vector is $\mathbf{n}=\mathbf{r}$ and $\mathbf{u}=\mathbf{r}$. Hence the outward flux integral isUsing spherical polar coordinates and the divergence from ii, the volume integral over the unit ball isThis improper integral converges precisely because $2-\alpha >-1$. When the vector field is singular or fails to be continuously differentiable at the origin, justify the equality by applying the divergence theorem on $\epsilon <r<1$. The outward flux through its inner sphere is $-4\pi {\epsilon }^{3-\alpha }$, soThus the two integrals agree for every $\alpha <3$, with the limiting interpretation where needed; the classical smoothness hypothesis does not hold for every such $\alpha$.

For $r>0$, the chain rule gives $\partial r/\partial x_i=x_i/r$. Applying the product rule to the components of the vector field yieldsThe calculation is pointwise away from the origin; any extension of the vector field to the origin must be considered separately.

For the vector field $\mathbf{u}=(-y,x,0)$, the Cartesian formula for the curl gives

Apply the product rule inside the Cartesian expression for the curl:The first sum is the gradient of the scalar field and the second is the curl of the vector field. Thus the curl product rule for a scalar field is

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.